Atmospheric.Thermodynamics - .NE by John A. Shanahan

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ATMOSPHERICTHERMODYNAMICS

ElementaryPhysicsandChemistry

Thistextbookpresentsauniquelyintegratedapproachinlinkingbothphysicsand chemistrytothestudyofatmosphericthermodynamics.

Thebookbeginsbyexplainingtheclassicallawsofthermodynamics,and discussesGibbsenergyandtheelementarykinetictheoryofgaseswithspecial applicationstotheatmosphere.Individualchaptersfocusonvariousﬂuidsystems, includingvaporpressureoverﬂatandcurvedsurfacesofpureliquidsandsolutions, andexaminetheverticaldependenceoftemperatureandpressureforenvironmental soundingandmovingairparcels.Recognizingtheincreasingimportanceof chemistryinthemeteorologicalandclimatesciences,achapterisdevotedto chemicalthermodynamicsandcontainsanoverviewofphotochemistry.

Althoughstudentsareexpectedtohavesomebackgroundknowledgeofcalculus, generalchemistryandclassicalphysics,thebookprovidesset-asiderefresherboxes asusefulreminders.Itcontainsover100diagramsandgraphstosupplementthe discussions.Italsocontainsasimilarnumberofworkedexamplesandexercises, withsolutionsincludedattheendofthebook.Itisidealforasingle-semester advancedcourseonatmosphericthermodynamics,andwillpreparestudentsfor higher-levelsynopticanddynamicscourses.

geraldr.north receivedaPh.D.inPhysicsfromtheUniversityofWisconsin in1966,andhasbeenaDistinguishedProfessorofAtmosphericSciencesatTexas A&MUniversityforover20years.Hisnotableresearchcareerincludesreceiving theOutstandingPublicationAward,NationalCenterforAtmosphericResearchin 1975,theExceptionalScientiﬁcAchievementMedalforNASAin1983,andthe JuleG.CharneyAwardfromtheAmericanMeteorologicalSocietyin2008.

tatianal.erukhimova receivedaPh.D.inPhysicsfromtheInstituteof AppliedPhysics,RussianAcademyofSciences,in1999,andisnowaLecturerin theDepartmentofPhysicsatTexasA&MUniversity.Herareasofresearchinclude large-scaleandmesoscaleatmospherictransportandmixing,atmosphericwave dynamics,atmosphericozone,andremotesensing.

ATMOSPHERICTHERMODYNAMICS

ElementaryPhysicsandChemistry

GERALDR.NORTHANDTATIANAL.ERUKHIMOVA

TexasA&MUniversity,USA

CAMBRIDGE UNIVERSITY PRESS

Cambridge, New York, Melbourne, Madrid, Cape Town, Singapore, São Paulo

Cambridge University Press

The Edinburgh Building, Cambridge CB2 8RU, UK

Published in the United States of America by Cambridge University Press, New York

www.cambridge.org

Information on this title: www.cambridge.org/9780521899635

This publication is in copyright. Subject to statutory exception and to the provision of relevant collective licensing agreements, no reproduction of any part may take place without the written permission of Cambridge University Press.

First published in print format

2009

ISBN-13 978-0-511-51677-1

ISBN-13 978-0-521-89963-5

eBook (EBL) hardback

Cambridge University Press has no responsibility for the persistence or accuracy of urls for external or third-party internet websites referred to in this publication, and does not guarantee that any content on such websites is, or will remain, accurate or appropriate.

Webothwishtodedicatethisbooktoourfamilies,whohave kindlyenduredourlongpreoccupationwiththeproject.

Preface

Thisbookisintendedasatextforundergraduatesintheatmosphericsciences.The studentsareexpectedtohavesomecalculus,generalchemistryandclassicalphysics backgroundalthoughweprovideanumberofrefreshersforthosewhomighthave lessexperienceorneedreminders.Ourstudentshavealsohadasurveyofthe atmosphericsciencesinaqualitativecourseatfreshmanlevel.Theprimaryaimof thebookistopreparethestudentforthesynopticanddynamicscoursesthatfollow. Weintendthatthestudentgainsomeunderstandingofthermodynamicsasitapplies totheelementarysystemsofinterestintheatmosphericsciences.Amajorgoalis forthestudentstogainsomefacilityinmakingstraightforwardcalculations.We havetaughtthematerialinasemestercourse,butinashortercoursesomematerial canbeomittedwithoutregretslaterinthebook.Thebookendswithtwochapters thatareindependentofoneanother:Chapter 8 onthermochemistryandChapter 9 onthethermodynamicequation.

Thisbookistheresultofteachinganintroductoryatmosphericthermodynamics coursetosophomoresandjuniorsatTexasA&MUniversity.Severalcolleagues havetaughtthecourseusingearlierversionsofthenotesandwegratefully acknowledgeProfessorsR.L.Panetta,PingYang,andDonCollinsaswellas thestudentsfortheirmanyhelpfulcomments.Inaddition,wehavereceiveduseful commentsonthechemistrychapterfromProfessorsSarahBrooks,GunnarSchade, andRenyiZhang.WealsothankProfessorKennethBowmanformanyfruitful discussions.WearegratefulforﬁnancialsupportprovidedbytheHaroldJ.Haynes EndowedChairinGeosciences.

Introductoryconcepts

Theatmosphereisacompressibleﬂuid,andthedescriptionofsuchaform ofmatterisusuallyunfamiliartostudentswhoarejustcompletingcalculus andclassicalmechanicsaspartofastandarduniversityphysicscourse.To complicatematterstheatmosphereiscomposedofnotjustasingleingredient,but severalingredients,includingdifferent(mostlynonreactive)gasesandparticlesin suspension(aerosols).Someoftheingredientschangephase(primarilywater) andthereisanaccompanyingexchangeofenergywiththeenvironment.The atmospherealsointeractswithitslowerboundarywhichactsasasource(and sink)offriction,thermalenergy,watervapor,andvariouschemicalspecies. Electromagneticradiationentersandleavestheatmosphereandinsodoingit warmsandcoolslayersofair,interactingselectivelywithdifferentconstituents indifferentwavelengthbands.

Meteorologyisconcernedwithdescribingthepresentstateoftheatmosphere (temperature,pressure,winds,humidity,precipitation,cloudcover,etc.)andin predictingtheevolutionoftheseprimaryvariablesovertimeintervalsofafew days.Thebroaderfieldofatmosphericscienceisconcernedwithadditionalthemes suchasclimate(statisticalsummariesofweather),airchemistry(itspresent,future, andhistory),atmosphericelectricity,atmosphericoptics(acrossallwavelengths), aerosolsandcloudphysics.Boththepresentstateofthebulkatmosphereand itsevolutionaredeterminedbyNewton’slawsofmechanicsastheyapplyto suchacompressiblefluid. Dynamics isconcernedprimarilywiththemotionof theatmosphereundertheinfluenceofvariousnaturalforces.Butbeforeonecan undertakethestudyofatmosphericdynamics,onemustbeableto describe the atmosphereintermsofitsprimaryvariables.Anessentialtoolneededinthis descriptionisthermodynamics,whichhelpsrelatethefundamentalquantitiesof pressure,temperatureanddensityasatmosphericparcelsmovefromplacetoplace. Suchparcelscontractandexpand,theirtemperaturesriseandfall;waterchanges phase,backandforthfromvaportoliquidtoice;chemicalconstituentsreact,

Introductoryconcepts

etc.Thekeytounderstandingthesechangesliesinapplicationsofthelawsof thermodynamicswhichrelatethesechangestoﬂuxesofenergyandotherless familiarfunctionswhichwillbeintroducedasneeded.

1.1Units

Theunitsusedinatmosphericsciencearethe StandardInternational (SI)units. TheseareessentiallytheMKSunitsfamiliarfromintroductoryphysicsand chemistrycourses.Theunitoflengthisthe meter,abbreviatedm;thatformass isthe kilogram,abbreviatedkg;andfortimetheunitisthe second,abbreviateds. Theunitsforvelocitythenarems 1 .Theunitofforceisthe newton (1kgms 2 , abbreviatedN).Tables 1.1–1.2 showtheSIunitsforsomebasicphysicalquantities commonlyusedinatmosphericscience.

Theunitofpressure,the pascal (1Nm 2 = 1Pa),isofspecialimportancein meteorology.Inparticular,atmosphericscientistslikethe millibar (abbreviated mb),butinkeepingwithSIunitsmoreandmoremeteorologistsusethe hectopascal (abbreviatedhPa,100Pa = 1mb).The kilopascal (1kPa = 10hPa)istheformalSI unitandsomeauthorspreferit. Oneatmosphere (abbreviated1atm)ofpressureis

1atm = 1.013bar

= 1013.25mb

= 1013.25hPa

= 101.325kPa

= 101325Pa

= 1.01325 × 105 Pa(1.1)

and1mb = 1hectopascal = 100Pa.Insomeoperationalcontextsandoftenin thepopularmediaonestillencounterspressureininchesofmercury(inHg)or millimetersofmercury(mmHg);1atm = 760.000mmHg = 29.9213inHg.

The dimensions ofaquantitysuchasdensity, ρ ,canbeconstructedfrom thefundamentaldimensionsoflength,mass,timeandtemperature,denotedby L, M , T, Temp respectively.Thedimensionsofdensity,indicatedwithsquare brackets [ρ ],are ML 3 .IntheSIsystemthe units arekgm 3 .Manyquantitiesare purenumbersandhavenodimension;examplesincludeargumentsoffunctions suchassineorlog.Theradianisaratiooflengthsandisconsideredheretobe dimensionless.

TemperatureinSIunitsisexpressedindegreesCelsius,e.g.20 ◦ C;orKelvin, e.g.285K.Wesay“285kelvins”andomitwritingthesuperscript“◦”when

Table1.1 Usefulnumericalvalues

Universal gravitationalconstant6.673×10 11 Nm2 kg 2

universalgasconstant(R∗ )8.3145JK 1 mol 1

Avogadro’snumber(NA )[grammole]6.022 ×1023 moleculesmol 1

Boltzmann’sconstant(kB )1.381 ×10 23 JK 1 molecule 1

protonrestmass1.673 ×10 27 kg

electronrestmass9.109 ×10 31 kg

Planck’sconstant6.626 ×10 34 Js speedoflightinvacuum3.00 × 108 ms 1

PlanetEarth

equatorialradius6378km polarradius6357km massofEarth5.983 × 1024 kg

rotationperiod(24h)8.640 × 104 s accelerationofgravity(atabout45◦ N)9.8067ms 2

solarconstant1370Wm 2

Dryair

gasconstant(Rd )287.0JK 1 kg 1 molecularweight(Md )28.97gmol 1 speedofsoundat0 ◦ C,1000hPa331.3ms 1 densityat0 ◦ Cand1000hPa1.276kgm 3

speciﬁcheatatconstantpressure(cp )1004JK 1 kg 1

speciﬁcheatatconstantvolume(cv )717JK 1 kg 1

Watersubstance

molecularweight(Mw )18.015gmol 1

gasconstantforwatervapor(Rw )461.5JK 1 kg 1

densityofliquidwaterat0 ◦ C1.000×103 kgm 3 standardenthalpyofvaporizationat0 ◦ C2.500×106 Jkg 1 standardenthalpyoffusionat0 ◦ C332.7kJkg 1 speciﬁcheatofliquidwater4179kJkg 1 K 1

STP T = 273.16K, p = 1013.25hPa

usingdegreeskelvin.Inoperationalmeteorologywesometimesﬁndtemperature expressedindegreesFahrenheit,e.g.70 ◦ F.

Eachsideofanequationmusthavethesamedimensions.Thisprinciplecan oftenbeusedtoﬁnderrorsinaproblemsolution.Theargumentoffunctionssuch astheexponentialhastobedimensionless.

1.2Earth,weightandmass

TheEarthisanoblatespheroid,withslightlylargerdiameterintheequatorialplane thaninameridional(pole-to-pole)plane.Thedistancefromthecentertothepoles

Table1.2 Selectedphysicalquantitiesandtheir units

QuantityUnitAbbreviation masskilogramkg lengthmeterm timeseconds forcenewtonN

pressurepascalPa = Nm 2 = 0.01hPa energyjouleJ

temperaturedegreeCelsius ◦ C

temperaturedegreeKelvinK speedms 1 densitykgm 3

speciﬁcheatJkg 1 K 1

Table1.3 GreekpreﬁxesappliedtoSIunits

PreﬁxNumericalmeaningExampleAbbreviation nano10 9 nanometernm micro10 6 micrometer µm milli10 3 millimetermm centi10 2 centimetercm hecto102 hectopascalhPa kilo103 kilogramkg mega106 megawattMW giga109 gigawattGW tera1012 terawatttW

Table1.4 SelectedconversionstoSIunits

QuantityConversion

energy4.186J = 1cal 1kWh = 3.6×106 J

pressure1atm = 760mmHg 1atm = 29.9213inHg distance1m = 3.281ft

temperature T (K) = T (◦ C) + 273.16

T (◦ F) = 9 5 T (◦ C) + 32 T (◦ C) = 5 9 (T (◦ F) 32)

Table1.5 Somerelationships betweenSIunits

QuantityEquivalent 1N1kgms 2 1J1kgm2 s 2 1Pa1Nm 2

is6356.91kmandtheradiusintheequatorialplaneis6378.39km.Abouttwothirds oftheEarth’satmosphereliesbelow10kmabovethesurface,hencetheatmosphere andtheoceans(depthaveraging4–5km)onlyformaverythinskinofabout1/60 theradiusofthesphere.

The weight ofamassistheforceappliedtothatmassbytheforceofgravity.It maybeexpressedasthemassinkilogramstimestheaccelerationduetogravity, g = 9.81ms 2 : W = Mg [weightandmass].(1.2)

Weightisexpressedin newtons,abbreviatedN;N = kgms 2 .Theacceleration duetogravityvariesslightlywithaltitudeabovesealevel gz = g0 (1 3.14 × 10 7 z ) z inmeters.(1.3)

Thereisalsoaslightvariation(< 0.3%)withlatitudeduetotheellipsoidal shapeoftheEarth(duetobothcentrifugalforceandtheequatorialbulge).In mostmeteorologicalapplicationsthesevariationsarenegligible.However,in calculationsofsatelliteorbitssuchvariationsareextremelyimportant.

Example1.1 Thedensityofwaterinoldfashionedunits(cgs)is

ρwater = 1gram cm3 .

ToexpressthisinSIunits,wecanmultiplyby

1 = 1kg 103 gram = unity(nodimension).

Weobtain:

ρwater = 1 kg 103 cm3 = 1 kg liter

Introductoryconcepts

Thisgivesusanintuitivemeasureofthekilogram.Nowwecanmultiplyby

Theﬁnalresultis

ρwater = 103 kgm 3 .

Physicsrefresher:verticalmotionofaparticle Theaccelerationduetogravityis g = 9.81ms 2 .Aparticlefallingfromaheight z0 withnoinitialvelocity,hasa velocity gt aftertime t .Afterthesametimeintervalitwillhavefallen 1 2 gt 2 meters. Thesearebothobtainedbysimpleintegration:

,since

=constant(1.4) and

Moreverticalmotionmechanics Theminimum work necessarytoliftaparticlea verticaldistance z againsttheforceofgravityisforce × distance = Mgz (Mg isthe weightorverticalforcenecessarytoliftthemasswithoutacceleratingit).Thiswork doneinliftingtheparticleisequaltothechangeinits potentialenergy Mgz .Ifthe particleisreleased,workwillbedonebythegravitationalforceappliedtothe particle.The kineticenergy oftheparticleduringitsfallis 1 2 Mv 2 .Theconservation ofmechanicalenergysaysthesumofthesetwoformsofenergyisconserved: E = PE + KE = constant,ormoreexplicitly

wherethesubscript0denotestheinitialtimeandthesubscript t denotesevaluationat alatertime.

TheconservationlawisderivedbyﬁrstwritingNewton’sSecondLaw:

Nowmultiplythroughby v d t andintegratewithrespectto t .Theleft-handside becomes

OntheothersideofNewton’sequationwehave

Equatingtheseexpressionsgivesouranswer(1.6).

1.3Systemsandequilibrium

Thermodynamicsisthestudyof macroscopic or bulk systemsofmassesand theirinterrelationsunderconditionsofsteadystate(nodependenceontime).By macroscopicwemeanthesystemcontainslargenumbersofindividualmolecules (withinafewordersofmagnitudeofa mole1 whichcontains6.02×1023 molecules). Wecallthesestates equilibriumstates iftheyarenotonlytimeindependentbut alsostableundersmallperturbations.Thermodynamicstatesaredescribablebya setofdimensionalquantitieswhichwerefertoas coordinates.Thermodynamicsis concernedwiththechangesinenergy-relatedquantities(certainofthecoordinates) whenthesystemundergoesatransitionfromonestatetoanother.A thermodynamic system isaregionofspacecontainingmatterwithcertain internallyuniform properties suchaspressureandtemperature.Wewillbeconcernedwiththeinterior ofthesystemandthevariables(coordinates)thatcharacterizeit.Forexample, amassofpuregas(onlyonechemicalspecies)containedinavesselmaybe characterizedbythepressureitexertsonthewallsofthevessel,thevolume ofthevesselandthetemperature( p, V , T ).Thesecomprisethecompletesetof thermodynamiccoordinates forthisparticularsystem.Formoregeneralsituations suchasmixturesofspeciesorphases,thecoordinatesnecessarytodescribethe statehavetobedeterminedexperimentally.Itisimportanttonotethatanindividual thermodynamicsystemisuniforminitsinterior.Therearenogradientsofpressure ortemperature,forexample,insidethesystem.2

1 ThemoleisanSIunitdeﬁnedasthenumberofcarbonatomsinamassof0.012kgofpurecarbon.The numberofmolesofasubstanceisthenumberofmultiplesofthisnumber(knownasAvogadro’snumber: NA = 6.02 × 1023 ).Informulastheunitisdesignatedas“mol.”

2 Notethatacolumnofairintheatmosphereisnotasimplethermodynamicsystembecauseitspressureand temperaturevarywithaltitude.However,itisconvenienttoconsiderthecolumnascomposedofthinslabs, eachofwhichcontainssubstancewithapproximatelyuniformtemperature,pressureandcomposition.Then eachindividualslabmaybeconsideredasasimplethermodynamicsystemformanypurposes.

1.3.1Examplesofthermodynamicsystems

Gasinavessel Supposeacontainerholdsagasofuniformchemicalcomposition.Let thewallsofthecontainerbethermallyinsulatingandletthevolumebeﬁxed.Inavery shorttimeafterﬁxingtheseconditionsthegaswillcometovaluesoftemperatureand pressurethatareuniformthroughoutandindependentoftheshapeofthecontainer. Thisisthesimplestthermodynamicsysteminastateofequilibrium.

Asecondcaseiswherethecontainer’swallsareheldataﬁxedtemperatureandthe pressureisallowedtovary.Equilibriumwillbeestablishedsuchthatthetemperature ofthegasbecomesequaltothatofthesurroundingwalls,thevolumeisgivenand thepressurecomestosomevaluethatwecanestimate.

Athirdcaseiswherethecontainerhasafrictionlessmovablepistonthatispushed uponexternallybyaﬁxedpressure(suchastheatmosphericpressure).Thismeans thatthepressureinthevesselisheldﬁxedalongwiththatofthetemperature.The pistonwillshiftinsuchawaytomakethepressureinsideequaltothatoutside,and thevolumewillchangeuntilalltheseconditionsaremet.

Ourgasmightnotbehomogeneous,butinsteaditmightbecomposedofamixture ofchemicallynoninteractinggases,suchasthoseinouratmosphere:nitrogen,oxygen andargon.Westillhaveathermodynamicsystemaslongasthecompositiondoes notvaryfromlocationtolocationorfromtimetotime.Ineachoftheabovecaseslet twoofthefollowingbeﬁxed:volume,temperature,orpressure.Thentheremaining variableisallowedtoﬁnditsequilibriumvalue.Notethatonceinequilibrium,the variablesorcoordinatesareuniformthroughoutthevessel.

Two-phasesystem Supposewehavealiquidofuniformchemicalcompositionsuch aswaterinourvesselandvacuumabovetheliquidsurface.Letthetemperatureand volumebefixed.Afterasufficientadjustmenttimesomeliquidwillhaveevaporated intothevolumeaboveitssurfaceandanequilibriumwillbeestablished(thefluxof watermoleculesleavingthesurfacebecomesequaltothefluxenteringandsticking tothesurface).Therewillbeagaspressureexertedonthewallsbythevaporthat evaporatedfromtheliquidsurface.Thisisatwo-phasesystemwithliquidandgaseous phases,butonlyone component (water)whichdepictsthenumberofdistinctchemical species.Thepressurethroughoutwillbeuniform(ignorethepressureincreaseas afunctionofdepthduetogravityintheliquid).Thetemperaturewillalsobe uniformthroughoutbothphasesofthesystem.Thistwo-phaseconfigurationisalsoa thermodynamicsystem.Thesystemcanbemadetopassthroughchangesinvolume, temperature,etc.,toestablishnewthermodynamicstatesofequilibrium.Notethat thetemperatureandpressureareuniformthroughoutbutthedensityvariesfromone phasetotheother.Asweshallseeinalaterchapterthereisanotherquantitythat isalsouniforminthetwo-phasesystemcalledthespecificGibbsenergy(chemical potentialinthechemicalliteraturewhenexpressedasmolarGibbsenergy).Itactsas anintensivevariable(seeSection 1.5)insuchmulticomponentsystemssimilarlyto pressureortemperature.

Aqueoussolutions Imagineavesselﬁlledwithwater(ataﬁxedtemperatureand pressure)andsomesaltisplacedintheliquid.Ifwecontinuetoputmoresaltinto thewatereventuallysomesaltwillremainincrystalformsinkingtothebottom (butignoregravityotherwise).Wewillhaveestablishedanequilibriumbetweenthe saturated salinesolutionandtheprecipitatedcrystallinesalt.Achangeintemperature willresultinanewequilibriumstatewithadifferentconcentrationofsaltinsolution (concentrationofaspeciesinsolutionisanotherthermodynamiccoordinate).Thisis anexampleofathermodynamicsystem.Variationsonthisincludeallowingthewater vaporabovetheliquidtobeinequilibriumwiththesalinesolution.Thepresence ofsaltinsolutionwillalterthevaporpressureovertheliquidsurface(aswell asthefreezingtemperature).Asthetemperaturechangesthevaporpressurewill change,etc.

Chemicalequilibrium Imagineagaseousmixtureinourvesselatﬁxedtemperature andpressurecomposedofOandO2 .Therewillbeareaction

whereMisabackgroundmoleculeusedtocarryawaymomentum(e.g.,O2 ,N2 or Arintheatmosphere).3 Someozonewilldecayandafterawhiletherewillbean equilibriumestablishedandthereactioncanbewritten:

O + O2 + M O3 + M.(1.11)

Theamountof reactants (theleft-handside)maybemorethantheamountof products (right-handside)foragiventemperature.Butasthetemperatureischangedthe balancemayshift.Thisisathermodynamicsystem.TheratioofO2 toO3 isnowa thermodynamiccoordinatealongwith T , p, V , Mtotal . Ofcourse,therearemanyothertypesofthermodynamicsystems,andwewill encounterseveraloftheminduecourse.

Everythingoutsidethesystemwhichmayaffectthesystem’sbehavioris calledthe surroundings.Inatmosphericscience,wecanoftenapproximatean infinitesimalvolumeofgasembeddedinthenaturalatmosphereashavinguniform interiorproperties.Whenappropriate,suchaninfinitesimalvolumeelementcanbe consideredasathermodynamicsystem.Inmanycasesthe“infinitesimalvolume element”mightbeasbigasaclassroomorsometimesassmallasacubiccentimeter dependingontheapplication.

Athermodynamicsystemcomposedofaverylargemassiscalleda reservoir andischaracterizedbyatemperature, TR .Ifaﬁnitesystemisbroughtintocontact withthereservoirthrougha diathermal membrane(onewhichallowsthepassage

3 Energyandmomentumcannotbeconservedsimultaneouslywhentwobodiesgotoonewithareleaseofenergy. Athirdbodyinthecollisioncanprovidethemeansofconservingboth.

O + O2 + M → O3 + M,(1.10)

Figure1.1Isothermsfor1kgofdryairtakenasanidealgas.Thevertical coordinateispressureinhPa,theabscissaisvolumeinm3 .Uppercurve,300K; lowercurve,200K.

ofthermalenergy,4 butnotmass),thesmallersystemwilladjustthevaluesofits coordinates(foragas, p, V , T )tonewvalues,whilethereservoirdoesnotchange itsstateappreciably(thisactuallydefineshowmassivethereservoirhastobe).The systemissaidtocomeintothermalequilibriumwiththereservoir(itstemperature approachesthatofthereservoir).Inthecaseofagaseoussystem,experiments haveshownthatthereisalocusofpairsofvalues(V , p)forwhichthesystemis inequilibriumwithagivenreservoir–inotherwords,acurve p = pT (V ) inthe V –p plane.Toputitanotherway,ifoursystemhasacertainfixedvolume,then whenitisbroughtintocontactwiththereservoiroftemperature T ,thepressurewill alwayscometothesamevalue, p = pT (V ).Aswedotheexperimentwithdifferent controlvolumeswecansweepoutthelocusofpointsinthe V –p plane.Thiscurve iscalledthe isotherm ofthesystemforthatreservoirtemperature(Figure 1.1).The isothermrepresentsaseriesofequilibriumstatesthatcanoccurwhilethesystem isincontactwiththereservoir(offixedtemperature).Forexample,thevolume mightbeforcedtoalterbyachangeinthewalldimension(e.g.,apistoncanhave differentpositionsinacylinderwhichcontainsthesysteminquestion).Inthiscase thepressurewillchangeasafunctionofvolumealongtheisotherm.Whilewecould inventanalgorithmbaseduponaseriesofreservoirsofdifferenttemperaturesto buildatemperaturescale,itwillsufficeforourpresentpurposessimplytousethe familiarthermometer.

4 Thermalenergyreferstothemicroscopicmotionofmoleculesinthesystem.Whenindiathermalcontact,the thermalenergyofmoleculesfromonesystemcanpassfromthesystemtoitsneighborthroughcollisions.In timethethermalenergiesofthetwosystemswillequalize.Moreonthisinlaterchapters.Thetransferofthermal energyislooselyreferredtoas heattransfer

Figure1.2Isothermandadiabatfor1kgofdryairtakenasanidealgas.Theupper curve(solidline)isthe300Kisothermandthedashedcurveistheadiabatpassing throughthe300Kisothermat V = 1m3 .Theverticalcoordinateispressurein hPa,theabscissaisvolumeinm3

Asystemcanalsobeinequilibriumwhenisolated(nomassorthermalenergy flowsintooroutofthesystem)fromothersystems.Wecallthisan isolatedsystem. Itcanhavecoordinatesjustasinthecaseofasystemincontactwithareservoir. Wecallthelocusofvaluesofpressureintheisolatedsystemfordifferentvolumes ofthesystem adiabats (Figure 1.2).Wecouldfindthetemperatureoftheisolated systematfixedvaluesof p and V bybringingitintocontactwithdifferentreservoirs untilwefindonewhichdoesnotcausethecoordinatesofthesystemtochange. Thesystemhasthesametemperatureasthisreservoir.Inthiswaywecouldmap outthelocusofpointsdefiningtheisothermwhichcrossestheadiabatatthepoint inquestion.Asasimpleralternative,wecouldinsertathermometer,whosemass issosmallthatitwillcometoequilibriumwiththesystem(whichnowactsasa reservoirwithrespecttothetinythermometer)withoutdisturbingthestateofthe systemappreciably.

Statesofthermodynamicequilibriummustnotinvolvetime.Theyaresteadyand onlyrequireaknowledgeofthethermodynamiccoordinatessuchastemperature, pressureandvolume.Whenthe“states”traversedbyasysteminvolvethetime wecannotusethermodynamicequilibriumstatestodescribethem.Conventional thermodynamicscannotbeusedtodescribewhatgoesoninstatesthatarenotin equilibrium.

Certainchangesofasystemcanbemadetooccurthroughasequenceof inﬁnitesimallynearbyequilibriumstates.Forexample,wemightbringthesystem intocontactoneatatimewithaseriesofreservoirsofinﬁnitesimallydiffering temperatures,andateachstepwewaitforequilibriumtobeestablished.Wecall

thisa quasi-staticprocess.Suchquasi-staticprocessescanbeapproximatedinthe laboratory.Fromamolecularpointofviewthegasintheinteriorofthesystem hastohavetimeduringeachinfinitesimalshiftoftheconstraintstoadjustto anewequilibriumwithitssurroundings.Inagasthisisroughlythetimefora typicalmoleculetomakeafewhundredcollisions,butoverafinitesizedvolumeit mightbemoreappropriatetousethetimeforsoundwavestotraversethevolume severalhundredtimes.Thismultiplepasstraversaltimeworksforpressure,but otherpropertiesmighttakeconsiderablylonger.Forexample,temperatureand speciesconcentrationssmoothoutmuchmoreslowlybecausethesedifferences aresmoothedoutbydiffusiveprocessessuchasthermalconduction.Stirringdue toturbulencecanspeedupthehomogenizationbuteventhentheadjustmentis slowerthanforpressuredifferences.Ateachinfinitesimalstep(waitingforthese adjustments)alongsuchasystempath,wecouldreversedirectionandretracethe samesteps.Thisisa reversibleprocess.

Notethatasystemmaygofromonethermodynamicstatetoanotherbyapath whichdoesnotinvolvesuchasequenceofthermodynamicstates.Wecallthisan irreversible changeinstate.Anexampleofanirreversibleprocessisthecaseof asystemwhichgoesfromstateAtostateBspontaneously,butnotfromBtoA. Aconcreteexampleisiftwobricks,onehotandonecold,arebroughtintocontact, theresultistwowarmbricks.Thisisanirreversibleprocess.Notethatitnever happensthatwhenwebringtwowarmbricksintocontactweendupwithawarm brickandacoldbrick(eventhoughenergyisconserved).

Reversibleprocessesdonotactuallyoccurinnature.Sowhystudythem?The reasonsareprettysimple.Firstofall,irreversibleprocessesarenearlyimpossibleto treattheoretically.Secondly,experiencehasshownthatapproximatingthenearly quasi-staticprocessesthatdooccurinnatureworksreasonablywellinmanycases whenwetreatthemasexactlyquasi-static.Weproceedthentoadoptthephilosophy usedbypractitionersformanyyears:wewillfreelyapproximatemanyprocessesin therealatmospherebyidealizedreversibleanalogiesinordertoobtainnumerical resultsthatcanbeusedinpracticalsituations.

1.4Constraints

Animportantconceptinthestudyofthermodynamicsystemsisthatofconstraints. Thisnotionisbestillustratedbyexample.Considerthegasinacylinderwhose volumeisdeterminedbythepositionofapistonasinFigure 1.3.Several constraintsareoperativeinthiscase.Mostobviousisthepositionofthepiston. Itconstrainsthevolumetohaveacertainvalue.Ifthepistonisremovedbya smallamounttheconstraintissaidtoberelaxed.Notethataforcemustbeapplied

GAS

ADIABATIC WALLS PISTON

Figure1.3Schematicdiagramofagasﬁlledcylinderwithadiabaticwallsanda movablepiston.

(actuallyrelaxed,thengraduallyreapplied)externallytoimplementthischangein theconstraint.Ifthepistonisremovedbyasmallamount,someagentmustperform work torestoreittoitsoriginalposition.Similarly,thewallsthatareimperviousto thetransferofthermalenergyformaconstraint.Ifaleakofthermalenergywereto occur,suchasonbringingthesystemintocontactwithatemperaturereservoirata slightlydifferenttemperature,thisconstraintwouldbesaidtohavebeenrelaxedand thethermodynamiccoordinatesofthesystemwillhavetobechangedtorestore theoriginaltemperature.Thermodynamicsystemsarealwayssubjecttocertain constraintsandtheirnatureandnumberareessentialingredientsinthedescription ofthesystemanditsstate.

Considertwothermallyisolatedchambersadjacenttooneanotherseparatedby apartition.OnonesideisgasAandontheotherisgasB.Letthechambershave thesametemperatureandpressure.Thepartitionformsaconstraintrestrictingthe twogasesfrommixing.Ifthepartitionisremoved,theconstraintisrelaxedand thetwosystemswillpassthroughnonequilibriumstatestotheirﬁnalwell-mixed equilibriumstate.Theirreversibleprocessfollowingremovaloftheconstraint representsonewhichforidealgasesinvolvesnochangesinpressureortemperature, butexternalworkmustbeperformedtorestoretheoriginalconditions.

1.5Intensiveandextensivequantities

Considerathermodynamicsystem.Theinteriorpropertiesofthesystemare uniform.Now,imaginesubdividingthesystemintotwoequalparts(say,two warmbricksincontact).Ifavariableisthesameforthetwoindividualparts (e.g.,pressure,temperature,chemicalcomposition,density,etc.),thevariableis an intensivevariable.Ontheotherhand,ifthethermodynamicvariableforeach subsystemisproportionaltothemassoftheconstituentsinthatsubsystem(e.g., volume,mass),wecallitan extensivevariable.

Figure1.4Schematicdiagramofagaseouspressurereservoirincontactwitha smallsystem.Themembranebetweenthesystemandthepressurereservoiris movablesothatthetwosystemscanadjusttheirvolumesinsuchawaythatthe pressuresequalize.

Example1.2 Anexampleofanisolatedsystemisa1kgmassofgaseousO2 , conﬁnedinaboxwiththermallyinsulatingwalls.Supposethevolumeis1m3 . Thismeansthedensityofthegasis1kgm 3 .Ifthetemperatureofthegasisgiven, say300K,thenthepressurewillbedetermined(thisisanexperimentalfact).The thermodynamiccoordinatesofthis(pure)systemare: V ,thevolume; M,themass; T ,thetemperature;and p,thepressure.

Example1.3 Athermodynamicsystemmightbeinthermalequilibriumwitha reservoir.InthecaseofthemassofO2 gasinaﬁxedvolumeof1m3 ,takethe gastobeinthermalcontactwithareservoirat350K.Thepressurewillbequite differentfromthelastexample.

Example1.4 Wemighthaveapressurereservoir.Considertheboxofgastobein contactwithareservoirwithaslidableinterface,suchthatthepressurescanequalize betweenthetwosystems.Letthesystemotherwisebeinsulatedthermallyfromthe reservoirandtherestoftheuniverse.Ifthegashasagiventemperatureinitially,it willexpandorcontractuntilitspressureequalsthatofthereservoir(pleaseletit happengradually).Thevolumeandtemperatureofthegasmaychangeinorderto establishequilibriumwiththepressurereservoir(seeFigure 1.4).

1.6Systemboundaries

Beforesettingupaprobleminthermodynamicsitisextremelyimportanttochoose thepartoftheuniverseyouwanttocallyoursystem.Itmightbeamassofmatteror itmightbeacertainvolumeinspace.Asintheatmosphericexamplesthemassor

volumemightbeinmotion.Ifweareconsideringamassinspacewithnoadditional matterallowedtoenterorleavethisﬁxedmasswesayitisa closedsystem.Inthe ﬁxedvolumecasemassmightenterorleave.Wecallthisanopensystem.

Calculusrefresher:theexponentialfunction Thefunction y (x ) whosederivative isitselfiscalledtheexponentialfunction:

Supposewetry y = a x .Then

Thefactorontherightmusttendtounityas x → 0.Itwillbemoreeasilyseenifwe let x = 1/N where N isaninteger.Alittlerearrangementyields

andthenumber a∞ isgiventhesymbolewhosenumericalvalueturnsouttobe 2.718281....Toseehowthelimitcomesaboutcalltheapproximatevalueof a∞ = eN .Simplecomputationgives, e5 = 2.48832, e10 = 2.59374, e100 = 2.70481, e1000 = 2.71692,and e10000 = 2.71815

Notethate0 = 1,e 1 = 0.367879 ...,andex iscalledtheexponentialfunction.We caneasilyderiveafewpropertiesof y = ex .Fromitsdeﬁnition,dex /d x = ex ,andwe canusethechainruletoshowthatdeα x /d x = α eα x .

Thefunctione α x decreases exponentially fromavalueofunityat x = 0toavalue

Figure1.5Theexponentialfunctionex asafunctionof x .

Figure1.6Thedecayingexponentialfunctione x asafunctionof x . ofe 1 = 0.367879 at x = 1/α ,whichiscalledthee-foldingdistance if x isa distanceorthee-foldingtime or timescale if x isatime,seeFigures 1.5 and 1.6.

1.7Thermodynamicsandatmosphericscience

Theplanofthisbookistopresentthesubjectofthermodynamicsinsucha wayastohelpusbetterunderstandtheatmosphere,butalsotogiveussome rulesandmethodsthatcanbeusedinthepracticalapplicationofatmospheric science.Thermodynamicsisahugesubjectmorethanacenturyold,treated byexcellenttextbooksinphysics,physicalchemistry,chemicalengineering, mechanicalengineering,etc.Wecannotpossiblycoverallthematerialinthese ﬁelds.Wecannotevencoverallthebasictheoryofthermodynamicsinashort courseintendedforstudentsmajoringinatmosphericsciences–especiallyat thesophomore/junioryearlevelforundergraduates,wherenotmuchscienceand mathematicscanberequiredprerequisites.Somecompromiseswillhavetobe made.Thismeansthosewhowanttodelvedeeperintosomeofthederivationswill havetocheckelsewhereamongthemanysourceslistedattheendsofchapters. Sometimeswewilllimitderivationsorjustiﬁcationstothepointthatitisclear thatenoughinformationistheretodeterminethatsuchandsuchaformulacanbe derivedbythemethodsalreadydiscussed.

Sowhatproblemsinatmosphericsciencecanbeaddressedbythermodynamics? Afterallwehaveseenalreadythatthermodynamicsconsistsofasetoflaws applicableunderconditionsthataresoidealizedthattheyarerarelyattainable eveninthelaboratoryletaloneinnature.Theprocessesthatoccurinnature arespontaneousandvirtuallyneverdoweﬁndasystem(perhapsourleading applicationconsistingofaparcelofair)intruethermodynamicequilibrium.The

answertothesequestionsisthatthermodynamicsaffordsusausefulframework inwhichsomeapproximatecalculationscanbeconducted.Overtheyearsthese approximateresultshaveproventobeuseful.

Thefollowingisalistofapplicationsthatwewillworkoninthisbook.

Gaseousmixtures Theseoccurinnaturesincetheatmosphereiscomposedmainly ofseveralrelativelyinertgases(attypicalatmospherictemperaturesanyway)and severalothergaseswhichoccurintraceamounts(i.e.,ozone,methane,carbondioxide, evenwatervapor).Wewillbeconcernedwithhowthepressuresoftheseindividual gasesadduptothetotalpressureandhowthedensityofalocalblob(orparcel)of thegasmightdifferfromthatofitssurroundings,whichmightgiverisetoparcels liftingthemselvesfromtheirpresentaltitudestosomewhereabove,justhowhigh andhowfastdependingontheparcel’sthermodynamicpropertiesandthoseofthe surroundings.

Liquid–vaporequilibria Waterexistsinallthreephasesintheatmosphere.We wouldliketoknowhowthepressureofwatervaporaboveliquidsurfacesvaries withtemperature.Wewouldliketohaveusefulwaysofdescribingtheamountof watervaporintheair(humidity)andhowthisaffectstheair’sbuoyancyandhow condensationleadstoreleaseofthermalenergyandthereforechangesinbuoyancy. Otherrelatedissuestreatableinthermodynamicsincludehowliquidwaterdroplets growinhumidenvironments.Doesthepresenceofairaffectthevaporpressureabove aliquidsurface?Doestheairdissolveappreciablyintheliquid?Doesthepresenceof saltdissolvedintheliquidaffectthevaporpressure?Doesthesizeofadropletaffect itsvaporpressure?

Dynamicsofairparcels Howcanwetellwhetheragivenenvironmentaltemperature proﬁle(functionofaltitude)leadstostableconditionsorunstableones(doestheair starttoturnoverspontaneously)?Whatistheroleofmoistureandcloudformation inthisprocess?Asparcelsrise,theyexpandandtheirtemperaturedrops(why?). Doesthismeantheyaredenserandtheymightreturn?Whataretheconditionsfor continuedrising?Whattemperatureproﬁlesarelikelytoleadtosevereweather?

Atmosphericchemistry Mostchemicalreactionsintheatmospherearebetweentrace gasessuchasozoneandso-calledairpollutants,butmanyoccurbetweennatural constituents.Whatarethecriteriaforareactiontoproceedonewayoranother?How arechemicalequilibriabetweenreactantsandproductsestablishedandhowdothese equilibriumconcentrationsvaryasthetemperaturevaries?

Notes

Acompletebibliographyisgivenattheendofthebook.Alluniversitylevelphysics bookscontainafewchaptersonthermodynamics;thenumerouseditionsofSears andZemanskiaswellasthoseofHallidayandResnickandtheonebyGiancoli aregoodexamples.Manygeneralchemistrybooksalsocontainagooddescription

Introductoryconcepts ofthesubject,forexample,Whitten,DavisandPeck(1996).Thelittlepaperback titledsimply Thermodynamics byEnricoFermihassomemarvelousdescriptions ofthermodynamicstates,equilibria,etc.

Anolderadvancedbookwhichcontainsspeciﬁcapplicationstometeorology andespeciallyapplicationstocloudphysicsisthatofIrebarneandGodson(1981). Twonewerbooksonapplicationsofthermodynamicstoatmosphericscienceand oceanographyarethosebyBohrenandAlbrecht(1998)andCurryandWebster (1999).Botharepitchedatahigherlevelthanthepresenttextandbothdelvemore deeplyintomanyaspectsofthesubject.

NotationandabbreviationsforChapter1 atmpressureunit,oneatmosphere

g , gz , g0 accelerationduetogravity,itsvalueasafunctionof altitude,itsvalueatthesurface(ms 2 )

G gravitationalconstant(Table 1.1)

h height

kB Boltzmann’sconstant(Table 1.1)

L dimensionlength mb,pressureunit,onemillibar(1mb = 1hPa)

M dimensionmass

M massofamacroscopicobjectorsystem(kg)

ME massoftheEarth(kg)

NA Avogadro’snumber(Table 1.1)

p pressure(Pa,hPa)

Paunitofpressure,1Pa = 1Nm 2

R, RG , Rd , Rw , R∗ gasconstants:nosubscriptindicatesforanunnamedgas, sometimesexplicitlyforaspeciﬁcgas,G;thesubscriptd fordryair,wforwatervapor,andthesuperscript*for theuniversalgasconstant(Table 1.1)

SIStandardInternationalsystemofunits

T dimensiontime,alsoperiodofarepeatingprocess,and temperature

Temp dimensiontemperature

v speed(ms 1 ),sometimeswithasubscriptindicating velocitycomponentalongacoordinateaxis(e.g., vx )

V volumeofasystem(m3 )

Problems

1.1Ausefulmathematicalmodeloftheverticaldependenceofpressureis p(z ) = p0 e z /H

where H iscalledthe scaleheight H isusuallybetween8and9km.Whatisthisin thousandsoffeet?Comparetothealtitudeoftypicaljetairﬂights.

1.2Whatistheratiooftheatmosphericscaleheight(Problem1.1)totheEarth’sradius?

1.3If p0 is1000hPa,whatisatypicalpressureinDenver(milehighcity)?Usethe expressioninProblem1.1with H = 8kmwith1mile = 1.6km.

1.4(a)ComputethecircumferenceoftheEarthattheEquatorinkm.(b)Howmanykm arethereperdegreeoflongitudeattheEquator?(c)Howmanykmarethereperdegree oflongitudeat30◦ N?

1.5Anumericalmodeloftheatmospherehashorizontalresolution(gridboxes)2◦ × 2◦ . WhatistheareaofoneoftheseboxesattheEquator,andat30◦ N?

1.6Usetheconservationofenergytoshowthatforaparticlefallingundergravityfrom restataheight z0 ,atheight z itsvelocityisgivenby

2 = 2g (z0 z )

where g istheaccelerationduetogravity(9.81ms 2 ).

1.7Newton’sLawofGravitysaysthattheforceonaparticleofmass M is

Usethefactthattheaccelerationofgravityatthesurfaceis g = 9.81ms 2 alongwith themassoftheEarth ME = 5.983 × 1024 kgandtheradiusoftheEarthof6365km tocomputethegravitationalconstant G (seeTable 1.1).

1.8AparticlefallsfromouterspacetotheEarth’ssurface.Farawayitspotentialenergyis zero.AttheEarth’ssurfacethepotentialenergyis GME M/RE =−MgRE (relative to r =∞)where ME istheEarth’smass, RE itsradiusand M isthemassinquestion (seeTable 1.1).WhatisthevelocityofthemasswhenitstrikestheEarth’ssurface? (Thisisexactlytheverticalvelocityitwouldhavetohaveifitweretoescapethe Earth’sgravityﬁeldafterbeingprojectedupwardsfromthesurface.Itiscalledthe escapevelocity.)

1.9Supposeaparticleisdroppedfromaheight h anditbounceselastically(i.e.,nokinetic energyislostinthecollision: vafter =−vbefore ).Howlongdoestheroundtriptake?

Gases

2.1Idealgasbasics

Gasesareaformofmatterinwhichtheindividualmoleculesarefreetomove independentlyofoneanotherexceptforoccasionalcollisions.Mostofthetimethe individualmoleculesareinfreeflightoutoftherangeofinfluenceoftheirneighbors. Gasesdifferfromliquidsandsolidsinthattheforcebetweenneighbors(onthe averageovertime)isveryweak,sincetheintermolecularforceisofshortrange comparedtothetypicalintermoleculardistancesfortheindividualgasmolecules. IfanimaginaryplateisheldverticallyinagasasshowninFigure 2.1,therewill beaforceexertedonthethinplatefromeachside.Theforcesonoppositesidesofthe insertedplaneareequal;otherwise,ifforcesontheopposingsideswereunbalanced, theplatewouldexperienceanacceleration.Theforceontheleftsideoftheplate iscausedbythereflectionofmoleculesastheyhittheleftfaceoftheplateand rebound.Theseimpulsiveforcesaresofrequentthattheresultingmacroscaleforce iseffectivelysteady.Theforceisperpendiculartothefaceandhasthesamevalue nomatterhowthefaceisoriented.Thiscanbeseenbyconsideringthecollisions withthewallandthetendencyfornomomentumtobetransferredparallelto theplanesurface.Theperpendicularcomponentoftheforceperunitareaonthe planeiscalledthe pressure.Tangentialcomponentsoftheforcecancelout(when averagedovermanycollisionswiththewall)andthereforevanishwhenaverages aretakenoveralargenumberofcollisionswiththesurface.Thepressurehasunits ofnewtonspermetersquaredorNm 2 ;1Nm 2 iscalledapascal,abbreviated Pa.AtmosphericscientistsusehPa(hectopascals1hPa = 100Pa)ortheequivalent mb(millibars).ThemoreappropriateunitwouldbethekPa,butthisisnotused muchinpractice.Anewtonistheforcenecessarytomaintainanaccelerationof 1ms 2 ona1kgmass.Theunitsofforcemaybedecomposedtokgms 2 .

Thestateofagasischaracterizedbythreequantities:itspressure, p;(mass) density, ρ (notethatknowledgeofdensityisequivalenttoknowledgeofthe

FORCES ON A VERTICAL PLATE IN A GAS

Figure2.1Illustrationoftheforcesexertedbyagasonathinverticallyoriented plate.Sincetheplatedoesnotfeelanetforce, FL =−FR .

volumeforasingle-phasesystemofagivenmass, ρ = M/V );andtemperature, T (inkelvins).Ingeneral,thereisamathematicalrelationshipbetweenthethree variables,orthermodynamiccoordinates,calledthe equationofstate.Itisimportant torememberthatthedensityisuniformthroughoutthevolumeofagasin thermodynamicequilibrium.

The numberdensity, n0 ,isthenumberofmoleculesperunitvolume ([n0 ] = moleculesm 3 ).Theequationofstateofan idealgas isgivenby

where kB iscalled Boltzmann’sconstant:

Boltzmann’sconstantisa universalconstant independentofthemolecular species.Almostallgasesbehaveasidealgasesiftheyaresufﬁcientlydilute. Theconditionforthisisthatthemoleculesspendalargefractionoftheirtime apartfromoneanothersothattheintermolecularforcesareactingonlyasmall fractionofthetimeforagivenmolecule.Thiswillbecomeclearerinthenextfew sectionswheresomeestimatesofintermolecularspacingsanddistancesbetween collisionsarecomparedtothesizesofthemolecules.Typicalintermolecularforces forneutralmoleculesareappreciableonlyoverdistancesoftheorderoftheradius

ofamolecule.Thisistobecontrastedwiththelong-rangeelectricalforcesofan ionicspecies(Coulomb’sLaw)wheretheforcesareoflongrange,varyingasthe inversesquareofdistance.

Example2.1 Agasisatstandardtemperatureandpressure(referredtoasSTP: p = 1atm = 1013hPa, T = 273.16K).Whatisitsnumberdensity?

Answer: n0 = p/(kB T ) = 2.69 × 1025 moleculesm 3 .Thisvalueiscalledthe Loschmidtnumber.

2.1.1Intermolecularspacing

Theapproximateintermoleculardistancecanbefoundbytakingthemoleculesat aninstantoftimetobeuniformlydistributedinspacewithnumberdensity n0 . Placeacubearoundeachmoleculeinthegas.Theneachmoleculesitsatthecenter ofacubeofsidelength d .Thenumberofthesecubesperunitvolumeis n0 .The volumeofoneofthemis d 3 = 1/n0 ;or d ≈ 1/n 1/3 0 = 3.34 × 10 9 m = 3.34nm (atSTP).Notethattheradiusofamolecule r0 isonlyafewtimes10 10 m = 0.1nm (severaltensoftimeslessthantheintermoleculardistance).Inaliquidorasolid intermoleculardistancesareontheorderofthemolecularsizes(seeTable 2.1).

Atomicrefresher TheBohratomhasradius a = h2 0 /π me Q 2 e ,where 0 = 8.85 × 10 12 Fm 1 (permittivityconstant), h = 6.63 × 10 34 Js(Planck’s constant), Qe = 1.60 × 10 19 C(electroncharge), me = 9.11 × 10 31 kg(electron mass).TheBohrradiusforahydrogenatomis a = 5.29 × 10 11 m = 0.0529nm. MosthighschoolorcollegechemistrybooksdescribetheBohrmodelofthe hydrogenatom.

2.1.2Meanfreepath

Theaveragedistanceamoleculetravelsinthegasbeforecollisioniscalledthe mean freepath.Toobtainanestimateofthemeanfreepathimaginethebackground gasparticlestobestationary.Takeourtestmoleculeofradius r0 tobemoving throughthelatticeofﬁxedpointsusedinthelastsubsection.Acollisionbetween ourprototypemoleculeandabackgroundmoleculewilloccurwhentheircenters arewithin2r0 ofeachother(Figure2.2).Wecanthinkofthetestmoleculehaving radius2r0 andthelatticecomposedofstationarypoints(seeFigure 2.3).Hence,as thetestmoleculemovesthroughthelatticeitsweepsoutacylinderofradius2r0 . Intime t thevolumesweptoutis (2r0 )2 π × v t (numberperunitvolumetimes volume).Thecross-sectionalareaofthesweeperissometimesgiventhesymbol σc andcalledthe collisioncross-section (seeTable 2.2).Thenumberofbackground

Figure2.2Atthemomentofcollisiontwosphericalmoleculeshaveadistance betweentheircentersof2r0 .

Figure2.3Asamoleculemovesadistance v t ,itencountersallthepointcenters inthevolumeofthecylinderwhoselengthis v t andwhosecross-sectionalarea is4π r 2 0 .Thenumberofpointcentersinthevolumeis n0 4π r 2 0 v t

moleculecentersinthiscylinderis n0 × σc v t (numberperunitvolume × volumeof thecylinder).Thenthenumberofcollisionsperunittime(thecollisionfrequency)is fcoll = n0 σc v [collisionfrequency].(2.3)

Wemaycalculatetheaveragedistancebetweencollisionstobethe[distanceper unittime] = [velocity] × [thetimebetweencollisions](thislastfactoristheinverse ofcollisionfrequency):

[meanfreepath,approximateform].(2.4)

TheGreekletter λ isusedinmosttextstodenotethemeanfreepath.Actually, itispossibletosolvetheproblemwhenalltheparticlesareinmotion,andthe derivationcanbefoundinbooksonthekinetictheoryofgases.Thesameformula occursforthemeanfreepathexceptforanadditionalfactorof √2 = 1.414 ... in thedenominator:

[meanfreepath,moreexactform].(2.5)

Table2.1 Somecharacteristiclengthsinthe kinetictheoryofgasesatSTP(allinunitsof nanometers, 10 9 m)

BohrIntermolecularMeanWavelengthof radiusspacingfreepathyellowlight 0.0533.352500

Table2.2 Somecollisioncross-sections σc forgasesof interestinnm2

Forinterspeciesestimatesusetheaverage.Datafrom AtkinsanddePaula(2002).

Foratypicalgas r0 = 2 × 10 10 m(0.2nm)andthus, σc = 4π r 2 0 ≈ 5 × 10 19 m2 (0.5nm2 ).Using n0 fromabovewehave λ ≈ 5.23 × 10 8 m(52.3nm),which isaboutanorderofmagnitudemorethantheintermolecularspacingatSTP (Table 2.1).

Example2.2 Takeasamodelfortheverticaldependenceofnumberdensity:

where z isthealtitudeabovesealeveland H iscalledthe scaleheight (typically 8kminmidlatitudes,butupto12kminthetropics).Ourexponentialmodelwas justcookedup,butitturnsouttobeaverygoodapproximation.Whataretypical meanfreepathsat z = 0, H ,2H ,takingSTPat z = 0?

Answer:

Then: λz =0 = 52nm, λz =H = 141nm, λz =2H = 384nm.

Fromthepreviousexampleweseethatthemeanfreepathisstillverysmall comparedtoourfamiliareverydaysizesofthings,especiallyweatherphenomena, evenataltitudesofseveralscaleheights(wellintothestratosphere).Whenthe dimensionsofthebodyofgas(say,astormoracoldairmass)arelargecompared

tothemeanfreepath,wecanignorethemolecularmotionsandtreatthegasasa ﬂuid,obeyingthemacroscopiclawsofﬂuidmechanics.Thisisusuallythecasein atmosphericscienceuptoaltitudesofabout100km.

Themeanfreepathalsogivesusanideaofthelengthscalesoverwhichtransport ofpropertiesoccurs.Ifapropertyistransportedviacollisionsbetweenmoleculesit hastohappenoverlengthscalescomparabletothemeanfreepath.Suchprocesses aresaidtobediffusive,andtheyareratherslowcomparedtosomeﬂuidmotions suchasconvection.

2.1.3Pressurefromkinetictheory

Anintuitivefeelingforthepressureofanidealgascanbegainedbyconsideringa gasenclosedinacubicalboxofedgelength L (seeFigure 2.4).Imagineamolecule goingleftandrightacrosstheboxandbouncingbackatthewalls.Aroundtriptakes atime2L/v0 where v0 isthespeedofthemolecule.Atareflectionitexperiences achangeofmomentum (m0 v0 ) = 2m0 v0 .Eachreflectionimposesanimpulsive forcetothewall(seethephysicsrefresherbelow).Thefrequencyofsuchreflections (bytheentireboxofmolecules)issolargethattheforceiseffectivelysteady(we shallseelaterthattypicalmolecularspeedsarehundredsofms 1 ).Therateof suchimpulsesbyanindividualmoleculeisthechangeinmomentumdividedby thetimeintervalbetweenreflections,2m0 v0 /(2L/v0 ) = m0 v 2 0 /L.Ifwesuppose thatonethirdofthemoleculesaregoingleft-right(theothersaregoingup-down andin-out),thenthenumbergoingleft-rightis n0 L3 /3.Thetotalforceonthe wallis

Figure2.4Amoleculemovinglefttorightwithvelocity v0 iselasticallyreﬂected bythewall.Afterthecollisionthevelocityis v0 .Thesidedimensionofthe cubicalboxis L

Thistotalforceonthewallshouldalsobethepressure × theareaofthewallwhich canbeexpressedas

Equatingthesewehave p = 1 3 n0 m0 v 2 0 [pressurerelatedto v 2

](2.8) wheretheoverbarindicatesstatisticalaveragingovertheprobabilitydistribution ofmolecularspeeds, v0 .Sinceweknowtheequationofstateforanidealgas(from experiments), p = n0 kB T ,wecanusethisalongwiththerelationshipjustfoundto identifythecorrespondingcoefﬁcientsandarriveatthenewrelationship: 1 2 m0 v 2 0 = 3 2 kB T [kineticenergyandKelvintemperature].(2.9)

Notethattheleft-handsideisjusttheaveragekineticenergy( 1 2 m0 v 2 0 )ofamolecule inthebox.Thisrelationshipsaysthatthe temperature expressedinkelvinsis proportionaltotheaveragekineticenergyofindividualmolecules.Thecoefficient ofproportionalityisthusdeterminedtobe 3 2 timestheBoltzmannconstant.Notethat whentheabsoluteorKelvintemperature T iszero,themoleculesareatrest(v 2 0 = 0). Allthermalmotionispresumedzeroatthistemperatureatleastintheidealgasas wehavedefinedit.Actually,thereisnosuchthingasanidealgasat0K(anyreal gaswouldhavebeenliquifiedorsolidifiedwellabove0K).Moreover,quantum mechanicstellsusthatthereismotionevenatthislowestoflowtemperatures. Fortunately,weneverencountertheselowtemperaturesinmeteorologyandthe IdealGasLawvirtuallyalwaysappliestothegasesofinterest.

Whilethederivationaboveishighlysimpliﬁedwithmanydetailsomitted,such asﬂightsthatarenotperpendiculartothewalls,distributionsofthemolecular velocities,collisionsbetweenthemolecules,etc.,thesedetailscanceloutinthe morerigorousderivation.Hence,theformulaanditsinterpretationarecorrect.

Physicsrefresher:impulseforce Whenaparticlereﬂectsfromarigidsurfaceit exertsaforceonthesurface.TheforceonthewallisfoundbyintegratingNewton’s SecondLawoverthetimeofthecollision: F τ = 1 τ t +τ/2 t τ/2 m0 d v (t ) d t d t where τ isthe(veryshort)timeintervalduringwhichthemoleculeisincontactwith thewall.Unfortunately,wedonotknowthevalueof τ ingeneral.However,over

longerperiodsoutsidetheinterval(t τ/2, t + τ/2)theintegrandvanishessothe integrationperiodcanbeextendedfrom t Trt /2to t + Trt /2where Trt isthelength oftimeforthemoleculetomakeitsroundtripbetweencollisionswiththewall.The forceexertedonthewallduringaroundtripofasinglemoleculeisthen

Iftheroundtriptime(Trt = 2L/v0 )isshortenough(comparedtothesluggishwall’s responsetime)thewallwillfeelanearlysteadyforceofthismagnitude.Nowifwe addthecollisionsofallthemolecules,asinthederivationofpressure,wecanbe assuredofasteadyforceperpendiculartothewall.

Thereareafewlooseendsthatmustbeaddressed.First,notallmoleculesare travelingstrictlyinthe x , y and z directions.Thiscanbedisposedofbynotingthatfor thewallperpendiculartothe x directiononlythe x componentofthemotionmatters. The y and z componentsdonotaffectthiswall.Dothecollisionsonebyonecancel their y and z componentsbeforeandafterthecollisionwiththewall?Afterall,the wallisnotasmoothsurfaceatthemolecularlevel.Theansweristhatoverthelong runandaveragingovermanyparticlesthiscancellationiscomplete.Lastly,the moleculesdonottraveluninterruptedfromoneendoftheboxtotheother.Theygo onlyonemeanfreepath(afewtensofnanometersatSTP)beforetheysuffera collisionwithanothermolecule.Thesolutiontothisproblemliesintheconservation ofmomentum.Afteracollisionthe x componentofmomentumisconservedforthe collidingpairanditisthemomentumchangeatthewallthatmatters,whetheritisthe samemoleculeornot.

InspeciﬁcapplicationssuchasmeteorologyitisusefultocasttheIdealGas Lawintoyetanotherformbymultiplyinganddividingbythemassofanindividual molecule, m0 :

where ρ isthemassdensity(ρ = m0 n0 = M/V )inkgm 3 ,andthegasconstant deﬁnedby R is kB /m0 (note:thisis not theuniversalgasconstantwhichistobe deﬁnedlater).Fordryair

Notethatthisdeﬁnitionofthegasconstantdependsonthemassofindividual molecules m0 .Dryairisamixtureofdifferentidealgases.Thevalueof Rd takes thismixtureintoaccountaswillbeexplainedshortly.

Table2.3 Thecompositionofdryair

Table 2.3 givesthecompositionofdryairbyvolumepercentage–thisistheratio ofthenumberdensityofasubstance n0 tothetotalnumberdensity.Thesametable alsogivesthepercentagebymass–theratio(×100)ofthemassoftheconstituent inasampletothemassofthewholesample.

Example2.3 Whatisthedensityofaparcelofdryairat270Katthe500hPalevel?

Answer:Use ρ = p/(Rd T ):

Example2.4 Whatisthe rootmeansquare (rms)speedofamoleculeofairatSTP?

Answer:Wecanwrite

Example2.5 Whatisthecollisionfrequencyof“air”atSTP?(air isinquotes becauseweimaginetheaircomposedofasinglespecieswhosemolecularweight is29.0,i.e.,29timesthemassofaproton).

Answer:Weusethecollisionfrequencyformula: ≈

σc v

× 1025 moleculesm 3 × 5 × 10 19 m2 × 485ms 1 = 6.52 × 109 collisionss 1 .

Example2.6 Whatisthetypicalnumberofcollisionswithawallperpendicularto the x -axisperunitareaperunittime?LettheconditionsbeSTP.

Answer:Thenumberofmoleculesimpingingonthewallfromtheleftis(n0 /4)(Av ) where n0 isthenumberdensity, A istheareaofthewall(1m2 ),and v istheaverage speedofthemoleculesasshowninSection 2.3. n0 = 2.69 × 1025 moleculesm 3

and v canbetakentoberoughlythe vrms ofthepreviousexample.Hence,there areabout1027 collisionspersecondwiththe1m2 wallatSTP.Sowheredidthe divisor6comefrom?Itcomesfromacarefulintegrationoveralltheangles,etc.A simpleminded(butcorrect)wayofobtainingitistonotethat 1 3 ofthemolecules aregoinginthe x direction,butonlyhalfofthesearegoinginthe+x direction. Withorwithoutthe6,thisisaverylargecollisionrate.

Mathrefresher:probabilitydensityfunction(pdf) Thequantity P (u ) d u isthe probabilityofﬁndingthevariable u tolieintherange (u , u + d u ).Theprobabilityof ﬁndingittohave any realvalueisunity;thus

The mean valueof u isdeﬁnedtobe

andits variance or meansquaredvalue isgivenby

Thevariable u iscalleda randomvariable.Intreatingrandomvariablesweconsider independent realizations ofthevariable(likedrawingvaluesfromahat).

2.2Distributionofvelocities

Obviouslythemoleculesinaboxarenotmovingparalleltothe x , y and z directions exclusively.Insteadmoleculeswillhaveinstantaneousvelocitycomponents vx , vy , and vz .Considerthe vx componentforanindividualmoleculeatagiventime. Thevalueof vx willtakeonarangeofvaluesfromonetimetothenextbecause ofcollisionswithothermolecules(itcanbethoughtofasarandomvariable). Computersimulationssuggestthatafteronlyahundredorsocollisionspermolecule theprobabilitydensityofvaluesof vx settlestoasteadyfunctionalform.Afterthis equilibration or thermalizationtime vx isfoundtobedistributedas:

Figure2.5Graphofthenormaldistributionfor σx = 1andfor σx = 2,where

σx = √R∗ T /M .Thelargervalueof σx (highertemperatureorlowermolecular weight)leadstoabroaderdistribution,butstillhasthesameareaunderit.Inthe caseofthe x componentofvelocity,thismeansthevelocityisexpressedinunits of280ms 1 at300K.Notethatthisdiffersfromthermsspeed(485ms 1 ),since weareonlyconsideringoneofthethreecomponentsofvelocity.

whichiscalledthenormaldistribution.Thenormal(sometimescalledtheGaussian) distributionoccursofteninnature.Itgenerallycomesaboutwhenthevariableis subjectedtoalonghistoryofrandomjoltsthatadduptoitscurrentvalue.After alongtime(manyadditiveincrementstothevalueofthevariable)itsprobability distributionapproachesthenormaldistribution.Thiscanbeprovedunderrather generalconditionsinmathematicalstatisticsundertheheadingofthe CentralLimit Theorem.ThenormaldistributionhasthefamiliarbellshapeshowninFigure 2.5 This probabilitydensityfunction (pdf)isnormalizedsuchthat

Theareaunderaportionofthecurvebetweentwovalues vx1 and vx2 isthe probabilityofagivenmoleculehavingits x componentofvelocitylyinginthat range.Obviouslytheprobabilityofitslyingintherange(−∞, ∞)isunity.The most probablevelocity istheoneforwhichthepdfismaximum–itiscalledthe mode ofthedistribution;themodeisthevalueof vx forwhichd P /d vx = 0.Theaverage valueof vx isgivenby

whichvanishesbecause P (vx ) isanevenfunctionontheintegrationintervaland itismultipliedbyanoddfunction, vx .

Themeansquareof vx (alsocalleditsvariance)isgivenby

andthiscanbeshowntobe

Example2.7 Theescapevelocityofamoleculeistheleastverticalvelocity vesc at whichthemoleculecanescapetheEarth’sgravitationalfield.Wecancomputethis velocitybyfindingthevelocitya(collisionless)moleculemighthaveuponfalling frominfinitytotheEarth’ssurface.Theprocedureistoequatethekineticenergyof theparticle 1 2 m0 v 2 esc tothepotentialenergyattheEarth’ssurface GMm0 /R.After cancellingthe m0 oneachsidewefind vesc = √2gR where g = 9.8ms 2 and R ≈ 6400km.Thefinalansweris vesc = 11.2kms 1 ,whichisindependentof mass(molecularspecies).

Therearemanyinterestingpdfformsthatariseinnature.Thesenexttwoexamples occuroften.Morecasescanbefoundinelementarystatisticsbooks.

Example2.8:uniformpdf

Afterperformingtheintegralsweﬁnd:

Example2.9:exponentialdistribution

Thisdistributionisgivenbytheformula

whichhasmeanvalue b andvariance b2

Wehavealreadyestablishedthevarianceforanidealgasfromtherelation (see(2.13))

Notethatthefactorof3seenbeforeisnotpresentbecauseoftheconsiderationof onlythe x componentofvelocityinsteadofallthreecomponents.

Thethreecomponentsofvelocityareactuallystatisticallyindependentofone another,andoneoftherulesofprobabilityisthatunderthesecircumstancesthe jointdistribution ofthethreevariatesisjusttheproductoftheindividualdensities:

whereexp(·) standsfore( ) .Andofcourse,thesquareofthevelocityvectorofa moleculeisgivenbythesumofthesquaresofitscomponents:

orwrittenmorecompactly

Theprobabilitydensityfunctionformolecularvelocities(2.23)iscalledthe Maxwell–Boltzmanndistribution (seeTable 2.4).

Thedistributionofvelocitieshasnodependenceondirection,onlyonspeeds (i.e.,itis isotropic).Wecangotosphericalcoordinatesinthevelocityspaceand replaced vx d vy d vz by v 2 sin θ d θ d φ d v .Sincethereisno θ or φ dependencein theintegrandwecanintegrateoverthemandthedifferentialbecomes4π v 2 d v . Thepdf(theremainingintegrand)becomesafunctionofspeedonly:

andtheintegralsnowrunfrom0to ∞.Thelastformulagivestheprobabilityof ﬁndingthespeedofamoleculeintheinﬁnitesimalinterval (v , v + d v ).Agraphof thisfunctionisshowninFigure 2.6.

2.2Distributionofvelocities

Table2.4 Comparisonofcharacteristicvelocityscalesfora Maxwell–Boltzmanndistribution;thevaluesareforthe“hypothetical”air molecule(M = 29)

Valueat

QuantityMathformFormula300K(ms 1 )

20040060080010001200

Figure2.6Graphofthedistributionofmolecularspeeds4π v 2 P (v ) forair molecules(M = 29)atatemperatureof300K.Thespeed v isexpressedin unitsofms 1 .Recallthattheescapevelocityis11.2kms 1 independentofmass. Thevalueoftheprobabilitydensityfunctionat v = 38σ is10 313 sm 1 ,which mighthelptoexplainwhytheEarthretainsitsatmosphere.

Itisinterestingtocomparetheescapevelocity(11200ms 1 )withthe distributionshowninFigure 2.6.Thevalueofthedistributionissome10 313 sm 1 . Thenumberofthesemoleculestoescapeevenoverthehistoryoftheplanet (4.7×109 years)isexceedinglysmall.Themedianofthedistributionmovesto higherspeedsifthetemperatureisraisedorifthemassofthemoleculesislowered. Forexample,Figure 2.7 showsthecaseofhydrogenatoms(M = 1)at350K,a valuecharacteristicofaltitudes ∼ 120km.Thevalueofthedensitydistribution attheescapevelocityis8 × 10 12 sm 1 ,andafternumericalintegrationofthe

0.00035 (v )

atomic hydrogen at 350 K 0.00005

v (ms–1)

Figure2.7Graphofthedistributionofmolecularspeeds4π v 2 P (v ) foratomic hydrogen(M = 1)atatemperatureof350K,avaluefortheupperatmosphere (∼ 120km).Thespeed v isexpressedinunitsofms 1 .Recallthattheescape velocityis11.2kms 1 independentofmass.Thevalueoftheprobabilitydensity functionat v = 11.2kms 1 is8 × 10 12 sm 1 ,andtheareaundertherestof thecurveis2×10 9 .ThisisavaluelargeenoughthatifHreachestheupper atmosphereitwillbedepletedoverplanetarylifetimes.However,Hiscontinually producedintheupperatmospherebyphotodissociation(seeChapter8)ofwater vapor.

densityfromthe11.2kms 1 toinﬁnity,weﬁndthattheprobabilityofthespeed beinghigherthantheescapevalueis2 × 10 9 .Thisisprobablylargeenoughfor Htoescape,butsmallenoughthatwatermoleculessteadilybeingdisassociated byhard(veryshortwave)solarradiationcanmaintainapresenceatveryhigh altitudes.

Example2.10 BycontrasttheMoonhasasmallerradius(0.24 rE =1737km)and mass(7.349×1022 kg = 0.01229ME )thanEarth.Thismeansthattheacceleration ofgravityis

gMoon = G MMoon /R2 Moon = 1.63ms 2 and vesc = 2gMoon RMoon = 2380ms 1 .

ThemaximumsurfacetemperatureontheMoonisabout400K.Figure 2.8 shows thedistributionofspeedsforthiscase.Theintegralfromtheescapevelocityto inﬁnityis2.73 ×10 7 ,easilylargeenoughfortheMoontoloseitsatmosphere overitslifetime.

Figure2.8ThedistributionofmolecularairmoleculesfortheMoonat400K.The escapevelocityis2380ms 1 .

2.3Fluxofmoleculesstrikingawall

Therearemanyderivationsofelementaryprocessesinkinetictheory.Wepresent onemoreheresincetheresultcomesupoften.Wewanttoknowthenumberof moleculesstrikingawall(perpendiculartothe x -axis)perunittimeandarea.This issimplythenumberdensitytimesthe x componentofvelocityaveragedoverthe velocitydistribution.Weproceedbyﬁnding n0 v x usingtheMaxwell–Boltzmann distributionforthe x component(theotherfactorsforthe y and z components integratetounity).Weconsideronlythepositivecomponentof vx :

with A = (m0 /(2π kB T ))1/2 and σ 2 = kB T /m0 .Theintegralcanbeevaluated togive

⊥

Andﬁnally:

.(2.28)

ﬂux/(⊥ area) = 1 4 n0 v [ﬂuxofmoleculeshittingawall].(2.29)

Ifweapplythistoleaksthroughasmallholeinthewalltheprocessiscalled effusion.Theformulaholdswhentheholeissmallerthanthemeanfreepathofthe moleculessothattheyflowthroughtheholewithoutcollisions;otherwisethegas actslikeafluidwhenpassingthroughtheopeningandonemustusefluidmechanics methodsratherthankinetictheory.

Table2.5 Gasnotation

p pressure(Nm 2 = Pa;100Pa = 1hPa = 1mb)

V volume(m3 )

ρ massdensity(kgm 3 )

α speciﬁcvolume(m3 kg 1 ), α = ρ 1

m0 massofanindividualmolecule(kg);forH, m0 = 1.67 × 10 27 kg

n0 numberdensity(moleculesm 3 )

kB Boltzmann’sconstant:1.381×10 23 JK 1 molecule 1

MG grammolecularweight;forhydrogen, MH =1gmol 1

MG thegrammolecularweightdividedby1000

Md dryaireffectivemolecularweight, Md = 28.97gmol 1

ME effectivegrammolecularweightofamixtureofgases

Mi bulkmassofconstituent i (kg)

NA Avogadro’snumber:6.022×1023 moleculesmol 1

ν numberofmolesofagas

R∗ universalgasconstant:8.3143JK 1 mol 1

Rd gasconstantfordryair:287JK 1 kg 1

R, RG gasconstantforaparticulargas,G(JK 1 kg 1 )

2.4Moles,etc.

The molecularweight, M ,ofapuregasisthesumoftheatomicweightsofthe atomsmakingupthemolecules.Themolecularweighthasdimensionsgramsper mole,denotedgmol 1 (seeTable 2.5).InkeepingwithSIunitsonemightchoose kgkmol 1 ,whichgivesthesamenumericalvalue.Forexample,themolecular weightofisotopicallypure(nodeuterium(2 H)ortritium(3 H)atomsinthegas)H2 is2andthatofCO2 is12 + 16 + 16 = 44.Thechemicalpropertiesoftheelementare determinedbythenumberofprotonsinthenucleus,whichisdesignatedthe atomic number.The atomicweight isdeterminedbythesumofthenumberofprotonsand thenumberofneutrons.Anelementcanhavedifferent isotopes,i.e.,thenumber ofneutronsmightvaryslightlyfromatomtoatom.Butthemostabundantisotope foundinnatureisusuallydominant,withonlyasmallpercentageoftheother isotopespresent.Ifwetakearandomsamplefromnaturethisleadstoaweighted averageoftheatomicweight,andthisisthevalueusedinmostcomputations. Forourpurposes,wecansimplyusethenumbersgiveninTable 2.6 whichtake intoaccountthedistributionsofnaturallyoccurringisotopes.Strictlyspeakingthe standardissetbythemostabundantisotopeofcarbonwhichisdeﬁnedtohavea molecularweightofexactly12.000kgkmol 1 .

Thenumberofmoleculesinagrammoleiscalled Avogadro’snumber

NA = 6.022 × 1023 moleculesmol 1 [Avogadro’snumber].(2.30)

Table2.6 Selectedatomicandmolecularweightsin gramspermole(equivalentnumericallytokgkmol 1 )

Ar39.948N14.0067

O15.9994

Thenumberofmolesofasubstanceisdenotedby ν .Itisthenumberofmolecules N underconsiderationdividedby NA :

Intermsofnumberdensity(n0 = N /V ),thenumberofmolesisgivenas ν = n0 V /NA ,where V isthevolumeoccupiedbythegasinm3 .Themass(ingrams!) ofanindividualmoleculeisrelatedtothesequantitiesby m0

MG /NA ,where MG isknownasthe(gram-)molecularweightofthegasG.(Avogadro’snumberof purehydrogenatomshasamassof1g.)Thegram-molecularweightistheoneused inmostexistingtables;hence,weuseithere.ButinkeepingwithourSIunits,we needtoexpress m0 inkilogramsinmostformulas.Thustheapplicableexpressionis

0 = MG NA [m0 ingrams].(2.32)

However,

TheIdealGasLawcanthenbewritten:

since ρ = n0 m0 .Now n0 = ν NA /V ,thetotalnumberofmoleculesperunitvolume. Alsonotethat m0 (inkilograms)is ˜ MG /NA .Then

iscalledthe universalgasconstant (R∗ = 8.3145JK 1 mol 1 ).

Notetherelationship:

Inotherwords,theBoltzmannconstantisthegasconstantpermoleculewhile R∗ isthegasconstantforamoleofmolecules.Thegasconstantforaspeciﬁcgasis relatedtotheuniversalgasconstantby

Againtheuseof MG insteadof MG abovesimplyconvertstoSIunits;thisis becauseoftheconventionaldeﬁnitionofthemoleintermsofgramsinsteadof kilograms.

Standardconditions (denotedSTP)foragasare0 ◦ Cand1atmofpressure (= 1013hPa).Itisusefultoknowthat1molofanidealgasoccupies22.4liters (1liter = 1000cm3 = 10 3 m3 )or0.0224m 3 atSTP.

Example2.11 Calculatethegasconstantfordryair.

Answer:Use(2.39)with Md = 28.97. Rd = 287Jkg 1 K 1 .

Example2.12 Calculatethegasconstantforwatervapor.

Answer:Sameasabove, Mw = 18.015, Rw = 461.5Jkg 1 K 1 .

Example2.13 CalculatethedensitiesofpuredryairandpurewatervaporatSTP. Answer:Usethegaslaw, ρ = p/RG T :

d (STP) = 1.013 × 105 Pa/(287Jkg 1 K 1 )(273.2K) = 1.292kgm 3 (2.42)

w (STP) = 1.013 × 105 Pa/(461.5Jkg 1 K 1 )(273.2K) = 0.803kgm 3 (2.43)

Watervaporatthesametemperatureandpressureislessdensethandryair.

Example2.14 Avesselcontains1.2kgofdryairatSTP.HowmanymolesofO2 , N2 ,andArarethere?

Answer:1200g = νd Md .Then νd = 1200/28.97 = 41.4molofdryair.Weknow that νO2 = 0.21νd , νN2 = 0.78νd , νAr = 0.0093νd .Inserting,weﬁnd: νO2 = 8.70 mol, νN2 = 32.3mol,and νAr = 0.41mol.

Example2.15 Howhighwouldamoleculewithupwarddirectedspeed485ms 1 gobeforeturningbackintheEarth’sgravityﬁeld?

Answer:Fromelementarymechanicstheheightofsuchaﬂightisgivenby convertingalltheinitialkineticenergyintopotentialenergy:

mv 2 0 = mgh

or h = v 2 0 /(2g ) ≈ 12000m = 12km.Thisisjustlargerthanthescaleheightofthe atmosphere.

Example2.16 Whatistheaverageforceexertedbyamoleculeofmass m0 making elasticreflectionsonthefloorundertheinfluenceofgravity?

Answer:Letthemoleculefallfromaheight h.Thetimeforitsfallis t = 2h/g . Itsspeeduponimpactis v0 = 2gh.Themomentumchangeonareflectionis 2m0 v0 = 2m0 2gh.Theforceexertedontheflooristhen2m0 v0 / t ; t isthe timeforaroundtripupandbackdownbythemolecule.Wefinallyobtain,

Inotherwords,theaverageforceontheﬂoorisjusttheweightofthebouncing molecule.Notethattheresultisindependentoftheinitialdroppingheight h.Isit anywonderthatatmosphericpressuremeasurestheweightofairinthecolumn aboveasquaremeter?

2.5Dalton’sLaw

Dalton’sLawdealswithamixtureofidealgases.Itstatesthatthe partialpressures oftheindividualconstituentgaseouscomponentsareadditive.Baseduponthe kinetictheoryderivationaboveitisnotsurprisingthatthepressureswouldbe additiveformixturesofidealgaseswithdifferentmolecularmasses m1 , m2 , .... Writingtheexpressionforthesumofpartialpressures,

Onceweacceptthisrulewecancomputetheeffectivegasconstantforamixtureof gases.Todosowesubstituteforthepartialpressureofthedifferentconstituents:

Aftermultiplyinganddividingthroughby ρ ,themassdensityofthemixture,

Afternotingthatthevolumesarethesamefor ρi and ρ ,wecanwrite

Notethatthecoefﬁcientsinthelastequationaremassfractions,notmolefractions. Itmightbenecessarytocalculatethe effectivemolecularweight ofamixture(as wehaveused Md = 29gmol 1 fordryair).Wecanstartwith

where M isthemolecularweight(kgmol 1 ).Nowreturnto(2.49)andset

Thisleadsto

Asachecklet Reff = 287Jkg 1 K 1 ,andweﬁnd Meff = 0.029kgmol 1 = 29gmol 1 .

Example2.17 Anargonatmosphere.Whatis RAr ?

Answer: RAr = R∗ / ˜ MAr = 8.31 × 103 JK 1 kmol 1 /39.94kgkmol 1 = 208.2JK 1 kg 1 .

Example2.18 WhatisthedensityofargongasatSTP?

Answer: ρ = p/(RAr T ) = (101325Pa)/(208.2JK 1 kg 1 × 273K) = 1.78kgm 3 .

Example2.19 Supposetheatmosphericdensityisgivenby ρ(z ) = ρ0 e z /H ,where z isaltitudeabovesealevel, ρ0 = 1.2kgm 3 and H isthescaleheight,nominally 10km.Whatisthemassofairabove1m2 atsealevel?

Answer:d m foraslabofthicknessd z is ρ(z )A d z where A isthehorizontalcrosssectionalareaoftheslab.Addingupalltheinﬁnitesimalslabsinthecolumn amountsto M = ∞ 0 Aρ0 e z /H d z = Aρ0 H = 1m2 × 1.2kgm 3 × 104 m = 1.2 × 104 kg.

Notes

Therearemanygoodclassicalthermodynamicsbooks.Onewhichisvery readableandincludesamixofelementarykinetictheoryaswellisSears(1953). AnothermorerecentbutveryreadableaccountisbyHouston(2001).Apurely thermodynamictreatmentisgiveninthephysicstextbyZemansky(1968).Modern physicalchemistrybooksareperhapsbestfordiscussionsofgasthermodynamics, forexampleAtkins(1994).Forareadablebutrigorousdiscussionofconstraints, etc.,seeReiss(1965).

NotationandabbreviationsforChapter2

a Bohrradius

F force(N)

h Planck’sconstant(Js)

h0 initialheight

kB Boltzmann’sconstant(Table2.5)

L lengthofboxedge(m)

λ meanfreepath(m)

me electronmass(kg)

m0 massofasinglemolecule(kg)

Meff effectivemolecularweight(gmol 1 )

MG grammolecularweightofagas(gmol 1 )

M bulkmassofanobject(kg)

n0 (z ) moleculardensityasafunctionofheight(moleculesm 3 )

N totalnumberofmolecules

Nnewtons

NA Avogadro’snumber

p, pG pressure,partialpressureforgasG(Pa)

P (vx , vy , vz ) jointprobabilitydensityfunctionforvelocitycomponents

P (z ) probabilitydensityfunction

r0 effectivemolecularradius(m,nm)

R∗ universalgasconstant(Jmol 1 K 1 )

Rd gasconstantfordryair(Jkg 1 K 1 )

Reff effectivegasconstantforamixtureofgases(Jkg 1 K 1 )

Rw gasconstantforwatervapor(Jkg 1 K 1 )

ρ density(kgm 3 )

t timeinterval(s)

T temperature(K)

σ standarddeviation

σ 2 variance

σc collisioncross-section(m2 )

(u , v , w ) velocitycomponents (vx , vy , vz )

v speed(ms 1 )

v meanspeed(ms 1 )

vesc escapevelocity(ms 1 )

v 2 meansquarevelocity(m2 s 2 )

v 2 x meansquareof x componentofvelocity(m2 s 2 )

V volume(m3 )

z elevation(m)

Problems

2.1Calculatethemassofa1m3 parcelofdryairatSTP.Calculateitsmassatthesame pressurebutat10 ◦ Cand20 ◦ C.

2.2Calculatethemassofa1m3 parcelofwatervaporatSTP.

2.3Whatisthepartialpressureofoxygeninadry1mcubeofairatSTP?

2.4Whatistheweightofthe1mcubeofdryairatSTP?Innewtons,inpounds?(Note: 1kgweighs2.2lbatsealevel.)

2.5Whatisthenumberdensityofavolumeofpureoxygen(O2 )atSTP?

2.6Express Rd intermsofhPainsteadofPa.

2.7Used P (v )/d v = 0toﬁndaformulaforthemostprobablespeedofamoleculeat STP.

2.8Computethe vrms forO2 ,N2 ,Ar,andH2 atSTP.Comparetotheescapevelocity.

2.9Themassofacertainairparcelis1kg,itstemperatureis0 ◦ Canditoccupiesavolume of1m3 .Itisknowntohave5gofwatervaporandtherestisdryair.Whatisthe partialpressureofwatervapor?Whatisthedensityofthismoistair?Comparetothe densityofdryairatthesameoverallpressure.

2.10Acylindricalcolumnofairhasradius1km.Thesurfacepressureis1000hPa.The entirecolumnisrisingataspeedof10cms 1 .Whatisthekineticenergyofthe column?

2.11Thecylinderofthepreviousproblemisrotatingaboutitsaxisofsymmetryatarateof 2π radiansperday(1day = 8.64×104 s).Whatisitsrotationalkineticenergy?(Hint: Themomentofinertiaofacylindricalslabis I = 1 2 mR2 ;kineticenergy = 1 2 I ω 2 where ω isangularvelocityinrads 1 .)

2.12Supposethenumberdensityofmoleculesinacolumnofairisgivenby n0 (z ) = n0 (0)e z /H .Whatisthetotalnumberofmoleculesinacolumnwithunitcrosssectionalarea?Whatarereasonablevaluesfor n0 (0) and H ?UseSTPat z = 0.

2.13Giventheconditionsofthelastproblemandareasonablevaluefor σc whatisthe approximatealtitude zH forwhichthemeanfreepath, λ,isequalto H ?

2.14Giventhatthemoleculesinacolumnofairaredistributedverticallyasinthelasttwo problemsandthatthetemperatureisconstantinthecolumnat T0 ,whatisthetotal gravitationalpotentialenergyinthecolumninJm 2 ?

2.15Supposethecolumnofairinthelastproblemisisothermalatatemperatureof300K. Whatisthetotalkineticenergyofthemoleculesinthecolumn?

2.16Comparethecollisionfrequencyof“air”moleculesatSTPtothefrequencyofyellow light.Notealsoforcomparisonthatthelifetimeofanexcitedstateofanatomisabout 10 8 s.

TheFirstLawofThermodynamics

Ofteninmeteorologywedealwithaﬁxedmassofagaswhosevolumeandother characteristicsmaychangeastheairmassmovesabout.Theparticularmassof gasmaybethoughtofasasmallparcelofmatterthatistransportedthroughthe environmentbynaturalforcesactinguponit.Wecouldalsoimaginemovingit virtuallyviaanabstractthoughtexperiment,forexampletodetermineitsstability undersmallperturbations.Asanairparcelrisesforwhateverreasoninthereal atmosphere,itwillalmostinstantaneouslyadaptitsinternalpressuretotheexternal pressureexertedbythelocalsurroundings,butthetemperatureandcomposition adjustmoreslowly.Inconvection,entrainmentofneighboringairalsospeedsup theprocessofequalizingthetemperaturebetweeninsideandoutsideair.Stillthere isahugeseparationofequalizationtimesbetweenpressureandtemperatureand/or tracegasconcentrations.Thistimescaleseparationhasmadetheparcelconcept ausefulandevenpowerfultoolintheatmosphericsciences.Wewillreturntoit often.

Thermodynamicsisconcernedwiththe stateofasystem (anexampleofwhichis theparcelalludedtoabovenowtreatedasanapproximatethermodynamicsystem) andthechangesthatoccurinitsstatewhencertainprocessesoccur(suchasits beinglifted).Inthecaseofaparcelcomposedofanidealgas,thestateiscompletely describedbythestatevariables p, V , M and T (actuallyinequilibriumonlythree variablesneedtobespecified,sincetheequationofstateintheformoftheIdeal GasLawcanbeusedtocalculatethefourthfromknowledgeoftheotherthree).In practiceinourapplicationsusingparcels,thetotalmass M(M = i Mi where i indicatesdifferentspeciessuchasO2 ,N2 ,etc.)isusuallyalsoheldfixed(unless otherwisespecified),makingonlythreestatevariables.Therearesomeotherstate variablesmoredirectlyrelatedtoenergeticsthatareconvenientforcertainpurposes andmuchofthischapterwillbeconcernedwiththefirsttwoofthem, internalenergy and enthalpy.

Figure3.1Illustrationofanexpansionofthesysteminthe x directionbya distance x .Theforceexertedbythesystemonthemovablewallis pA,where A istheareaofthewall.Theworkdonecanthenbeexpressedas pA x = p V where V isthechangeinvolumeofthesystem.

Themostbasicenergeticquantitytoconsideristhe work performedbythegas onitssurroundingsduringanexpansionofthesystem’s(e.g.,theparcel’s)volume (seeFigure 3.1).Ifthesystem(whoseshapewetaketobethevolumedeﬁnedby thecylinderofFigure 3.1)expandsinthe x directionbyadistance x ,thenthe forceexerted on theenvironmentis F = pA,where A istheareaofthemovable wallinthecylinderand p isthepressurethesystemexertsonthewall.Theamount ofworkdonebytheparcel on theenvironmentinthisinﬁnitesimalprocessis 1

workdonebythesystem = F x = pA x = p V

where V istheinfinitesimalchangeinvolumeassociatedwiththeexpansionin the x direction.Theexpansionneednotbesolelyalongthe x -axisasshowninthe figure,butcanbeadistortioninanyoralldirections.Theformula W = p V still holdsforinfinitesimal V

Ifthepressureisnotconstantduringtheexpansion,wemustsumthe contributionsfromasequenceofinﬁnitesimalexpansions

VA →VB =

[expansionworkdonebythesystem].(3.1)

Onacoordinateplanewithpressureandvolumeasordinateandabscissa,thework performedistheareaunderthecurveof p versus V (Figure 3.2).Thismakesit clearthattheamountofworkdonebythesystemonitsenvironmentdependson thepathtaken p = p(V ),whichdeﬁnesacurveintheplane.Onemightimagine acyclicprocessinwhichthesystemexpandsfrom VA to VB alongonepathand returnsalonganotherpath.Theareabetweenthetwocurvesrepresentsthenetwork

1 Inthistextweusethesignconventionthatpositive W meansworkdone by thesystemonitsenvironment. Sometextbooksusetheoppositesignforworkdonebythesystem.

Area = Work

Figure3.2Graphofpressure p versusvolume V foranexpansionofagaseous systemfrom VA to VB alongaspeciﬁedcurve p(V ).Theareaunderthecurveis theworkperformedbythesystemontheenvironment.

Work

Figure3.3Graphofpressure p versusvolume V foranexpansionofagaseous systemfrom VA to VB alongaconstantpressurepath(isobaricpath).Thearea underthecurveistheworkperformedbythesystemontheenvironment.

donebytheparcelonitssurroundingsduringthecyclicprocess.Whentheparcel’s volumedecreases,itdoesnegativework.

Example3.1 Considerasystemcomposedof2kgofdryairat0 ◦ C.Letthesystem expandisobaricallyfromitsinitialvolumeof1m3 to3m3 .Howmuchworkis donebythesystem?(SeeFigure 3.3.)

Answer:Theamountofworkdonebythegaseoussystemonitsenvironmentis givenbytheformula

where VA = 1m 3 and VB = 3m 3 , pA = ρ Rd TA .Thedensityisgivenby2kg/1m3 , TA = 273K, Rd = 287Jkg 1 K 1 ;hence pA = 1.57 × 105 Pa.Finally, W = 3.13× 105 J = 313kJ(kilojoules).

Example3.2 Howmuchdoesthetemperaturechangeduringtheaboveprocess?

Answer: TB = pA /(ρB Rd ).Wecancompute ρB = mass/VB = 2kg /3m 3 = 0.667 kgm 3 .Then TB = 820K.Obviouslyanisobaricprocessleadingtoatriplingof volumeisveryunlikelyintheatmosphere.

3.1Reversibleandirreversiblework

Intheprecedingweassumedthattheworkdonebythesystemwasalonga well-definedpath p(V ).Actuallythisisaratherstrongassumption–thatat eachinfinitesimaladjustmentthecurve p(V ) exists.Weareimplicitlyassuming thatweareinastateofthermodynamicequilibriumateachstep–inother wordsthesystemhastimetocometoequilibrium(i.e.,uniformtemperature throughout,etc.)beforethenextinfinitesimalstep.Inrealprocessessuchas thecompressionofapistoninaninternalcombustionengine,thegasinthe chambermightbehighlynonuniformandlocallydisturbedbysuchthingsas shockwavesduringthenextchangeinvolume(perhapstheequationofstatedoes notevenholdduringthisintervaloftime).Foran irreversiblechange suchas intheinternalcombustionengine,anamountofworkwillbedone,butitmay notbecalculableusing p d V .Inmoreadvancedbooksonthermodynamicsitis shownthatwhenthesystemdoeswork(forexamplebyexpansion) p d V isthe maximumwork thatcanbedone.Butwhenthesystemiscompressed,thereversible (calculable)work( p d V )istheminimalworkdonebythesystemduringthe compression.Inthehighcompressionenginetheamountofworkdoneisseldom morethan75%oftheestimatebasedonthereversibleassumption.Theunfortunate mechanicalengineersimplycannotwininthefaceofirreversibleprocesses. Luckily,mostnaturalprocessesofinteresttotheatmosphericscientistarebetter behaved.

Theidealizationof reversiblework allowsustodocalculationsusing p d V even thoughinrealityitneverquiteworksthatway.Inmostapplicationsthatfollowin thisbooktheassumptionofreversibleworkisadequate.

3.2Heatingasystem

Intheexpansionprocessundergonebyaparcel(ﬁxedmass)inmoving from(VA , pA )to(VB , pB ),theparcelwilldosomeworkontheenvironment, VB VA p(V ) d V .Duringthisprocesssomeenergymightbetransportedintothe systembecauseofatemperaturedifferencebetweentheinteriorofthesystemand thesurroundingenvironment.Thisenergytransportedintothesystemis thermal energy 2 asdescribedinChapter2.Thermalenergyconsistsofallthemodesof energyassociatedwithindividualmolecules:translationalkineticenergy,rotational energyforpolyatomicmoleculesandvibrationalenergy(potentialandkinetic) forpolyatomicmoleculesthatexperienceinternalstretchingoscillations.These individualenergytermseachcontributetothethermalenergyofamolecule(but onlythetranslationalkineticenergycontributestopressure).Infact,theenergyon theaverageissharedequallybetweenthedifferentmodes(translationalkinetic, etc.),althoughatatmospherictemperaturesthevibrationalmodesarenotexcited becauseofquantumthresholdeffects.3

Thistransportofheatiseffectedatthemolecularlevelbythecollisionsof individualmolecules.Ifthereisagradientoftemperature,moleculesfromthe warmerregionwillpenetrateadistanceoftheorderofameanfreepathbefore sufferingacollisionintothecoolerregion(andviceversa),causingthecoolerregion towarm;moleculesmovingtheotherwaycausethewarmerregiontocoolthrough individualcollisions.Therandommotionsofmoleculescrossingtheboundary bringthenewsoftheirdifferent“temperature”viaa randomwalk process(each stepforwardorbackwarddeterminedbytheproverbialflipofacoin).Thenews andconversionarebroughtaboutslowlybutsurely.Thedistanceadvancedby thespreadingedgeofa“warmfront”atthemolecularlevelisproportionalto thesquarerootofthetimeelapsed.Thisisinstarkcontrasttothepropagationof pressuredifferenceswhichmoveviaasound-likewave(distanceofadvanceofthe pressurefrontbeingproportionaltotime).Toobtainanideaofthiscontrastconsider aone-dimensionalgas(x directiononly)andletaninstantaneoushotspotdevelop at x = 0(perhapsafirecrackerexplodes).Itispossibletosolvethisproblem analytically,butthedetailsneednotbegivenhere.Thebasicideaisthatheatflux

2 Inmosttextsthisthermalmotionisreferredtoas heat asthoughitwereamaterialsubstancemovingaround inspace,butsomeauthors(e.g.,BohrenandAlbrecht 1998)shuntheuseofthenoun heat infavoroftheverbs heating or cooling asatransportprocessinvolvingtheenergizingofneighboringmoleculesbytheiraggregate beingincontactwithanaggregateofmoleculesofadifferenttemperature.Wewillusetheterm heat tomean theintegralovertheheatingratewithrespecttotime.Justkeepinmindthatheatisnotaﬂuidﬂowingaboutin themedium.

3 Theenergylevelsinquantummechanicsarediscreteandthedisturbingcollisionsneedtohaveasufﬁcient energytransfertoeffectatransitiontothenexthigherenergylevel.Typically,rotationallevelsarecloselyenough spacedforthemtobeexcited,butvibrationalthresholdsaremuchhigher,requiringveryhightemperaturesfor excitation.

Figure3.4Spreadinmetersofalocalizedpulseofthermalenergydue(only) tomolecularthermalconductivityinairatSTPafter2handafter12h.The functionalformisthenormaldistributionwithstandarddeviation σ = √2Dt , where D = κH /ρ cp ≈ 2 × 10 5 m 2 s 1 and t istimeinseconds.

isproportionalto κH d T /d x ,where κH isthe thermalconductivity ofair(≈0.024 Jm 1 s 1 K 1 ).Thepulsespreadsoutintheshapeofanormaldistributionasshown inFigure 3.4.Thestandarddeviationofthespreadoftheelevatedtemperaturesis onlyabout3mafter12hours.Thismeansthattheconceptofparcelintegrityfor objectsoftheorderofseveralhundredmetersissafefordaysiftheonlystirring mechanismismoleculardiffusion.Thereareothermechanismsthatcanshorten thetimeofmixingdependingontheconditions,butthesearestillusuallyslow comparedtotheadjustmentoftheinteriortotheexteriorpressure.Notethatsound wavestravelatseveralhundredmeterspersecond.Theadjustmentofpressures shouldbeaccomplishedinseveralhundredpassesofsoundwavesbackandforth acrosstheparcel–stillveryfastcomparedtomolecularandeveneddy(turbulent) transportprocesses.Thesoundwavesareeventuallydissipatedintothermalenergy.

Thermalenergyorheataswehavebeendiscussingitcannowbecontrastedwith the work beingdonebyasystemduringaprocess.Thethermalenergyisatthe molecularlevelanditmigratesfromplacetoplaceviagradientsinthetemperature (sayfromthesystemtoareservoir),whileworkisatthetrulymacroscopic level.Workisperformedwhenoneofthemacroscopicdimensions(sometimes called conﬁgurationcoordinates),saythepositionofapiston,isalteredaﬁnite (macroscopic)amount.

Returningtooursystem,heatcanbetransportedintoitbecauseofsmall differencesintemperaturebetweenthesystemanditsenvironment.Theamountof heattakenintothesystemduringaﬁniteprocessistraditionallygiventhesymbol Q . Wesay 4 QA →B asthesystemmovesfromthestatedenotedAtothestatedenotedB.

4 Notethatinournotationifapositiveamountofheatis absorbed bythesystemthen QA →B isassigneda positivevalue.Thisisthesignconventionfollowedbyvirtuallyalltextbooks.

Considerfirstthesimpleheatingofaparcelwherethevolumeisheldfixed(an isochoric process).Theparcelisheatedbyanamount QA →B anditstemperature undergoesachange T = TB TA .Theheatenergyabsorbedandthetemperature changearerelatedwiththecoefficientofproportionalitybeingthemasstimesthe specificheatcapacitycv (unitsJkg 1 K 1 ); Cv = Mcv ,where Cv (unitsJK 1 )is calledthetotalheatcapacity:

[heatingatconstantvolume].(3.3)

Ingeneral, cv mightbeafunctionoftemperature,butforanidealgasitisnot. Fordryairthevalueof cv is717Jkg 1 K 1 .Inthisprocessnoworkisdoneby theparcelonitsenvironment,sincethevolumeoftheparceldoesnotchange. Alltheheatgiventothesystemgoesintoits internalenergy.Fromthemolecular relation 1 2 mv 2 = 3 2 kB T werecallthattheaveragekineticenergyofthemolecules isproportionaltotheKelvintemperature.Hence,achangeininternalenergyis equivalenttoachangeinthekineticenergy(foranidealmonatomicgas).Aswewill seelaterthekineticenergyoftranslationstillhasthesamerelationtotemperature formulti-atomicgases,buttheinternalenergyinthemulti-atomiccaseismodiﬁed (nextsection).

Inchemicalapplications(alsochemicaltextsandhandbooks)itiscommonto usethemolarspeciﬁcheatforasubstance, c v .Inthiscasethetotalheatcapacity is Cv = ν c v where ν isthenumberofmolesinthesystem.Inthisformulation:

Example3.3 Asealedroomwithwallsmadeofperfectlyinsulatingmaterialhas dimensions4m×4m ×3m.Theconditionsare p = 1000hPa, T = 300K.Whatis themassofairintheroom?Howmanyjoulesarerequiredtoraisethetemperature by1K?

Answer: M = pV /Rd T = (105 Pa × 48m 3 )/(287Jkg 1 K 1 × 300K ) = 55.7kg.

Q = cv M T = 717Jkg 1 K 1 × 55.7kg × 1K = 4.0×105 J = 400kJ.

Example3.4 Howmanykilowatthoursofenergyareexpendedinthelastexample?

Answer:1kWh = 3600kJ.Sotheresultis0.11kWh.Notethatthecostof1kWh isafewcents(US).

Example3.5 Inthelastexample,considertheeffectofthinwalls.Letustakethe wallstobewoodand1cmthick.Whatistheamountofheatnecessarytobring thesewalls(andﬂoor)up1◦ C?

Answer:Thespeciﬁcheatofwoodis1760Jkg 1 K 1 ,andthedensityofwoodis 600kgm 3 .Thewalls,ceilingandﬂoorhaveanareaof72m2 makingavolume of0.72m3 .Themassofthismaterialis600kgm 3 × 0.72m3 = 432kg.Thetotal heatcapacityofthesolidmatteris760kJ,nearlytwicethatoftheaircontained.

QA →B = Mcv T

3.2.1InternalenergyandtheFirstLaw

Whenasystemundergoesatransitionfromonestatetoanother,energypasses fromthesystemtoitsenvironmentorviceversa,sinceenergyinallitsformsis conserved(eveniftheprocessisirreversible).Thechangeininternalenergy, U , forasystemisdeﬁnedtobe

Thedifferentialformis

wherethebarcrossingthedofd Q andd W istoremindusthatbothofthese differentialsdependonthepath(itisnotaperfectdifferentialasinmultivariable calculus,whereasdU is).InthelastexpressiondU istheinﬁnitesimalchangein internalenergy.Thelastequationisastatementoftheconservationofenergy.The FirstLawofThermodynamicsactuallygoesmuchfurtherandstatesthattheinternal energy U isafunctiononlyofthestateofthesystem.Fortheidealmonatomic gasthisisobviousfromoursimplekinetictheorymodelsincetheinternalenergy isthetotalkineticenergysummedoverallthemoleculesinthesystemandthisis proportionaltotheKelvintemperature.Foragivenmass,beingafunctionofthe statemeansitsvalueisuniquelydeterminedateachpointina V p diagram.Note thatneither Q nor W arefunctionsofstate.

3.3Idealgasresults

3.3.1Internalenergyofanidealgas

Inadditiontotheidealgasequationofstateanotherpropertyisnecessarytodeﬁne anidealgas.Wemustspecifyitsinternalenergyasafunctionofthethermodynamic coordinates.Thiscanbeaccomplishedbylaboratoryexperimentstoyield:

where N isthenumberofmoleculesinthesystem.Therearemanyalternative formsofthisrelationbecauseofthedifferentwayswecandescribeanidealgas:

where kB isBoltzmann’sconstant(Table1.1), M isthemassofgasinthesystem, and f isaconstantequalto3foranidealmonatomicgas(e.g.,Ar),and f = 5 formostdiatomicgasesatroomtemperature(e.g.,air).Theinternalenergycan bedeterminedbyaseriesofexperimentsinvolvingadiabaticprocessesinwhich

U =−WA →B iseasilymeasured.Theconstant f dependsupontheinternal structureofthemolecules;itisknownasthe numberofdegreesoffreedom inthe molecule.Foradiatomicmoleculethatisverystiff(doesnotstretchandcontract underthetemperatureconditionsbeingconsidered)suchasO2 andN2 nearroom temperature,thevalueof f is5becausetherearetwomoredegreesoffreedomdue totheabilityofthemoleculetorotateinatwo-dimensionalplane,butnotaboutthe axisjoiningthetwoatomicconstituents(itsmomentofinertiaistoosmallabout theaxisjoiningthetwoatoms).Athightemperatures(notnaturallyoccurringinthe loweratmosphere)thenumberofdegreesoffreedomgoesupbytwobecauseof molecularvibrationduetothespring-likebinding.Hence,fornormalatmospheric conditionssuchasencounteredinthetroposphereandstratosphere:

f = 3idealmonatomicgas(e.g.,argon)(3.10)

f = 5N2 ,O2 nearSTP(3.11)

f = 6CO2 ,H2 Oandothernonlinearmolecules.(3.12)

Itisnotdifﬁculttoseewhythespeciﬁcheatshouldbelargerformoleculeswith larger f .Considertheconstantvolumeheatingcase.Ifheatingoccursinabox ofmonotonicgas,alltheenergymustgointoincreasingthelinear(translational) kineticenergyofthemolecules.Ifthemoleculeisspatiallyextendedsuchasa diatomicmolecule,itcanrotateaswellastranslate.Theaddedenergycangointo rotationalenergyaswellastranslationalenergy.Hence,theheatcapacity(amount ofheatnecessarytoraisethesystem’stemperatureby1K)willbelarger.Basically theheatenergy(thatatthemolecularlevel)mustbesharedamongallthedegrees offreedom,butonlythelinearkineticenergygoesintocausingpressuresinceit carriesmomentumtothewalls(oracrossboundaries).

Aremarkabletheoremprovedintheclassicalstudyofstatisticalmechanicsshows thatinequilibriumtheenergywillbesharedequallybetweeneachoftherotational andtranslationalmodes(andvibrationalmodeswhenapplicable): theprinciple ofequipartitionofenergy.Inthecaseofadiatomicmoleculeonlytworotational modesareavailable,therotationabouttheaxisjoiningthetwoatomsdoesnot count.Inthecaseofatriatomicmoleculeinwhichtheatomsarenotinastraight line(e.g.,CO2 ),allthreerotationalmodesareinvolved: f = 6.Moleculesactually canvibrate(stretchingandcontractinglikemassesjoinedbyaspring)aswell,and atsufﬁcientlyhightemperaturesthesemodescanenterandraise f evenmore,but asremarkedabove,thisvibrationaldegreeoffreedomisnotimportantinmost

applicationsofthermodynamicstotheatmosphere.Ontheotherhand,suchmodes ofvibrationandrotationplayanimportantroleintheabsorptionandemissionof infraredradiationasitpassesthroughair.Insummary,eachmoleculeontheaverage possesses 1 2 kB T foreachofitsmechanicaldegreesoffreedom.For1molofsuch moleculesat300Kthisis 1 2 R∗ T = 2.5kJforeachdegreeoffreedom.Hencefor argonitis7.5kJmol 1 andforO2 andN2 itis12.5kJmol 1 .

Example3.6 Findtheinternalenergyofa1kgmassofdryairatSTP.

Answer:Wecanuse U = ( f /2)ρ Rd TV = ( f /2)MRd T0 ,where M isthemassof thegasinthesystem(here1kg)and f is5.Then U = 1.96 × 105 J = 196kJ.

Example3.7 Comparetheriseinpotentialenergyduetoliftingthe1kgparcelto 9km,approximatelyonescaleheight.

Answer:Thechangeinpotentialenergyis Mgh = 1.08 × 105 J = 108kJ.Hence thegravitationalpotentialchangeiscomparabletotheinternalenergyforaliftof onescaleheight.

3.3.2Heatcapacities

Ifasystemcomposedofanidealgasabsorbsheatataconstantvolume,its temperaturewillincrease.Sincethevolumeisheldconstant,thesystemcando noworkonitsenvironmentintheprocess,therefore

Differentiatingtheequationfortheinternalenergyofanidealgas(3.9)weobtain

Notethat R asusedinthisequationisforaparticulargassuchasdryair.The heatcapacityatconstantvolume, cv ,isproportionaltothenumberofdegreesof freedom.

Anotherimportantprocessistheheatingofthegasatconstantpressure.Inthis case

Wealsohave

Thisleadsto

If U isa functionofstate astheFirstLawclaims,5 thenitisgivenby Mcv T , andwehaveanidentity:

whichisaveryimportantrelationforidealgases,holdingindependentlyofthe valueof f .Thislasttellsusthatforanidealgas

That cp isalwaysgreaterthan cv hasaneasyinterpretation:someoftheheat absorbedintheisobariccaseis“wasted”bytheexpansion(workdonebythe systemontheenvironment)ratherthanbeingdevotedtoraisingthetemperature.

Example3.8 Dryair:whatare cv , cp ,usingidealgasrules?

Answer:Wehave cv = 5 2 Rd = 717.5Jkg 1 K 1 ,and cp = ( f /2 + 1)Rd = cv + Rd = (717 + 287) = 1004Jkg 1 K 1 .

Example3.9 Howmuchheatisrequiredtoraisethetemperatureofa1kg parcelofairatconstantpressure(constantaltitudeintheatmosphere)by1◦ C?

Answer: 1004J.

Example3.10 Amassof2kgofdryairisheatedisobaricallyfromatemperature of300Kto310K.Howmuchheatisrequired?

Answer:

5 Asubtlethingisabouttohappenhere:eventhoughtheprocessisoccurringatconstantpressurefromone temperaturetoanother,wecanusetheformula U = Mcv T .Thisisbecause U isafunctionofstate.No matterhowwegofrom TA to TB ,thechangein U isthesame.Hereweseeoneofthemostpowerfultools inthermodynamics,theinvokingofastatefunction’sbeingafunctiononlyofitsstate(notpath).Remember, however,thatformanysubstancesotherthantheidealgas U mightdependonmorethanjusttemperature; nevertheless,itisonlyafunctionofstate.

Example3.11 ForExample3.10,whatisthechangeininternalenergy?

Answer:

Example3.12 ForExample3.10,whatistheamountofworkdone?

Answer:UsetheFirstLaw:

Example3.13 Acolumnofdryair1kmthick(approximatethicknessofthe atmosphericboundarylayer)andunitcross-section(1m2 )isheatedbysunlight atarated Q /d t = 250Js 1 .Iftheheatingtakesplaceatconstantpressure ( p = 1000hPa, T = 300K),byhowmanydegreesperdayisthecolumnheated? (Assumetheairaboveandthegroundbelowareinsulatedfromthewell-mixedair intheboundarylayer.)

Answer: M = 1.16 × 103 kg,

whered Q /d t istherateofheating(Js 1 ).Thend T /d t = 2.14 × 10 4 Ks 1 = 18.5Kday 1 .Thisofcourseisalargerateofincreasefornormal conditions.Theassumptionof250Wistheproblem.Thenextexampleshowsa morerealisticcaseinwhichtheheatingisaccomplishedbyblackcarbonparticles intheair.Muchoftheheatingintheatmosphericboundarylayercomesfromthe absorptionofinfraredradiationaswellassolarradiationbywatervapor.

Example3.14:blackcarbonaerosolinair Supposethereare100blackcarbon particlespercubiccentimeterintheair(10 8 particlesm 3 ).Letustaketheradius ofoneoftheseparticlestobe1 µm.Thismeansthecross-sectionalareaofan individualparticleis3.14 × 10 12 m2 .Thetotalareaofinterceptingcarbonin a1m3 blockofairis3 × 10 4 m2 .(Notethatthisisonlyatinyfractionof the1m2 cross-sectionalareaofthecubeofair.)Ifthesunisstraightoverhead itsﬂuxis1370Wm 2 .Theheatingrateofthiscubicmeterofaerosol-loaded airisthen0.43W.Iftheairisatsealevelitsdensityisabout1.2kgm 3 ,and

cp = 1004Jkg 1 K 1 .Aftersomearithmeticweﬁndthattheairwillexperience anincreaseoftemperatureof1.29Kh 1 .

Calculusrefresher:thenaturallogarithm Thenaturallogof x isdenotedby y = ln x (Figure 3.5)andisdeﬁnedby ey = x

Wecandeduceafewvalues,forexample,for x = 1, y = 0,as x →∞, y →∞ and y isonlydeﬁnedfor x positive.As x → 0, y →−∞.Moreover, d x d y = ey = x .

Wecanturnthisintod y = d x /x ,andthenintegrate: y 0 d y = x 1 d x x whichleadstoanalternativedeﬁnitionofln x :

Nowafewproperties.Firstthewonderfulidentity: x = eln x whichfollowsfromthe deﬁnition.Thiscanbeusedtoderiveanumberofusefulthings: ab = eln a eln b = eln a +ln b = eln ab ⇒ ln ab = ln a + ln b. x a = eln x a = (eln x )a = ea ln x ⇒ ln x a = a ln x .Finally,wealreadyfoundthat

Figure3.5Thenaturallogarithmln x asafunctionof x

3.3.3Adiabaticprocessesandpotentialtemperature

Manyprocessesinmeteorologyinvolveparcelsofairmovinginsuchawaythat noheatisexchangedwiththeenvironmentduringpassagefromonelocationto another,i.e., Q = 0.Theparceleffectivelyissurroundedbyathermallyinsulating blanket.Akeypointhereisthatittakesalongtimefortemperaturedifferences todiffuseintoaparcelfromoutsidecomparedtotherelativelyshorttimeforthe pressuresinsideandoutsidetoequalize.Thismeansthatinmanysituationswe canregardtheprocessasbeing adiabatic,thatis,isolatedfromdiathermalcontact. Inthatcaseforaninﬁnitesimaldisplacement

Thenextstepsbecomesmootherifweusetheidentity cv = cp R,leadingto:

NextweusetheIdealGasLaw

andtakenaturallogstoobtain

Takingthedifferentialsweﬁnd:

Thetermd (ln MR) disappearsbecauseitisconstant.Nowwecansubstituteinour originaladiabaticequationtoeliminatethe V dependence:

andﬁnally,

Afterdividingthroughby cp anddeﬁning

wecanintegratefrom(

ortakingtheanti-log:

Inatmosphericsciencetheformula(3.36)isveryimportant;oneusuallyseesitin theform

where θ (θ = T0 at p = 1000hPa)iscalledthe potentialtemperature.Weoften seeitinthefollowingform:

Thelastequationiscalled Poisson’sequation.Itgivesthetemperaturethat aparcelofdryairwouldhaveifitwerebroughtadiabaticallytoapressure of1000hPa.Nomatterwheretheparcelliesasapieceoftheenvironment,its potentialtemperatureiswelldefined.Iftheparcelmovesadiabaticallyitspotential temperaturewillnotchange.Whenwefindaquantitydescribingthesystem(defined hereastheparcel)whichdoesnotchangeastheparcelmovesabout(inthiscase adiabatically),werefertoitasa conservativeproperty.Inpracticeparcelsdo moveadiabaticallyinconvectivemotionstoagoodapproximation.Moreover,

6 Takingtheanti-logmeansraisingeachsidetobethepowerofe: y = f (x ) ⇒ e y = e f (x ) .Nowusing x = eln x meanstheanti-logofln x isjust x

Table3.1 Importantformulasforidealgases inadiabaticprocesses

VariablesFormula

175200225250275300325

Figure3.6Dryadiabatforaparcelofairwhosepotentialtemperatureis300K. Theabscissais T inK,andtheordinateis ln (p/p0 ) = z /Ha .Thevalue z /Ha is theheightabovesealevelinunitsofthescaleheight, Ha ,whichistypicallyabout 8kminmidlatitudes.

parcelsmovinghorizontallymovealongconstant θ surfaces(isentropicmotion) (seeTable 3.1).

Tip Itmightbedifﬁculttoremembertheformof(3.38)(Poisson’sequation),sinceit willappearinavarietyofforms:isthe κ withanegativesign?Is p0 ontopornot? etc.Justremember:letaparcelrise,indoingso θ staysconstant,sothatas T goes down,somust p.Thiscanhelpustoobtainthecorrectformula.

Figure 3.6 showshowthetemperatureofaparcelisloweredifitisliftedin anatmospherewhosepressurefollows p(z ) = p0 e z /Ha ,whichisareasonable approximationtotheactualbehaviorofthepressure.Here Ha iscalledthescale heightoftheatmosphere.Typicallyinmidlatitudes, Ha ≈ 8km.

Example3.15 Airinajetplanenearthetropopauseistakenintotheplaneand compressedadiabaticallyfromanoutsidepressureof300hPato1000hPa.The temperatureoutsideis255K.Whatmustbedonetotheairtobringittoaninside temperatureof300K?

Answer:InadiabaticallycompressingthegastheKelvintemperatureisraisedbya factorof1.411to360K.Theairmustnowbecooledatconstantpressureto300K bytheairconditioningsystem.Recallingthat3600J=1Wh,thisrequires60kJ = 0.0167kWhforeachkilogramofairbroughtintotheplanefromtheoutside.

Example3.16 A1kgparcelofdryairislocatedatthe500hPalevelinthe atmosphere.Itstemperatureis246K.Whatisitspotentialtemperature?

Answer: θ = (1/2) 0.286 T .Weﬁnd θ = 300K.

Example3.17 A1kgmassofdryairislocatedat500hPa.Itspotentialtemperature is θ = 300K(asabove).1kJofheatisabsorbedbytheparcelatconstantpressure. Whatisthechangeintheparcel’spotentialtemperature?

Answer:First,computethechangeintemperature, T = Q /(Mcp ) = 1kJ/ (1.004kJkg 1 K 1 × 1kg) = 0.996K.Next,usePoisson’sequationat500hPa:

Example3.18 A50kgparcelofdryairhastemperature300Katthesurface (1000hPa).Itisliftedadiabaticallytothe700hPalevel.Whatisitspotential temperature?

Answer: θ = 300K.Whatisitstemperatureatthe700hPalevel? T = θ p 1000mb κ = 300K × (0.7)(0.286) = 270.9K.(3.40)

Example3.19 Airat300Kisforcedupamountainslopeadiabaticallythrougha verticalheightof2km.Supposethepressureisgivenby p(z ) = p0 e z /H , H = 10km.Howmuchisthetemperaturechanged?Bywhatratioisthevolumeofsuch aparcelchanged?

Answer: p(2km) = p0 e 2/10 = 0.819 p0 T /300K = (0.819)0.286 T = 283K. VB /VA = (TB pA )/(TA pB ) = 1.15.

Example3.20:otherformsoftheadiabaticcurve Returningtotherelation(3.27), use R = cp cv ,thendividethroughby cv .Weobtain

where γ = cp /cv iscalledtheratioofspeciﬁcheats.Foranidealdiatomicgas(and airactslikeone) γ = 1.400.Wecanintegrateasinthetextandobtain

andﬁnally

Anotherformusefulinphysicsandengineeringcanbederived:

3.4Enthalpy

Inmanymeteorologicalandchemicalapplicationstheinternalenergyisnot themostidealstatefunctionfordescribingenergeticchangesduringtransitions. TheclassicalformoftheFirstLawisespeciallyusefulfortransitionsinwhich thevolumeisheldﬁxed(d V = 0)sincethevolume-worktermvanishes,but inatmosphericapplicationsmostchangesinthestateofaparceloccureither adiabaticallyorisobarically.Hence,itbecomesconvenienttointroduceanew functionofstatecalledthe enthalpy, H ,deﬁnedby

Takethedifferentialtoobtain

AftersubstitutingtheearlierformoftheFirstLawintermsoftheinternalenergy weobtain

Veryoftenatmosphericprocessestakeplaceataﬁxedpressure(altitude).These includeheatingofaparcelbysolarradiationataparticularaltitude,condensation heating,andcontactheatingatthesurface.Inthiscasetheenthalpyisavery convenientfunctiontodescribetheparcel’sthermodynamicstate.Notethatthe changeinenthalpyunderaconstantpressureprocessisjust

Integratingthelastequation:

where F (V ) isanarbitraryfunctionofvolumeappearinghereasanintegration constant.Wecanﬁndthisfunctionbynotingthat U = Mcv T andusingthe deﬁnitionof H ,(3.45),toobtain

H = Mcp T [enthalpyforanidealgas].(3.51)

Inotherwords,thearbitraryfunction F (V ) isidenticallyzerofortheidealgas. Theadiabaticprocessisexpressedas

andfortheidealgas,

and

whichwillquicklyleadustoPoisson’sequation(3.36).

Calculusrefresher:partialderivatives Thermodynamicfunctionsnearlyalways involvemorethanonevariableaswehaveseenalready,e.g., V (T , p).The“partial” of V (T , p) withrespectto p holding T constantisdeﬁnedby

Inmostﬁeldsthesubscript T followingthelargeparenthesesisomitted,butin thermodynamicsitisconventional(anduseful)toretainthisreminderofwhich variableisbeingheldconstant.Sometimesespeciallyinmathematicsandphysics,the partialderivativeisdenotedbyasubscript.Forexample,let f beafunctionof x and y ,then ∂ f /∂ x = fx ,etc.Yousimplytaketheordinaryderivativebutholdallvariables constantexcepttheonebeingvaried.Forexample,taketheidealgas: V = MRT /p, then (∂ V /∂ p)T =−MRT /p2 .

Thisisagoodtimetosearchoutyouroldcalculusbookandreviewthechapteron partialdifferentiation.Animportantresulttorememberisthatifwegotosecond partialderivativessuchas fxx or fxy ,the order doesnotmatter:

The differential of f (x , y ) is

Ifwedividethroughbyd x andsetd y tozeroweobtain

Supposethefunction f (x , y ) isheldconstant.Then

= 0(3.58) andweﬁnd

Thenotationinthelastequationwillbeencounteredoften.

Example3.21 A1kgparcelisheatedatthesurface(p =1000hPa)atarated Q /d t = 20Wkg 1 (W = watts).Whatistherateofchangeofenthalpy?

Answer:Notethatd p/d t = 0.Then,d H /d t = d Q /d t .

Example3.22 Aparcelmovesalonganisobaricsurface(constantpressure)andis heatedatarated QM /d t = 10Wkg 1 .Whatistherateofchangeof T alongthe pathofmotion?

Answer: (d H /d t ) = Mcp d T /d t = d QM /d t ;d T /d t = (d QM /d t )/cp = 10Wkg 1 /1004J 1 kg 1 = 0.01Ks 1 .

3.5Standardenthalpyoffusionandvaporization

Enthalpyisaveryusefulfunctionindescribingtheenergytransfersinprocesses involvingachangeofphase(e.g.,liquidtovapor).Enthalpyisespeciallyuseful sincetheseprocessesoftentakeplaceatconstantpressure.Anexampleisthe evaporationof1molofwater.Inthiscasethesystem(thevolumecontaining thewater)isheatedbymaintainingasmalltemperaturedifferentialwithits surroundingsatconstantpressureandconstanttemperature.Theheat(nowwe cancallitenthalpy)absorbedtoeffectthistransitionisoftencalledthe latentheat

Table3.2 Standardenthalpiesoftransitionforselected compounds

Thestandardenthalpiesoffusionandvaporizationare evaluatedatthefreezingandboilingpoints(K) respectivelyat1atmofpressure.Unitsinthistablefor theenthalpiesarekJmol 1 ,butbecarefulinusing tablessincetheunitsofenergymightbeinkcal(4.18kJ = 1kcal).InthetableTf isthefreezingpointandTb is theboilingpointat1atmofpressure.

ofvaporization intheolderliterature,butinkeepingwithcurrentconventionitis calledthe enthalpyofvaporization.Manytablesgivevaluesintermsofmolesrather thankilogramsofthesubstance.Tomakeusefulstandardizedtables,conventions havebeenadopted.Inthecaseofevaporationforexample, vap H ◦ indicatesthat 1molofthesubstanceisbeingconsidered(theoverbar)andthesuperscript ◦ indicatesthatitisatastandardtemperature(unitsshouldbeindicatedinthetable).

SeeTable 3.2.

Bydeﬁnition,the standardenthalpyforvaporization, vap H ◦ ,istheheat transferredtothesystematconstantpressurepermoleintheprocessofvaporization ofthesubstancefromitsliquidtoitsvaporform.Thestandardquantityisdeﬁned attheboilingpoint(373Kforwater)at1atmofpressure.

Similarlythe standardenthalpyforfusion istheheattransferredbythesystemto thesurroundingsatconstantpressurepermoleintheprocessoffusingthesubstance fromitsliquidtoitssolidform.Itislabeled fus H ◦ .Byconventionthestandard quantityisevaluatedatthefreezingpointat1atmofpressure.

Example3.23 Wehave36gofliquidwaterat373Kand1atmofpressure.We wishtoevaporatethewaterandraiseitstemperatureto473Katconstantpressure. Whatisthechangeinenthalpyforthetwosteps?(cpvap ≈ 2kJkg 1 K 1 .)

Answer:36gis2.0molofwater.Wemustthengivethesystem2mol × 40.656 kJmol 1 = 81.312Jforstep1.Step2isaheatingatconstantpressure. H2 = Mcp T = 0.036kg × 2kJkg 1 K 1 × 100K = 7.2kJ.

Example3.24 Twogramsofliquidwaterareevaporatedinto1kgofdryairat1 atmconstantpressure.Thetemperatureis300K.Whatisthechangeinenthalpy ofthesystem?

Species

Answer:Thedryairisirrelevant.Wemustdoseveralstepstoaccomplishourgoal. TousethevalueinTable3.2wemustheatthe2gofwatertoitsboilingpoint.The changeinenthalpyforthisis H1 = Mcliq T = 2g × 4.18JK 1 g 1 × 73K = 610J.Nextthewaterisevaporated: H2 = (2/18) mol40.7kJmol 1 = 4.52kJ. Instep3wemustcoolthevaporbackdowntoitsstartingtemperature: H3 = 2kJK 1 kg 1 × 0.002kg ×( 73K) =− 292.0J.Combiningallthreesteps: H1 + H2 + H3 = 4.84kJ.

Example3.25 Inthelastexample,whatisthetemperatureanddensitychangefor theoriginal1kgofairwhichholdsthe2gofliquidwater?

Answer: T = H M 1 c 1 p = 4.84kJ/(1kg1004Jkg 1 K 1 ) = 4.5K.Thechangeindensityis ρ =−p T /Rd T 2 = 105 Pa × 4.5K /(287Jkg 1 K 1 )(300K)2 = 0.019kgm 3 .Thisrepresentsa1.6%change inthedensity,enoughtocauseimportantbuoyancyeffects.

Intheseexampleswemadeliberaluseofthefactthattheenthalpyofasystem dependsonlyonitsstate,notonthepaththroughwhichthestateisfound.This freedomallowsustousestandardtables.

Notes

Mostofthegoodthermodynamicsbooksreferredtoinearlierchaptersworkwell forthisone.

NotationandabbreviationsforChapter3

cv , cp speciﬁcheats(heatcapacityperkg)atconstantvolume,pressure (Jkg 1 K 1 )

c v , c p molarspeciﬁcheats(Jmol 1 K 1 )

Cv , Cp heatcapacitiesatconstantvolume,pressure(JK 1 )

(d H )p changeinenthalpyatconstantpressure(J)

d H /d t ,d Q /d t timerateofchangeofenthalpy,heattransferrate(Js 1 )

d Q ,d W differentialsforheat,work,thebaremphasizespath dependence(J)

V changeinvolume(m3 )

fus H

◦ (X ) standardenthalpyoffusionofsubstance X (Jmol 1 )

vap H changeinenthalpyduringvaporization(J)

vap H ◦ (X ) standardenthalpyofvaporizationofsubstance X ,theoverbar indicates1molofsubstance,thesuperscript ◦ indicatesat standardconditions(usually25 ◦ C)(Jmol 1 )

x displacementin x f numberofdegreesoffreedomofamolecule

fx partialderivativewithrespectto x

F force(N)

γ ratioofspeciﬁcheats cp /cv (dimensionless)

H enthalpy(J)

Ha scaleheightoftheatmosphere

kB Boltzmann’sconstant

κ = R/cp (dimensional)

κH thermalconductivity(JmK 1 )

M bulkmass(kg)

n0 numberdensity(moleculesm 3 )

ν numberofmoles

p pressure(Pa,hPa)

d QM /d t rateofheatingperunitmass(Jkg 1 mol 1 s 1 )

R, RG , Rd , R∗ gasconstant(Jkg 1 K 1 ),gasconstantforagasG,fordryair, universalgasconstant(JK 1 mol 1 )

ρ density(kgm 3 )

T temperature(K)

θ potentialtemperature(K)

V volume(m3 )

VA , VB initialandﬁnalvolumes

WVA →VB workingoingfrom VA to VB

W , Q workdonebythesystem,heattakenintothesystem

Problems

3.1Suppose p(z ) = p0 e z /H .Evaluatethefollowing.

(a)d p/d z at z = H /3.

(b) ∞ 0 p(z ) d z .

(d) ∂ p/∂ H at z = H /2.

3.2Let p = ρ RT .Evaluatethefollowing.

(a) ∂ p/∂ T .

(b) ∂ρ/∂ T .

3.3The compressibility ofasubstanceisdeﬁnedby

where X isthevariablebeingheldconstant.Wecancompressthegasisothermally(κT ) oradiabatically(κθ )(θ isthepotentialtemperature).Calculatebothforanidealgas.

3.4The coefﬁcientofexpansion isdeﬁnedby

Compute β foranidealgas.

3.5Showthatforanygas

Hint: d V =

,seetheCalculusrefresherinthischapter.)

3.6Findtheinternalenergyof1kgofdryairatSTP.

3.7Supposetheatmospherehasitspressuregivenby p(z ) = p0 e z /H with p0 = 1atm and T (z = 0) = 273K.Nowsupposea1kgparcelisliftedadiabaticallyonescale height H .Howmuchworkdoestheparceldoontheenvironmentintheprocess? Whatisthechangeinitsspeciﬁcinternalenergy, u ?

3.8 Isobaricprocess A1kgparcelofdryairhastemperature285Kandpressure 1000hPa.Itisheatedbycontactwiththedrygroundtoatemperatureof295K. (a)Whatis Q ?(b)Whatisthechangeoftheparcel’sspeciﬁcinternalenergy?(c) Whatisthechangeintheparcel’sspeciﬁcenthalpy?

3.9 Isothermalprocess 1kgofdryairat300Kand1000hPaisexpandedisothermally (prettyunusualintheatmosphere)fromavolumeof2m3 totwicethatvalue.(a)What istheworkdonebythegasinthisexpansion?(b)Whatistheheatabsorbed?(c)What isthechangeinenthalpy?

3.10 Isochoricprocess 1m3 ofdryairat1000hPaand290Kisenclosedinarigidbox. Whatisitsdensity(kgm 3 )?3gofliquidwaterareevaporatedintothebox.What istheincreaseoftemperatureinthisboxwhosevolumeisheldﬁxed?Whatarethe changesininternalenergyandenthalpy?Sketchadiagramofthechangeinthe V –p plane.

3.11 Adiabaticprocess Aparcelofmass1kgisliftedadiabaticallyfrom800hPa,where itstemperatureis270K,to600hPa.Whatisthenewtemperature?Whatarethe changesininternalenergyandenthalpy?Sketchadiagramofthechangeinthe V –p plane.

3.12 Isobaricprocess Aparcelofessentiallydryairisheldatafixedaltitudewherethe pressureis950hPa.Itisheatedbyinfraredradiationbeingabsorbedbysomewater vaporintheparcel.Theheatingrateis20Jkg 1 s 1 .Whatistherateofchangeofthe temperature,specificenthalpyandspecificinternalenergy?Howdoesthepotential temperaturechangeperunittime?Sketchadiagramofthechangeinthe V –p plane.

3.13Anaircolumniscomposedofdryairandthedensityoftheairisgivenby ρ(z ) = ρ0 e z /H ,where ρ0 = 1.25kgm 3 and H = 10km.

(a)Whatisthemassofair(kg)lyingabove1m2 ?

(b)Howmanyidealized“air”moleculesareabovethe1m2 ?

(c)Thenapproximatelyhowmany“airmolecules”arethereintheentireEarth’s atmosphere?

3.14Thespeedofsoundinaircanbecomputedfromtheformula vsound = √1/ρκX ,where κX isthecompressibilityholdingtheparameter X constant.Comparethesoundspeeds (taking ρ = p/RT )when X = κT and X = κθ .Thelatterﬁtsthedata.Doyourecall fromphysicswhytheadiabaticcompressibilitygivesthecorrectanswerinsteadof theisothermalcompressibility?SeeProblem3.3above.

3.15Supposetheatmospheresatisﬁes p(z ) = p0 e z /H andthatitisisothermal(T (z ) = T0 ). Whatisthepotentialtemperature θ asafunctionof z ?Sketchagraph.

TheSecondLawofThermodynamics

TheSecondLawofThermodynamicsdealswithchangesintheconditionsof asystemanditssurroundingsundertransitionsfromonethermodynamicstate toanother.Wewillintroduceanewfunctionofstatebyconsideringasimple quasi-staticseriesofchangesforanidealgas.Firstrecallthatsomequantitiesare functionsofthestateonly,whileothersdependonthepathtakenbetweentwo states.AccordingtotheFirstLaw,theinternalenergy U isafunctiononlyof statewithchangesingoingfromstateAtostateB,denoted U ,dependingonly ontheinitialstatecoordinates(e.g., pA , VA , TA )andtheﬁnalstatecoordinates (pB , VB , TB ).Recallthattheenthalpy H isalsoafunctionofstate.Changesinother quantitiessuchastheamountofworkdonebythesystemonthesurroundingsin the(quasi-static)transitiondependuponthepathtakenbetweenthetwostates.In thiscasetheworkdonebythesystemonthesurroundingsmaybewritten,

andthefunction p(V ) isaspecificcurveinthe V –p planedefiningthepathtaken ingoingfromAtoB.Thepathmustbequasi-staticandreversible,otherwisethe path p(V ) maynotevenbedefined,norcouldweuse pd V incomputingdW .The changeininternalenergydependsonlyonthedifference

nomatterwhatthepath(eveniftheprocessisirreversible).Forexample,inthecase oftheidealgastheinternalenergy(andenthalpy)dependonlyonthetemperature, independentofthehistoryofthesystem.Formorecomplexsystems U and H mightdependonotherstatevariablesaswell,butnotonthehistoryofthesystem. Notethattheheat(inducedbyatemperaturedifferencebetweenthesystemandits environment)takenintothesystemduringthetransitionmustalsodependonthe

Figure4.1IllustrationoftwopathsjoiningpointsAandBina V –p statediagram.

pathsince

andtheworkdonebythesystemsurelydependsonpathaswehaveseenearlier(see Figure 4.1).Nextweexplorequasi-statictransitionsofanidealgastoseewhether theremightbeanotherfunctionofthestateofthesystem.

Considerthecaseofan idealgas wherethemass M isﬁxed.Wecanusethe deﬁnitionoftheenthalpytoobtain:

Ifwemultiplythroughbythe integratingfactor 1/T weobtain

UsingtheIdealGasLaw, V /T = MR/p andsolvingford Q /T :

whereasbefore κ = R/cp .Thelastexpressionsaysthatd Q /T isa perfect differential.Thatis,itschangeingoingfromAtoBdoesnotdependonthepath chosenforthesequenceofquasi-staticinﬁnitesimalsteps.Anecessaryandsufﬁcient conditionforthedifferentialtobeperfectisthatintegralsaroundarbitraryclosed loopsresultinnochangeinthefunction.

Calculusrefresher:perfectdifferential Suppose F isdeﬁnedby:

where A(x , y ) and B(x , y ) arewell-behavedfunctions.Itmighthappenthatthereexists afunction f (x , y ) suchthat

ifso

istheexactdifferentialof f and

whichmeansthat F isindependentofthepathalongwhichtheintegralistaken joiningtheupperandlowerlimits.Itmightbeeasiertoseethisifweintroducea parameter t whichdeﬁnesthecurve

(forexample,aparabola y = x 2 couldbewritten x (t ) = t , y (t ) = t 2 )thisdeﬁnesa curveinthe x –y plane(thepath).Theintegralcanbewritten:

Example4.1 Considerthefunction

Then fx (x , y ) = y cos xy and fy (x , y ) = x cos xy .Then

Example4.2:integratingfactor Considertheﬁrst-orderlineardifferentialequation forthefunction y (t ):

where b isaconstantand g (t ) isagivenfunction.Nowmultiplythroughbythe integratingfactor ebt .Wecanshowthat

andtheleft-handsideisrenderedaperfectdifferential,byvirtueofourknowingthe properintegratingfactor.Nowweareinpositiontosolvethedifferentialequation byintegratingeachside:

Then

Example4.3 Considerthedifferentialexpression

Wewanttoconsiderintegralsofthisdifferentialfromthepoint(0,0)tothepoint (1,1)alongsomedifferentpaths.Firstconsiderthestraightlinepath, y = x ,which connectsthetwopoints.Then,

Nextconsidertheparabolicpath, y = x 2 ,d y = 2x d x ,whichalsopassesthrough thetwopoints:

Weobtainthesameanswerforthetwodifferentpaths.Infact,itispossibletowrite

Inotherwords,d Z (x , y ) isaperfectdifferential.Integratingfromonepointinthe x –y planetoanotheralwaysyieldsthesameanswer. Ontheotherhand,supposewehadthefunction

Thisdifferentialisnotperfect,butwiththeaidoftheintegratingfactor, y 2 ,wecan createtheperfectdifferential,d Z .

Example4.4 Inclassicalmechanicsweencounterthesameconceptwith conservativeforces,whichimpliestheexistenceofapotentialenergyfunction. Ifaforcecanbedescribedbyapotentialenergyfunction V (x , y , z ) andiftheforce, F(x , y , z ) canbewrittenasthegradientoftheforce:

Orwecanwrite

HencetheworkdoneingoingfromAtoB:

Inthecaseoftheidealgasmentionedabove,wecanusePoisson’sequation,

andinsertinginto(4.6)weobtain:

where S iscalledthe entropy.Inchemistryandphysicstheentropyisusually denoted S andtheentropyperunitmass s;insomeoldermeteorologycontextsit isdenoted φ .Theimportantthingisthatentropyisafunctiononlyofstateand thatitschangeingoingfromonestatetoanothercanbecalculatedbychoosinga reversiblepathjoininginitialandﬁnalstates.

Foraninﬁnitesimaldisplacementalongareversiblepath(forageneral thermodynamicsystem,notjustanidealgas),

Theextrasubscriptrevisaddedtoremindthereaderthatthecalculationofd Q /T mustbealongareversiblepath.Recallthatareversiblepathisoneinwhichthe infinitesimalstepsalongthepatharequasi-staticandsuchthateachcanbereversed torestorethesystemtoitspreviousstate.Inperformingthecalculation,weshould findsuchanimaginarybutrealizablepathjoiningthetwostatesforthepurposesof calculation.Ofcourse,innaturethetransitionmightbe(andoftenis)anirreversible orspontaneousonefromstateAtostateB,buttherealwaysexistsareversiblepath joiningthetwostatessothatthechangeinentropycanalwaysbecalculated.As indicatedinthelastequation,theveryexistenceofsuchafunctionofstateas definedaboveconstitutesonestatementoftheSecondLawofThermodynamics.

Inthecaseofdryparcelsofidealgas(nearlyalwayssatisﬁedinthedry atmosphere)wecanusetheformula

andintermsofspeciﬁcquantities(kg 1 ):

whereanarbitraryintegrationconstanthasbeensettozero,sinceitisneveractually needed.Intheselastformulas s referstotheentropyperunitmass(unitsJK 1 kg 1 ) orthe speciﬁcentropy.Wewilluse s (unitsJK 1 mol 1 )toindicateentropyper mole.Foraparcelofdryair,knowingthepotentialtemperatureisequivalent toknowingitsentropy.Hence,thechangeinentropyforaparcelundergoinga reversibletransition(throughaseriesofquasi-staticchanges)canbecomputed simplybycomputingthechangeinitspotentialtemperature

Itisobviousthatforareversibleadiabaticprocess,thechangeinentropyvanishes, since d Q = 0alongthereversiblepath.Forso-called diabatic processes(ones inwhichsomeheatisexchangedbetweenthesystemanditssurroundings),the formuladeﬁningthechangeinentropycanbeused(d S = d Qrev /T ).

Example4.5 Showthatd Qrev /T = Mcp dln θ reallyworksforanidealgas.First write

Substitutingford H and V :

Afterdividingthroughby T wesimplyrepeatthestepsleadingto(4.6).

Example4.6 Computethechangeinentropyforaparcelofmass M atpressure level p0 beingheateddiabatically(e.g.,byradiationheating)fromtemperature T0 to T .

Answer:Sincetheheatingisdoneataﬁxedlevel,thepressure p remainsconstant atavalue p0 .Wecanproceedtocomputethechangeinentropybyusing

Example4.7 Howmuchdoesthepotentialtemperatureoftheparcelchangeinthis heatingifittakesplaceat500hPa?

TheSecondLawofThermodynamics

Answer:Since θ = T (p/p0 ) κ , p = constant = 500hPa,and p0 = 1000hPa,we have θ = T (p/p0 ) κ = (T T0 ) (500/1000) 0.287 ,foranisobaricheating process.

Example4.8 Aparcelofidealgasistakenfrom (V0 , p0 ) to (2V0 , 1 2 p0 ) byﬁrst(path a)expandingisobaricallyto2V0 ,thenthepressureisreducedatconstantvolume to 1 2 p0 (pathb).Whatisthechangeinentropy?

Answer:Alongpathawecompute Sa = d Q /T = Mcp

d T /T = Mcp ln2.Alongpathb, Sb = d Q /T = d U /T = Mcv T

/T = Mcv ln2.Now Sa + Sb = M(cp cv ) ln2 = MR ln2.

Example4.9 Usingthesameinitialandﬁnalstatesasinthepreviousexample, computethechangeinentropybutalongtheisothermalpathjoiningthetwostates (path c ).

Answer:Alongcwehave d Q /T = d W /T = (

)

= (MRT0 /T0 ) 2V0 V0 d V /V = MR ln2,whichisthesameasinthealternative reversibletwo-steppathchoseninthepreviousexample.

Example4.10:changeinphase Computethechangeinentropyfor1kgofwater beingevaporatedtogasat373Katconstantpressure.

Answer:Firstnotethat1kgofwateris55.56mol.Thechangeinentropyis Q /(373K).Sincethepressureisheldconstant, Q = ν vap H ◦ = 55.56mol × 40.66kJmol 1 = 2259kJ.Thechangeinentropyisthen S = 2259kJ/(373K)= 6.06kJK 1

4.2TheSecondLawofThermodynamics

ThereareseveralequivalentstatementsoftheSecondLawofThermodynamics. Forourpresentpurposesitsufﬁcestostatethelawasfollows.

ThereexistsafunctionSoftheextensiveparametersofthesystemthatisa functionofstateandwhosechangesfromonestatetoanothercanbecalculatedby

andsuchthat

foranisolatedsystem.

Thesubscriptisagainareminderthatthecalculationmustbeconductedalong areversiblepath.

4.3Systemsandreversibility

Itistimetopauseandreviewsomedefinitions.Considera system whichis embeddedinits surroundings.Togetherwesaythesecomprisethe universe.The systemisincontactwithitssurroundingsbyvariousmovableorstationarywalls andmembraneswhichmight(ormightnot)allowfluxes(heatormass)tocross.Ifno masscrosseswesaythesystemisclosed,whereasifmasscrossesintooroutofthe volumeconfiningthesystemwesayitisanopensystem(sometimesthisiscalled a controlvolume).Sofarinthisbookwehaveonlyconsideredclosedsystems,but wewillencounteropensystemsaswell.Inequilibrium,boththesystemandthe surroundingshavethermodynamiccoordinates.Areversiblechangeisonewhichis quasi-static(takeninsmallslowstepsinsuchawaythatequilibriumismaintained; thatis,thereisenoughtimebetweenstepsforthepressureandtemperatureto homogenizethroughoutthevolumeofthesystem)andwhichcanbereversedat anypointreturningbothsystemandsurroundingstotheirformercoordinatevalues withouttheexpenditureofanyadditionalwork(moreonthisbelow).

Oftenasystemincontactwithsurroundingsundergoesaspontaneoustransition whenaconstraintisreleasedorrelaxedtosomenewconﬁguration.Suchatransition isirreversible.Undermanyofthesespontaneoustransitionstheinternalenergydoes notchange,buttheentropydoes.

Example4.11: Freeexpansion Considerachamberisolatedfromtheoutsideby adiabaticwalls.Insidethechamberisawallseparatinghalfthevolumeoneach side.Thereisanidealgasononesideofthepartition,vacuumontheother.Suddenly thepartitionisremoved(slippedoutsidewayssothatnoworkisdone),suchthat thegasexpands(irreversibly)toﬁllthewholevolume.Whatarethechangesin internalenergy,enthalpyandentropy?

Answer:TheinternalenergymaybecalculatedfromtheFirstLaw.Thesystemdoes noworksincethevacuumexertsnobackpressureduringtheexpansion.Alsono heatistakenintothesystembecausethewallsareimpermeabletosuchatransfer. Therefore,theinternalenergydoesnotchange: U = Qfreeexp. Wfreeexp. .Note thatwewereabletoapplytheFirstLaweventhoughthepathwasirreversible. Thefactthattheinternalenergyisinvariantmeansthatinthefreeexpansion,the temperaturedoesnotchange(trueforanidealgas–inarealgasthereissome temperaturechange,eventhoughnochangein U occurs).Sincethetemperature doesnotchangewecanseethattheenthalpydoesnotchangefortheidealgas.The changeinentropymustbecomputedbyanalternativereversiblepath.Thereare many,butwecanchoosetheonealonganisothermfrom

Alongtheisotherm,d Q = d W = p d V = MRT d V /V ,therefore

Wecouldalsocomputethechangeinentropyforthesurroundings(theadiabatic walls).Itiszero.Hence,thechangeinentropyfortheuniverseis MR ln2which isapositivenumber.

Inthefreeexpansionprocessintheexampleabove,theentropyofthesystem experiencedanetincrease,whilethatofthesurroundingsdidnotchange.This isanexampleofthesecondpartoftheSecondLaw. Foranirreversibleprocess theentropyofthesystemanditssurroundings(takentogether,theuniverse)will alwaysincrease.Notethatforaninfinitesimalincreaseinvolumethatisanadiabatic freeexpansion,theentropyoftheuniverseincreases.Thismeansthatsomequasistaticprocessesareirreversible.Anirreversibleprocesscouldbedefinedasonein whichtheentropyoftheuniverseincreases.Sohowcouldanadiabaticexpansion everbereversible?Thewayoutisthatonecanimaginemakinginfinitesimal(and reversible!)expansionsatconstantvolumethenatconstantpressureinastair-step proceduretoapproximatetheadiabaticexpansioncurveinthe V –p plane,muchas anintegralcanbeapproximatedbysummingrectangularboxeswhoseupperedges approximatethecurvebeingintegrated.Inthiswayareversibleapproximationcan befoundtotheadiabaticexpansioncurve.

4.4Additivityofentropy

Theentropyforasetofthermodynamicsystemsisthesumoftheindividual entropiesoftheconstituentsubsystems.Hence,theentropiesofthesystemand itssurroundingsare additive.Thesame additivityprinciple appliestotheinternal energyandtheenthalpyandanyotheroftheextensiveparametersdescribingthe compositesystem.Theextensiveparametersvolumeandmassalsosatisfythis principle.Whenweaddupseveralsystemstoformalargersysteminwhichthe entropies,internalenergiesandenthalpiescanbeaddedup,wecallthisa composite system.Wecanexpressthisinequationform:

Example4.12:acolumnofair Wecanthinkofacolumnofairasacomposite system.Itspressureandtemperaturemightbevaryingwithaltitude z butwecan

addupthesecontributionsfromindividualslabs:

where ρ(z ) ismassdensity, u (z ) isthespeciﬁcinternalenergy,etc.

Ingeneral,iftwosystemsarebroughtinto“contact”wecansaythesumofthe internalenergiesofthesystemwillremainthesame.Bringingtwosystemsinto contactamountstoremovingorrelaxingaconstraint.Forexample,iftwogasesA andBatthesamepressureareinachamber,butseparatedbyapartition,theremoval ofthepartitionwillallowthegasestomix.Nochangeininternalenergywilloccur ifthechambercontainingbothsubsystemsisinsulatedfromitsenvironment.On theotherhand,thechangeinthetotalentropymustbezeroorpositive.Processes (eitherreversibleornot)inwhichthesumoftheentropiesofasystemandits surroundingsyieldanegativechangeintotalentropydonotoccurinnature.This principlecanbeofgreatutilityindeterminingwhichwayaprocesswillproceed whenaconstraintisrelaxedorremoved.Recallthatremovalofaconstraintmeans thatinordertorestorethesystemtoitsoriginalstateadditionalworkmustbe performedbysomeexternalagent.ThislastisreallytheessenceoftheSecond LawofThermodynamics.Solutionstoproblemsofthistypemaynotalwaysbe facilitatedbyuseoftheinternalenergyalone,sinceitmayremainﬁxedwhenthe constraintisremoved.Butrestorationoftheoriginalconditionsdoesrequirework, andhencechangesintheinternalenergyaloneareinsufﬁcienttodescribewhathas happened.Itturnsoutthattheentropychangeprovidesthisadditionalinformation.

Example4.13 HowcouldwerestorethegasinExample4.11toitsoriginal conditionandhowmuchworkwouldberequired?

Answer:Firstwewouldhavetobringtheidealgasintocontactwithareservoir oftemperature T ,thenwewouldperformanisothermalcompressionofthegas fromvolume2V0 to V0 .Asweperformthecompressionofthegas,inﬁnitesimal temperaturedifferenceswilldevelopbetweenthesystemandthereservoir:heat willtransferfromonesystemtotheothertomaintaintheﬁxedtemperature.The workperformed(bythesystem)intheisothermalcompressionisjust MRT ln2. NotethatinthecaseofmixingtwogasesAandBmentionedabovewewould needtorecompresseachtoitsoriginalvolumebyanisothermalrouteinorder torestorethemtotheiroriginalstates.Wemightinthiscaseuseamembrane

TheSecondLawofThermodynamics

whichallowsmoleculesAtopassintoanewadjacentchamberbutnotB.But supposeAandBarethesamegas.Doestheentropyincreasewhenthepartitionis removed?No.

4.4.1Expressionforentropyofanidealgas

Itispossibletoderiveananalyticalexpressionfortheentropyofanidealgas. Toproceedconsidertheexpressionforanidealgasdryairparcel.Recallthatthe entropy S isgivenby

Usingtheexpressiondeﬁningthepotentialtemperaturewehave

Whilethisisaperfectlygoodexpressionfortheentropy,itisnotintermsof extensivevariables M, U , V .Makinguseofthedeﬁningequationsforanideal gas: p = MRT /V , U = ( f /2)MRT andtheidealgasproperty cp = ( f /2 + 1)R, weeventuallyﬁnd

[entropyofanidealgas](4.53)

whichholdsforanyidealgas.Fordryair, R = Rd , f = 5.Toexpressthisinmolar formreplace MR with ν R∗ .Weimmediatelyseethatsuchafunction S doesexist forthesimpleidealgassystem.Wealsoseethat S isextensive(proportionalto M),andtheextensivearguments U and V areexpressedasratiosofstandardstates withthesamemass.Inadditionwecanseethat S (U , V ) isanincreasingfunction oftheinternalenergy U andthevolume V

AmoredirectwayofderivingtheentropyofanidealgasistostartwiththeFirst Law(nowwithd Q replacedwith T d S ;hencetheexpressionthatfollowswillhold forreversiblepathsinthe V –S plane):

d U = T d S p d V [differentialfor U intermsofentropy].(4.54)

Dividingthroughby T andrearranging: d S = 1 T d U + p T d V .(4.55)

Using U = ( f /2)MRT and p = (MRT /V ):

Then,

Thelastequationisequivalenttotheexpression(4.53)with S0 = 0.

4.4.2Expressionfortheinternalenergyofanidealgas

Wecansolvefor U usingthelastequation:

Thisexpresses U = U (S , V , M) forasingle-componentsystem.The thermodynamicpotentials aretheintensiveparametersofthesystemthatcomefromtaking partialderivativesof U withrespecttoitsarguments:

Theﬁrstleadsto T = U /(( f /2)MR) whichdeﬁnes U fortheidealgas.The secondleadstotheidealgasequationofstate, pV = MRT . Whileweworkedouttheanalyticalexpressionsforthecaseofanidealgas, thedifferentialrelationsford U andd S aregeneralandholdforanysubstance. Thereforethelasttwoequationsforthepartialderivativesalsoholdgenerally.We thenlearntwoimportantfacts.(1)Theinternalenergyisalwaysanincreasing functionof S (likewise, S isanincreasingfunctionof U ).(2)Forconstantentropy theinternalenergyisadecreasingfunctionof V .

4.4.3Enthalpyofanidealgas

Byaseriesofstepssimilartothoseofthelastsubsectionwecanﬁnd:

(S , p) = H0 p p0

[enthalpyofanidealgas].(4.61)

Thisexpressionshowsexplicitlythatforasimpleidealgas,theenthalpyisafunction oftheentropyandthepressure.

Inthegeneralcase(notjusttheidealgas)wehaveforasingle-componentsystem d H = T d S + V d p [differentialof H intermsofentropy](4.62) andtherefore

Animportantcorollaryoftheexpressionsford U ,d S andd H isthatthey arenaturalfunctionsofcertainpairsofvariables(ﬁxedmass).Forexample, U = U (S , V ), S = S (U , V ) and H = H (S , p).

4.5Extremumprinciple

Sincetheentropiesofthesystemanditssurroundingstogetheralwaysincrease inanatural(spontaneous)process,wecanseethattheﬁnalstateafterconstraints havebeenreleasedwillbeoneinwhichtheentropyisamaximum.Forexample, theequilibriumconcentrationofgaseousreactantsandgaseousproductsina chemicalreactionwillreachitsequilibriumvaluewhentheentropyofthesystem ismaximized(seeChapter 5).

Example4.14 Considertwoisolatedparcelsofdryairwithmasses M1 and M2 at thesamepressure p1 = p2 (samealtitude)butatdifferentinitialtemperatures T 0 1 and T 0 2 .Thetwosubsystemscomeintothermalcontact.Whatistheﬁnaltemperature afterthisirreversibleprocesstakesplace?

Answer:Theenthalpyisconstantduringthismixingprocess

where TF istheresultingtemperatureofthemixture.Wecansolvefor TF :

250255260265

Figure4.2Entropyversusﬁnaltemperatureforcomponentoneofamixture subjecttotheconstraintthatenthalpybeconserved.

where x = M1 /M, M = M1 + M2 .Thelastequationistheanswertoour problem,butletuslookfurthertoseewhathappenstotheentropy.Thepotential temperatureofsubsystem i is θi = (p0 /pi )κ × Ti ; p0 = 1000hPa.Theentropyof thesystemisthesumoftheentropiesoftheindividualsubsystemscomprisingthe wholesystem

Consideraparticularexampleforwhich M1 = 2kg, M2 = 3kg, x = 0.4, T 0 1 = 250K, T 0 2 = 260Kandforsimplicity cp = 1.Theresulting TF = 256K. Figure 4.2 showsaplotof S asafunctionof T1 .Ourassumptionbasedonintuition andexperiencethatbothmassescometothesametemperatureof256Kisjustiﬁed byitsbeingthevalueoftemperaturethatmaximizestheentropyofthecombined system.

4.5.1Carnotcycle

TheCarnotcycleisthemostimportantclosedloopprocessinthermodynamics.We illustrateithereforanidealgas.Thisisafoursteploopprocessasillustratedin Figure 4.3.Thebranch ab isalongahotisothermattemperature Th duringwhich heat Qh istransferredtothesystemfromthehotreservoir.Thenextstepisan adiabaticexpansion bc tothecoolertemperature Tc .Thethirdstep, cd isisothermal andanamountofheat Qc (> 0) isexpelledfromthesystemtothecoolerreservoir at Tc .Finally,thereistheadiabaticcompression da,whichcompletesthecycle

Figure4.3Carnotcyclefor1kgofdryair(takenasanidealgas)inthe V –p plane.Thecycleproceedsasfollows.Step ab isanisothermalexpansionfrom a to b,attemperature Th ,drawinginheat Qh .Step bc isanadiabaticexpansion from b to c .Step cd isanisothermalcompressionattemperature Tc expellingheat Qc fromthesystem.Finally,step da isanadiabaticcompressionfrom d to a .In thiscase Va = 0.50m 3 , T2 = 300K, Vb = 1.50m 3 , T1 = 200K.Thenallother intersectionsaredetermined,e.g., pa =1720hPa, pb = 574hPa.

backtothestartingpoint a .Wecanlisttheproducts:

andsinceitisaclosedloop, d S = 0,wehave

whichleadsto

andinadditionwecanﬁndfromtheaboveformulas:

IntheCarnotcyclewefollowconventionandcalltheheattransferredtothe systemalongthehotisotherm(a → b) Qh andtheheatrejectedalongthecooler isotherm Qc .Thiswayboth Qh and Qc arepositivenumbers.Notethat Qh > Qc :moreheatisalwaysdrawnfromthehotreservoirthanisrejectedtothecold one.Thedifferenceistheworkextractedfromthesystemduringthecycle.It wouldbenicetousealloftheheatfromthehotreservoirtocreatework,but thisisimpossible,sincesomeheatisalwaysrejectedtothecoldreservoir.We canneverturnallofourheatextractedfromthehotreservoirintowork.This isamajorconsequenceoftheSecondLaw.TheFirstLawsaysenergymustbe conserved,buttheSecondLawgoesfurthertotellusthatwenotonlycannot obtainsomethingfornothing,wecannotevenobtainallofthesomethingtobe usedforwork.

The efﬁciency oftheprocessistheworkperformedoverthewholeloopdivided byheatextractedfromthehotreservoir:

Figure 4.4 shows(nottoscale)thesamecycleintheentropy–temperatureplane, referredtoasthe S –T plane.Inthisplanethecycleisamoresimplegeometric ﬁgure,arectangle.Considertheareaenclosedinaclosedﬁgureinthe S –T plane

Figure4.4SchematicdiagramoftheCarnotcycleofFigure4.3exceptinthe S –T plane.Thegeometricalﬁgureisarectangleindependentofthesubstance(i.e.,this diagramisnotrestrictedtotheidealgas).Theareaenclosedistheworkdoneby thesystem.

TheSecondLawofThermodynamics (suchasFigure 4.4):

areaenclosed = T (S ) d S [area-energyinthe S –T plane].(4.76)

TheFirstLawofThermodynamicsstatesthatd U = T d S p d V .Sincetheloop integralofd U vanishes,wefindthattheareaenclosedinthe S –T planeaboveis exactlytheworkdoneintheloopprocess.Inmostcasesinatmosphericscience itiseasiertodealwithareasinthe S –T planethaninthe V –p plane.Thisis becausethevolumeofaparcelisnoteasilyobserved,whereasitspressure(same astheenvironmentalpressureandapproximatelyequivalenttoaltitude)iseasily observed.Similarlyforadiabatictransformations s = cp ln θ isfixedandina diabaticheating p isusuallyfixed.

NotethattherectangularshapeoftheclosedfigureforaCarnotcycleinthe T –S planedoesnotdependonthesystembeingcomposedsolelyofanidealgas. Isothermsandadiabatsarestraightlinesforanysysteminthe S –T diagramin Figure 4.4.ThismeansthattheCarnotcycleisusefulindescribingcycleseven ofacompositesystemcomposedofanysubstances.Aparticularexampleisthe caseofacompositesystemconsistingofaliquidinequilibriumwithitsvapor, bothenclosedinachamber.Thislastconfigurationissimilartoasteamengine whereinvaporiscondensedandlaterevaporatedasseparatelegsofthecycle.It isshowninmoreadvancedthermodynamicsbooksthattheCarnotcycleisthe mostefficientreversiblecycleintermsofobtainingworkfromthesystemby extractingheatfromahotreservoirandrejectingpartofittoacoldreservoir. Irreversiblecyclesarealwayslessefficientthanreversibleones.Thermodynamic loopdiagramsforatmosphericprocesseswillbeexploitedinlaterchaptersas weanalyzetheenergeticsinvolvedinparcelsundergoingtransitionsinthereal atmosphere.

Example4.15: Carnotcycleforanysystem The S –T depictionoftheCarnotcycle (Figure 4.4)worksforanythermodynamicsystem.Wecanshowafewproperties thatholdforthegeneralcasethatwerederivedaboveonlyfortheidealgascase. Forexample,considertheworkdoneinthecycle.TheFirstLawtellsusthatsince ( U )loop = 0fortheclosedloop, Wloop = Qh Qc .NextconsidertheSecond Lawwhichsaysthat ( S )loop = 0;hence Qh /Th = Qc /Tc .Theefﬁciencyisstill Wloop /Qh anditcanbewrittenas1 Qc /Qh = 1 Tc /Th ,exactlyasfortheideal gas.Finally,wecanverifythattheareaenclosedintherectangleis Wloop .Thearea isgivenby ( T )( ab S ) = (Th Tc )(Qh /Th ) = Qh Tc Qh /Tc = Qh Qc .

4.6Entropysummary

Theextensivethermodynamicvariableswehavediscussedsofararethevolume V ,themass M,internalenergy U ,theenthalpy H andtheentropy S .Thereare alsotheintensiveparameters,pressure p,andtemperature T .Itisbesttothinkof theinternalenergyasafunctionoftheentropy,volumeandmass.Notethateach isanextensivevariable:

ThefollowingformoftheFirstLawactuallyincorporatestheSecondLaw(for ﬁxedmass):

anditshowsexplicitlythattheinternalenergyisbestcharacterizedbythesetwo variables, S and V .Notethatthislastexpressionshowshowtheinternalenergy changesintermsofotherstatevariables(S and V ).Theassociatedintensive parametersaregivenbypartialderivatives:

Theenthalpycanbewritten:

andtheFirstLaw(combinedwiththeSecond)isexpressed(forﬁxedmass)as

alongwiththecorrespondingpartialderivativeexpressions.Wecanalsothinkof theentropyasthedependentvariable:

4.7Criteriaforequilibrium

Wereturnnowtotheextremumprinciple.Ifweconsiderasystemandits surroundings,andweallowaspontaneouschangetooccur(releaseaconstraint), weknowthat

wherethesubscriptsrefertothesystemandtothesurroundingswhiletheequality signholdsforareversibleprocess.Notethattheentropychangeforthesurroundings canbewritten

Herewehavetakend Qsurr tobethesameasd Q rev surr sincethesurroundingsmight beassumedtoundergoareversiblechange(becauseofitslargemass)whilethatin thesystemmightnotnecessarilybereversible.Butwecantaked Q rev surr =−d Qsys sincetheheatgainedbythesurroundingshastobesuppliedbythesystem.The d Qsys neednotbereversibleinthisproblem.Wenowcanwrite

orrearrangingtoobtaintheimportantformula

Ssys ≥ d Qsys T [equalforreversible,largerforirreversible].(4.88)

Theequalitysignappliesonlyiftheprocessisactuallyreversible.Asanexample, considerthe(irreversible!)freeexpansionstudiedinExample4.11.Inthatcase, d Qsurr = 0whilewefoundthatd Ssys > 0. Considerwhathappenstoasystemundergoingaspontaneoustransitionbecause someconstrainthasbeenrelaxedorremoved.WecanusetheFirstLaw(which holdsforreversibleorirreversibletransitions)towrite

Ifthetransitionoccursatconstantvolumewecanwrite

Thesecanalsobeexpressedas

wherethesubscriptsindicatewhichvariablesaretobeheldconstant.Inwords,if theinternalenergyandthevolumeareconstrainedtobeﬁxedinthetransition,the entropyofthesystemwillincreaseinthetransition.Theotherthreeinequalities canbeexpressedsimilarly.

4.8Gibbsenergy

Thereisanotherthermodynamicstatefunctionwidelyusedinapplicationsto atmosphericscience,the Gibbsenergy (sometimescalledthe Gibbsfreeenergy orjustthe freeenergy).Itprovestobeusefulforprocesseswhichoccuratconstant pressureandconstanttemperature.WecanusetheGibbsenergytohelpusin decidingthedirectionofachemicalreactionandindeterminingtheequilibrium phasesorconcentrationsofchemicalspeciesinequilibrium.TheGibbsenergy isparticularlyusefulforopensystems(thoseinwhichmasscanenterorleave thesystem)andforsystemsinwhichtheinternalcompositionmightchange duetochemicalreactions.Wewilltakeupsomeofthesecaseslaterinthis chapter.

TheGibbsenergycanbedeﬁnedas

G = H TS [deﬁnitionofGibbsenergy](4.94) where H isenthalpy, T temperatureand S isentropy.Thedifferentialof G canthen bewritten:

Substitutingford H :

Alongareversiblepathwecantaked Q = T d S whichleadsto d G = V d p S d T [differentialforGibbsenergy].(4.97)

Thislastexpression(whichcombinestheFirstandSecondLaws)tellsusthat G isanaturalfunctionof T and p, G (T , p).Hence,inachangeinwhichthemass,

TheSecondLawofThermodynamics pressureandtemperatureareheldfixed,theGibbsenergywillnotchange.This actuallyhappensinaphasetransition.Forexample,considerachamberwitha movablepistonheldatfixedtemperature.Letthechambercontainaliquidwith itsownvaporinthevolumeaboveit.Ifthepistoniswithdrawnisothermally andquasi-staticallysomeoftheliquidwillevaporateintothevolumeabovethe liquidsurface.Thepressureisjustthevaporpressureandisconstantsinceit dependsonlyonthe(fixed)temperature.Weshouldnotethatthepressureinthe liquidisthesameasthepressureinthevapor(weignoregravityhere).Different positionsofthepiston(leadingtodifferentvolumesofthevapor)correspondto thesametemperatureandthesamepressure(inbothliquidandvapor).Along thislocusofpointsinthestatespace(saythe V –p diagram)forthiscomposite systemtheGibbsenergyisconstant.Wewillreturntothetwo-phaseproblemin Chapter 5.

Returningtothegeneralproblemweseefrom(4.97):

Asindicatedinanearlierchapter,thereactionsoftracegasesintheatmosphere occuratconstantpressureandtemperature.Inthiscasetheatmospherewhich containsordersofmagnitudemoreneutralbackgroundmolecules(nitrogen,oxygen andargon)thanthe(usuallytrace)reactantsactsasamassivethermalandpressure bufferholdingthetemperatureandpressureconstant.Hence,thereactionsamong tracegasesintheatmosphereoccuratﬁxedpressure(altitude)andtemperature (thatofthebackgroundgas).Inthesecaseswhere T and p areheldconstantonly theconcentrationofthespeciesisallowedtochange.Thisistheperfectsetupfor useoftheGibbsenergy.

4.8.1Gibbsenergyforanidealgas

Beginwiththedeﬁnitionof G

where κ = R/cp asbefore.Then

The speciﬁc Gibbsenergyis G normalizedbythemass,sometimesdenoted g (T , p):

Anotherformthatisusefulespeciallyinchemistryisthe molar Gibbsenergy. InsteadofspecifyingthespeciﬁcGibbsenergyasperunitmassitmaybemore convenienttoexpressitasapermolequantity.Intheexpressionfor G notethat forheatingatconstantpressure:

(4.103) or

(4.104) where cp isinJkg 1 K 1 , M isinkg, ν isthenumberofmoles,and c p isin Jmol 1 K 1 .Alsorecallthat R∗ istheuniversalgasconstant(8.3145Jmol 1 K 1 ). Then

(4.105) andusing G (T , p) = G (T , p)/ν wehave

Intheseexpressions, κ isthesamedimensionlessnumbersince R/cp = R∗ /c p . Acompositesystemconsistingofseveraldistinctsubsystemsofidealgases i = 1, , n leadstoanexpressionfortheGibbsenergy:

4.8.2EquilibriumcriteriafortheGibbsenergy

Aswithentropyandinternalenergy,thereisaconditionforequilibriumforthe Gibbsenergyanditisoftenmoreusefulthantheothers:

whichafterdividingeachsidebyd S simpliﬁesto

Nowforaprocessatconstanttemperatureandpressure,

[equilibriumcriterionforGibbsenergy].(4.110)

Thislastisanimportantresult,sincesomanyprocessestakeplaceatconstant temperatureandpressure.Theinequalitiesderivedearlieraresomewhatlessuseful sinceinapplicationsitismoredifﬁculttocontrol H , S or U .Thelastequation statesthatinaspontaneousprocessthe(possiblycomposite)systemwilladjust itscoordinatesinsuchawayastolowerthevalueofthesystem’sGibbsenergy; inthissenseitbehaveslikeapotentialenergyfunctioninmechanicswherea systemtendstowardminimumpotentialenergy(seeFigure 4.5).Equilibriumwill establishitselfattheminimumof G (T , p),muchasitdidinthelastchapterfora maximumof S (U , V ) onlythistimewehaveafunctionwhosedependentvariables aremoreunderourcontrol(ormoretothepointthosefoundinnaturallyoccurring circumstances).

Figure4.5InaspontaneoustransitiontheGibbsenergyisaminimum.

4.9Multiplecomponents

Inallthethermodynamicfunctionswehavestudiedsofarwehaveignoredthefact thatthechemicalcompositionofthesystemmightchange.Infact,thefunctionsof statemightbesummarizedby:

(4.111)

(4.112)

(4.113)

(4.114)

where ν1 , ... , νK indicatethenumberofmolesofeachchemicalspeciesinthe system.Thesemolarindicatorsarethermodynamiccoordinates.Wecanwrite

Wecanwritemorecompactly

wherethe G i arethemolarGibbsenergiesfortheindividualcomponentsinthe mixture.1 Notethatasimilarexpression(butwithdifferentialsoftheirnatural variablesservingascoefﬁcients)holdsford U andd H withthesamevaluesof G i

expresseshowmuchthecompositeGibbsenergychangespermoleofspecies1 beingaddedtothesystem.Aswithotherintensiveparametersinsubsystemsin contact(suchas T1 and T2 whenthesubsystems1and2areindiathermalcontact, or p1 and p2 whenthepressuresareallowedtobeunconstrained)thespeciﬁcGibbs energies G i willtendtowardequalitywhenthenumberofmolesofthedifferent speciesareallowedtovary(e.g.,bychemicalreactionsorphasechanges).

Wecanobtainsomeinsightintothisequalizingofthe G i byconsideringasystem atconstantpressureandtemperatureinwhichtherearetwochemicalspecies,A andB.Wehavethereaction:

1 The G i aredenoted µi inthechemicalliteratureandarecalledthe chemicalpotentials ofthesystemcomponents.

TheSecondLawofThermodynamics

ThenumberofmolesofBbeingcreated, δνB =−δνA .Andsince

Weﬁndthatatequilibriumwhered G = 0:

ThespeciesAandBmightbedifferentphasesofthesamesubstance.Weagain ﬁndtheequalityofthetwomolarGibbsenergiesforeachphasewhenequilibrium isestablished.Wewillﬁndthistobeofgreatutilityinthenextchapter.

Nextconsiderasystemcomposedoftwosubsystemsofequalvolume,oneis filledwith νA molesofidealgasspeciesAtheotherwith νB molesofB.Further, supposethetwogaseshavethesamepressure p andtemperature T .Nowsuppose thetwosubsystemsarebroughtintomaterialcontactwiththevolumebeingthe sumoftheoriginalvolumes,thepressuresandtemperaturesalsobeingthesame. WhatisthefinalGibbsenergy?Whatarethefinalenthalpy,internalenergy,and entropy?

TheinitialGibbsenergyis:

wherewehaveusedthesamespeciﬁcGibbsenergy G (p

T ) foreachoftheideal gasesAandB.Oncethegasesaremixedintothelargervolume,thetotalpressure willbethesame,butthepartialpressureswillbeonlyhalfasmuchsincethey occupytwicethevolumebutatthesametemperature.Hence,

Takingthedifferenceandusingtheformula(4.101)weget

Thisillustratesthatthespontaneousprocessofmixingtwoidealgasesleadsto adecreaseintheGibbsenergy.

Fortheinternalenergyandenthalpy,thejobiseasy.Thechangeoftheinternal energyis

since T = 0.Thesameholdsforenthalpywiththesubstitution c V → c p . Asexpected,duringthemixingofidealgasestheGibbsenergydecreases,while theenthalpyandinternalenergiesdonotchange.

Tocalculatetheentropychangewechooseareversibleisothermalpath.Weuse

with H = T = 0.Hence,

Themixingoftwoidealgasescausesanincreaseoftheentropyaswelearned earlier.

SupposenowthatthegasesAandBareidentical, M

.Then,from(4.126)wegettheincreaseoftheentropyafter mixing:

Doesthismakesense?Inthebeginningeachsubvolumecontainsthesamenumber ofmolesofidenticalgases.Whatchangesafterthemixingofthegases?Nothing. Thenthechangeinentropyshouldbezero.Sowegettwodifferentanswersfor thesameproblem.ThishasbecomeknownastheGibbsParadox.Thereason thisparadoxarisesisthatinclassicalphysicswecannotconsiderthemixingof twoidenticalgasesasalimitingcaseofthemixingoftwodifferentgases.Ifwe startourconsiderationfordifferentgases,theyhavealwaystobedifferent.Itis impossibletogettheanswerfortheentropychangeofthemixingoftwoidentical gasessimplybyequatingthemassesandthegasconstantsinequation(4.126). Inclassicalphysicstheexchangeofcoordinatesbetweentwoidenticalparticles (gasmoleculesinourcase)correspondstoanewmicroscopicstateofthesystem (twogasesinthecylinder),althoughnothingchangeswithsuchanexchangeat themacroscopiclevel.Thisparadoxdoesnotexistinquantumtheory,wherethe exchangeoftwoidenticalparticlesdoesnotcorrespondtoanewmicroscopicstate ofthesystem.Therefore,whentwoidenticalgasesaremixed, theentropydoesnot change.

Notes Asidefromthebooksalreadymentionedinearlierchapters,abeautifultreatment ofthermodynamicsfromanaxiomaticpointofviewisgivenbyCallen(1985). Thermodynamicsanditsapplicationsinengineeringhasalonghistory.Agood introductorylevelengineeringbookisthatbyÇengalandBoles(2002).Bothof

Emanuel’sbooks(1994, 2005)aswellasthebookbyCurryandWebster(1999) discussthethermodynamicsofconvectionphenomena.

NotationandabbreviationsforChapter4

c v , c p speciﬁcheats,theoverbarindicatesquantitiesexpressedper mole(Jmol 1 K 1 )

d Qrev inﬁnitesimalabsorptionofheat,subscriptindicatingthatthe changebereversible(J)

d SU ,V inﬁnitesimalchangeinentropyduringwhich U and V are heldconstant(JK 1 )

F force(N)

g Gibbsenergyperkilogram(Jkg 1 )

G GibbsenergypermoleormolarGibbsenergy(Jmol 1 )

G Gibbsenergy(J)

H enthalpy(J)

κ = R/cp (dimensionless)

M bulkmass(kg)

µi chemicalpotentialofspecies i ,sameas G i (Jmol 1 )

ν , νA , νB numberofmoles,numberofmolesofspeciesA,B

p pressure(Pa)

p(V ) pressureasafunctionofvolume;expressionforacurvein the (p, V ) plane

R gasconstantforaparticulargas(Jkg 1 K 1 )

s entropyperunitmass(lowercaseindicatesperunitmass) (JK 1 kg 1 )

S , SA , SB entropy,entropyofstateA,stateB(JK 1 )

Ssys , Ssurr , Suniverse entropyforthesystem,surroundings,universe(sys+surr)

θ potentialtemperature(K)

U internalenergy(J)

UA , UB internalenergyatstatesA,B(J)

WA →B , QA →B workdonebythesystem,heattakenintothesystemin goingfromstateAtostateB(J)

Problems

4.1Aparcelisliftedadiabaticallyfrom z = 0to z = H ,whatisitschangeinentropy?

4.2Computethechangeinentropyforanidealdrygasofmass M whichisheatedat constantvolumefrom T1 to T2 .Take M = 1kg, T1 = 300Kand T2 = 310K.

4.3Aparcelisliftedisothermallyfrompressure p0 to p1 .Finditschangeinpotential temperature.Take p0 = 1000hPaand p1 = 500hPa, T0 = 300K.

4.4A1kgparcelat500hPaand250Kisheatedwith500Jofradiationheating.What isthechangeinitsenthalpy?Whatisthechangeinitsentropy?Itspotential temperature?

4.5Aquantity18gofwateris(a)heatedfrom273Kto373K,(b)evaporatedtogasform, and(c)heatedto473K.Allstepsareperformedatconstantpressure.Computethe changeinentropyforsteps(a),(b),and(c).Note:theheatcapacityforwatervaporis ≈ 2kJK 1 kg 1 .

Usethesedatainthenexttwoproblems:2kgofanidealgas(dryair)isattemperature 300K, p = 1000hPa.

Step1:thevolumeisincreasedadiabaticallyuntilitisdoubled.

Step2:thepressureisheldconstantandthevolumeisdecreasedtoitsoriginalvalue.

Step3:thevolumeisheldconstantandthetemperatureisincreaseduntiltheoriginal stateisrecovered.

4.6(a)Sketchtheprocess(steps1,2and3)inthe V –p plane.

(b)Whatarethevolumeandtemperatureattheendofstep1?

(c)Whatisthechangeofenthalpy H ,internalenergy U ,andentropy S ,during step1?

(d)Howmuchworkisdonebythesysteminstep1?

4.7ContinuingProblem4.6.

(a)Howmuchworkisperformedinstep2?

(b)Whatisthetotalamountofworkinallthreesteps?

(c)Whatistheentropychange S instep2?

(d)Whatarethetotalchangesin U , H , S duringallthreesteps?

4.8Adryairparcelhasamassof1kg.Itundergoesaprocessthatisdepictedinthe V –p planeinFigure 4.6.Calculatethechangeofentropy,enthalpyandinternalenergyfor thisairparcel.

Figure4.6DiagramforProblem4.8.

Figure4.7DiagramforProblem4.9.

4.9Findthechangeofinternalenergyandenthalpyforthecyclicprocessshownin Figure 4.7.StartingfrompointAdescribehowthetemperaturechangesduringthis cyclicprocess.

4.10Showthattheworkperformedbyasystemduringareversibleisothermalcycleis alwayszero.

4.11ShowthatfortheCarnotcycleofanidealgasholds(4.74)

Hint: Divideoneofthefourequationsforworkdonealongthedifferentlegsofthe cyclebyanotherone.

4.12AtropicalstormcanbeapproximatedasaCarnotcycle.Airisheatednearly isothermallyasitﬂowsalongtheseasurface(typically27 ◦ C).Theairislifted adiabaticallyintheeyewalltoaheightabovethetropopausewhereitbeginsto coolduetolossofinfraredradiationtospace.Thetemperaturewherethisoccurs isabout 73 ◦ C.Finallytheairdescendstothesurfaceadiabatically.Calculatethe thermodynamicefﬁciencyofthis“heatengine.”2

4.13Showthattheworkdoneina(reversible)Carnotcycleistheproductoftheentropy differencebetweenthetwoadiabatsandthetemperaturedifferencebetweenthetwo isotherms(seeFigure 4.4).Theresultholdsforanysystem,notjustanidealgas.

4.14Airisexpandedisothermallyat300Kfromapressureof1000hPato800hPa.What isthechangeinspeciﬁcGibbsenergy?

2 KerryEmanuel(2005)explainsthissimplemodelatbeginner’slevelinChapter10ofhisbook DevineWind

Airandwater

Innaturewaterpresentsitselfinsolid,liquidandgaseousphases.Energytransfers duringtransformationsamongthesephaseshaveimportantconsequencesin weatherandclimate.Thesystemofredistributionofwaterontheplanetconstitutes the hydrologicalcycle whichiscentraltoweatherandclimateresearchand operations.Waterisalsoanimportantsolventintheoceans,soilsandincloud droplets.Thepresenceoftinyparticlesintheaircaninﬂuencetheformationof clouddropsandtherebychangetheEarth’sradiationbalancebetweenabsorbed andemittedand/orreﬂectedradiation.Theseandothereffectsleadusintothe fascinatingroleofwaterintheenvironment.Ofcourse,thermodynamicsisan indispensabletoolinunravelingthisverychallengingpuzzle.

5.1Vaporpressure

Westartwithadiscussionoftheequilibriumgaspressureinachamberindiathermal contactwithareservoirataﬁxedtemperature, T0 .Thechamberistohaveavolume thatisadjustable,asshowninFigure 5.1.Inthefollowingletthechamberhaveno airpresent–onlythegasfromevaporationoftheliquid.Wearetochooseavolume V suchthatthereissomeliquidpresentatthebottomofthechamber(wesayherethe bottomofthechamber,butotherwiseweignoregravity).Therearegasmolecules constantlystrikingtheliquidfromaboveandsticking.Therateatwhichparticles entertheliquidphasewillbeproportionaltothenumberdensityofmoleculesin thegasphase(recallfromourdiscussionofkinetictheorythattheﬂuxofmolecules perunitperpendicularareacrossingaplaneis 1 4 n0 v where v ismeanspeedofthe vapormolecules(seeChapter 2)).Moleculesintheliquidphasemusthaveatleast acertainminimumverticalcomponentofvelocityinsidethecondensedphaseto escapetheliquidsurface(theyhavetoovercomethepotentialenergynecessary toleavethesurface).Ifwewaitfortheequilibriumtoestablishitself,therateof moleculesleavingtheliquidsurfacegoingintothevolumeabovewillexactlyequal

Figure5.1Quasi-staticcompressionofavaporatconstanttemperature.InA→ B,thereisnoliquidpresent.AtpointB,liquidbeginstoformonthebaseofthe chamber.InB → C,thereisliquidinequilibriumwiththevapor.

therateofmoleculesarrivingandsticking.Iftherateofdeparturesshouldexceed therateofstickingarrivals,thenumberdensityofgasmolecules n0 wouldsteadily increaseuntiltheratesequalize.

Ifthevolumeofthechamberisdecreasedslightly,theequilibriumwillhave tobere-established.Thesteadystateequalityofarrivalanddepartureratescan onlybemaintainedforthesamenumberdensityinthegasphaseasbeforesince thetemperatureisheldﬁxed.Indecreasingthevolumewemustcondenseanet amountofvapormoleculesintotheliquidphaseundertheseconditions.Asthe excessnumberofstickingmoleculesfallsintoapotentialenergywellwhenthey entertheliquidtheirvelocitiesintheliquidincrease(pictureaslowlymoving marblerollingoffthetable’sedge,whereitskineticenergysuddenlychangesfrom nearzerotoalargevalue).Thisexcesskineticenergyofthemoleculesentering theliquidisquicklysharedwiththeotherwatermoleculesintheliquid,slightly raisingitstemperature.Thistinyexcesstemperatureoverthatofthereservoirin contactwiththesystemisquicklywipedout(withaheat(enthalpy)transferto thereservoir),maintainingtheisothermalconstraint.Asthevolumeisdecreased, thereisanotherformofenergybeingaddedtothesystem,becauseduringthe compression,workisbeingperformedonthesystembythepiston(maintaining constantpressure).

Considerthesituationina V –p diagram(Figure 5.2).Wewishtotraceanisotherm forthissystem.WestartatpointAwhereallthematterinthechamberisinthegas phase.WecompressthegasisothermallyuntilwereachpointBwhereliquidbegins tocondenseontheﬂoorofthechamber(nodroplets,please,becausesurfacetension

B to C

A to B

criticalisotherm

Figure5.2Pressureversusvolumediagramforamixtureofliquidandvapor.We startatpointAwherethesystemisallvapor,andcompressisothermallytopoint B,whereliquidbeginstoappear.IngoingfromBtoCamixtureofliquidand vaporispresent.Alongthissecondstagetheisothermisalsoanisobar.Theportion ofthecurvetotheleftofCrepresentstheliquidphase.Thecriticalisothermis shownbytheboldline. T1 < T2 < T3 <...< T7 .Thedashedlineboundsthe areawheretheliquidanditsvaporareinequilibrium.

ofthecurveddropletsurfaceswouldcomplicatetheenergeticshere;butalsono gravityinthisexperiment–thewatercouldcongregateontheceilingforallwe carehere).Asweisothermallyandquasi-staticallycompressfurther,thepressure remainsconstantasmorematterisconvertedfromgastoliquidphase.Heatreleased fromthecondensationandfromtheworkperformedduringthecompressionmust betransferredfromthechambertothereservoirinordertomaintainthesame temperature.Thechangeininternalenergyiscomposedoftwocontributions,the workdonebythepistononthegasandtheheatassociatedwiththematterbeing convertedfromvaportoliquid.Thechangeinenthalpyofthesystemdoesnot dependonvolume,soitschangeonlyinvolvesthecondensationcontribution.The changeinenthalpyinmovingalongfromBtoCinFigure 5.2 is

where M isthe(positive)amountofmattercondensedintheprocessand L = Hvap iscalledthe enthalpyofvaporization (latentheatofevaporation inold fashionedterminology).Forwaterat0 ◦ C, L = 2.500 × 106 Jkg 1 ,anditisnearly independentof T (error < 1%,overtherangeofinterestinatmosphericscience). Whenthevolumeisreducedtosuchanextentthatwehaveonlyliquidinthe chamber,thefurtherdecreaseinvolumerequiresaverylargeincreaseinpressure,

becauseliquidsarealmostincompressible(seethesegmentofthecurveC → Dof theisotherm).

Ifweplotisothermscorrespondingtohighertemperatures(T1 < T2 < T3 ), weseethatthelengthofthehorizontalportionoftheisothermdecreases.This meansthatwithincreasingtemperaturethevolumeintervalforwhichtheliquid andvaporcancoexistinequilibriumdecreases.Thishappensuntilwereachthe so-called criticalisotherm,wherethisintervalshrinkstoapoint.Thetemperature correspondingtothisisothermiscalledthe criticaltemperature.Atthecritical temperature Tc , ∂ 2 p/∂ V 2 = 0,an inﬂectionpoint.Theisothermswithtemperature wellabovethecriticaltemperaturearehyperbolae,becausethesubstanceatvery hightemperaturesbehaveslikeanidealgas.

5.2Saturationvaporpressure

Theequilibriumpressureofwatervaporaboveaﬂatsurfaceofliquidwaterina chambersuchasshowninFigure 5.1 iscalledthe saturationvaporpressure.Itis independentoftheshapeofthevolumeinthecylinder(sinceitonlydependsonthe numberdensity ns ofthevapor).Thesaturationvaporpressure(usuallydenoted es ) ishoweveraverystrongfunctionoftemperature T .Thisisintuitivelyreasonable sinceanincreaseintemperaturewillincreasetheproportionofliquidmolecules havingvelocitiesabovethethresholdtodepartfromthesurface.Moredepartures willrequiremorearrivalratestomaintainequilibrium.Thisinturnwillrequirea largernumberdensitywhichisproportionaltothevaporpressure.Notethatthe ﬂuxofmoleculesmovingdownperpendicularlyis 1 4 ns v (seeChapter 2).

Figure 5.3 showsagraphofthesaturationvaporpressureofwateroveraﬂat liquidsurfaceversusthetemperatureindegreesCelsius.Manyaspectsofweather andclimatedependonthisveryrapidincreasewithtemperature.Asaroughbut

Figure5.3Saturationvaporpressureforwateroveraﬂatliquidsurfaceversus temperatureindegreesCelsius.

usefulruleofthumb,thesaturationvaporpressuredoublesforevery10 ◦ Cincrease intemperature(atleastintherangeofinterestforatmosphericscience).Evenso, atmoderatetemperaturesthesaturationvaporpressureisverysmallcomparedto atmosphericpressurenearthesurface(usually5to30hPacomparedto1000hPa).

Doesthepresenceofdryairaffectthesaturationvaporpressureofwater?Perhaps theaddedpressureoftheairontheliquidsurfacesqueezesmorewatermolecules intothevaporphase.Butonthecontrary,someairdissolvesintheliquidandthereby mighthindertheﬂuxofmoleculesoutoftheliquidsurface.Botheffectsarepresent buttogethertheirimpactislessthan1%ofthesaturationvaporpressure.

5.3VanderWaalsequation

Aswelearnedearlier,theapproximationofanidealgasworkswellifwecanneglect theintermolecularforces.Thisisvirtuallyalwaysthecaseforthemajorconstituents ofairatEarth-likeconditions.Butasagasnearsitscriticaltemperatureandthe liquidorsolidstatecancoexistwiththegasphase,thedeparturefromidealityis important.AsweseefromFigure 5.2 theidealgasequationofstatedescribesthe behaviorofrealgasesinlimitingcasesofhightemperaturesandlowpressures. Isothermsforanidealgasarerectangularhyperbolae(p ∝ 1/V ).Asmallpressure decreaseleadstoalargeincreaseinvolume(BtoAinFigure 5.2).However,the idealgasequationofstateisnolongeragoodapproximationwhenthetemperature ofthegasisbelowitscriticalpoint,andthevolumeisintherangewherethe isothermsbecomehorizontal(seetheﬂatsegmentCtoBinFigure 5.2);i.e.,there isamixtureofliquidandgasinequilibriumtogether.

Averyusefulequationthatdescribesthebehaviorofmanysubstancesovera widerangeoftemperaturesandpressureswasderivedbyvanderWaals.Thevan derWaalsequationfor1molofgasis:

p + a v 2 (v b) = R∗ T [vanderWaalsequation](5.2)

where a and b areconstants(differentfordifferentsubstances)and v isthevolume permoleofthegas(i.e.,the reducedvolume orthe speciﬁcvolume).Theterm b in(5.2)isduetotheﬁnitesizeofthemolecules,whiletheterm a /v 2 isduetothe effectoftheattractivemolecularforces.For a = b = 0,thevanderVaalsequation reducestotheIdealGasLaw(5.2).

UsuallythevanderWaalsequationiswrittenintheform p = R∗ T v b a v 2 .(5.3)

Figure 5.4 showsanexampleofisothermscalculatedusingthevanderWaals equation.IfwecompareFigures 5.2 and 5.4,weseethatvanderWaalsisotherms

Figure5.4VanderWaalsisotherms.Theisothermwiththeinﬂectionpointisthe criticalisotherm.Theequilibriumvaporpressure e issuchthattheshadedareas areequal.

reproducemanyfeaturesofrealgasbehavior.AsshowninFigure 5.4 forlarge v andlow p thereisalargeincreaseinvolumewithasmalldecreaseinpressure.Fora liquid(small v andhigh p)thereisasmalldecreaseinvolumewithalargeincrease inpressure.Thereisacriticalisothermwithtemperature T = Tc indicatingapoint ofinflection(∂ 2 p/∂ v 2 = 0).Theisothermswithtemperatureshigherthan Tc are verysimilartothoseinFigure 5.2.However,theisothermswithtemperatureless than Tc lookverydifferent:theyarenothorizontalintheregionwheretwophases, waterandvapor,coexist.ConsideroneparticularisothermABCDEFGderived fromthevanderWaalsequation.Letuscompressthegasuntilsaturationoccurs atpointFontheisotherm.Thenwithafurtherdecreaseofthevolumethereisno increaseinpressure,whichcorrespondstothehorizontalstretchFB.Instead,the vanderWaalsisothermshowsanincreaseinpressure(partofthediagramFE). Alongthisbranchofthecurvethevaporis supersaturated.Vaporcantheoretically existforthesevalues,butifasmallimpurityispresentsuchasadustparticle,ora scratchonthewall,thevaporwillbegintocondenseonthissiteandthesystemwill collapsetotheflathorizontallineBFinFigure 5.4.Inotherwordsthisstateofthe vaporisunstable:anydisturbancecausesittomigratetoastableconditionwhich containstwosubsystems,vaporandliquid.So,ifweplotthevanderWaalsisotherm foragiventemperature,wewillnotfindtheflatportion(BDF)whichweknow shouldbethere(fromexperiment).Wehavetoputitin“byhand.”Buthowdowe decidetheproperpressurevalueatwhichtoinsertthisflatportion?Therule(first discoveredbyMaxwell)isthattheareasboundedbythecurvesBCDBandDEFD havetobeequal.Letussketchaproof.ConsiderthecycleFEDCBDFinFigure 5.4

(a“ﬁgure8”onitsside).FromtheFirstLawweknowthatforanisothermalprocess theworkdoneduringaclosedcycleisequaltotheamountofheatabsorbedbythe system, W = Q ,sincethechangeininternalenergyiszeroforacyclicprocess. Wealsoknowthattheloopintegral d Qrev /T = 0(ourprocessiscontrivedtobe reversible).Foranisothermalprocesswecantaketemperatureoutoftheintegral andget Qloop = 0.Since Qloop = 0,wealsohave Wloop = 0.Ifthehorizontal linewerenotsuchthattheareasareequal,ourimaginary(butrealizable)process wouldviolatethelawsofthermodynamics(either d U = 0or d S = 0orboth).

Anexcellentdiscussionofunstablestates(supersaturated,etc.)canbefound inadvancedbooks,especiallythediscussioninCallen(1985),wherethecase isillustratedwiththevanderWaalssystem.1 Thecriterionforstabilitycanbe expressedintermsoftheconcavityorconvexityofthethermodynamicfunctions:

orfortheGibbsenergy:

Ifthegraphsfor S (U , V , M) and G (T , p, M) havethewrongsignofconcavitythe branchofthecurvewherethecriteriafailwillbeunstable.

5.4Multiplephasesystems

Weproceedwiththecaseofwaterinbothitsliquidandvaporformsinequilibrium inacontainer.Thisisaone-component(onlyonechemicalspeciesispresent) systemwithtwophases(liquidandgas)inequilibrium.Experiencetellsusthat thetwophasescancoexistinequilibriuminthisconﬁguration.Infact,wehave seenthatforagivenmassofthesubstancethereisarangeofvaluesofvolumefor whichtheequilibriumexistswithtransfersofmassfromonephasetotheotheras thevolumeischanged(atconstantpressureandtemperature).Thisisthehorizontal lineCBinFigure 5.2.Letthetemperatureandpressuresbe T0 and p0 alongCB.In the T –p planethislineisasinglepoint(T0 , p0 ),seeFigure 5.5.Ifweweretomake aninﬁnitesimalchangeinthetemperaturereservoirto T0 + T (seeFigure 5.6), thenwewouldmovetoahigherhorizontallineinFigure 5.2,therebyoperatingat

1 CallengivesanexpressionfortheentropyofavanderWaalsgas: S (u, v ) = ν R∗ ln (v

where c isthemolarheatcapacityatconstantpressure.Forwatervapor,

, and c = 3.1.

Figure5.5Apointinthe T –p planeinwhichliquidandvaporareinequilibrium.

Figure5.6Asthetemperatureisincreasedfrom T0 to T0 + T ,thesaturation vaporpressurewillincreasefrom p0 to p0 + p.

(p0 + p, T0 + T ).Aswechangefromoneﬂatlineinthe V –p plane,wetrace outanewcurveinthe T –p plane.Letuscallit pequil (T ). Alongthiscurve, p = pequil (T ),thetwophasescanexistinequilibrium.Infact, if T and p lieonthecurve(i.e., p = pequil (T ))thenthevolumecanbevaried isothermallyandisobaricallycausingmasstotransferfromonephasetotheother untiloneofthephasesisexhausted.ThepointinFigure 5.5 liesbetweenpointsB andbinFigure 5.7.Thevariationcanbethoughtofasintooroutofthe T –p plane alongthe V (volume)axis.

Theupshotofallthisisthatwhenthetwophasesaretogetherinequilibrium therewillbeauniquecurveinthe T –p plane.Thislineisofgreatinteresttous. Forexampleitsslopetellsushowmuchthesaturationvaporpressurewillincrease forasmallchangeinthetemperature.

Watercanformiceasitssolidphase.Itturnsoutthatasingle-componentsystem suchaspurewatercancoexistinallthreephasessimultaneouslyonlyatasingle pointinthephasediagramcalledthe triplepoint. Thetriplepointforwateris 273.16Katapressureof6.11hPa.Atpressuresbelow6.11hPaiceandvaporcan

Figure5.7Aschematicphasediagraminthe T –p planeofthephasesofwater. BelowthelineABCthephaseisvapor.TotheleftofABDthephaseissolid. AboveDBCthephaseisliquid.Theselinesarecalledphaseboundariessince alongthemtwophasescancoexist.ThepointBistheso-calledtriplepointsince allthreephasescancoexistatthispoint.Thedashedlineabindicatesatmospheric pressure.Theboilingpointisatb.

coexist,butnoliquidwaterwillexistinequilibrium.Therearemanyotherphases ofwaterinitsdifferentcrystallineforms,andtheseareimportantinhighpressure situationswelloutsidetherangeencounteredinatmosphericapplications.

5.5Phaseboundaries

Considerapointrepresentingastateinthe T –p plane.Atsuchanarbitrarypointthe Gibbsenergyforeachofthesubsystemphaseswillhavegivenvalues Ml gl (T , p) and Mg gg (T , p),where Ml and Mg arethemassesoftheliquidandgaseous subsystemsand gl (T , p) and gg (T , p) arethespecificGibbsenergiesforeach(by useoftheterm specific wemeanthateachisperunitmass).TheGibbsenergy forthewholesystemisthesumofthemassweightedspecificGibbsenergies. Notethatwhenthetwophasescoexistinequilibrium,thepressure,temperature andspecificGibbsenergiesofeachphasearehomogeneousthroughout(e.g.,the specificGibbsenergyoftheliquidequalsthatofthevapor).Onlythedensitydiffers fromonephasetotheother.Toshowthis,supposeweareonaphaseboundaryin the T –p plane(thecurveofequilibriumstatesdescribedinthelastsection),andwe changethevolumeofthesystemslightly, V .Inthattransition,thetotalsystem Gibbsenergy, G (T , p) doesnotchangesince p, T ,and M donotchange.This meansthat

Ml gl (T , p) + Mg gg (T , p) = 0(5.6) wherewehavemadeuseofthefactthatthespeciﬁcGibbsenergiesdonotchange because p and T areﬁxed.Nowinthelastequationwecannotethat Ml =− Mg

becausethemasshastobeconserved.Thisleadstotheinterestinganduseful conclusionthatalongaphaseboundary, gl (T , p) = gg (T , p) [alongaphaseboundary].(5.7)

Inthe T –p phaseplanedifferentregionsrepresentdifferentphasesofthistwophasesystem.Aphaseboundaryexistsalongwhichtheliquidandgaseousphases cancoexistinequilibrium.WehaveshownabovethatthespeciﬁcGibbsenergies foreachindividualphaseareequalalongthephaseboundaryinthe T –p plane. Thisresultwillallowustocalculatetheslopeofthephaseboundaryinthenext section.

Gibbsphaserule Inthelasttwosectionswediscussedmultiplephasesandin particularthecaseofwateranditsthreephases.Ingeneraltheremightbemorethan onecomponentaswell(forexample,amixturewithdifferentphasesforeach).The intensivevariablesintheproblemarethetemperatureandthepressure(commonand equalforallthecomponents).Wealsoknowthatwhenthesystemisinequilibrium, thespeciﬁcGibbsenergiesforagivencomponent G i (T , p) willbeequalforthe phasesofthatcomponent.Inlookingbackatthewaterproblemweseethatthereare regionsinthe T –p diagramwhereboth T and p canbevariedindependently.These areregionswherethereisasinglephasepresent.Thelinesinthediagram(Figure 5.7) representalocusofpointswheretwophasesarepresentinequilibrium.Finally,the triplepointisthesinglepointwherethreephasesarepresentinequilibrium.Thisis thesituationwhenthereisonlyonecomponentpresent(water).

The numberofdegreesoffreedom denotedhereas F (differentfromthesamename usedinkinetictheory)referstothenumberofwaysoneoftheintensivevariables (T , p, G 1 , G 2 , ... , G c ,where C isthenumberofcomponents)canbevaried independently.Forexample,intheregionsawayfromphaseboundariesinFigure 5.7 both T and p maybevariedindependently(twodegreesoffreedom),butonaphase boundary,onlyoneofthesevariablesisindependentsincethephaseboundaryis deﬁnedbyafunction, p = p(T ) (onedegreeoffreedom).Atthetriplepointthe numberofdegreesoffreedomiszero.

Itispossibletoderiveaformulaforthenumberofdegreesoffreedomfora multi-component,multi-phasesystemanditisworthpresentinghere.Firstnotethat thenumberofmolarconcentrationvariablesis C ,butonly C 1areindependent, sinceweareinterestedonlyinmolefractions.Thenumberofphasesis P .Sothetotal numberoftheseintensivevariablesis P (C 1).Buttherearesomerelationsbetween thesevariablesbecausesomeoftheGibbsenergiesarerelatedtooneanother.For eachindividualcomponenttheGibbsenergiesofthephaseshavetobeequal.For eachcomponentthereare P 1relations.(Forexample,iftherearetwophasesthere isonlyonerelation,say G 1 = G 2 ,etc.)Thisreducesthenumberofindependent

intensivevariablesby C (P 1).Westillhavethepressureandtemperaturethatare independenttoadd.Thusweobtain

leadingtothe Gibbsphaserule:

F = C P + 2 [numberofdegreesoffreedom:Gibbsphaserule].(5.9)

Weseethatforasingle-componentsystemwithonephase, F = 2(theregions betweenphaseboundariesinFigure 5.7).Whentherearetwophasespresent(ona phaseboundaryline), F = 1.Whenallthreephasesarepresent, F = 0,thetriple point.

Othersystemsofinterestincludethecaseofwaterwithadissolvedsolutesuchas salt.Thiswouldbeatwo-componentsystemwithtwophases(theliquidsolutionand thesaturatedvaporinequilibriumwithitabove).Thiscasewillbediscussedlater.

5.6Clausius–Clapeyronequation

HavingestablishedthatthespeciﬁcGibbsenergiesforliquidandgaseousphases arethesamealongthephaseboundary,wecannowproceedtocalculatetheslope ofthephaseboundaryinthe T –p plane.Thisslopeistherateofchangeofthe vaporpressurewithrespecttotemperatureasthesystemisallowedtomovealong thephaseboundary.Thisslopemeasurestherateatwhichthesaturationvapor pressureincreasesforincrementalchangesintemperature–animportantquantity inmeteorology.

Firstconsidersuchareversiblechangeofthecompositesystem(gasandliquid inequilibrium)alongthephaseboundary.Wehave:

Then(usinginﬁnitesimalnotationinsteadof )

wherethesmallletters s and v referto speciﬁc entropyandvolumesrespectively, andhenceforthwedenotethesaturationvaporpressureas es .Rearranging:

First,noticethat vg vl (e.g.,foronegrammoleofvapor22.4 × 103 cm3 versus18cm3 ofwater)sothat vl canbeneglected.Thespeciﬁcvolumeofan

idealgasis RT /p,where R isthegasconstantfortheparticularspecies(herewater vapor).Thedifferenceofspecificentropiescanbecalculated.Thisdifferenceisthe changeinspecificentropyasweconvertaunitmassofliquidintogasformata fixedtemperature: Hvap /T = L/T ,where L istheenthalpyofvaporization(per kilogram).Wearriveatthe Clausius–Clapeyronequation:

[Clausius–Clapeyronequation].(5.13)

Beforeproceedingtointegratethisequationtoﬁndanexpressionfor es (T ),it shouldbenotedthattheprocedurejustemployedisverygeneralandcanbeapplied tomanyotherproblems.Whilewewillnotpursueithere,itisperhapsclearthat theequilibriumwespeakofcouldbethatofchemicalspeciesinsteadofphases, oritcouldbeacombinationofboth.Inphysicalchemistrytextsthetechniqueof equilibriumboundariesutilizingtheGibbsenergycanbefoundtoleadtosuch diverserulesasthetemperaturedependenceofreactionratecoefﬁcients.

5.7IntegrationoftheClausius–Clapeyronequation

Toproceedwedivideeachsideoftheequationby es andmultiplythroughbyd T . Theleft-handsidewillbeafunctiononlyof es andtheright-handsidewillbeonly afunctionof T .Thisallowsustointegrate:

Nextwechoosethelowerlimittobe273.2Ksothat es (0) = 6.11hPa.Thevalue 6.11hPahastocomefromobservations–thermodynamicscannottellusthevalueof suchaconstant.Afterall,this integrationconstant shouldbedifferentfordifferent substances(e.g.,comparethisvaluetothevaporpressureofmercuryat20 ◦ C whichis0.16Pa).Insertingtheothernumericalvaluesleadsto:

wherewehaveinsertedthenumericalvaluesfor R = Rw and L = Lvap .This equationcanalsoberearrangedtogivethehandyformula:

es = 2.497 × 109 e 5417/T (hPa) [integratedformofthe Clausius–Clapeyronequation].(5.17)

Figure5.8Saturationvaporpressureoverice(dashedline)andliquidwater (solidline).ThecurveswerecomputedwiththeClausius–Clapeyronequation with Lvap = 2.5 × 106 Jkg 1 and Lsublime = 2.83 × 106 Jkg 1 .

TheresultinggraphisshowninFigure 5.8.Ifweconsiderthevaportobein equilibriumwithanicesurfacewemustuse Lsublime = 2.83 × 106 Jkg 1 andthe resultofthisisshowninFigure 5.8 asthedashedline.Notethatbelow T = 273K thesaturationvaporpressureislargeroverliquidthanoverice.Thismeansthatif thereisanicesurfaceinthechamber,theicesurfacewillnotbeinequilibriumwith thevaporinthechamber,whichisatthesaturationvalueforaliquidsurface.The upshotisthattheicevolumewillgrowinsizeattheexpenseoftheliquidmass. Eventuallythevaporpressureinsuchachamberwillbecomethatofthesaturation vaporpressureoverice.Thiseffectisimportantinacloudattemperaturesbelow freezing(0 ◦ C)inwhichicecrystalsareembeddedinaﬁeldof supercooled water droplets.Thetermsupercooledisappliedsincewaterdropletscanbebelowthe freezingpoint(0 ◦ C)withoutactuallyfreezing.Wewillseelaterthatthepresenceof animpurityinthedroplet,suchassilverchloride,cancausethedroplettofreezeat slightlyhighertemperatures.Thepointatwhichallthesupercooleddropletsfreeze is 40 ◦ C.

Example5.1:comparingvaporpressureovericeandliquidwater Aplotofthe saturationvaporpressureoveraﬂatliquidwatersurfaceisshowninFigure 5.8 Ifweconsiderthevaportobeinequilibriumwithanicesurfacewemustuse Lsublime = 2.834 × 106 Jkg 1 .Thenthevaporpressureovericeisapproximately

Thecorrespondingformulaforvaporpressureoveraliquidsurfaceis

wherethetemperatureinbothformulas 2 isinkelvinsandthevaporpressureisin hPa.Moreaccurateformulascanbederivedbytakingintoaccountthetemperature dependenceof Lsublime ,etc.

TheseareshowninFigure 5.8 withthevaporpressureoverliquidasthesolid lineandoverthesolidsurfaceasthedashedline.Notethatbelow T = 273K,if thereisapieceoficeinthechamber,itwillgrowinsizesincetheevaporationrate fromtheicewillbelessthantheevaporationratefromaﬂatliquidsurface.

Anothereffectwhichisimportantinsomeapplicationsisthedifferencein saturationvaporpressuresfordifferentisotopesofwater,H2 O18 versusH2 O16 . Bothoftheseisotopesareradiologicallystable(theydonotdecay)andbothare foundinnature.Theheavierisotopemakesuponlyabout0.20%ofoxygenatoms. Thevibrationfrequenciesofthewatermoleculesareaffectedslightlybythesmall amountoftheheavierisotope.Thisleadstoaverysmallchangeinthesaturation vaporpressureofwaterhavingmoreorlessoftheheavyisotopepresentintheliquid. Thissmalleffecthasatemperaturedependenceanditleadstoslightlydifferent evaporationratesoverwarmversuscooloceanwaters.Thisleadstoadifferent concentrationoftheisotoperatiointhewatervaporoverthesedifferenttypesof oceanwater.Theratiooftheheavyisotopicwatertothelighteronecanbemeasured inicecoresandinothermaterial.Theapplicationisinpaleoclimatologywheresnow depositedonpolariceﬁeldsleavesinitslayeringarecordoftemperaturesignatures ofpastclimates.Thisisaveryactiveareaofcurrentresearch–notjustforthewater isotopesbutforthoseofmanyotherelements.

5.8Mixingairandwater

Whenthereisamixtureofairandwatervaportheeffectivemolecularweightof thegaschangesslightly.WecanuseDalton’sLaw(Section 2.5)toﬁndtheeffective valueofthegasconstant, Reff .Firstweﬁndtheeffectivevalueofthemolecular weightwhensomewatervaporispresent.Theresultis 3

2 Emanuel(1994)containsextendeddiscussionsoftheserelations.

3 Thealgebrarequiredbeginshereandcontinuesonthenextpage.

where w ,the mixingratio,isgivenbytheratioofthemassofwatervaportothe massofdryairinaparcel:

Enteringthecalculationwastheratioofmolecularweightsofwatertoair:

The mixingratio w isusuallygiveninunitsofgramsofwatervaporperkilogram ofdryair(ofcourse,inequationssuchasthosejustaboveitwouldbein (kgvapor)(kgair) 1 ).

Theeffectivevalueofthegasconstantisthen

Theresultsaboveallowustowritetheequationofstateforthemoistairas

where Rd = 287Jkg 1 K 1 and

iscalledthe virtualtemperature.Notethatthevirtualtemperatureisalwayslarger thantheactualtemperature,butthattheyseldomdifferbymorethan1K(w is seldomgreaterthan4×10 2 (kgvapor)(kgdryair) 1 .

Theuseofvirtualtemperatureallowsthemeteorologist(whoseinterestis buoyancy)tocorrectthedensitytolowervalueswhenwatervaporispresentwhile retainingthesimplicityoftheIdealGasLawfordryair(Rd = 287Jkg 1 K 1 ).This

Footnote3 continued

workstoaverygoodapproximationinpracticalsituations.Rememberthatwater vaporhasalowermolecularweightbuteachwatervapormoleculeattemperature T hasthesameeffectonpressureasanairmolecule.Thusifthereisamixtureof watervaporandairwherethetotalpressureisthesame,thedensitywillbelower thanforthesamevolumeofdryairatthesametemperatureandpressure.

The saturationvapormixingratio isdenoted ws anditisastrongfunctionof temperature.Keepinmindthatthesaturationvapormixingratioisalsoafunction oftheairpressure(equivalenttoaltitude)becauseitistheratioofwatermassto airmass.

The relativehumidity isgivenby

Orintermsofpercent:

The dewpoint temperature, TD ,isthetemperatureatwhich

Inotherwords,foragivenvalueof w ,itisthetemperatureforwhichthatvalueof mixingratio, w ,isequaltothesaturationmixingratio, ws .Wereachthedewpoint bycoolingatconstantpressuretothetemperaturewhere w inaparcelreachesits saturationvalue.

Toﬁndtherelationshipbetweenthepartialpressureduetovaporandthepartial pressureofthedryairinaparcel,wewrite:

Takingtheratioandrearranging:

[mixingratiointermsofvaporandairpressure](5.33) where

Andofcoursetheformulaholdsforthesaturationcaseaswell ws = es p [saturationmixingratio].(5.35)

Strictlyspeaking, p aboveis pdry whichisnot patmo = e + pdry ,butusuallythe differenceisnegligible.

Wecanalsowrite

andsincewearetolowerthetemperatureto TD isobaricallywecansay

Thenifeither w or e areknownwecanﬁnd TD bysolvingoneoftheseequations for TD .

Example5.2 Watervaporismostlydistributedintheboundarylayerofthe atmosphere(lowest1–2km).Assumetheatmosphereisisothermalat300Kand thatthesurfacehumidityis95%.Ifallthevaporisdistributeduniformlyinthe ﬁrst1.5km,howmuchwaterliesaboveagivensquaremeterofsurfaceinkgm 2 ? Alsoexpresstheresultinmmequivalentofliquidwater.

Answer:FirstcomputethesaturatedwatervaporpressurefromtheClausius–Clapeyronequation: es = 36hPa.Thenthesaturationmixingratioisgivenby ws = (0.622) × (0.036) (kgwater)(kgair) 1 = 0.022(kgwater)(kgair) 1 .Thus, thewatervapormixingratiois95%ofthis,0.021(kgvapor)(kgair) 1 .Themass ofairinthe1.5kmcolumnis1500m3 × 1.2kgm 3 = 1800kg.Multiplyingyields 38kgofwatervaporinthe1m2 column.Theequivalentdepthofliquidwateris Mwater /(ρliqwater × 1m 2 ) = 38mm.

Example5.3 : dryline Thisisafairlysharpboundaryoftenfoundintheareaeastof theRockyMountainsrunningnorth–south.Theboundaryseparatesdryaironthe westfrommoistairontheeast.Thedryaircancomefromwindsfromthesouth (Mexico)ordryairdescendingfromtheRockies.Totheeasttheairmaybemoist becauseofsoutherlyﬂowfromtheGulfofMexico.Reductionsofdewpointofas muchas18 ◦ Ccanbefoundingoingfromeasttowestacrossthedryline.Iftheair columnispushedeastwardstheheavierdryaircanwedgeunderthelightermoist airandsometimesleadtocloudinessorevenrain.

Anothermeasureofmoistureencounteredinatmosphericscienceisthe speciﬁc humidity, q.Itisthenumberofgramsofwatervaporperunitmassofairplus

Airandwater

watervapor(usuallyexpressedinkilograms).Inapplications, q isnumerically closeenoughtothemixingratio w thatoneseldomneedstodistinguishbetween them.

Example5.4 Compare q and w ifthepressureis800hPaand q = 0.010(kgvapor) (kgmoistair) 1 .Wecanwrite(using p = pd + e ):

Notethatthepressureof800hPareallydidnotmatter.

Example5.5 Moisteningthetropicalboundarylayer(temperature300K).Suppose theairabovetheseasurfaceisstill(i.e.,ignoreadvection)andabsolutelydry. Supposeevaporationtakesplacesteadilyatarateof1myr 1 .Howlongdoesit taketheboundarylayertocometo80%humidity?

Answer:Theevaporationrateis1myr 1 .Then

At300Kand80%RH,theamountofwatervaporabove1m2 intheboundarylayer (1.5km)is32kg.Hence,

Example5.6 Canitsaturateintime?Supposeadryparcelisintroducedtothe tropicalboundarylayerat30 ◦ N.Itﬂowsalongthesurfaceinthetradewindsuntil itreachestheEquator.Doesithavetimetosaturate?

Answer:Letthetradesﬂowtowardthesouthwestat10ms 1 .Themeridional distancetotheEquatoris30×100km = 3 × 106 m.Alongadiagonalat45◦ this isenhancedto4.24×106 m.Thetimerequiredforthispassageisapproximately 4×106 s = 46days.Accordingtothepreviousexamplethisappearstobesufﬁcient timefortheparceltosaturate.

Example5.7 Thetemperatureis20 ◦ Candthevaporpressureis10.0Pa.Whatis thedewpointtemperature?

Answer:Solve(5.37)for TD :

Figure5.9Relativepositionsofthetemperature T ,dewpoint TD ,andwet-bulb temperature Tw ofanairparcel.

5.9Wet-bulbtemperature,LCL

Thethermodynamicstateofanairparcelisdeterminedbyitstemperature,pressure andmixingratio.However,insomeapplicationsitisnotconvenienttousethe mixingratio.Forexample,themixingratioisnoteasilymeasureddirectly.There isanotherindicatorofatmosphericmoisturecalledthewet-bulbtemperature(Tw ) whichcanbemeasuredmoredirectly.Thewet-bulbtemperatureliesbetweenthe dewpointtemperatureandtheairtemperature(seeFigure 5.9).

Beforeintroducingthewet-bulbtemperature,letusturnagaintothedewpoint temperaturewhichisalsoameasureofmoistureintheair.Recallthatitisthe temperaturetowhichairmustbecooledatconstantpressureinordertobecome saturatedwithrespecttoaplanesurfaceofwater.Asweperformthecoolingwe mustkeepthemixingratiooftheairﬁxed.Atthistemperaturetheactualmixing ratiobecomesequaltothesaturationmixingratio:

Attemperatureshigherthanthedewpointtheaircontainssomemoisture,but lessthanthesaturationvalue.Asthetemperatureisdecreasedtothedewpoint condensationoccurs.Sinceatthedewpointthemixingratioisequaltothesaturation mixingratio,itisevidentthatthedewpointtemperatureisalwayslowerthanor equaltotheairtemperature.Ifthetwoareclosetogether,therelativehumidityis high.Ifthedewpointisfarbelowtheairtemperature,therelativehumidityislow. Thedewpointisagoodindicatorofhumandiscomfort.Whentemperaturesarehigh itismorecomfortablewithalow TD ratherthanwithahighdewpointtemperature becauseofthehigherrelativehumidity.Whentherelativehumidityishighthe rateofevaporationfromamoistsurfaceislow(itisactuallyproportionalto100 minustherelativehumidityinpercent).Meteorologistsoftenrefertothedifference betweentemperatureanddewpoint,whichiscalledthe dewpointdepression. Thefrostpointisdeﬁnedsimilarlytothedewpoint.Thefrostpointisthe temperaturetowhichairmustbecooledatconstantpressureinordertobecome

saturatedwithrespecttoaplanesurfaceofice.Notethatthemixingratioisconstant duringaconstantpressurecoolingofaparcel.

Thedewpointtemperatureisdifﬁculttomeasuredirectly.Itiseasytomeasure the wet-bulbtemperature.Thisisthetemperatureofawetsurface(nominallyawet clothwrappedaroundthebulbofathermometer)thatisimmersedintheambient air.Thewetsurfaceevaporatesmoistureintothesurrounding(typicallylessthan saturated)airandinsodoingthetemperatureofthewetsurfaceislowered(as perspirationintodryaircoolstheskin).Thewetsurfacewillcometoanequilibrium temperatureafterashorttimeiftheairnearthewetsurfaceiscontinuallyventilated withthedryerambientair.Thisequilibriumtemperatureofthewetclothiscalled thewet-bulbtemperature.Ifthesurroundingairisfullysaturated,therewillbeno netcoolingsincetherateofevaporationwilljustequaltherateofcondensationonto thewetsurface,leavingthewet-bulbtemperaturetobethesameasthedewpoint temperature.Forunsaturatedairthewet-bulbtemperaturealwaysfallsbetweenthe dewpointtemperatureandthedry-bulbtemperature(thedifferencebetweenthe dry-bulbandwet-bulbtemperaturesiscalledthe wet-bulbdepression).Notethatan evaporatingclouddropletorraindrophasatemperatureatthewet-bulbtemperature.

Thewet-bulbtemperaturecanbemeasuredwitha slingpsychrometer,which consistsofathermometerwithwetgauzecoveringitsbulb.Thishand-helddeviceis swungaroundfromashortchaintomaintaintheproximityoffreshambientairatthe wetsurface.Withouttheswinging,stagnantsaturatedairwouldaccumulatearound thewetbulbandraiseitstemperaturetoanerroneouslevel.Watermoleculesleaving thewetsurfacediffuseawayfromthebulbthroughthethinboundarylayerofair surroundingit.Atthesametimeheatisbeingconductedfromthewarmerambient airtowardsthecoolerwetsurfacethroughthesamethinboundarylayer.Equilibrium isestablishedbetweentheenthalpyfluxduetoevaporationcarriedbyout-flowing moleculesandthein-flowingenthalpyflux.Aformulacanbederivedfortherelative humiditygiventhewet-bulbanddry-bulbtemperatures(see Wet-bulbderivation in theboxbelow).Forpracticalusetherelationshipiscommonlyexpressedintables. Itisinterestingthatthegeometricalconfigurationofthewetclothsurroundingthe bulbdoesnotmatterbecausethosefactorscancel.Morediscussionofthewet-bulb temperaturecanbefoundinExercises7.8and7.9inChapter 7.

Thesaturationmixingratiodependsontemperatureandairpressure,thusit isafunctionofheight.Whenaparcelisliftedadiabatically,itstemperatureand pressurebothdecrease.Thetemperaturedependenceoftheparcelislinearwith height(wewillseeinthenextchapterthatitis10Kkm 1 ).Hence,astheairrises 1kmthetemperaturewillfallabout10K.Thesaturationvaporpressurebecomes halfitssurfacevaluebecauseofthisdecrease(remembertheruleofthumbabout thedoublingofvaporpressureforevery10 ◦ C).Themixingratio w0 staysthesame forthisascent,whiletheairpressure p(z ) andthevaporpressure e (z ) intheparcel

Lifting Condensation Level

eachfallabout12%(usingascaleheight H oftheatmosphereofabout8km):

Sothevaporpressuregoesdownby ≈12%whilethesaturationvaporpressuregoes downby50%.Thisshowsthatthesaturationvaporpressureisdecreasingmuch fasterthantheactualvaporpressureintheparcel.Asthiscontinues(thesame percentagesforeachkilometerofascent)thecurveswillcrossandcondensation willoccur.Thelevelatwhichaninitiallyunsaturatedparcelreachesitssaturation levelwhilebeingliftedadiabaticallyiscalledthe liftingcondensationlevel (LCL). Ifweknowthetemperature,pressureandmixingratioofanairparcel,wecanﬁnd theLCL,wherecondensationstartstooccuralongtheascent.

Ifweknowthemixingratio,wecaneasilyfindthedewpointandwet-bulb temperatureandviceversa.Figure 5.10 showstheLCLforaparcelatinitial temperature T ,pressure p,anddewpoint TD .Itislocatedattheintersection ofthelinewithconstantpotentialtemperature(theunsaturatedparcelascends dryadiabatically)andthelineofconstantsaturationmixingratiostartingatthe dewpointtemperature(thisisbecausethemixingratioisfixedatitsinitialvalue duringthisnoncondensingpartoftheascent).Thephysicalsignificanceofthese parameterswillbemoreclearwhenwestarttoworkwiththermodynamicdiagrams inChapter 7.

Wet-bulbderivation Intheprecedingweindicatedthatthegeometricalfactors cancelinthewet-bulbdepression.Toshowthiswecanderivetherelationship betweentherelativehumidityandthewet-bulbdepression.Forsimplicitywetakethe bulbtobeasphereofradius R0 ,butintheenditdoesnotmatterwhatthegeometrical

Figure5.10Illustrationoftheliftingcondensationlevel(LCL).

Figure5.11Watervapormoleculesdiffuseawayfromthewetsurfaceasheatis conductedtowardsthewetsurface.Whentheseﬂuxesmatch,thetemperatureatthe wetsurfacewillbethewet-bulbtemperature.

shapeis.Therearetwoﬂuxesfromthespherethathavetobecomputed:(1)theﬂux ofwatervapormoleculesleavingthesphereduetoevaporationtowardlessmoistair faraway;(2)theheatconductedtothespherebecauseofthewarmerairfaraway (Figure 5.11).

Thewatervapormoleculesdiffuseawaywithaﬂux Fw (r ) where r isthedistance fromthecenterofthesphere, r ≥ R0 .Theﬂuxatanyvalueof r passingthrougha surroundingsphereis

where D isthediffusioncoefficientforthisprocess(knownfromtables), nw (r ) isthe numberdensityofvapormoleculesatradius r .Thisfluxmustbeindependentof r sinceotherwisetherewouldbeasourceoffluxotherthanthewetbulb’ssurface. Wecanthenset Fw (r ) = F0 = constant.Thisallowsustointegrateeachsidefrom R0 to ∞ afterdividingthroughby4π r 2 D .Thisproduces:

Wecanrecognizethat

where nsat (Tw ) isthesaturationnumberdensityfor T = Tw .Ifwedividethroughby nsat (TD ) wehave

whereRH = n∞ /nsat (TD ) istherelativehumidityawayfromthewetbulb.Sofarwe donotknowthevalueof F0 intheaboveformula.Anothermeasurementisrequired. Wemustnowturntocondition(2),theheatconduction.Heatflowsfrominfinity towardthewetbulbbyheatconduction.Thefluxofheatcrossingasphericalsurface ofradius r is

where κH isthe(availablefromtables)thermalconductivitycoefﬁcientandthesame argumentisusedasabovefortheconstancyoftheﬂuxpassingthroughspheresof differentradii.

Asabovewecanintegrateafterdividingthroughbythefactor4π r 2 κH .Weget:

Thereisoneothercondition,thatthefluxofheathastoberelatedtothefluxof vapormolecules.Eachmoleculeofwatervaporleavingtoinfinitycoolsthewetbulb by = L/NA where L istheenthalpyofvaporizationpermoleofwatervaporand NA isAvogadro’snumber.Thissaysthat

Nowwecansubstitutefor H0 anddividethetwoequationsabovetoobtain:

Usingthepsychrometerwemeasureboth TD and Tw .Wecanthencalculate nsat (TD ) and nsat (Tw ) fromtheClausius–Clapeyronrelation.TheonlyunknownaboveisRH, whichcannowbecalculated.Thecomputationistedious,hence,thetables.Notethat thegeometricfactors4π R0 cancelledout.Itcanbeshownthatnomatterwhatthe geometricalshapeofthewebbulb,thesegeometricalquantitieswillcanceloutand thewet-bulbtemperatureisindependentoftheshapeofthebulb.Sowhyswing thepsychrometeraround?Thereasonistohavefreshenvironmentalairwithina millimeterortwoofthewetbulbtoinsurethatthemoistairatinﬁnityis representativeandnotcontaminatedbythewetbulb’sevaporation.

5.10Equilibriumvaporpressureoveracurvedsurface

Sofarourdiscussionofthesaturationvaporpressurehasbeenrestrictedtothat overaplanesurfaceofwater.However,inatmosphericphysicswealsoencounter situationswherethesurfaceiscurved.Thisisthecase,forexample,inclouddroplet formation.Clouddropletshaveapproximatelysphericalshape,whichmeansthat growthofadropletimpliesanincreaseofthesurfaceareaofthedrop.Increasing

surfaceareaofaliquidrequireswork(considerblowingasoapbubble).Growthofa dropletthenrequiresconsiderationof surfacetension.Thustoﬁndtheequilibrium vaporpressureoveradropletwehavetoincludesurfacetensionintheenergy balance.Theenergyrequiredforanincreaseofsurfacearead A is σ d A,where σ isthesurfacetension(surfaceenergyperunitarea),inJm 2 (thevalueforwater is0.0761Jm 2 at0 ◦ C).Theenergyrequiredtocreateasphericaldrop4 ofradius a is σ 4π a 2 .

Considertheformationofaclouddropletfrompurewatervapor(noaerosols orotherimpuritiespresent).Suchaprocessiscalled homogeneousnucleation as opposedto heterogeneousnucleation,whensmallaerosolparticlestakepartinthe dropletformation.Supposethatinitially,at t = t1 ,wehavewatervaporofmass M atpartialpressure e andtemperature T .TheGibbsenergyofthesystem, Ginitial ,is

initial = Mgv (5.52) where gv isthespeciﬁc(perunitmass)Gibbsenergyofthewatervapor.Itdepends onthevaporpressureandthetemperature.Supposethatatsomelatertime, t = t2 , anembryonicdropletstartstoform.Itgrowsbyoccasionalstickingcollisionsby watervapormoleculesandatsomemomentdevelopsaradius a andmass Mw . Thesphericalsurfaceareaofthedropletis A = 4π a 2 .Thetotalmassisconserved, soif Mv isthemassofwatervaporremainingaftercondensation,then

ThetotalGibbsenergyofthesystemattime t2 , Gﬁnal ,is

Theﬁrsttermontheright-handsideof(5.54)istheGibbsenergyofthewater vapor,thesecondistheGibbsenergyoftheliquid,andthelasttermisdueto surfacetension.

ThechangeoftheGibbsenergyduetothedropletformationcanbefoundby subtracting(5.52)from(5.54)andtakingintoaccount(5.53):

Thenextstepistoﬁndthedifference gw gv .Weknowthatataconstanttemperature (inourcasethetemperatureisﬁxed)thechangeintheGibbsenergyisd g = v d p,

4 Theexampleofasoapbubbleishelpful.Soapywaterhasahighersurfacetensioncoefﬁcientthanpurewater sothesurfacetensionisveryimportant.Alsoonemustrememberthatthebubblehastwicethesurfacearea becausethereisaninsidesurfaceaswellasanoutsidesurface,incontrasttothewaterdroplet.

Figure5.13Schematicdiagramillustratingthenotationforequilibriumvapor pressureoveradroplet, e ,andthatoveraﬂatsurface, es .

where v isthespeciﬁcvolume(v = (density) 1 ).Then,forthevaporwehave d gv = vv d e ,fortheliquidd gw = vw d e ,andthedifferenceisequalto

since vv vw .Substitutingspeciﬁcvolume vv fromtheIdealGasLaw,weobtain

and,aftertheintegration,

Wecanﬁndtheconstantofintegrationbytakingintoaccountthatinequilibrium (seeFigure 5.13),alongthephaseboundarywhere e = es (T ), gw = gv (see(5.7)). Then,

Substituting(5.59)into(5.55)gives

Themassofasphericalwaterdropletwithradius a anddensity ρw is Mw = 4 3 πρw a 3 ,andthesurfaceareais A = 4π a 2 .ThenthechangeintheGibbsenergy duetothedropletformation, G = Gﬁnal Ginitial ,is:

From(5.61)weseethatinthesubsaturatedair,when e < es ,ln (e /es ) isnegative, and G isalwayspositive(seeFigure 5.14).FromSection4.8.2weknowthat equilibriumoccurswhentheminimumoftheGibbsenergyisachieved;or,inother

Figure5.14ThechangeofGibbsenergyduetotheformationofawaterdroplet asafunctionoftheradiusofthedroplet.

words,thesystemtendsspontaneouslytowarditsequilibriumstatebyhavingthe Gibbsenergydiminishmoreandmoreuntilitreachesitsminimum.Weconclude thattherearenofavorableconditionsforclouddroplet(a > 0)formationin unsaturatedair.

Thesituationisdifferentwhentheairissupersaturated,whichmeansthat e > es andln (e /es ) ispositive(seeFigure 5.14).Inthiscase G increasesastheradius a increases,thenitreachesamaximumatsomeradius,andthendecreaseswiththe furtherincreaseof a .Iftheclouddroplethasaradiuslessthanthecriticalradius a ∗ , itwilldisappearbyevaporation.If,however,theclouddropletreachesthecritical radius a ∗ ,thenitwillcontinuetogrow.Wecanﬁndanexpressionforthecritical radiusbyequatingthederivative ∂ G /∂ a tozero.Theresultis

whichiscalledKelvin’sformula.Itallowsonetoﬁndtheradius a ∗ ofadroplet whichisinequilibriumwithairwiththevaporpressure e .Thisequilibriumstate isunstable.Thisbecomesevidentifweconsideraslightgrowthofthedroplet. Thisgrowthmightbeduetothecondensationofwatervaporinthevicinityofthe droplet,whichmeansthattherelativehumiditydecreasesjustabovethesurface. Sincetheairinthevicinityofthedropletbecomesdrier,thereisadiffusiveﬂuxof moistairtowardthedroplet,andthecondensationprocesscontinues,leadingtothe furthergrowthofthedroplet.Ontheotherhand,ifthedropletevaporatesslightly, therelativehumidityjustabovethedropletsurfaceincreases,thewatervaporstarts todiffusefromthedroplet,andthedropletwillcontinuetoevaporatetomaintain therelativehumiditycorrespondingtoradius a ∗ .

5.10Equilibriumvaporpressureoveracurvedsurface 125

Wecanrewrite(5.62)inordertodeterminetheequilibriumvaporpressure e over adropletwithradius a ,

wheretheparameter b isdeﬁnedby

Ifwesubstitute σ =0.076Jm 2 , ρw =1000kgm 3 ,and Rw = 461.5Jkg 1 K 1 , weobtain

(5.65) (inmeters).At273K, b = 1.21nm(typically b a ).Formula(5.63)showsthat theequilibriumvaporpressureoverasphericaldropletisnotequaltothesaturation vaporpressureasdeterminedoveraplanesurfaceofwater.Thishappensbecause ofthesurfacetension.Iftheradiusofthedropletgoestoinﬁnity, a →∞,which correspondstoaplanesurface,weobtaintheresultforaﬂatsurface e = es (see Figure 5.13).

Example5.8 Waterdropletsareinequilibriumwithsurroundingvaporata temperatureof2 ◦ C.Calculatethevaporpressureandrelativehumidityoverthe dropletwithradius0.008 µm.

Answer:Thesaturationvaporpressureat2 ◦ Cis: es = 2.497 × 109 exp( 5417/ 275.2) (hPa)=7.06hPa.From(5.63)weobtain e = 7.06 × exp(3.3 × 10 7 / (275.2 × 0.008 × 10 6 )) = 8.2hPa.RH = 116%.Asupersaturationof16%is requiredforthecreationofaclouddropletwithradius0.008 µmbyhomogeneous nucleationat2 ◦ C.

Toobtaintherelativehumidity,dividebothsidesof(5.63)by es :

ThisrelationshipisillustratedinFigure 5.15 foratemperatureof5 ◦ C.Asshown intheﬁgure,theformationofadropletwithradius0.01 µmrequiressupersaturation of112%.Atthesametime,inrealcloudstherelativehumidityrarelyexceeds101%. AsweseefromFigure 5.15,1.0%supersaturationisrequiredtoformadropletwith radiuslargerthan0.1 µm.Largedropletsinacloudsimplycannotformbyrandom collisionsofmoleculesontheirsurfaces.Fromtheabovediscussionitfollows thattheprocessofhomogeneousnucleationisunlikely.Innatureclouddroplets formbyheterogeneousnucleationonatmosphericaerosols.Considerthenhowthe

Figure5.15Therelativehumidityasafunctionofdropletradius.Thecurveis plottedforatemperatureof5 ◦ C.Thisgraphshowstherelativehumiditynecessary toformadropletofthesizeindicatedontheabscissa.

equilibriumvaporpressureoverthedropletchangesifthedropletcontainsdissolved electrolytes(usuallyfromaerosols,e.g.seasalt).Wewillconsider hygroscopic particles (thosethataresolubleinwater).Themostcommonofthesearesodium chloride(NaCl)andammoniumsulfate((NH4 )2 SO4 )whichdissolvewhenwater condensesontothem.Inthiscaseawaterdropletcanbetreatedasasolutionwith thewaterasasolventandthesaltasasolute.

Chemistryrefresher:Raoult’sLaw Theratiooftheequilibriumvaporpressure overasolution, e ,totheequilibriumvaporpressureoverpuresolvent, e ,isequalto themolefractionofthesolventinthesolution, f , e /e = f .Thepresenceofsaltin solutionalwayslowersthevaporpressure.

Consideraclouddropletthatcontains nsalt moleculesofsaltand nw molecules ofwaterperunitvolume.If e istheequilibriumvaporpressureoverthesolution (seeFigure 5.16)then,accordingtoRaoult’sLaw,

Figure5.16Illustrationofthenotation.Left:saturationvaporpressure e overa solutiondroplet.Center:saturationvaporpressure e overapurewaterdroplet. Right:saturationvaporpressureoveraﬂatsurface es .

where e istheequilibriumvaporoveradropletcomposedofpuresolvent.The secondexpressioncomesafterdividingnumeratoranddenominatorby nw .Ifwe takeintoaccountthatforadilutesolution nsalt nw andexpandthedenominator in(5.67)inageometricseries,5 weﬁnd(retainingonlythelinearterm),

In(5.68)theeffectofdissociationoftheionsisnotyettakenintoaccount.Ifthe saltdissociatesinto i ions,sayNa+ andCl giving i = 2,thenthenumberofmoles ofsoluteindividualsinthedroplet,

,is

where Msalt and Msalt arethemassandmolecularweightofthesaltrespectively (besuretousekgpermolehereforthemolecularweight).Thedegreeofionic dissociation i ≈ 2forsodiumchlorideand i ≈ 3forammoniumsulfate.The numberofmolesofwaterwithmolecularweight Mw inthemass Mw is

where a istheradiusofthedroplet.6 Theratioofmolesisequaltotheratioof numberdensities:

Aftersubstituting(5.71)into(5.68)weobtain

5 Thegeometricseriesfor 1 1+ = 1 + 2 ,where | | < 1.

6 Weassumethatdissolvingthesaltparticledoesnotchangethevolumeofthedroplet.

wherewehaveintroducedanewparametertosimplifythenotation

Aftersubstituting ρw = 1000kgm 3 , Mw = 18gmol 1 weobtain

CombiningKelvin’sformula(5.66)and(5.72)weobtaintheformulafor equilibriumvaporpressureofasolutiondroplet:

Tobetterunderstandthisformulaconsideralimitingcase.Weagainlettheradius ofthedropletgotoinﬁnity, a →∞.From(5.75)weobtaintheknownresult:the equilibriumvaporpressureovertheplanesurfaceofwater(a =∞)isequaltothe saturationvaporpressure, e = es .

Considernowadropletwitharadius a b.Thenonecanexpandtheexponential functionin(5.75)inaTaylorseries 7 andget

Thesecondtermontheright-handsideof(5.76)isresponsibleforthesurface tension,thethirdtermisduetothepresenceofthesaltinthedroplet.Agraphical illustrationofformula(5.76)isgiveninFigure 5.17 for10 19 kgofsodiumchloride. Thecurveshowingthedependenceofrelativehumidityonradiusofthesolution dropletiscalledtheKöhlercurve.Thedashedcurvecorrespondstohomogeneous nucleation(5.63)withnosaltpresent.Thesetwocurvesdifferconsiderablyfor smallvaluesoftheradiusofthesolutiondroplet.Themostimportantresultisthat withanembeddedsolublesaltnucleusinthedropletamuchlowersupersaturation isrequiredforthedroplettobeinequilibriumwithitsenvironmentthaninthecase ofapurewaterdropletofthesamesize.Forexample,withsupersaturationofjust 0.1%adropletwitharadiusslightlylargerthan0.1 µmcanbeformed.Withavery smallradiusthedropletcanbeinequilibriumwiththesurroundingairevenwitha relativehumiditylessthan100%.Thisispossibleonlybecauseofthepresenceof hygroscopicparticles.Withrelativehumidity90%asolutiondropletwithradius 0.05 µmcanformifthereare10 19 kgofsodiumchloride(notshowninthegraph).

Forasmalldropletradius,theKöhlercurveincreasesmonotonicallyuntilit reachesamaximumatradius a = a ,andthenitdecreasesmonotonically.Consider

Figure5.17Relativehumiditywithwhichasolutiondropletformedon10 19 kg ofNaClisinequilibriumwithwatervaporasafunctionofthedropletradius.The temperatureis5 ◦ C.

adropletwitharadiuslessthan a .Iftherelativehumidityincreases,suchaparticle growstoadjusttothenewequilibriumconditions.Theequilibriumisstable,since slightfluctuationsincondensationorevaporationdonotleadtofurthergrowingor shrinkingofthedroplet.Ifthedropletgrowsslightlybycondensation,therelative humidityincreases,thediffusivefluxisdirectedawayfromthedropletandthe dropletevaporatestoreturntoequilibrium.Withslightevaporation,therelative humidityinthevicinitydecreases,whichcausesdiffusivefluxtowardthedroplet, andthedropletagainreturnstoequilibriumwiththesurroundingair.Suchdroplets withradiuslessthan a arecalledhazeparticles.

Thesituationisdifferentifthedropletreachestheradius a .Nowtheequilibrium isunstable.Withanyfurthergrowthofthedroplettherelativehumiditydecreases andthedropletcontinuestogrow.Thisisonemechanismforclouddroplet formation.

Example5.9 Howbigisasaltparticlewhosemassis10 19 kg?Usingthedensity of2165kgm 3 ,wecancalculatethatasphericalparticleofthismasswouldhave aradiusof0.022 µm.Howmanymoleculesareinthisparticle?Thenumber

ofmolesis M/M NaCl = 10 19 × 1000/58.44kgmol 1 = 0.017 ×10 16 mol. ThenumberofmoleculesisAvogadro’snumbertimesthenumberofmoles: 1.03×106 molecules.

5.11Isobaricmixingofairparcels

Whentwoparcelsofdryairatthesamepressure(altitude)arebroughtintocontact andmixed,theﬁnaltemperatureis

Ifthemixingratiosoftheparcelsare w1 and w2 ,wecanarriveatthesamekindof linearrelationship

Itispossiblethatalthoughthecombinations (w1 , T1 ) and (w2 , T2 ) areneitherone individuallysaturated,themixedairparcel (wf , Tf ) issaturated.Twoclearmoist airparcelscancomeintocontactwithafoggyresult.

Example5.10 Supposetwoparcelsofequalvolumesandpressures,butdiffering temperatures(273Kand293K)comeintocontactneartheground(p = 1000hPa). Leteachhaverelativehumidity90%.Wecanuse

Theﬁnaltemperaturesare:

Wecancomputetheinitialmixingratios:

Theﬁnalmixingratiois

Butthesaturationmixingratioat Tf is

Wefindthatthefinalmixingratioisabovethesaturationvalueforthefinal temperature,hencewehavesupercooledwatervaporandthereforefog.

Theﬁnaltemperaturewillactuallybeslightlyabove Tf becausesomeheating willoccurduringthecondensation.

Notes

MoreadvancedtreatmentsofwaterandaircanbefoundinBohrenandAlbrecht (1998),IrebarneandGodson(1981),andCurryandWebster(1999).Discussions ofclouddrops,etc.,canbefoundinFleagleandBusinger(1980),RogersandYau (1989),Houze(1993)andEmanuel(1994).

NotationandabbreviationsforChapter 5

a dropletradius(m)

a ∗ criticaldropletradius(m)

e , es vaporpressure,saturationvaporpressure(Pa)

e vaporpressureoverasolution(Pa)

= Mw /Md = 0.622(dimensionless)

gl , gg speciﬁcGibbsenergyforliquid,gas(Jkg 1 )

gv , gw speciﬁcGibbsenergyforvapor,liquidwater(Jkg 1 )

G Gibbsenergy(J)

h speciﬁcenthalpy(Jkg 1 )

H (r ) ﬂuxofheatcrossingthesurfaceofasphere(Js 1 )

κH thermalconductivitycoefﬁcient(JmK 1 s 1 )

L = Hvap theenthalpy(latentheat)ofevaporation(Jkg 1 )

LCLliftingcondensationlevel

Mv , Md , Me grammolecularweightofvapor,dryairandeffective (gmol 1 )

Ml , Mg bulkmassofliquid,gas(kg)

ns numberdensityofvapormoleculesatsaturation (moleculesm 3 )

nsat numberdensityofvapormoleculesatsaturation (moleculesm 3 )

nw numberdensityofwatermoleculesinvapor (moleculesm 3 )

n0 numberdensity(moleculesm 3 )

NA Avogadro’snumber(moleculesmol 1 )

q speciﬁchumidity(kgwatervapor/kgofmoistair)

r relativehumidity

Rw thegasconstantforwatervapor(Table 1.1)

Reff effectivegasconstantforamixtureofspecies (Jkg 1 K 1 )

R∗ universalgasconstant(Jmol 1 K 1 )

sl , sg speciﬁcentropyforliquid,gas(JK 1 kg 1 ))

S entropy(JK 1 )

σ surfacetension(Jm 2 )

T Kelvintemperature(K)

TD dewpointtemperature(K)

Tv virtualtemperature(K)

Tw wet-bulbtemperature(K)

θ potentialtemperature(K)

v meanmolecularspeed(ms 1 )

vl , vg speciﬁcvolumeforliquid,gas(m3 kg 1 )

w mixingratio(kgwatervaporperkgofdryair)

ws saturationmixingratio(kgwatervaporperkgofdryair)

Problems

5.1Akilogramofwaterisvaporizedat0 ◦ Candat1000hPaatmosphericpressure.(a) Calculatethechangeinenthalpyofthewatersubstanceinthetransitionand(b)the changeofentropyfortheprocess.

5.2Whatisthevirtualtemperatureforakilogramofairat T = 283K,relativehumidity 50%at1000hPa?

5.3Thenormaltemperatureofhumanbloodis37.2 ◦ C.Ifapersonisliftedinaballoon theairpressuredecreases.Therewillbeapressure(altitude)wherethebloodbegins toboil.WhatisthatpressureinhPa?Atabouthowmanymetersisthatabovesea level?

5.4Itisamuggynightattheoldballpark.Thetemperatureis30 ◦ Candthehumidity is85%.Whatisthechangeindensity(%)oftheairfromadrynightatthesame temperatureandpressure(1000hPa)?

5.5Assumetheatmosphereisisothermalat303K(verytropical),whichgivesascale heightof H = 8.87km.Thesurfacehumidityis80%.Thevaporisdistributed uniformlyintheﬁrst1.5km,andtheairisdryabovethat.Forsimplicitytakethe airpressuretobeuniforminthislowest1.5km.Howmuchwater(invaporform)lies aboveagivensquaremeterofsurface(kgm 2 )?Expresstheresultinmmequivalent ofliquidwater.Comparetothesituationwhenthetemperatureis273K.

5.6Asystemconsistsofdryairmixedwithwatervaporatatemperatureof20 ◦ C.The pressureofthemixtureis990hPa.Therelativehumidityis50%.

(a)Whatisthesaturationvaporpressure?

(b)Whatisthepartialpressureofthewatervapor?

(c)Whatisthedensityofthemixture.Compareitwiththedensityofdryairatthe same T and p

(d)Whatare w and ws ?

(e)Ifthesystem(parcel)isliftedadiabaticallyto500hPa,whichisconserved e or w ?

5.7Calculatetheequilibriumvaporpressureoversphericaldropletswithradii0.01,0.1, 1,10 µmattemperature273K.Plottherelativehumidity(withrespecttoaﬂatwater surface)asafunctionofradius.

5.8Whatsupersaturationisneededforthedropletswithradius0.5 µmtobeinthe equilibriumwithwatervaporattemperatureof10 ◦ C?

5.9Anammoniumsulfate((NH4 )2 SO4 )particleofmass10 20 kgofradius0.07 µm ispresentintheairattemperature0 ◦ C.Findtherelativehumiditynecessaryfor heterogeneousnucleation.

5.10Findtheexpressionforthecriticalvalue(maximumoftheKöhlercurve)ofthedroplet radiusandrelativehumidity.Calculatethesevaluesforadropletcontaining10 16 kg ofNaClat0◦ C.

Proﬁlesoftheatmosphere

Thepropertiesoftheatmospherethatvarywithaltitudeincludethepressure, temperatureandthecompositionofconstituentssuchaswatervapor.Thischapter providessomeinsightintothesedependencieswithsimplederivationsthathold underidealizedconditions.Thestagewillbesetforthefollowingchapterwhich providesmethodsforanalyzingtheconditionsatthetimeofobservation.

6.1Pressureversusheight

Atmosphericpressuredropsoffdramaticallywithheightabovethesurface.Thisis indicatedbythegraphinFigure 6.1 whichshowsthedependenceofpressureon altitudefortheUSStandardAtmosphere.1

Bybalancingtheverticalcomponentsofforceonaslabofairatanarbitrary height z itispossibletoderiveaformulafortheaveragepressureasafunction ofheight, p(z ).Consideracolumnofairwithcross-section1m2 .Inthecolumn

Figure6.1Dependenceofpressure p (hPa)onaltitude z (km)fortheUSStandard Atmosphere.

1 TheUSStandardAtmosphereisamodeloftheatmosphericproﬁleofvariousproperties.Itisusedprimarilyin aviationandsatellitedragcalculations.Itattemptstogiveaglobalaverageofconditions.Morecanbelearned aboutitontheinternet.

(z +dz)A p(z)A

(d )g

Figure6.2Diagramofacolumnofairofcross-sectionarea A withaslabof thicknessd z atheight z .Thepressureandgravitationalforcesontheslabare indicated.

wepictureathinhorizontalslabofairwhosebottomsurfaceislocatedatheight z abovesealevelandwhosethicknessisd z (seeFigure 6.2).Themassofmaterialin theslabis(densitytimesvolume):d M = ρ Ad z ,where A = 1m2 isthehorizontal cross-sectionalareaoftheslab.Theweightoftheslabofgasis (d M)g = (ρ Ad z )g .

Beneaththeslabisapressureforcepushingupwards: p(z )A.Abovetheslabisa pressureforcepressingdownwards,

Equatingthenetpressureforceontheslabtothegravitationalforce,

Aftercancellationsweobtainthe hydrostaticequation: d p d z =−ρ g [hydrostaticequation].(6.3)

WecanusetheIdealGasLawtowrite:

(6.4)

whereweindicateexplicitlythatbothtemperatureandpressurearefunctionsof altitude z .Inthelaststepweusedtheidealgasequationofstate.Thehydrostatic equationhasmanyuses,butitisparticularlyusefulifthedependenceof T on z is known.Thismayoftenbethecase,atleastapproximately.Ifitistruewecanwrite:

Thenintegratingeachsidefromthesurfaceuptolevel z leadsto:

wherewehaveindicatedthe“dummy”integrationvariableby z todistinguishit fromtheupperlimitoftheintegral.Finally,

Iftheintegralscanbeperformed,wehaveananalyticalexpressionfor p(z ).Even ifthe z dependenceof T isknownonlygraphicallyorintabularform,theintegral canbeperformednumerically.

Example6.1 InExample2.16wefoundthataballofmass m bouncingelastically onthefloorgivesanaverageforceonthefloorof mg .Wecanalsoderivethe hydrostaticequationforaballbouncingonthefloor,butreflectingbackelastically onaceilingonlyashortdistanceabove.Lettheballleavethefloorwithvertical velocity v0 andwhenitgetstotheceiling h,itsvelocitywillbe v1 .Therateof momentumtransfertotheceilingis2mv1 /T ,where T isthetimeforaroundtrip ceilingtofloorandback.Wecanshowthat T = 4h/(v0 + v1 ).Theaveragepressures exertedbythereflectingballattheceiling ph andatthefloor p0 are

where A istheareaonthesurfaceoftheﬂoororceiling,andthedifferenceinthe pressuresis

Wethenhave

Finally, p/ z =−ρ g .Ofcourse,wearetopicturealargenumberofballs (molecules)bouncingupanddownsoastomakeasteadypressure.

Example6.2: constantdensitycase Considerthecaseofconstantdensityasa functionofheight.Thisproﬁleismoreliketheoceanthantheatmosphere.Taking

z = 0tobeatthebottomoftheﬂuid,wecanintegratethehydrostaticequation directlyfrom0tothetopoftheﬂuid H :

Integratingfrom0to z weﬁndthat

Thepressureiszeroatthetopoftheﬂuidanditincreaseslinearlywithdepthbelow thesurfaceuntilthebottom(z = 0)isreached.

Example6.3 : pressureintheocean Whatisthepressureat5kmbelowthesurface intheocean?

Answer:Thedensityofwateris103 kgm 3 .Now ρ gh = 5 × 107 Pa ≈ 500atm. Forreference,itmightbenotedthat1atmofpressureisequivalenttothatatabout 10mdepthofwater.

Example6.4 : isothermalatmosphere Takethetemperaturetobeconstant(T = T0 ) withrespectto z .Thenwecanintegrate:

where p0 isthepressureat z = 0andthe scaleheightH isgivenby

Notethestraightforwardphysicalinterpretation:largetemperaturesleadtolarge scaleheights(swelling);large g leadstosmallerscaleheights.Moreover,larger R (smallermolecularweight)leadstolarger H

Itturnsoutthattakingthescaleheight H tobeaconstantisarather goodapproximationtothepressuredependenceonaltitude,ascanbeseenin Figure 6.3 whereinacalculatedproﬁle(dashed)issuperimposedontheUSStandard Atmospherevalues.Inthecalculationavalueof H = 7.89kmwasusedalongwith pz =0 = 1013hPa.Thismodelﬁttotheempiricalresultisverygoodconsidering thatthetemperatureisveryaltitudedependentfortheUSStandardAtmosphereas showninFigure 6.4.

Figure6.3Thepressureversusaltitudecurvewithacomputedproﬁle(dashed) superimposed.Thecalculationwasconductedwiththeconstantvalueof H = 7.89 kmandavalueof pz =0 = 1013hPa(toagreewiththeStandardAtmosphereatthe surface).

Figure6.4ThetemperatureversusaltitudecurvefortheUSStandardAtmosphere. Thealtitudeofthediscontinuityinslopeatabout11kmiscalledthe tropopause.

6.2Slopeofthedryadiabat

Considernextthedryadiabat.Thisisthetemperaturedependenceofaparcelof dryairasitisdisplacedupwards 2 underadiabaticconditions(noheatingduetothe temperaturedifferentialwiththeenvironment).Forasmallverticaldisplacement oftheparcelwecanwritethechangeinenthalpyfortheparcel(treatedasa thermodynamicsystemundergoingareversibletransformation):

2 Someexternalmeansmustcausethedisplacementsuchasbuoyancyortheriseofairduetoupslopewind.

Dividingeachsideby M d z ,

andusingthehydrostaticequation(d p/d z =−ρ g )and ρ = p/RT weobtain:

Finally,

Thisverysimpleandelegantresultdoesnotdependontheactualtemperature proﬁleoftheatmosphere.Wecanevaluatethisformulatoﬁnd:

Notethattheadiabaticlapserate d isdeﬁnedtobeapositivenumber.Figure 6.5 showsthetemperatureandsizeofaparcelbeingliftedadiabaticallyfromthesurface to10km.

Example6.5: dryadiabaticatmosphere Considerthepressureproﬁleofan atmospherewhoseverticaldependenceoftemperatureisthatofadryadiabat.This

Figure6.5Illustrationofthesizeofaparcelasitisliftedadiabatically.Thevolume ofthesphericalparcelwascalculatedfrom(3.43).

Proﬁlesoftheatmosphere isanatmospherethatisthoroughlymixedintheverticaldimension.(Forexample, theturbulentboundarylayerwhichoccupiesthelowest1to2kmoftheaircolumn. Wewillshowthislater.)Itstemperaturefallsofflinearlyasd T /d z =−g /cp .For theseconditions θ isthesamethroughout.Suchanatmosphericproﬁleiscalled isentropic.Wehavethehydrostaticequation

andPoisson’sequation

Wetake θ tobeaconstantinPoisson’sequation.Thus,

andﬁnally,

whereinthelastweused κ = R/cp . Notethattheformulafailswhen z >θ cp /g ≈ 30km.

Theisentropic(θ = constant)proﬁleisoftenobservedinthedaytimeboundary layeranduptotheLCLwheretheairiswellmixedvertically.Abovethislevelthe lapserateissmallerbecauseofwarmingduetocondensationofwatervaporinthe parcelintodroplets.

Whendoesthehydrostaticapproximationnotwork?Thiscanhappeninunusual circumstances,butﬁrstconsidertypicalconditions.Iftheforceofgravityisnot balancedbytheverticalgradientofthepressureﬁeld,theparcelmustbeaccelerating vertically.Supposetheimbalanceis1%oranaccelerationof ≈ 0.1ms 2 .After only10saparcelstartingatrestwouldhaveaverticalvelocityof1ms 1 ,a

veryrareoccurrenceexceptinathunderstorm.Verticalvelocitiesaretypically oftheorderof0.01ms 1 ,whichsuggeststhatlargeimbalancesareveryrare. Argumentscanalsobeconstructedfromthree-dimensionalconsiderationsthatat synopticscales(scalesthatmatchthetypicalobservingstationsonaweathermap, afewhundredkilometers)oneﬁndsthathorizontalmotionsaretypicallyonthe orderof1to10ms 1 andverticalmotionsareoftheorderofcentimetersper second.Theseargumentscanbefoundintheﬁrstfewchaptersofmostdynamics books.

6.3Geopotentialheightandthickness

Themechanicalpotentialenergyperunitmassofaparcel(calledthe gravitational geopotential)is (z ) = gz where z isitsheightabovesomereferencelevel(typically sealevel).3 Wecanwriteforthechangeinpotentialfromoneleveltoanother:

Thisistheamountofworkperformedinliftinga1kg parcelfrom zbelow to zabove (notcountingbuoyancyforces,justgravity).Thegeopotentialcanbeturnedaround slightlytobeconsideredafunctionofthepressureleveloftheparcel.Soinstead of (z ) wecanthinkof (p).Thisisjustthegravitationalpotentialenergyper unitmassofaparcelatpressurelevel p.Nowthechangeingravitationalpotential energyingoingfromonepressureleveltoanotheris

where zpabove and zpbelow aretheelevationsabovethereferencelevelforwhichthe pressuresare pabove and pbelow .The geopotentialheight, Zp ,isdeﬁnedtobethe heightinmetersofthepressurelevelforagivenvalueofthepotentialenergyper unitmass:

= (p)/g .(6.31)

The geopotentialheight, Zp1 (x , y ),istheelevationofthesurfaceforagiven pressure p = p1 .Theheightofthisconstantpressuresurfaceisafunctionof x and y (longitudeandlatitude)overtheEarth’ssurface.Allmeteorologistsarefamiliar withthe500hPaheightﬁeld,sinceitissoimportantinweatherforecasting.

Example6.6 : heightﬁeldofanisothermalatmosphere Supposethetemperatureis everywhere T0 .Whatisthe500hPaheightﬁeld?

3 If z werelargeenoughwewouldhavetotakeintoaccountthe z dependenceof g = g (z ) (duetotheweakening ofthegravitationalforcewithdistancefromtheEarth’scenter)anduse z g (z ) d z ,butthisisseldomimportant instudiesofweatherandclimateofthetroposphere.

Answer:Foranisothermalatmospherewehave p = p0 e z /H (6.32)

oraftertakingnaturallogsofeachside, z = H ln p0 p .(6.33)

Foratemperatureof300K,thescaleheight H = Rd T0 /g is8786m.Ifthereference pressureisﬁxedat1000hPa,usingln2 = 0.693,wehave Z500 = 6090m.

Notethatbecauseitismuchcolderatthepoles,the500hPaheightﬁeldis loweratthepoles(T ≈ 250K)thanattheEquator(TEquator ≈ 300K),orabout 5000mversus6000m.Roughlyspeaking,theheightﬁeldscalesinverselyto T , butrememberthisisforanisothermalatmosphere.

Inoursolutionwealsohadtospecifythesurfacepressure.Ifthereismore atmosphericmassaboveonepositionontheEarth(asmeasuredbysurface pressure),thiswillliftthegeopotentialheightﬁeld.Sotheheightﬁeldisdetermined bytheamountofmassabovethereferencesurface(thissetsthesurfacepressure) atapointandthethermalstructureoftheairabovethatpoint.Keepinmindthat thesurfacepressure(proportionaltothemassabovethesite)variesnomorethana fewpercentfromtimetotime.Itfallsasmuchas8%or9%intheeyeofthemost intense(hurricanestrength)tropicalstorms.

Theverticallyaveragedtemperatureinalayercanbedeﬁnedas

= pabove pbelow T dln p pabove pbelow dln p (6.34)

wheredln p = d p/p.(Notethatd z ∝−d p/p atleastlocallyaccordingtothe hydrostaticequation.)Thisjustiﬁestheuseofdln p asourintegrationincrement. Afterapplyingthehydrostaticequationweﬁnd:

.(6.35)

Finally, zabove zbelow = RT g ln pbelow pabove = H ln pbelow pabove [thickness](6.36) andfordryair H = 29.3T isthe(local)scaleheight.Thequantity z = zabove zbelow iscalledthe thickness ofthelayerlyingbetweenthetwopressuresurfaces.

6.4Archimedes’Principle

Clearlythethicknessisameasureofthelocalverticallyaveragedtemperatureof thelayer.

Example6.7 Howthickisthe500to600hPalayerwhentheaveragetemperature is280K?

Answer: z ≈ 1500m.(6.37)

Example6.8 The1000hPato500hPathicknessisoftenusedinweatherdiscussions todescribetheaveragetemperatureinthelowerpartofthetroposphere.Whatis theaveragetemperatureforasealeveltemperatureof295Kandalapserateof 6Kkm 1 ?

Answer:Use T = (1/ z ) z 0 (T0 z ) d z and(6.36).Thenthetemperature T = 278K.Thethicknessis ≈ 5647m.

Itisimportanttonoticethatthicknessisameasureoftemperature.Coldlayers arethin,warmlayersthick,indirectproportiontotheaverageKelvintemperature inthelayer.

6.4Archimedes’Principle

Inafluidsuchastheatmospherealowdensityparcelembeddedinadenser environmentwillrise.Considerafluidofuniformdensity.Figure 6.6 showsan arbitraryvolumeisolatedinsideaboxoffluidofuniformdensity(pictureathin filmenclosingaportionofthesameuniformfluid).Thefluidincludingtheenclosed parcelisatrest.Therefore,allcomponentsoftheforcesactingontheenclosed interiorportion(parcel)mustsumvectoriallytozero.Weneedonlyconsider

Figure6.6Diagramofanirregularvolumeinafluid.Theweightofthemassin thevolumeis ρe Vg ,where ρe isthedensityofthefluid, V isthevolumeand g is theaccelerationduetogravity.Thepressureforcesofsurroundingfluidindicated bythearrowspointinginwardexactlybalancethedownwardpointingforcedue togravity.Butifafluidoflowerdensity,say ρ ,issubstitutedinthesameirregular volume,therewillbeanetupwardsforceonthismass.

Profilesoftheatmosphere theverticalcomponents,sincethehorizontalcomponentsbalancewithoutany gravitationalcontribution.Theweightoftheinteriorvolumeindicatedbythe downwardpointingvector(seeFigure 6.6)hasmagnitude Mg = ρe Vg .The subscripteindicatesthedensityoftheuniformfluid(theenvironment).Thepressure forcesexertedbythesurroundingfluidmustaddupvectoriallytoavectorpointing upwardswithmagnitude Mg .Nowsupposetheisolatedvolumeiscarvedoutand refilledwithmatterofadifferentdensity,say ρ .Theforceexertedonthisvolume (parcel)isupwardsandofmagnitude ρe Vg ,buttheweightoftheparcelisonly ρ Vg .Thereisthusanetupwardforceof

Ifthedensityoftheparcel ρ islessthantheenvironmentaldensity,theforce willbeupwards.Thisisthe buoyancyforce ontheparcelandtheformulais called Archimedes’Principle.Oftenwewanttoknowtheforceperunitmassor acceleration.WecanuseNewton’sSecondLaw(F = Ma where a isthevertical acceleration):

Wemayuse p = pe andforanidealgas ρ = p/RT ;hence

andsimplifying

Notethatif T > Te ,theforceisupward;andif T < Te ,theforceisdownward. (Proofthatwarmairrisesaswelearnedinelementaryschool!)

Example6.9 Considerathunderstormshaft.Letthetemperatureinsidetheshaft nearthesurfacebe5Kwarmerthanthatoutside ≈ 300K.Thenthevertical accelerationofaparcelintheshaftisapproximately0.163ms 2 .Iftheacceleration isconstantthevelocityaftertime t is v = at andthealtitudeis z = 1 2 at 2 .Theparcel reaches4kminabout3.7minatwhichtimeithasaverticalvelocityof36ms 1 .

Ofcourse,thisanalysisisverysimpliﬁedbecausetheeffectsoflatentheatrelease areignored.Inadditionitwasassumedthatthedifferentialtemperaturebetween theinterioroftheparcelandtheenvironmentremainedconstantduringtheascent. Inensuingsectionswewillseehowtheseeffectsaretakenintoaccount.

6.5Stability

Aslabofair(parcel)sitsatrestinanenvironmentwithacertaintemperatureprofile. Theforcesarebalancedontheparcel.Whathappensiftheparcelisnudgedupwards ordownwardsbyatinyamount?Dotheforcesinthenewpositionbalanceordothey impartarestoringforceorperhapsarepellingforce?Firstconsideraparcelatpoint AinFigure 6.7.Notethatastraightlinesegmentslantedupwardsintotheupper leftquadrantdirectionpassesthroughthepointatA.Thisstraightlinesegmentisa dryadiabatpassingthroughA.IfaparcelisliftedadiabaticallyatA,itmovesalong thisadiabatanditfindsitselfwarmerthantheenvironmentalcurvewhichliesto itsleft.Inthiscasetheparcel,beingwarmerthantheenvironment,willexperience abuoyantforceupwards.Hence,aparcelatpointAunderasmallperturbation upwardswillexperienceanetbuoyantforcetocontinuegoingupward.Ifthesame parcelisnudgeddownwardsfromAitwillexperienceadownwardbuoyantforce. WesaythepointAis unstable withrespecttodryconvection. NextconsiderthepointC.AnanalysissimilartotheaboveshowsthatthepointC is stable withrespecttoinfinitesimalperturbations.ThepointBis neutral.Infact, wecanquicklyseethatallpointsbetweenthesurfaceandBalongtheenvironmental curveareunstable.Wesaythisisan unstablelayer.Similarly,thepointsaboveB onthecurveasshownforma stablelayer

Figure6.7Asoundingcurveoftheatmosphere:temperature T versusheight z . Alsoindicatedarestraightlinesegmentsrepresentingdryadiabatspassingthrough thepointsA(unstable),B(neutral)andC(stable).

Obviously,thekeytotheseanalysesisthelocalslopeoftheenvironmentalproﬁle comparedtotheslopeofadryadiabat.Clearly

Theverticalderivativeofthepotentialtemperature(seeSection 3.3.3),d θ/d z , providesanevensimplerrule.Thederivativeof θ withrespectto z canbecalculated fromitsdeﬁnition, θ = T (p/p0 ) κ (with κ = Rd /cp ):

Thehydrostaticequationcanbeusedtosimplifythesecondterm:

Thisgives

Inthelastequationweintroducedthelocalenvironmentallapserate:

/d z . Thislastexpressionimmediatelytellsusthat

Figure6.8Schematicofthreesoundingsona θ –z diagram.Theonewithd θ/d z > 0isstable;theneutralsoundinghasd θ/d z = 0;theunstableonehasd θ/d z < 0.

Inthe θ versus z diagram,thedryadiabatsareverticallines(Figure 6.8).If θ is increasing withaltitude,thelayeris stable.If θ isdecreasingwithaltitude,the layerisunstable.WewilltalkmoreaboutstabilityinChapter 7,whenwework withthermodynamicdiagrams.

6.6Verticaloscillations

Consideralevel z0 inastablelayeroftheatmospheresuchaspointCinFigure 6.7. Sincethelayerisstable,aparceldisplacedupwardswillexperiencearestoringforce tendingtopushitbacktothepoint z0 .Similarlyadownwardsdisplacementbelow z0 willresultinanupwardsrestoringforce.Theacceleration a ofaparcelslightly displacedalongadryadiabatatthispointisgivenbyArchimedes’Principle:

where Te (z ) istheenvironmentaltemperatureproﬁleor sounding; Ta (z ) isthelocal adiabatpassingthroughthecurve Te (z ) atheight z0 (thisispointCinFigure 6.7); T0 isthevalueoftheenvironmentaltemperatureatthepointofintersection, z0 , T0 ≡ Te (z0 ).Bothenvironmentalandadiabaticcurvescrossatthispoint.Recall thatd Te /d z =− e andd Ta /d z =− d .Thentheenvironmentalandadiabatic temperaturesnearthepoint z0 canbewrittenas:

wherewehaveusedtheapproximatesign(≈)toindicatethatweareusingonly thetangenttotheenvironmentalcurve;ofcourse,theadiabaticlapseratecurveis astraightline,sotheapproximationisnotnecessaryinthesecondequation.For

notationalconveniencelet x = z z0 .Thenwecanequatetheaccelerationtothe secondtimederivativeof x :

Thisisthefamiliar harmonicoscillator equation,whosesolutionis

where A and B areconstantsdependingontheinitialconditions,4 and(inunits rads 1 )

Thisfrequencyiscalledthe Brunt–Väisälä frequency.Thefrequencyincyclesper second(Hz), f = ω/2π ,is

Physicsrefresher:oscillatornotation Theangularfrequencyofalinearoscillator isdenotedby ω .Itsunitsareradianspersecond.Thecorrespondingfrequency f is givenby ω/2π inunitsofcyclespersecondorhertz.Theperiodoftheoscillationis P = 1/f = 2π/ω .

Whentheatmosphereisstable( d > e )aparcelwilloscillate;iftheatmosphere isunstable,then ω 2 < 0andthetrigonometricfunctionsbecomeamixtureof exponentialsatleastoneofwhichisgrowingintime.Toseethisgobacktothe differentialequationfortheoscillator(6.54)andinsteadof ω 2 insert λ2 > 0.Then thesolutionsbecome Aeλt + Be λt .(Youcaninsertthisinthedifferentialequationto satisfyyourself.)Thismeansthatinsteadofoscillating,theparcel“runsaway.”

Anotherimportantobservationisthatforsmallamplitudeoscillations(whenthelinear formulaisvalid)thefrequencyisindependentoftheamplitudeofthedisplacement.

4 Weusuallyarenotinterestedintheinitialconditionsoftheoscillation,butratherthe(angular)frequency, ω

Analternativeexpressionfor ω intermsofthepotentialtemperatureisoften useful.Referringtothelastsubsectionweﬁnd:

ω 2 = g θ d θ d z [squareoftheBrunt–Väisäläfrequency].(6.59)

Thisparticularrepresentationshowsexplicitlythatifd θ/d z ispositive(stable atmosphericlayer),thenthequantity ω isarealnumberandtheoscillationswill occur.Iftheslopeisnegative,thenthefrequencybecomesanimaginarynumber, whichmeansthattheparcel’sdisplacementwillgrowexponentiallyintime,either upordowndependingontheinitialperturbation.Ofcourseintherealatmosphere theparceldoesnotaccelerateallthewaytoinﬁnity,butinsteaditsmotionislimited bythefailureofourlinearanalysiswhichassumedsmalldeviations.Unstablelayers leadtooverturningandmixingoftheparcelswithinthelayer.

Example6.10 Supposetheenvironmentallapserateis0Kkm 1 (isothermal atmosphere)and T0 = 300K,thenwhatistheoscillationfrequency?

Answer:

whichcorrespondstoaperiodof5.7min.

Itistypicalofatmosphericverticaloscillationsthatthecharacteristictimeisof theorderofafewminutes.

6.7WhereistheLCL?

InthissectionwederiveanapproximateformulathatshowsthattheLCL(lifting condensationlevel,seeSection 5.9)isdeterminedasafunctionofthetemperature andmixingratio, T0 and w0 ,ofaparcelatthesurface.Considerwhathappensto thesaturationvaporpressureforthisparcel.Astheparcelrisesitstemperaturefalls fromitssurfacevalue,

where Ta (z ) isusedtodenotethetemperatureoftheparcelasitrisesadiabatically toaltitude z abovethesurface.Thesaturationvaporpressureonlydependsonthe temperatureinsidetheparcel.WecanusetheintegratedformoftheClausius–Clapeyronequation(5.15)toevaluatethesaturationvaporpressure es (Ta (z )) inthe parcelasafunctionofaltitude(alongthedryadiabat):

where Rw = 461.5Jkg 1 K 1 isthegasconstantforwatervapor,and L isthe enthalpyofvaporization(latentheat)forwater(L = 2.500 × 106 Jkg 1 ).Wecan simplifythelastequationbynotingthat

whichleadsto:

Astheparcelisliftedadiabatically(bysomeexternalmechanism)itsmixing ratio, w = w0 ,willremainﬁxedsinceitisaconservativequantity.Theexternal airpressurecanbewrittenasanexponentialfunction, p0 e z /H ,toagood approximation,where H istheatmosphericscaleheight.Asaparcelrises adiabaticallyitsinternalpressurewilladjusttothatoftheexternalpressureat eachaltitude z .Since w0 = e (z )/p(z ) wecanobtainaformulaforthevertical dependenceoftheactualvaporpressure e (z ) intheparcel

wheretheatmosphericscaleheight, H ,canbetakentobe Rd T0 /g with T0 the temperaturenearthesurface.Thevalueof Hw (≈ 1.7km)ismuchsmallerthan typicalvaluesof H .AnexampleisshowninFigure 6.9 where H wastakento be8.3km.Atthesurface e (z ) es (z ),butthegapnarrowsastheparcelislifted adiabatically.Thevalueof e (z ) willcatchupwith es (z ) as z increases.Thesaturation mixingratiochangesastheparcelrises,asshowninFigure 6.10.

Considerthe T –z diagramshowninFigure 6.11 toseehowthetemperature oftheparceldecreaseslinearlywithaltitudeasitrises.Atthesametimethe temperatureversusheightfor ws = constantalsodecreasesbutmoreslowly.The intersectionofthetwocurvesistheLCL.Thislatestviewistheoneusuallyshown inthermodynamicdiagramswhicharethesubjectofthenextchapter.

Byequatingtheformsfor es (T (z )) tothatof e (z ) weﬁndaformulafor zLCL

LCL = ln (es (T0 ) /w0 p0 ) 1/Hw 1/H .(6.66)

Wecansimplifythisfurtherbynotingthattheargumentofthelogarithmreduces to ws /w0 = 1/r where r istherelativehumidityatthereferencelevel(surfaceor sealevel):

LCL = ln (1/r )/(1/Hw 1/H ).(6.67)

Figure6.9Anexampleofhowthesaturationvaporpressureshrinksfasterthanthe vaporpressureinaparcelrisingfrom z = 0.Theinitialmixingratiois0.007kg/kg andtheinitialsaturationmixingratiois0.015kg/kg.Theinitialtemperatureis 20 ◦ C.Thescaleheightoftheatmosphereistakentobe8.3kmandthescale heightforthesaturationvaporpressureis1.66km.Wherethecurvescrossisthe liftingcondensationlevel(LCL).

Figure6.10Illustrationofhowthemixingratio(w (z ))andthesaturationmixing ratio(ws (z ))changeastheparcelisliftedfrom z = 0.Notethat w (z ) = w0 isaconstantofthemotion,but ws (z ) decreases.Wherethecurvescrossisthe liftingcondensationlevel(LCL).Thevaluesoftheparametersarethesameasin Figure 6.9.

ThisformulaallowsustoﬁndanapproximatevaluefortheLCLforagiven valueof w0 and T0

Example6.11: profileofwatervaporintheatmosphere Considertheprofileof es (z ) in(6.64).Letusimaginethatthewatervaporintheatmosphereissaturated.Wecan usetheresultsabovetocomputetheratioof es (T (z )) tothesaturationvalueatthe surface, es (T (0)).Lettherelativehumiditybe100%allthewayup.Theprofileis showninFigure 6.12 fortheparametervaluesofthepreviousfigures.Notethatthe scaleheightforthe(saturated)watervaporis ≈ 1.66km,whichisatypicalheight oftheatmosphericboundarylayer.Thisisessentiallytheexplanationofwhywater vaporisconfinedtothelowestfewkilometersoftheatmosphere.

Figure6.11Theverticalascentofaparcelfrom z = 0asafunctionoftemperature.Thecurvelabeled (T T0 )/ d isthetemperatureoftheparcelduring dryadiabaticascent; T0 isthetemperatureatthesurface.Thecurvelabeled ws (T , p(z )) = constantisthesaturationmixingratio.Wherethecurvescross istheliftingcondensationlevel(LCL).

Figure6.12Theratio es (T (z ))/es (T (0)) forwatervaporembeddedinanatmospherewithtemperatureproﬁle T (0) d z .Thescaleheightinthisexampleis Hw = Rw T 2 0 /L d ≈ 1.66km.Thevaluesoftheparametersarethesameasin Figure 6.11.

6.8Slopeofamoistadiabat

Aswasdiscussedintheprevioussection,anairparcelliftedadiabatically experiencesadecreaseintemperatureatthedryadiabaticrateuntilthevaporinthe parcelbecomessaturated,deﬁningtheLCL.Astheparcelcontinuestorise,water vaporwillbeconvertedtodropletswithanaccompanyingwarmingoftheparcel. ThisassumesthepresenceofcondensationnucleiasdiscussedinChapter 5.This isvirtuallyalwaysthecase(butthenumberdensityofcondensationnucleidiffers signiﬁcantlyfromplacetoplaceespeciallyfromabovelandsurfacestoabove oceansurfaces).Thisadditionalheatingfromcondensationcausesthetemperature todecreaseatalowerrate(asafunctionofaltitude)comparedtothedryadiabatic process.Theslopeofthemoistadiabatcanbefoundusingargumentssimilarto thoseintheprevioussection.

Thechangeinenthalpyforaparcelofmass M andvolume V undergoinga smallverticaldisplacementisgivenby

(6.68) whereweusetheapproximatesignbecauseweareneglectingsomesmallterms. Forexample,wedonotincludethecontributionsfromwatervaporintheparcel Mv c v p aswellasthatoftheliquidwaterdroplets.Thecompositeparcelriseswith noheatingfromtheoutsidesuchthatd Q = 0,butifwetreatthedryairasthe onlyconstituentofoursystem,therewillbeheatingaswatervaporisconverted todroplets.Wecanwrited Q =−ML d ws .Notethatthisisapositivenumberfor risingairsinced ws < 0(ws decreasesfortherisingparcel).Thislastisbecausethe saturationmixingratiodecreaseswithaltitudeastemperaturesareloweredalong thelift.Takingintoaccountthehydrostaticequation(6.3)anddividingbothsides oftheequation(6.68)by M,werewrite(6.68)as

or,afterdividingbothsidesby

Identifying

Thesecondtermontheright-handsidehastwopartssince ws = ws (T , p) = es (T )/p.Expandingthederivative:

Next,insertthehydrostaticexpressionford p/d z ,usingthedryairdensity.Then evaluatethepartialderivativeof w

(T , p):

Nowevaluate ∂ ws /∂ T usingtheClausius–Clapeyronequation(whichintroduces Rw ,thegasconstantforwatervapor):

Figure6.13Relativeslopeofadryadiabat(lapserate d )andthatofamoist adiabat(lapserate m )startedfromthesamepointona T –p diagram.

Theninsertingthisinto(6.71)andrearranging,wehave

Thisrathercumbersomeformulacannotbesimpliﬁedfurther.Thenumeratorin largeparenthesesisoftheorderof1.3andthedenominatorisoftheorderof2.8 whentypicalnumbersareinserted.Thismeansthatthemoistadiabaticlapserateis lessthanthedryadiabaticlapserate(Figure 6.13).5 Thisishardlyasurprisesince theparcelisheatedasitrisesatsaturationbecauseofthecondensation.

6.9Liftingmoistair

Aswehavejustlearned,anadiabaticallyrisingparcelcontainingsomemoisture willeventuallyreachitscondensationlevel.Abovethispoint,withfurtherlifting, waterdropletsformandgrowattheexpenseofvaporintheparcel.Asvaporis condensed,theparcelofairisheated,causingtemperaturechangesalongthevertical path,andtherebyaltering(actuallyincreasing)thebuoyancyoftheparcel.Ifthe waterdropletsarelostduetoprecipitation,itisreferredtoasa pseudo-adiabatic process.Thepreﬁx pseudo indicatesthatthisprocessisirreversible(becauseofthe precipitation)andcannotstrictlybeconsideredasanadiabaticprocess.Another possibilityisthatanascendingairparcelretainsthewaterdropletsaftertheyare formed.Suchaprocessiscalleda moist adiabaticprocess.Temperaturedifferences betweentheinsideoftheparcelandthatoutsideareapproximatelypreserved becausethemolecularthermalconduction(andeventheeddytransport)istoo slow.Consistentwiththeotherapproximationswehaveadopted,thechangesin theparcel’sthermodynamiccoordinatesinaliftingprocesscanbetakentobe

5 MoredetailedderivationsretainingthecontributionsduetoliquidandvaporcanbefoundinthebookbyBohren andAlbrecht(1998)andthatbyIrebarneandGodson(1981).Thesecomplicationsaddgreatlytothetediumof thederivation,buttheapproximateresultisinsufﬁcientagreementtowarrantomittingthemhere.

reversibleandadiabatic.Theactualdifferencebetweenthepseudo-adiabaticandthe moistadiabaticprocessesisnegligibleforawiderangeofproblemsinatmospheric science.Inthisbookwewillnotdistinguishbetweenpseudo-adiabaticandmoist adiabaticprocesses.Thepathofasaturatedairparcelwillbecalleda moistadiabat Ithasalreadybeenshownthatthepathofadryairparcelcanbeconveniently describedintermsofthepotentialtemperature,whichisconstantduringadry adiabaticprocessand,consequently,servesasaperfect tracer ofadiabaticmotion foradryairparcel.Wecallitatracersincewecouldlabeltheparcelwithits(ﬁxed) potentialtemperatureandfollowitaround.Inthissectionwewillintroduceanew variablethatisconservedduringbothdryandmoistadiabaticprocesses,andcan beusedtotraceamoistairparcel.

Westartwiththedefinitionofentropyasappliedtoanidealgas.Supposethe parcelissaturated.Inaninfinitesimalliftatiny(positive)massofwater(per kilogramofair)equalto d ws iscondensed.The(positive)changeinentropyfora moistairparcelwithmass M isthen ML d ws /T .Ontheotherhand,thechange inentropyforadryparcelofidealgasis Mcp d θ/θ (seethediscussionsurrounding equation(4.34)).Wecanequatethesetwoexpressionsfortheinfinitesimalentropy change,andaftercancellingthecommonmassfactorwehave:

Toagoodapproximation6

Thenwehave:

Uponintegratingeachside:

6 Toshowthisconsiderthequantity

WecanusetheClausius–Clapeyronequationtoshowthattheﬁrstterminparenthesesismuchlargerthanthe secondterm(recallthatforevery10Kofincreaseintemperaturethereisadoublingofvaporpressure;then d ws /ws ∼ 1,andd T /T ∼ 10/300.Compare:1 10/300).Wemaythensubstituted (ws /T ) ford ws /T toa goodapproximation.

Proﬁlesoftheatmosphere

where θ(T , p) = T (p0 /p)κ .Thislastrelationship(6.80)formsanimplicit functionalrelationshipthatdeﬁnesacurveinthe T –P plane.Therelationship canalsobewritten

Thecoefﬁcientinfrontoftheexponential, θe ,iscalledthe equivalentpotential temperature.Theequivalentpotentialtemperatureisconservedalongthepathofa moistparcel.

Foradryadiabat θdry (T ) = constant,butforthe moist adiabat,d θ/d T > 0.If wesolvefor θe from(6.81),weobtain

Toseethephysicalsigniﬁcanceof θe letuslifttheparceluntilallitswateris condensedout(thismeans p → 0or z →∞).Inthislimit ws → 0in(6.82)and theequivalentpotentialtemperature θe becomesequaltothepotentialtemperature θ .Inotherwords,toﬁndanequivalentpotentialtemperature,oneshouldlifttheair parceluntilallmoistureiscondensedandprecipitatesout,thencompressthedry parceladiabaticallydownwardsuntilitreaches1000mb.Thetemperaturetheparcel attainsatthe1000hPalevelistheequivalentpotentialtemperature θe .Thewhole processissupposedtooccurwithoutexchangingheatwiththeenvironment.Note that θe isauniquelabelthatcanbeattachedtoanyairparcel,givenitsvaluesof T , w and p ataparticularlevel.

Iftheparcelisinitiallysaturatedandhastemperature T0 atlevel p0 ,theequivalent potentialtemperature θe canbecalculatedbysubstitutingitstemperature T0 ,its potentialtemperature θ(T0 , p0 ),andthesaturationmixingratio ws (T0 ) into(6.82). Iftheparcelisinitiallyunsaturated,thenthetemperature,potentialtemperature, andsaturationmixingratioaretobecalculatedattheliftingcondensationlevel (LCL).Sincethemixingratio w isequaltothesaturationmixingratio ws atthe LCL,theformulaforequivalentpotentialtemperatureforanunsaturatedparcel becomes

Weemphasizeagainthattheequivalentpotentialtemperatureisconservedduring bothdryandmoistadiabaticprocesses,whilepotentialtemperatureisconserved onlyduringdryadiabaticprocesses.Thisisthereasonforusing θe :itcanserve asagoodtracerforamovingairmass.Imagine,forexample,amoistﬂowthat

passesoveramountain.Ifairintheﬂowisinitiallyunsaturated,itcouldbe liftedbyconvectionontheupslopesideofthemountaintotheLCL,wherethe condensationprocessstarts.Duringtheliftingmoistureisremovedbyraining outontheupslopeside.Thenastheairdescendsbacktothesurfaceonthe leesideofthemountain,itwillbemuchwarmeranddrierthanontheupslope side.ThisistheoriginoftheChinookwind(moreonthisinChapter 7).So,the temperature,potentialtemperature,andmixingratiovaryduringbothascentand descentofairparcels.Atthesametime,theequivalentpotentialtemperatureis thesameatthestartingandendingpoints;itisconservedduringthiscomplicated process.

Thereisyetanothergoodtracerofmoistair:theso-calledwet-bulbpotential temperature.The wet-bulbpotentialtemperature, θw ,isthetemperatureanair parcelwouldhaveifcooledfromitsinitialstateadiabaticallytosaturation, andthenbroughtto1000hPabyamoistadiabaticprocess.Thisalgorithmof ﬁndingthewet-bulbpotentialtemperaturedependsonwhetherornottheparcelis initiallysaturated.Iftheparcelisinitiallysaturated,itshouldbecarriedalonga moistadiabattothe1000hPapressurelevel.Iftheparcelisinitiallyunsaturated, itshouldbeliftedﬁrsttotheLCLandthentakenmoistadiabaticallytothe 1000hPalevel.Whendescending,anairparcelmayneedadditionalwatervapor tomaintainsaturation.Thewet-bulbpotentialtemperature,liketheequivalent potentialtemperature,isconservedduringbothdryandmoistadiabaticprocesses. So,inthecaseoftheChinookwinditisthesameontheupslopeandleesidesofthe mountain.

Thelastusefulcharacteristicofmoistairthatweintroduceinthissection isthe saturationequivalentpotentialtemperature θs .Consideranunsaturated parcel.Thesaturationequivalentpotentialtemperatureistheequivalentpotential temperaturetheparcelwouldhaveifitstartedoutcompletelysaturated.The saturationequivalentpotentialtemperature θs canbedeﬁnedas:

Itisimportanttounderstandthedifferencebetween(6.82)and(6.84). θ and ws in(6.82)arethepotentialtemperatureandsaturationmixingratioofsaturated airattemperature T ,whereasthesamevariablesin(6.84)arecalculatedatthe temperature T ofunsaturatedair.Thesaturationequivalentpotentialtemperatureis notconservedduringanunsaturatedprocess.Forsaturatedair, θe isequalto θs .The reasonweintroduce θs isthatitisausefulcharacteristicofairﬂowwhenanalyzing airstability(wewilldiscussitbrieﬂyinChapter 7).

6.10Moiststaticenergy

Wecanﬁndaverysimpleformfortheenthalpyofmoistair,

wherethesecondtermrepresentsthecontributionduetotheenthalpyof vaporization,andthethirdistheenthalpyofthevapor.Ifthewaterweretocondense, thesecondtermwouldcontributetoraisingthetemperature.Neglectingthethird term(whichisverysmallcomparedtotheothers)thespeciﬁcenthalpycanbe written

wherewehaveneglectedthemassofwatervaporcomparedtothatofthedryair. Ifweconsideraparcelofairbeinglifted,thereisactuallyanothertermdue tothegravitationalpotentialenergyperunitmass gz ,where g isthegravitational accelerationand z istheelevationabovesomereferencelevel.Kineticenergycould alsobeaddedbutweneglectithere.Thesumofenthalpyandgravitationalpotential energyisconservedalongaverticalpath.Thissumiscalledthe moiststaticenergy. Thetermstaticisusedbecauseweneglectedkineticenergy.Astheparcelislifted, themoiststaticenergy(callit hmse )isconserved.NotethatbelowtheLCLthis meansthat

whichisthedryadiabaticlapserate.AbovetheLCLwecanobtainfurther informationaboutthemoistlapserate.Forexample,theformulafortheslope forthemoistadiabat(6.75)canbederivedfromit.7

6.11Proﬁlesofwell-mixedlayers

Veryoftenalayeroffinitethicknessiscausedtobewellmixedbyturbulent processes(stirring).Forexample,inthefirstkilometerortwoabovetheground theairisturbulentduetomechanicallydriveneddiesthatareinducedbythelarger scaleairflowsinteractingwiththesurfacefeatures.Iftheatmosphericprofileis stable,buildings,treesandotherprotuberancesaboveaflatboundarywillcause irregularitiesoftheairflow.Moreover,iftheairisunstable,convectiveirregularities willaddtothemechanicalturbulence.Thiskindofturbulent,well-mixedlayermay

7 SeeBohrenandAlbrecht(1998).

6.11Profilesofwell-mixedlayers 159 persistuptoafewkilometerswhereitgentlychangestothemoreorderlylarger scaleflow.Thelayersabovethisboundarylayerarecalledthe freeatmosphere. Inamixedlayerasawholewedonothaveastrictthermalequilibrium. Thatistosaythelayerwillnotreachauniformtemperatureasafunctionof height.Themechanicalstirringoverridesthetendencyforthelayertocometoa uniformtemperatureduetothermalconduction(duetomolecularoreddietransport processes).Thereasonforthisisthatasparcelsrisetheirtemperaturesarelowered becauseofadiabaticexpansion.Observationsshowthatforsuchwell-mixedlayers, especiallynearthegroundandongustydays,thetemperatureprofileapproaches thedryadiabaticlapserate.Anexampleisthelayerbetween850hPaand1000hPa showninFigure 7.13.

6.11.1Well-mixedtemperatureproﬁle

Aheuristicproofoftheadiabaticlapserateinawell-mixedlayercanbeconstructed byassumingthattheatmosphereissubdividedintohorizontallayers,eachlabeled byanindex, i .Nowsupposeapieceofoneofthelowerlayersiscarriedtoahigher layerandinturnthesameamountofmassfromtheupperlayeriscarriedbelowby themechanicalstirringmechanism.Astheparcelfrombelowisliftedadiabatically andthenbroughtintocontactwiththelayerabove,itwillbeinthermalcontactwith otherparcelsinthathorizontallayer.Theliftedparcelandtheothersintheupper levelwillreachatemperature(isobaricmixing)betweentheoriginalenvironmental temperatureofthelayerandthatoftheadiabaticallyliftedparcel.Theadjustment toequilibriuminthisupperhorizontallayerrepresentsanincreaseinentropyof thatlayerwhichcanbetreatedasathermodynamicsystem.Thecollectionofall thelayerscanbethoughtofasacollectionofthermodynamicsystemswhichwe allowtointeractinthispeculiarway.Eachtimeaparcelisliftedorloweredand broughtintocontactwithalayeratadifferentpressurelevel,theentropyofthat systemincreasesandfurthermoretheentropyoftheentirecollectionincreases.As themixingproceedsinthisway,eachsteppreservingthemassatanindividuallevel andpreservingthetotalenthalpyofthesystem,thesystemwillcometoaprofile inwhichfurthermixingwillnolongerincreasetheentropyofthecollection.This finalstatemustbetheoneinwhichtheentropyishomogeneousthroughout.This isthestatewithconstantpotentialtemperature(recall S = Mcp ln θ )andthisis theadiabaticprofile.

Mathematicalderivation Wecanmakeourargumentmorecompellingbyusingan analyticalapproach.First,taketheentropyofthewholesystemoflayerstobe

Wewishtopreservethetotalenthalpyofthecompositesystem:

Wewouldliketoﬁndtheextremumof S subjecttotheconstraintthat H beheld constant.Aconvenientwaytodoitisthroughtheuseofa Lagrangemultiplier, λ (almostallcalculusbooksnowadaysdiscussthistechnique).Weproceedby writing

andsetthepartialderivativestozero:

Thisprocedureﬁndsthesetof

thatwillmake S extreme.Weﬁnd

Inotherwords θi doesnotdependon i or z ;itisaconstant. Ofcourse,itmustbekeptinmindthatthemathematicalproofdoesnoteasethe assumptionswemadeaboutadiabaticliftingandloweringandtheassumptionsabout horizontal(constantpressure)exchangesofheatbetweentheparcelbeingmovedand itsenvironmentatthesamelevel(pressure).Ontheotherhand,thefactthatsucha simpleargumentreproducestheproﬁleseeninnaturesoregularlysuggeststhatour assumptionsarereasonable.

6.11.2Watervaporinawell-mixedlayer

Wehaveremarkedinearlierchaptersthatthemixingratio, w ,isaconservedquantity underverticalmotionsbelowtheLCL.Wecangothroughthesameargumentas abovetoshowthatthemixingratioofwater(orthatofanyotherinertchemical species)shouldbecomeuniforminthelayer.Basically,whenwebringaparcelinto alayerinwhichthebackgroundisdifferentfromthemixingratiointheparcel,the twowillmixinsuchawaythatthenewmixingratiowillliebetweenthatofthe parcelandthatofthewholelayerinproportiontothemasses.Thismixinginan individuallayerwillincreasetheentropyofthatlayer.Eachexchangeofparcels willcauseanincreaseintheentropyoftheentirecompositesystemuntilfurther

exchangesdonotincreasetheentropy.Thisﬁnalconﬁgurationwilloccurwhenthe entirelayerorcompositeoflayersisatauniformmixingratio, w .

Auniformvalueof w inalayerexhibitsacharacteristicshapeofthedewpoint temperatureinathermodynamicdiagram.Itturnsoutthedewpointcurvewilllie exactlyparalleltooneofthesaturationlinesonthechart.

Oceanmixing Therearelayersintheoceaninwhichthemixingtheorygivenabove fortheatmosphereworks.Themostimportantanalogousconservedquantityinthe oceanisthesalinity.Thisquantity,alongwithpotentialtemperature,isuniformin thedeepestlayersoftheoceanwhereenoughtimehaselapsedsincetheirisolation toleavethesewatermasseswellmixed.

Notes

ManyofthesubjectsinthischapterarecoveredindynamicsbookssuchasHolton (1992).ThethermodynamicdetailsarediscussedinmoredetailbyBohrenand Albrecht(1998)andIrebarneandGodson(1981).

NotationandabbreviationsforChapter6

A horizontalareaofaslab(m2 )

g accelerationduetogravity(9.81ms 2 )

d , m , e lapserate, d T /d z ofdryairascendingadiabatically,ofmoist adiabat,oftheenvironment(Km 1 )

h heightaboveareferencelevel

H scaleheight

Hw ascaleheightforwatervapor(m)

κ = R/cp (dimensionless)

L = Hvap enthalpyofvaporization(latentheat)(Jkg 1 )

ω , f angularfrequency(rad 1 ),frequency(Hz)

p, p(z ), p0 pressure,asafunctionof z ,atareferencelevel(hPa) (z ), 1 , 2 geopotentialheightasafunctionofheight,attwolevels(meters, onchartsoftenindecameters,dm)

ρ , ρ0 , ρe density,atareferencelevel,oftheenvironment(kgm 3 )

T , T (z ), T0 temperature,asafunctionof z ,atareferencelevel(K)

Te (z ), Ta (z ) temperatureoftheenvironment,ofanadiabat(K)

T verticalaveragetemperatureinalayerofair(K)

θ , θe , θs , θw potentialtemperature(K),equivalentpotential,saturation equivalentpotential,wet-bulbpotential

w , ws mixingratio,saturationmixingratio(kgwatervaporperkgdry air)

z , z verticaldistance,incrementofit(m)

Zp potentialenergyperunitmassduetogravity(m)

Problems

6.1Supposethetemperatureoftheatmospherehasthedependence T = T0 e z /z0 .Find anexpressionforthepressure p(z ).

6.2A1m3 parcelofmoistair(r = 75%, T = 303K, p = 1000hPa)isembeddedin surroundingdryair.Whatistheverticalaccelerationofthisparcel?

6.3Supposeaparcelhasaverticalaccelerationof0.12ms 2 (seepreviousproblem).If itstartsatrestatthesurface,whatisitsverticalvelocityafter5s,10s,30s?How longdoesittaketoreachthetopoftheboundarylayer(≈ 2km)?

6.4Atacertainlevelofthe(dry)atmosphere z ,thetemperatureis303Kandthelocallapse rateis12Kkm 1 .Isthislayerstable?Supposea1m3 parcelisdisplacedupwardsby 0.5kmadiabatically.Whatisitsaccelerationduetobuoyancy?Howwilltheanswer changeiftheparcelisdisplacedisothermally?

6.5Supposetheatmospherehasitstemperatureequalto300Kandpressure1000hPaat z = 0.Thetemperatureproﬁlefallslinearlywithalapserateof6Kkm 1 upto10km. Above10kmthetemperatureisconstant.Whatisthepressureasafunctionof z ?

6.6UsetheresultsofProblem6.1tocomputethepotentialtemperatureasafunctionof height z

6.7FindthedrylapseratenearthesurfaceforMars.ThemeanradiusofMars RMars = 3.40×106 mis0.530 × REarth ;massofMars = 0.107MEarth ;universalgravitational constant G = 6.67 × 10 11 Nm2 kg 2 ;forCO2 , cp ≈ 0.76kJkg 1 K 1 .

6.8Supposeanatmosphericproﬁleisgivenby T (p) = a + b ln p/p0 ,0 < p ≤ p0 .Find anexpressionforthegeopotentialheight Z (p) asafunctionofpressure, p

6.9Whatisthethicknessofthe1000to900hPalayerifthemeantemperatureis300K?

6.10Whatistheaccelerationofadryairparcelwhosetemperatureis300Kembeddedin anenvironmentof285K?

6.11ComputetheBrunt–Väisäläfrequencyfordryairinalayerwhered θ/d z = 1Kkm 1 , θ = 300K.GivetheanswerinHzaswell.Computetheperiodoftheoscillations.

6.12Considerthedifferentialequation:

Showthat x = A cos ω t + B sin ω t isasolutionforconstantvaluesof A and B

6.13Relatingthelastproblemtobuoyantoscillationsofadryairparcel,ﬁndthecoefﬁcients A and B fortwosituations,usingd θ/d z =1Kkm 1 and θ ≈ 300K:(a) x (0) = 10m, v (0) = 0ms 1 ;(b) x (0) = 0m, v (0) = 1ms 1 .

Thermodynamiccharts

Atmosphericscientistsmakeuseofavarietyofchartsintheiranalysisofweather conditions.Inthepreviouschapterwelearnedtoidentifywhetheraparcellocatedat aparticularaltitudeisstableundersmallperturbations.Thestabilityisdetermined bythelocalslopeoftheenvironmentalcurvecomparedtothatofanadiabatpassing throughthesamepoint.Thediagramusedinthosestudieswasthetemperature versusthealtitude.Onsuchadiagramaplotoftheobservedenvironmental temperatureversusaltitudecouldbecomparedwithplotsofhypotheticaladiabatic trajectoriesofparcels.Welearnedthatcomparingthelocalslopescouldrevealthe stabilityofairlocatedatapoint(altitude)ontheenvironmentalcurve.

Whilethetemperatureandthealtitudeareconvenientfordeterminingthelocal stability,thetemperatureandthelogarithmofthepressurearemoreappropriate coordinatesforcomputingenergeticquantitiesofinterestwithoutgivingupthe convenientstabilityrules.Sincetheatmosphereisverynearlyinhydrostatic equilibrium,thealtitude,temperatureandlogpressurearecloselyrelatedthrough thehydrostaticequation,d p/p =−(g /RT )d z .Usingln p0 /p (p0 isastandard pressure,usually1000hPa)astheverticalcoordinateinsteadofaltitudewillallow ustorelatetheenergeticsofaparcel’smovementsinanunstableenvironment.

Toseehow T versusln p isrelatedtothermodynamicdiagramsusedearlierin thisbook,consideraclosedloopintegralenclosinganareainthisplane:

where v = 1/ρ isthespeciﬁcvolumeofaparcel.Wecanwrite

Thermodynamiccharts

since d (pv ) = 0.Thismeanstheareaenclosedbytheloopintegralinthe T versus ln p planeisproportionaltothenegativeoftheworkdone by thegasintheparcel intraversingtheloop.Thismeanstheareaisproportionaltothework on thegas bytheenvironment.Thisprovidesaconnectiontothethermodynamicdiagrams wearefamiliarwith.Wewillseethephysicalsigniﬁcanceandapplicabilityofthis connectioninthenextsection.

7.1Areasandenergy

Consideraparcelperchedatanunstableequilibriumpoint(altitude)inthe atmosphere.Aslightupwardsnudgewillcausetheparceltoaccelerateupwards becauseoftheincreasing(fromzero)buoyantforce.Astheparcelrisesawayfrom itspreviousbutprecariousequilibriumlevelitwillgainkineticenergyandlose buoyantpotentialenergy.Wecanobtainaformulaforthekineticenergyofthe parcelasafunctionofdistanceabovetheinitiallevel.Weconsideronlyadiabatic motion,whereintheairparcelmoveswithoutheatexchangewiththesurrounding environmentalair.

AccordingtoArchimedes’Principle,theupwardforceperunitmass(equalto acceleration)ontheparcelis

where M isthemassoftheparcel, ρa (z ) isthedensityoftheairintheparcel asitmovesverticallyalonganadiabaticpath,and ρe (z ) isthedensityofthe environmentalairjustoutsidetheparcelatlevel z .Notethatfordisplacements forwhich ρa islessthan ρe therewillbeanupwards(positive)buoyantforceon theparcel.Theworkdoneperunitmassontheparcelbythebuoyancyforcein movingtheparcelfrom z0 to z is z z0 (F /M) d z ,whichisalsothechangeinthe kineticenergyperunitmassoftheparcelinthisdisplacement.Inotherwords, thepositivebuoyancyforcecausestheparceltoincreasespeedasitmovesin theverticaldirection.Using K todenotethekineticenergyperunitmass,we obtain

Substitutingthehydrostaticequation,d p/d z =−ρ g ,andtheidealgasequationof state, p = ρ RT ,yields:

Theresultisthatthekineticenergyofaparcelisproportionaltotheareainthe closedloopdeﬁnedbydoublyintersectingenvironmentalandadiabaticcurvesin a T –ln p diagram:

Itisworthrememberingthattheabovederivationisvalidforbothdryandmoist adiabaticprocesses.Astheparcelrisesadiabatically,itskineticenergygoesup,if thereisapositiveareaenclosedbytheparcel’spathandenvironmentalcurve.

Thesamefactis,ofcourse,trueforthe v –p diagram.Thechangeofkinetic energyperunitmassin v –p variablesis

wherewehavesubstituted v = va ve

Wehaveseenthatthe v –p and T –ln p diagramshavethearea-energyproperty: theareaenclosedbytheenvironmentalcurveontheleftandanadiabatic(parcel) curveontherightisproportionaltothekineticenergyperunitmassassumedbya parcelbeingforcedupwardsbybuoyancyinsuchanunstableatmosphericproﬁle.

Figure 7.1 showsadiagramofanenvironmentalsoundingcurvewithadry adiabatleavingthesurfaceandrejoiningthesoundingatahigherlevelinthe atmosphere.Theoretically,aparcelleavingthesurfacealongthedryadiabatwill haveakineticenergyonreachingtheenvironmentalcurveproportionaltothe shadedareaboundedbythetwocurves.

Ofcourse,theideaoffrictionlessmotionoftheparcelishighlyidealized. Exchangeofmomentumtransmittedbysmalleddiesbetweentheparcelandits environmenttendtoslowtheparcelandalteritsmotionfromtheidealconditions offrictionlessmotion.This entrainment processalsoexchangesotherproperties suchaschemicalcompositionandenthalpy.Nevertheless,theidealizedkinetic energyparameterhasprovenusefulindiagnosingandpredictingtheconsequences ofunstablesituations.

SOUNDING T

Figure7.1Schematicdiagramofasounding(leftborderoftheshadedarea)along withadryadiabatrisingfromthesurface(rightborderoftheshadedarea).The shadedareaisproportionaltothekineticenergyacquiredbyaparcelinrisingfrom thesurfacetotheintersectionofthetwocurves.

Wewillreturntoenergeticconsiderationsafterintroducingtheskew T diagram andpassingthroughaseriesofchartexercisesdesignedtobuildfamiliaritywith thecharts.

7.2Skew T diagram

Whileintheprevioussectionwesawthatthe T –ln p diagramhastheveryuseful area-energyproperty,experiencehasshownthatarelateddiagramisevenmore usefulwhilepreservingthearea-energyproperty.Whileseveralsuchdiagramshave beenproposedovertheyearsanddiscussionsofthemcanbefoundontheinternet, byfarthemostwidelyusedistheskew T –log p chartwhichwewillrefertosimply astheskew T chart.Useofthediagramsavestime,avoidstediouscalculations, andprovidesaneasyvisualmeansofsummarizingtheverticalstructureofthe atmosphericthermal,stability,energeticandmoisturecharacteristics.

Datafroma radiosonde (aninstrumentedballoonlaunchedevery12hoursata networkofthousandsoflocationsovertheglobe)areplottedonthediagramto formthe sounding orenvironmentalcurve.Themainparametersderivedfrom theradiosondearethepressure,temperature,humidity,altitudeandhorizontal componentsofwindvelocity.

Theskew T diagramdiffersfromthe T –ln p diagraminthattheabscissaisrotated abouttheorigin(T0 ≈−50◦ C,ln1000)byabout45◦ downwardsinaclockwise directionasillustratedinFigure 7.2.

Theresultingcoordinateplane(ordiagramorchart)(Figure 7.3)isshownin theformusedbypracticingmeteorologistsandresearchers.Theabscissa X onthis diagramisproportionalto (T + β ln (p0 /p)),where β isanadjustablecoefﬁcient setonceandforallforconvenience(fortheusualskew T chart,thevalueis closetounitymakingtheangleofrotation45◦ ).Theordinate Y isproportionalto ln (p0 /p) whichisverynearlyproportionaltothealtitude,makinginterpretation

Figure7.2Intheskew T diagramthe T -axisisrotatedabout45◦ clockwise.The originalisothermswereverticalwhiletherotatedonesaretiltedasshown.The isobarsarehorizontalbeforeandaftertherotation(onlytheabscissaisrotated).

Figure7.3Illustrationofaskew T –log p diagram.Theisolinesarelabeled.

ofthiscoordinateeasier.Tobesurethatthispairofcoordinatesisviablefromthe pointofviewofenergetics,considertakinganintegralaroundaclosedloopinthe plane:

Thistellsusthataclosedloopinaskew T diagramhasthesamevalueasthat sameloopina T –ln p (unskewed)diagram.Thelargertheareaenclosedthemore energywillbeinvolvedinarelatedclosedloopprocess.Theclosedloopprocess canthenberelatedtotheconversionofbuoyantpotentialenergyintokineticenergy ofconvectionaswediscussedinthelastsection.

Intheskew T diagramthelinesofconstantpressure(isobars)arehorizontal whiletheisothermsarenolongerverticalbutaretiltedtotheright.Examination ofFigure 7.3 showsmanylinesbesidestheisobarsandtheisotherms.Letustake onecurveatatime.

Dryadiabat Wecanobtainanequationforthedryadiabatbytakingthelogarithmof Poisson’sequation,

Thisrelationshipshowsthatthedryadiabatsarenotexactlystraightlinesonaskew T diagram.Hence,thedryadiabatsareslightlycurved(solid)lineswhichrunfrom thelowerrighttotheupperleftofthediagramandarenearlyperpendicularto theisotherms.The45◦ anglebetweentheskewedisothermsandisobarsandthe resulting90◦ anglebetweenisothermsandadiabatsmakesiteasierfortheobserver toseethedifferencebetweenthesoundingandtheadiabats.Thispropertyhasled tothewideadoptionoftheskew T diagram.ThepressureisinhPa(sameasmb), thetemperatureisindegreesCelsius.Thedryadiabatsarelabeledbythepotential temperatureassociatedwiththem.

Saturationmixingratio Theselinesaredrawnontheskew T diagramasdashedlines runningtowardtheupperright.Theunitsaregkg 1 ,whichindicatetheamountof wateringramsperkilogramofairatsaturationattheparticulartemperatureand pressure.Thevalueforeachsaturationmixingratiolineisshownonthebottomof thediagram(onsomechartsontheinternetitmightbeshownontheupperright).

Moistadiabat Theseareshownasdashedlinesrunningtowardtheupperleft.They canbecomputedfromtheinformationinthepreviouschapter,butthisisunnecessary sincethechartsalreadyprovidetherelationship.Moreaboutthesebelow.

Figure7.4Exampleofasoundingonaskew T –log p diagram.

Thethermodynamicstateofaparcelofaircanberepresentedasapointinthe diagram.Forexample,aparcelhavingtemperature 10◦ Candpressure600hPa isseentohavepotentialtemperature ≈30◦ C.Tospecifytheparcel’sproperties completelyweneedtoknowitswatervaporcontent, w ingkg 1 .Supposeourparcel whichislocatedat600hPaand 10◦ Ccontainswatervaporwithamixingratioof 0.5gkg 1 .Thesethreequantities p, T , w aresufﬁcienttodeﬁnethethermodynamic stateoftheparcel.Onecouldequallywellspecify p, θ ,andRHsinceonetriplet canbefoundfromtheother.

NextconsiderthesoundingplottedinFigure 7.4.Therearetwosoundingcurves plottedonthechart:thetemperatureplotandthedewpointplot.Bothproﬁles arebasedonradiosondemeasurements.Thetemperaturelineisalwaystothe rightofthedewpointline.Onecanreadthevaluesofpotentialtemperature, equivalentpotentialtemperature,saturationmixingratio,andactualmixingratio ofanairparcelsituatedatanyparticularpressureandtemperature.Aparcelof aircanbemovedhypotheticallyindifferentdirectionsonthechart.Asitmoves

Thermodynamiccharts

itsthermodynamiccoordinateschange.Forexample,consideraparcelonthe environmentaltemperaturecurveat400hPa;ahorizontalmove(OA)isanisobaric change(thismightresultfromcoolingduetoradiation).Amovealonganisotherm (OB)showshowtheparcel’spropertieschangeunderanisothermaldisplacement.A movealongadryadiabat(OC)indicatesthechangesthataparcelwouldexperience ifitwereliftedadiabaticallytoadifferentpressurelevel.Asaparcelislifteddry adiabatically,thechartshowsthatthetemperatureofsuchaparceldecreases(it crossesisothermsofdecreasingtemperatures).Thiscoolingisagainagraphical expressionofPoisson’sequation.Tointroduceskew T charts,letuswalkthrough somechartexercises.

7.3Chartexercises1

Exercise1 Anairparcelhasatemperatureof253Katthe600hPapressurelevel.

(a)Finditspotentialtemperatureandsaturationmixingratiousingbothskew T –log p chart andformulas.

Answer:(Figure 7.5)Ontheskew T diagramthatweuse(seeFigure 7.5),the temperatureisindegreesCelsius;forthatreasonwefirsthavetoconvertKelvins todegreesCelsius(T =−20◦ C).Findthepointonthediagramwheretheabscissais equalto 20◦ Candtheordinateisequalto600hPa(pointA).Tofindthepotential temperatureoftheparcel,followadryadiabattotheintersectionwiththe1000hPa level(pointB).ReadthetemperatureatpointB.Thisisthepotentialtemperatureof theparcelwhichisequalto293K.Tofindthesaturationmixingratio,frompointA followthelinewithconstantsaturationmixingratiototheintersectionwiththeabscissa (dashedline).Readthevalueofsaturationmixingratio(1.3gkg 1 ).

Figure7.5DiagramforExercise1.

1 Intheseexercisesweusereducedaccuracy, TSTP ∼ = 273Kand pSTP ∼ = 1000hPaasSTPexceptwhen TSTP appearsintheintegratedClausius–Clapeyronequation.

Usingformulas,(i)ﬁndthepotentialtemperaturefromPoisson’sequation,(ii) ﬁndthesaturationmixingratiofromtheintegratedformoftheClausius–Clapeyron equation(5.17)andformula(5.35).

(b)Movetheparcelalongthedryadiabat(dryadiabatically)tothe850hPalevel.Whatis thenewtemperatureoftheparcel?Whatisthenewsaturationmixingratio?Whatis theparcel’spotentialtemperatureat850hPa?

Answer:FrompointAfollowthedryadiabattoapressurelevelof850hPa(point C).Toﬁndthetemperature,followtheisothermfrompointCtotheintersectionwith theabscissa.Readthetemperature(T = 7◦ C).Toﬁndthesaturationmixingratio, frompointCfollowthelineofconstantsaturationmixingratiototheintersection withtheabscissa.Readthevalueofsaturationmixingratio(7gkg 1 ).Theparcel’s potentialtemperatureatthe850hPalevelisthesameasat600hPa,sincethedescent wasconducteddryadiabaticallywithconservationofpotentialtemperature.

Exercise2 Anairparcelhasatemperatureof298Kat1000hPalevel.Itsmixing ratiois14gkg 1 .Findtherelativehumidityusingthechartandformulas.

Answer:FindthepointonFigure 7.6 withabscissa25◦ Candordinate1000hPa correspondingtotheinitialconditionsoftheparcel(pointA).

Weknowthemixingratio,hence,toﬁndtherelativehumiditywehavetoﬁnd thesaturationmixingratio(seeExercise1).ThesaturationmixingratioatpointA is20gkg 1 .Thustherelativehumidityis

Usingtheformulas,thetemperatureoftheparcelisknown,thereforefromthe Clausius–Clapeyronequationweﬁndthesaturationpressuretobe32hPa.Then, calculatethesaturationmixingratiofromformula(5.35).Finally,ﬁndtherelative humidityfrom(5.28).

Exercise3 UsingthesameinitialconditionsasinExercise2,ﬁndthedewpoint oftheparcelusingbothchartandformulas.

Figure7.6DiagramforExercises2and3.

Thermodynamiccharts

Answer:Bydefinition,thedewpointisthetemperaturetowhichanairparcelmust becooledatthesamepressurelevelinorderforittobesaturated.Inourcasethe parcelisunsaturated,itsmixingratiobeinglessthanthesaturationmixingratio.So, tofindthedewpointmovetotheleftfromtheinitialconditionoftheparcel(point A)alongtheisobarcorrespondingtothe1000hPalevel.Stopattheintersection oftheisobarwiththelineofconstantsaturationmixingratiocorrespondingto 14gkg 1 (pointB).Thisshiftalongtheisobarcorrespondstocoolingtheparcelat thesamepressureleveluntilitsmixingratiobecomesequaltoitssaturationmixing ratio.Therefore,thetemperature(theabscissa)atpointBisthedewpoint.Forthis problem TD = 19 ◦ C.Itisimportanttounderstandthatthedewpointandmixing ratiooftheparcelreflectequivalentinformation:ifyouknowthedewpoint,you canfindthemixingratioatthesametemperatureandpressure,andviceversa.Now letuscalculatethedewpointusingformulasinsteadofdiagrams.Tofindthedew point,wehavetoequatethesaturationmixingratiototheactualmixingratioof theparcel:

Thisgivesusthedewpointtemperature TD = 19 ◦ C.

Exercise4 Anairparcelisliftedadiabaticallyfromthe1000hPalevelwhere theparcelhasatemperatureof20◦ Canddewpoint6◦ C.FindtheLCL(lifting condensationlevel).Whatarethetemperatureandpotentialtemperatureofthe parcelatthislevel?

Answer:Findtheparcel’sinitiallocationonFigure 7.7 correspondingto20 ◦ C abscissaand1000hPaordinate(pointA).Thesaturationmixingratioat20◦ Cis 14.5gkg 1 .Sinceweknowthedewpoint,wecaneasilyﬁndtheactualmixing ratio:themagnitudeofthesaturationmixingratioatthedewpoint(6◦ Catthe 1000hPalevel,pointB)istheactualmixingratio,whichisequalto5.7gkg 1 in ourcase.Sincetheactualmixingratioislessthanthesaturationmixingratio,the airparcelisunsaturated.Therefore,theparcel,whenliftedadiabatically,followsa dryadiabaticlinepassingthroughpointA.Duringascentthepotentialtemperature isconstant,aswellasthemixingratiooftheparcel(thereisnocondensationand latentheatrelease).Atthesametime,thesaturationmixingratiodecreasessince thetemperaturedecreases.So,atsomepressurelevelthesaturationmixingratio andtheactualmixingratiobecomeequaltoeachother,whichmeansthatadiabatic liftingeventuallyleadstosaturationandcondensation.Thelevelatwhichtheairin theparcelhascooledbyadiabaticexpansionsufﬁcientlytobecomesaturatedisthe

LCL.TheLCLcanbefoundattheintersectionofthedryadiabatstartingfromthe initialparcel’stemperatureandpressure(AConthegraph)andconstantsaturation mixingratiolinestartingfromthedewpoint(BC).InourcasetheLCL(pointC)is at810hPa.Thetemperatureatthislevelis3◦ C.Thepotentialtemperatureis293K attheLCL.Thepotentialtemperatureisconservedduringthelifting;itisthesame asthatatthe1000hPalevel.

WecancalculatetheLCLwithoutusingcharts,butthisismoredifﬁcult.We havetoﬁndatwhatpressurelevelthedryadiabatthatstartedatpointAintersects withthelineofconstantsaturationmixingratiothatstartedfrompointB.Wecan, forexample,expresstemperatureintermsofpotentialtemperatureandpressure usingPoisson’sequationandsubstituteitintotheClausius–Clapeyronequation.As aresult,wehavetosolveatranscendentalequationforpressure.Chartsarequicker.

Exercise5 ContinueExercise4.Iftheparcelisliftedadiabaticallytothe450hPa level,whatisitsﬁnaltemperature?

Answer:At810hPatheparcelhasbeensaturated.So,itsfurtherliftingisalonga moistadiabat.FollowthemoistadiabatstartingfrompointCtothe450hPalevel. Thetemperatureatthispointis 29◦ C.

Exercise6 Anairparcelatthe800hPalevelwithtemperature 10◦ Cissaturated (mixingratio2.2gkg 1 ).Computetheequivalentpotentialtemperature θe using boththeskew T –log p chartandtheformulafor θe .

Answer:Lifttheparcelfromitsinitiallocation(pointAonFigure 7.8)alongthe moistadiabattoinﬁnity(approximately200hPaonthegraph,moistanddry adiabatsarenearlyparalleltoeachotheratthisandhigherlevels).Allvaporthat theparcelinitiallycontainedhascondensed.Thenmovetheparcelbacktothe

Figure7.7DiagramforExercises4and5.

Figure7.8DiagramforExercises6and7.

1000hPalevel(pointB)alongthedryadiabat.ThetemperatureatpointBisthe equivalentpotentialtemperature (286K).

Fromformula(6.82)weget

Exercise7 ContinueExercise6.Lifttheparceltothe425hPalevel.Howmuch wateriscondensedduringtheascent?

Answer:BringtheparcelfrompointAtothe425hPalevelalongthemoistadiabat (pointC).Theascentisalongthemoistadiabatsincetheparcelwassaturatedat theinitialtemperatureandpressure.AtpointCthesaturationmixingratioisequal to0.10gkg 1 .Hence,theamountofwatercondensedoutduringtheascentis 2.2gkg 1 0.1gkg 1 = 2.1gkg 1 .

Figure7.9DiagramforExercise8.

Exercise8 Airat800hPahasatemperatureof10◦ Candadewpointof0 ◦ C. Determinethewet-bulbtemperatureandthewet-bulbpotentialtemperature.

Answer:Thewet-bulbtemperature,bydeﬁnition,isthetemperatureanairparcel wouldhaveifcooledadiabaticallytosaturationatconstantpressurebyevaporation ofwaterintoit.Toﬁndthewet-bulbtemperatureonFigure 7.9 youhavetoperform thefollowingsteps.(1)Findtheliftingcondensationlevel(seeExercise4).Itisat 690hPa(pointC).(2)DrawamoistadiabatstartingatCdowntotheintersection withthe800hPaisobar(pointD).Readthemagnitudeofthetemperatureatpoint D,whichisthewet-bulbtemperature(5◦ C).Thewet-bulbpotentialtemperature canbefoundbyextrapolatingthemoistadiabatlinestartingattheLCL(pointC) tothe1000hPalevel(288K).

Exercise9 Therelativehumidityoftheairatthe950hPapressurelevelis47%. Plottherelativepositionsoftemperature,dewpointandwet-bulbtemperatureon thechart.

Answer:TheRH = 47%meansthatairisunsaturated.Therefore,thedewpoint (pointBonFigure 7.10)istotheleftofthetemperature(pointA)atthe950isobar level.Thewet-bulbtemperatureisalwayshigherthanthedewpoint.Thishappens becausethedewpointistheresultofcoolingtosaturationatconstantpressurewith constantmixingratio.Instead,thewet-bulbtemperaturecharacterizescoolingto saturationoftheairparcelbyevaporatingwaterintotheparcel,whichraisesthe actualmixingratioasthecoolingproceeds.So,thewet-bulbtemperature(pointC) isbetweenthedewpointandtheactualtemperature.

Exercise10:Chinookwind TheChinookwindisthewarmdrywindthatisthe resultofwinddescendingeastwardsfromtheRockyMountains(Figure 7.11). Chinookwindscancauselargetemperaturechangesoccurringonlyoverafew

hours.“Chinook”isaNativeAmericanwordthatmeans“snow-eater,”reﬂecting thewarmingeffect,whichcouldbeaccompaniedbysubstantialmelting.Itiscalled FöhninEurope.Lettheairhaveatemperatureof10◦ Cand5gkg 1 mixingratio atpressure950hPaattheupslope(western)sideofthemountain.Whenpassing overthetopofthemountainatthe600hPalevel,assumethat80%ofthemoisture isprecipitatedout.Theairreturnstothe950hPalevelontheeasternsideofthe mountainafterbeingheatedbythecondensation.Comparethetemperature,relative humidity,potentialtemperatureandwet-bulbpotentialtemperaturesonbothsides ofthemountainatthe950hPalevel.

Answer:Findtheinitiallocationoftheparcelonthechart(pointAonFigure 7.12). Atthispointthesaturationmixingratioisequalto8gkg 1 .Sinceweknowthe mixingratiooftheparcelatpointA,wecanﬁndtherelativehumidityRH = 62%.ThepotentialtemperatureatpointAis287K(seeExercise1).Theparcelis unsaturatedatpointA.Thus,whenlifting,itfollowsadryadiabatuntilitreaches theLCL(seeExercise4).TheLCLisatthe860hPalevel.Theintersectionof thedryadiabatstartingatpointAandthelineofconstantsaturationmixingratio of5gkg 1 isatpointL.Toﬁndthewet-bulbpotentialtemperature(seeExercise8),

Figure7.10DiagramforExercise9.

Figure7.11Exercise10,illustrationoftheChinookwind.

Figure7.12DiagramforExercise10.

drawamoistadiabatstartingfromLbacktothe1000hPalevel.Readtheabscissa, whichyields9◦ C.Withfurtheruplifting,theparcelfollowsthemoistadiabatto thetopofthemountainatthe600hPalevel(pointB).Atthispointtheparcel hassaturationmixingratioequalto1.6gkg 1 .Nowwecanﬁndtheamountof watercondensedoutduringtheascenttothetopofthemountain.Itisequaltothe differencebetweenmixingratiosatpointsAandB,5gkg 1 1.6gkg 1 = 3.4 gkg 1 .Weknowthat80%ofthismoisture,whichis2.7gkg 1 ,isprecipitated out.Therefore,thereis3.4gkg 1 2.7gkg 1 = 0.7gkg 1 ofliquidwateratthe topofthemountain.Whendescendingontheothersideofthemountain,theair parcelfollowsthemoistadiabatagainsinceitissaturated.Whendescending,the parcelwarmsandexpands,sowaterevaporates.Eventually,atsomelevel,allthe liquidwaterevaporates,theparcelisnolongersaturatedanditsfurtherdescentis alongadryadiabat.Howcanwecanﬁndthis“threshold”pressurelevelwhenthe parcelreachesitsnewsaturationmixingratiocorrespondingtotheevaporationof allliquidwater?Weknowthatatthetopofthemountainthereis1.6gofliquid waterperkilogramofdryair.Wealsoknowthataftertheprecipitationthereis still0.7gofliquidwaterperkilogramofdryair.Hence,thenewsaturationmixing ratio,whenallliquidwaterevaporates,is1.6gkg 1 + 0.7gkg 1 = 2.3gkg 1 . Theparcelintersectsthelinewith2.3gkg 1 saturationmixingratioatthe660hPa level(pointC).Thisisthe“threshold”pressurelevelbecausethereaftertheparcel becomesunsaturatedandfollowsitsdryadiabattothe950hPalevel(pointD). Theparcel’snewtemperatureatpointDis17◦ C,whichis7 ◦ Cwarmerthaninthe beginning.Duringdryadiabaticdescentthemixingratioisconstant,soatpointD theparcelhasthesamemixingratioasatpointC,whichis2.3gkg 1 .Thesaturation mixingratioatpointDis12.6gkg 1 .Therefore,therelativehumidityoftheair

ontheothersideofthemountainis18%(comparewith62%atthebeginning). ThepotentialtemperatureatpointDis294K(comparewith287KatpointA).The wet-bulbtemperatureisagain9 ◦ C;itisconservedduringtheprocess.

7.4Stabilityproblem:examplesounding

Figure 7.13 showsatypicalsounding.Inthisexampletheairnearthesurface isunsaturatedsincethedewpointislessthantheairtemperature.Whenlifted, theairfollowsadryadiabatuntilitreachestheLCLasdepictedinthegraph. Furtherupliftingisalongamoistadiabat(indicatedbythedashedline).Theﬁrst intersectionofthismoistadiabatwiththesoundingcurveoccursatthe leveloffree convection (LFConthegraph).IfaparcelisliftedtoaheightlowerthantheLFC, itreturnstowardthesurfacebecauseitexperiencesnegativebuoyancysinceitis

Figure7.13Illustrationofahypotheticalsounding.Solidcurvescorrespondtoair temperatureanddewpointsounding,thedashedcurveshowstheactualpathofa parcelliftingfromthesurface.

7.4Stabilityproblem:examplesounding 179 alwayscoolerthantheenvironmentalongitspath.If,however,theparcelreaches theLFC,itbecomeswarmerthanitssurroundings.So,theLFCisthelevelwhere theparcelbecomespositivelybuoyant.Thepositivebuoyancycarriestheparcelup tothe levelofneutralbuoyancy (LNB),wheretheparcel’spathintersectswiththe (measured)temperaturesounding.

InChapter6wediscussedstabilitycriteriaforanunsaturatedparcel.Letusapply thesetodifferentlayersofthesounding.Bylayerwemeanathinslabofairalong thesoundingacrosswhichthereisanapproximatelyconstantlapserate.Consider thestabilityoflayersdepictedonFigure 7.13 (Figure 7.14 enlargesthepartofthe chartweareinterestedin).

LayerABisstablesinceitslapserateislessthanthedryadiabaticlapserate. LayerBCisalayerexhibitingaslight inversion.Aninversionoccursinalayer whenthetemperatureincreaseswithheight–suchalayerisobviouslystable.The layerCDisalsostable,itslapseratebeinglessthanthedryadiabaticlapserate. ThelayerDEisneutral;itisparalleltothedryadiabat,sotemperaturedecreases atthesamerateaswithadryadiabaticprocess.

Nowconsiderthecasewhenthetemperaturedecreaseswithheightatarateless thanthedryadiabaticlapseratebutgreaterthanthemoistadiabaticlapserate, d > > m (forexample,layerAB).AnairparcelinlayerABisnegatively buoyantifliftedashortdistancebutcouldbecomepositivelybuoyantifan imposedverticalmotionisstrongenoughtobringthisparceltoitslevelof freeconvection(LFC).Forexample,theairmightbepushedupamountainside orliftedbymechanicallyinducedoverturning(turbulence).Suchasituationis called conditionalinstability.Thelayerisstablewhenairisunsaturated,butcould becomeunstablewithexternallyimposedverticalmotion.Wecantestthelayerfor

Figure7.14TheenlargedareaofinterestofthesoundingshowninFigure 7.13.

Thermodynamiccharts

conditionalinstabilitybycalculatingtheverticalgradientofsaturationequivalent potentialtemperatured θs /d z ,ratherthanbycalculatingthelapserate.Inthecaseof conditionalinstabilityd θs /d z < 0.Indeed,onecanseefromthegraph(Figure 7.14) that θs (A) >θs (B).If = m (AandBareonthesamemoistadiabat),then θs (A) = θs (B),andd θs /d z = 0.

Tosummarize,ifthetemperatureintheparticularlayerdecreasesatarategreater thanthedryadiabaticlapserate,thislayerisunstableinanycaseforbothsaturated andunsaturatedparcels.Ifthetemperaturedecreasesataratelessthanthemoist adiabaticlapserate,thislayerisabsolutelystable;thesaturationequivalentpotential temperatureincreaseswithheightinthiscase.Theformulasbelowlistthestability criteria:

Thereisanothertypeofinstabilitycalled potentialinstability.Potentialinstability occurswhenthelayerisliftedasawhole,forexamplebyconvectionassociated withamovingfrontorwithaﬂowpassingoveramountain.Whenmoving,the saturationconditionsand,consequently,pathsaredifferentforthebottomand topofthelayer,whichcanchangetheinitialtemperaturegradient.Considerthe invertedlayerBConthesamesounding(Figure 7.15).Thereasonwechosean inversionlayeristhattheeffectwewanttodemonstrateismorepronouncedinthis case.Imaginethatanupliftingﬂowmovesthislayerasawhole200hPahigher. Whathappens?Thebottomofthelayer(pointB)isalmostsaturatedinitially. Therefore,whenlifted,itquicklyreachesitsLCL(labeledasLCLB onthegraph) andfollowsamoistadiabatthereafter(pointB1 ).Thesituationatthetopofthe layer(pointC)isdifferent.Atthebeginning,airatthetopofthelayerhasalow relativehumidity,itsdewpointisfartotheleftofitstemperature.Whenuplifted, theairatpointCreachesitsLCL(labeledLCLC onthegraph)andthenfollows amoistadiabat(pointC1 ).NowconsiderthelapserateoftheB1 C1 layer.Itis largerthanthedryadiabaticlapserate.So,theabsolutelystablelayerBCbecomes unstablewhenuplifted.Thisisthecaseofpotentialinstability.Thecriterion forthisinstabilityisanegativegradientoftheequivalentpotentialtemperature inthelayer,d θe /d z < 0.Indeed,theequivalentpotentialtemperatureremains constantduringtheliftingeverywhere:bothaboveandbelowtheLCL.Whenthe airreachestheLCL,theequivalentpotentialtemperaturebecomesequaltothe saturationequivalentpotentialtemperature, θe = θs .Afterthat,fromthecriterion

Figure7.15Illustrationofpotentialinstability.ThestablelayerBCbecomes unstableafteruplifting.

forconditionalinstability(7.15)weobtaind θe /d z < 0forpotentialinstability.One canseefromFigure 7.15 that θe (B)>θe (C).Thisisacommonoccurrenceinthe southeasternUSAaswarmmoistairisadvectedfromthesouth,overridingdrier airadvectedfromthewest.Ifliftingoccurs,suchaconﬁgurationcanleadtosevere weatherconditions.

7.5Convectiveavailablepotentialenergy(CAPE)

Inprevioussectionsweanalyzedthestabilityofthedisplacementofasmallparcel intermsoftemperaturelapserate.Inthissectionwewillcontinuetoanalyze stability,butintermsofenergy.Wehavealreadyshownthatwhenthereisapositive areaintheclosedloopbetweenenvironmentalandadiabaticcurvesona T –ln p diagramor,inotherwords,ifaparcel(afteranudge)ispositivelybuoyant,the parcel’skineticenergyincreases.Consideraparcelbeinginitiallyunsaturatedin aconditionallyunstableatmosphere.Wedenotetheparcel’sinitiallocationbyA intheexampleofatemperaturesoundingshowninFigure 7.13.Whenlifted,the parcelﬁrstfollowsadryadiabatuntilitreachestheLCL.Withfurtherlifting,it followsamoistadiabat.Iftheupwardmotionisstrongenoughtobringtheparcel toitsLFC,theparcelbecomespositivelybuoyant.Figure 7.16 showsthesame soundingasFigure 7.13.Thepositivearea(shadeddarkonFigure 7.16)between theparcel’spathandthesoundingboundedbytheLFCandtheLNBiscalled the convectiveavailablepotentialenergy (CAPE).CAPErepresentsthemaximum kineticenergythatapositivelybuoyantparcelcanacquirebyascendingwithout exchangingmomentum(eddyfriction),heatandmoisturewithitsenvironment.

Figure7.16ThesamesoundingasinFigure 7.13.Darkandlightgrayareas representconvectiveavailablepotentialenergy(CAPE)andconvectiveinhibition energy(CIN)correspondingly.

Wecancalculatetheidealchangeofkineticenergyperunitmassduetopositive buoyancybyintegrating(7.5)fromLFCtoLNB.Theamountofkineticenergy releasedinthissituationis

CAPEisausefulmeasureofthunderstormseverity,sinceitallowsustoestimate thevalueofmaximumpossibleverticalvelocity.Indeed,ifaparcelhaszerovertical velocityattheLFC,thenfrom(7.17)

Inthisconsiderationwehaveneglectedtheeffectofwatercondensation,which reducesbuoyancyslightly.ValuesofCAPEgreaterthan1000Jkg 1 implythe possibilityofstrongconvection.Eveniftheﬁnalverticalvelocityislessthanthe maximumvalue,theenergyisstilldissipatedinturbulencewithinthecloud.

LetusreturnnowtoFigure 7.13.BeforetheparcelstartingatpointAreaches itsLFC,ithastoovercomeapotentialenergybarrierbetweentheLCLandthe LFC,whereaparceliscoolerthanitsenvironmentandnegativebuoyancytends toreturntheparceltowardthesurface.This negative areabetweentheparcel’s pathandenvironmentboundedbytheLCLandtheLFCiscalledthe convective inhibitionenergy (CIN).ItisshownasthelightgrayareainFigure 7.16.CIN controlswhetherconvectionactuallyoccurs.Itisameasureofhowmuchenergy isrequiredtoovercomethenegativebuoyancyandallowconvection.ToﬁndCIN wehavetointegrate(7.5)fromtheLCLtotheLFC,namely

IftheCINisgreaterthan100Jkg 1 ,asigniﬁcantsourceofliftingisneededtobring theparceltoitsLFCinordertocreatefavorableconditionsfordeepconvection.

Figure7.17ThesamesoundingasinFigure 7.13.Foraparceloriginatingatpoint B,CAPEiszero.

Figure7.18Illustrationofahypotheticalsounding.

Ifwe,forexample,areinterestedintheCAPEofanairparcelstartingatpointB ratherthanatpointAonthesamesoundingdiagram(Figure 7.17),thentheCAPE iszero.ThepathoftheparcelstartingatpointBisshownbyadashedlineon Figure 7.17.Thisparcelisalwayscoolerthanitslocalenvironment.Itisimportant tonotethatthevalueofCAPEdependsontheinitialparcellocation.

ConsiderthesoundingshowninFigure 7.18.Aparcelstartingfromthesurface willexperiencenegativebuoyancy.TheareacorrespondingtotheCINisshown inlightgrayinFigure 7.19,whichsimplyenlargesthepartofFigure 7.18 weare interestedin.TheareashadedindarkergraycorrespondstotheCAPE.Tobecome positivelybuoyant,aparcelstartedfromthesurface(pointAonthegraph)has toovercomethis“lightgray”area.Imaginenowthatweexpectthesurfacetobe warmedinthenextcoupleofhours.Then,insteadofpointA,theparcelstartsfrom pointA1 (Figure 7.20).Itdoesnotexperiencenegativebuoyancyanylonger;itsLFC coincideswithitsLCL,andtheseareexcellentconditionsforseverethunderstorm activity.If,onthecontrary,weexpectthesurfacetobecooled(Figure 7.21,point

Figure7.19ThesamesoundingasinFigure 7.18.CAPEandCINfortheparcel startedfromthesurface(pointA)areshownindarkandlightgraycorrespondingly.

Figure7.20ThesamesoundingasinFigure 7.18.Illustrationofahypothetical surfacewarming.FortheparceloriginatingatpointA1 ,thereisnoCIN.

A2 ),thenthesituationisreversed.CINbecomeslarger,andCAPEissmallerthan theprevioussituation.Thismeansthattheconditionsforathunderstormareno longerfavorable.

Fortunatelyfortheforecaster,thevaluesofmanyoftheparametersdiscussed above(CAPE,CIN,etc.)areprintedrightontheskew T chartsthatarepublishedat manysitesontheinternet.Hence,notediouscomputationsofareasarenecessary bytheuser.

Figure7.21ThesamesoundingasinFigure 7.18.Illustrationofahypothetical surfacecooling.FortheparceloriginatedfrompointA2 ,CAPEdecreases,butCIN increasesincomparisonwiththeparcelstartedfromA.

NotationandabbreviationsforChapter7

CAPEconvectiveavailablepotentialenergy(Jkg 1 )

CINconvectiveinhibitionenergy(Jkg 1 )

F , F /M force,perunitmass(Nkg 1 )

h, u speciﬁcenthalpy,internalenergy(Jkg 1 )

K kineticenergy(J)

Levap latentheatofevaporation(Jkg 1 )

R gasconstant(JK 1 kg 1 )

ρa parceldensityforadiabaticchange(kgm 3 )

ρe environmentaldensity(kgm 3 )

Va , Ve volumesofaparcelalonganadiabatandoftheenvironment(m3 )

w , ws mixingratio,saturated(gwatervaporperkgdryair)

X , Y abscissa,ordinate

Problems

7.1RefertothesoundinginFigure 7.22

(a)Estimatethemixingratioatthesurface.

(b)Estimatethesaturationmixingratioatthesurface.Whatistherelativehumidityat thesurface?

(c)Whatisthedewpoint?

(d)WhatisthepressurelevelattheLCL?

(e)Whatisthewet-bulbtemperatureatthesurface?

(f)IsCIN > 0?

Figure7.22AsoundingfromLakeCharles,LA,at00Z,June1,2007.Takenfrom theUniversityofWyomingwebsite.

(g)IstherealargeCAPE?

(h)Whatisthemixingratioat800hPa?

(i)Whatisthesaturationmixingratioat800hPa?

7.2RefertothesoundinginFigure 7.23

72764 BIS Bismarck

00Z 01 Jan 2007

Figure7.23AsoundingfromBismarck,ND,at00Z,January1,2007.Takenfrom theUniversityofWyomingwebsite.

Thermodynamiccharts

(a)Whereisthetropopause?

(b)DescribetheairmassoverBismarckonthisday.

(c)Describethehumidityasafunctionofaltitude.

(d)Arethereanytemperatureinversionsasafunctionofaltitude?

(e)IsthereanyCAPEorCIN?Stable?

7.3RefertothesoundinginFigure 7.24.

(a)Whereisthetropopause?

(b)DescribetheairmassoverBismarckonthisday.

(c)Describethehumidityasafunctionofaltitude.

(d)Arethereanytemperatureinversionsasafunctionofaltitude?

(e)IsthereanyCAPEorCIN?Stable?

00Z 01 Aug 2007

Figure7.24AsoundingfromBismarck,ND,at00Z,August1,2007.Takenfrom theUniversityofWyomingwebsite.

72764 BIS Bismarck

Figure7.25Ablankskew T chartforProblem7.4.TakenfromtheUniversityof Wyomingwebsite.

7.4Theairat1000hPaand11◦ Chasdewpoint 0.5 ◦ C(seeFigure 7.25).

(a)Findthemixingratio,relativehumidity,andthepotentialtemperatureusingboth theskew T chartandformulas.

(b)Findtheliftingcondensationlevelusingthechart.

(c)Findtheequivalentpotentialtemperatureusingthechart.

(d)Whatarethemixingratioandthepotentialtemperatureiftheparcelrisesto 900hPa?

(e)Whatistheequivalentpotentialtemperatureiftheparcelrisesto600hPa?

7.5Consideraparcelofmoistairthatrisesfromthesurfacewhere p = 1000hPato400hPa. Assumeallofthecondensedwaterisprecipitatedoutduringtheascent.Theparcelthen descends(unsaturated)backtothesurface.Iftheinitialtemperatureis20◦ Candits initialdewpointis0◦ C,ﬁndthefollowing(useFigure 7.26).

(a)Howmuchwateriscondensedduringtheascent?

(b)Thetemperatureoftheparcelanditsdewpointtemperaturewhenitreturnstothe surface(1000hPa).

Figure7.26Ablankskew T chartforProblem7.5.TakenfromtheUniversityof Wyomingwebsite.

Thermochemistry

Applicationsofthermodynamicsformtheheartof physicalchemistry.Withthe FirstLawofThermodynamicswecanﬁndsomeelementaryapplicationsof thermodynamicstoprocessesrelevanttotheatmosphere.Asanexample,consider theelementaryreaction

wheretheginparenthesesindicatesthatthechemicalspeciesisinthegaseousstate (thesolidphaseisindicatedbys,liquidbylandaqueoussolutionbyaq).

AtthemolecularlevelamoleculeofCO2 strikesanH2 moleculeandthe rearrangementcollisionoccurswithacertainprobabilitydependingonvelocities, spatialorientationofthecolliders,etc.Therateatwhichareactionproceedsin asystemistheproductofthelikelihoodofacollisionbetweentheimportant partiesandtheprobabilityofrearrangement,giventhecollision.SometimesaCO moleculebumpsintoanH2 Omoleculeandthereversereactionoccurs.Thatthe reactionmightgobothwaysisindicatedbytheequation

Onceequilibriumisestablished(rateofreactionsproceedingtotherightequals therateofthosegoingtotheleft)inasuitableenclosure,wecanconsiderthematter involvedtobeathermodynamicsystemwhichcanbetreatedbythemethodsof equilibriumthermodynamics.Thenumberofmolesofthespecies νCO2 , νH2 , νCO and νH2 O becomethermodynamiccoordinatesorfunctionsalongwiththoseweare alreadyacquaintedwith, M, p, V , U , H , S , G ,and T .Infact,wewanttoknowhow thesecoordinates(equilibriumconcentrations)varyasafunctionofthetemperature ifthepressureisheldconstant.Fixedpressureistheusualconditionforgasphase reactionsintheatmospheresincetheycanbetakenasoccurringinasmallparcelor volumeelement,whosepressureinsidequicklyadjuststothatoutside(whichinour

CO2 (g ) + H2 (g ) → CO(g ) + H2 O(g )

CO2 (g ) + H2 (g ) CO(g ) + H2 O(g ).(8.2)

casedependsonaltitude).Intherealatmospheretherearemanychemicalspecies invariousstatesofequilibrium.Thetaskoftheatmosphericchemistisoftentosort outwhichreactionsareimportant,whatthesourcesofthevariousspeciesare,what isthefeasibilityandenergeticsofchemicalreactionsandhowfastthereactions proceed.

8.1Standardenthalpyofformation

Considerageneralchemicalreaction

whereAandBarecalledthe reactants,CandDarethe products,and a , b, c and d areintegers(sometimesrationalnumbers)insertedtobalancetheequation.

Supposetheingredientsontheleft-handsideoftheequation(the reactants)are placedinaclosedcontainerthatisimpermeabletomattercrossingitsbounding surface.Furthermore,letthereaction(8.3)proceedfromlefttorightatconstant pressure.Ifnoheatisallowedtoenterorleavethesystemduringthe(irreversible) process,theﬁnaltemperaturewillbedifferentfromthatbeforethereactionbegan. Ifthetemperatureofthesystemgoesup,wesaythereactionis exothermic.Ifthe temperaturegoesdown,itis endothermic.Chemistshavefoundaconvenientway ofcharacterizingtheenergeticsofsuchreactions.Supposethereactiongoesfrom lefttorighttocompletion(noreactantsremaining),thentheheatrequiredtorestore thesystemtoitsoriginaltemperatureatconstantpressureisitschangeinenthalpy duringtheirreversibleprocess, H .

Inordertoﬁndtheheatofreactionforaparticularchemicalprocessitis necessarytostartwiththeso-called standardenthalpyofformation oftheindividual compounds.Thesearebasedupontheenthalpyneededtoformthecompoundfrom thestateoftheindividualatomicspeciesmostcommonlyfoundinnature.For example,theconventionfortheelementoxygenistostartwiththegaseousform O2 ,notO.SimilarlythebasestateaccordingtotheconventionfornitrogenisN2 andforhydrogenitisH2 .ForargonitistheatomicformArandforcarbonitisC. Thestandardenthalpyofchemicalreaction,whenreactantsintheirstandardstate areconvertedtoproductsintheirstandardstates,isequaltothedifferencebetween standardenthalpyofformationofproductsandreactants:

Theoverbarindicatesthat1molofthesubstanceisconsidered,thesuperscript ◦ referstothestandardstate,whichisat1atmand25 ◦ Cbyconvention(seeTable 8.1). If H ◦ isnegative,heatisreleasedandthereactionisexothermic.Exothermic

Table8.1 Standardenthalpiesofformationforselected compounds( H ◦ inunitsofkJmol 1 )

Thesymbolinparenthesesafterthecompoundindicateswhether itsphysicalstateisliquid,solidorgas.Allvaluesrelateto298K.

reactionscanproceedspontaneouslyintheatmosphere.Iftheoppositeistrue, H ◦ ispositive,thereactionisendothermic,andanexternalsourceofenergyis neededforthereactiontoproceed.

Theexothermicreactionscanbesigniﬁcantforthethermalbudgetofthe atmosphere.Theclassicalexampleisthereactionleadingtotheformationofozone. Theheatreleasedinthisprocessdominatestheformofthetemperatureproﬁlein thestratosphere.

Example8.1 Themainmechanismofozoneformationinthestratosphereisthe recombinationofatomicoxygen:

whereMisamoleculeinthebackgroundgaswhichisneededtocarryoffthe excessmomentuminatwo-bodies-to-onemolecularcollision.Findhowmuch heatisreleasedbythisreaction.

Answer:Toﬁndhowmuchheatisliberated,weneedtocalculatetheenthalpyof thereaction:

Theminussignindicatesthatthisisanexothermicreaction.Therefore,withthe reactionofozoneformation(8.5)106.4kJpermoleisliberated.Thisliberatedheat warmsthestratosphericairandraisesitstemperaturewhichreachesamaximum atabout50kmaltitude.NotethattheconcentrationofOin(8.5)isdeterminedby thephotodissociationofO2 ,O3 andotherspecies.

Example8.2 Supposewewantedtoknowthechangeinenthalpyforthereaction:

Weform

H ◦ ispositive,whichmeansthatthisreactionisendothermic,andheatisabsorbed duringtheprocess.

8.2Photochemistry

Furtherexamplesofendothermicreactionsincludethe photochemicalreactions. Inthiscasetheadditionalsourceofenergynecessaryfortheendothermic reactiontoproceedissolarradiationwhichcanbreakthechemicalbondsof atmosphericspecies.Inthisbookwewillconsideronlyonephotochemicalprocess: photodissociation. 1

Physicsrefresher Solarradiationconsistsofelectromagneticwaves. Electromagneticradiationhasadualwave-particlenature.Thismeansthat electromagneticradiationexhibitsbothwave-likeandparticle-likeproperties.Inits waveformelectromagneticradiationcanbethoughtofasagroupofsuperimposed wavessometimesreferredtoasan ensemble propagatinginvacuumwiththespeedof light c = 2.998 × 108 ms 1 independentofwavelength.Eachwaveinthisensemble canbetreatedasasimplesinusoidalfunction(seeFigure 8.1)withacertain wavelength,frequency,andamplitude.Thewavelength, λ,isthedistancebetween twosuccessivepeaksofthewave.Theunitsof λ aremeters.Thefrequencyofawave, f ,isthenumberofcyclesthatpassanobserverinasecond.Theunitoffrequencyis thehertz(1Hzisoneoscillationpersecond).Theproductofwavelengthand frequencyforanindividualwaveisequaltothespeedoflight(speedisdistance dividedbytime): c = λ × f .Fromthisequationonecanseethatwaveswithhigher frequencieshaveshorterwavelengths,andwaveswithlowerfrequencieshavelonger wavelengths.

Whenradiationinteractswithatomsormolecules,itcanbeabsorbedoremitted onlybycertaindiscreteamountsofenergy.Inotherwords,electromagneticradiation is quantized.Thewavesmaybethoughtofasabeamofparticlescalledphotons carryingdiscreteamountsorpackagesofenergy.Theenergyofaphotonof

1 Interestedreadersarereferredto BasicPhysicalChemistryfortheAtmosphericSciences byPeterV.Hobbs (2000)formoreinformationonphotochemicalreactions.

Figure8.1Asinusoidalwaveofagivenwavelength.

Hence,theshorterwavelength(higherfrequency)photonsaremoreenergetic.The electromagneticspectrumfromradiowavestogammaraysisillustratedin Figure 8.2.Themostenergeticphotonsaregammarays.Asonemovesvertically downthespectrumfromgammaraystowardsradiowaves,theenergydecreasesand

Figure8.2Electromagneticspectrum.

sodoesthefrequency,whilethewavelengthincreases.Anarrowbandofthespectrum correspondstothevisiblelight.Photonsinthevisiblerange(approximately 400–700nm)canbedetectedbyahumaneye.Ultravioletradiationhasshorter wavelength(higherfrequency)thanthevisiblepartofthespectrum.

Todescribetheradiationpenetratingtheatmosphereitisusefultointroducethe ideaofan energyﬂux.Theenergyﬂux(energypassingperunitareaperpendicular tothebeam,perunittime)isgivenby

where n0 isthenumberofphotonsperunitvolume(numberdensity asinagas). Theenergyﬂuxofsolarphotonsatthetopoftheatmosphereis1370Wm 2 . Thisparameteriscalledthe solarconstant.Solarphotonspropagatingthroughthe atmospherecanbeabsorbedand/orscatteredbyatmosphericconstituents.Consider the attenuation ofaphotonﬂuxatwavelength λ duetophotonabsorptionassuming normalincidenceforsimplicity(thesunisat zenith,directlyoverhead(Figure 8.3)).

Wedenoteaphotonenergyfluxatwavelength λ as Fλ ;itsdimensionisenergyper unitarea,perunittime,andperunitwavelength.Ifatthetopoftheatmospherethe fluxperunitwavelengthis Fλ(top),thefluxatheight z , Fλ(z ),isdescribedby

λ(z ) = Fλ(top) exp( τ(z )) [attenuationofaverticalsolarbeam].(8.12)

Figure8.3Schematicdiagramofasolarbeamcomingfromdirectlyoverhead withattenuationofthebeam’sintensityindicatedbyshading.

F = energyﬂux = n0 hcf (8.11)

Fλ(top)

Fλ(z) z

Thisequationfollowsfromtheradiativetransfertheory. 2 Thecoefﬁcient τ in theexponentiscalledtheopticaldepth:

Theopticaldepth τ isproportionaltotheverticallyintegratedcolumndensity ∞

z N (z )d z where N isaconcentrationofatmosphericspeciesabsorbingat λ.The integrationfrom z to ∞ reflectsthepaththephotonstravelfromthetopofthe atmospheretoheight z .Thecoefficientofproportionality σ iscalledtheabsorption cross-section.Thisparameterdescribestheabilityofaparticulargaseousspecies toabsorbphotons;itismeasuredinm2 (oftenintheliteratureascm2 ).Absorption cross-sectionscanbemeasuredinthelaboratory.Whentheopticaldepthgets closetounity,thefluxisattenuatedbyafactorofroughlythree(e ≈ 2.7).For example,for λ between240and300nm(ultravioletrange) τ reachesunitydueto theabsorptionbyozoneapproximatelyatheightsof30–38km.Thismeansthat thesolarphotonsinthisrangeareabsorbedbyozoneinthestratosphereanddonot reachthetroposphere.Atshorterwavelengths,between175and200nm,radiation isabsorbedbyoxygenatheightsof40–80km.Atwavelengthsgreaterthan310 nm,mostphotonspenetrateintothetroposphereandreachthesurface.Ifthesun haszenithangle = 0,thencos hastobeaddedintheformulafortheflux attenuation(Figure 8.4):

Thelargerthezenithangle,thestrongertheattenuationatagivenheight z

Thephotonsintheultravioletandvisiblerangesareenergeticenoughtobreak moleculesapart.Thisprocessiscalledphotodissociation.Photodissociationplays averyimportantroleinthetroposphereandthestratosphere.Forexample,akey reactioninthetroposphereisthephotodissociationofozonebyultravioletradiation:

+ hf → O2 + O(8.15) wherethenotation hf denotesaphotonwithfrequency f .Thisnotationemphasizes thattheenergycarriedbythephotonisthefrequencytimesPlanck’sconstant.The formationoftroposphericozoneisduetophotodissociationofNO2 :

2 Abeamisattenuatedinadistanceintervald z byanamountproportionaltotheincomingbeam’sﬂuxandto theamountofattenuatingmaterialintheinterval.Theresultisd Fλ =−AFλ d z where A isproportionaltothe amountofattenuatingmaterialperunitvolume.Integrationofthisequationleadstoexponentialdecayalong thepath,knownasBeer’sLaw.

Figure8.4Schematicdiagramofasolarbeamcomingfromtheupperrightand passingthroughaslabofmatterwiththicknessd z .Thepathlengthintheslabis d z / cos ,where isthesolarzenithangle.

TheatomicoxygenthenleadstotheformationofozonebyrecombinationwithO2 : O + O2 + M → O3 + M.(8.17)

Theformationoftheozonelayerisalsocausedbyphotodissociation.Inthe stratosphere,ultravioletradiationwith λ ≤ 240nmphotodissociatesmolecular oxygenO2 creatingatomicoxygen:

+ hf → O + O.(8.18)

Ozoneisthenformedbyrecombinationofatomicandmolecularoxygen(reaction (8.17)).

Example8.3 Considerthephotodissociationofanoxygenmoleculethatcreates twogroundstateoxygenatoms:

+ hf → O + O.(8.19)

Whatphotonenergyisrequiredforthisreactiontoproceed?Whatpartofthe electromagneticspectrumcorrespondstothisenergy?

Answer:Firstletusexaminethestandardenthalpyofthisreaction:

where NA isAvogadro’snumber.Toﬁndtheminimumenergyofaphotonrequired tobreakoneO2 moleculewehavetoequatethestandardenthalpyofthisreaction tozero.Onlyphotonswithenergy hf greaterthanthisminimumenergyareableto

breakanO2 moleculeapart: hf ≥ 498.34/NA kJ.Thisinequalitygivesthevalue ofthesmallestfrequencyrequired: f ≥ 498.34 × 103 /(6.022 × 1023 × 6.62 × 10 34 ) = 12.49 × 1014 Hz,orthelargestwavelength. λ ≤ 0.24 × 10 6 m = 240nm. Therefore,radiationwith λ ≤ 240nm,whichcorrespondstotheultravioletpartof thespectrum,isneededforthereaction(8.19)toproceed.

Wecanﬁndtheenergyofaphotonnecessaryforacertainreactiontoproceed withoutexaminationoftheenthalpy,ifweknowtheenergyofdissociationofa chemicalbond.

Example8.4 DuringthedaytimeanimportantsourceofNOinthestratosphereis thedissociationofNO2 molecules:

Findthemaximumwavelengthofelectromagneticradiationrequiredforthis reaction,iftheenergyofdissociationofanNO2 moleculeis5.05 × 10 19 J.Energy ofdissociationisoftengiveninelectronvolts:3

1electronvolt(eV) = 1.6 × 10 19 J(8.22) or 5.05 × 10 19 J = 3.16eV(8.23) (thisreactionisalsoimportantinpollutedurbanair,sinceitisasourceof troposphericozone).

Answer:Photonswithenergy hf ≥ 5.05 × 10 19 Jareneededtodissociatean NO2 molecule.Then, f ≥ 5.05 × 10 19 /(6.62 × 10 34 ) = 7.6 × 1014 Hz.Finally, λ ≤ 2.998 × 108 /7.62 × 1014 = 0.39 × 10 6 m = 390nm,whichisattheboundary betweenthevisibleandultravioletpartsofthespectrum.

8.3Gibbsenergyforchemicalreactions

Earlierweshowedhowtoﬁndtheenthalpyforphasechangesandchemical reactionsbymanipulatingvaluestakenfromstandardtables.Inthissectionwe

3 Theunitelectronvolt(eV)istheenergyanelectronhasafterbeingacceleratedacrossapotentialdifferenceof 1volt.Thisisthepreferredunitinatomicandnuclearphysics.Thebindingenergyofanelectronintheground stateofahydrogenatomis13.6eV.

willworkwiththeGibbsenergy,whichisusefulindeterminingthefeasibility ofachemicalreactionandtheabundancesofspeciesinchemicalequilibrium situations.Usingenthalpytodeterminewhetherareactionwillproceedislimited, sinceenthalpydependsonentropy S andpressure p aswellastheconcentrationsof thevariousspeciespresent.Ifwearetoexaminewhetherareactionwillproceed,we willﬁndithardtoholdtheentropyconstant,especiallyinnature.Ontheotherhand, theGibbsenergyisusefulwhenthetemperatureandpressureareheldconstant. Thisisoftenthecaseintheatmospherewhenthereactionoccursbetweentrace gasesatacertainaltitude(pressure)andthetemperatureisconstantbecausethe reagentsarebufferedthermallybythesurroundingbackgroundgasmolecules.

The standardGibbsenergy isintroducedsimilarlytothestandardenthalpyof thereaction.ThestandardGibbsenergyofachemicalcompound, G ◦ ,isthe changeoftheGibbsenergywhen1molofacompoundisformed(theoverbar isanindicationof1molbeingconsidered).Conventionally,thestandardGibbs energyofcompoundsintheirmoststableformistakentobezero.Thesuperscript ◦ indicatesthestandardstate,whichisat1atmand25 ◦ C.

Forthegeneralchemicalreaction

thestandardGibbsenergyisthedifferencebetweentheGibbsenergiesofproducts andreactants:

InChapter4welearnedthatiftemperatureandpressureareheldconstant,thenas thesystemtendsspontaneouslytoitsequilibrium,itsGibbsenergywilldecreaseto aminimum.Applyingthisequilibriumcriteriontochemicalsystems,weconclude thatif G ◦ ofthereactionisnegative,thereactantsintheirstandardstateare areconvertedtotheproductsintheirstandardstate.If,ontheotherhand, G ◦ is positive,thenanadditionalsourceofenergyisneededforthereactiontoproceed.

Example8.5 CalculatethestandardGibbsenergyofformationat25 ◦ Cand1atm forthereaction:

Canitproceedspontaneously?

Table8.2 StandardGibbsenergyfor selectedcompounds( G ◦ inunits kJmol 1 ),allvaluesrelatetothe standardconditions298Kand1atm ofpressure

Answer:ThestandardGibbsenergyofthisreaction(Table8.2)

Since G ◦ isnegative,thereaction(8.26)canproceedspontaneously.Notethat thereisnoinformationabouthowlongthereactionwilltaketocomplete.

Example8.6 SupposewearelookingforsomeeffectivemechanismofOH productionintheatmosphere.WesuggestthattherecombinationofH2 OandO2 canworkasasourceforOH:

Beforewestartthelaboratoryexperimentstocheckouridea,wecancalculatethe Gibbsenergyofthisreaction:

AftersubstitutingthenumbersfromTable 8.2,weget G ◦ = 281.3kJmol 1 . ThepositivevalueofstandardGibbsenergymeansthatthesuggestedmechanism forOHformationisthermodynamicallyimpossibleintheatmosphere.

8.4Elementarykinetics

Wehaveseenhowtoestimatetheenergeticsandfeasibilityofchemicalreactions proceedingonewayortheotherusingthemethodsofequilibriumthermodynamics. Butequilibriumthermodynamicscannottellushowrapidlyareactionwillproceed.

Thisisthebusinessofchemicalkineticswhichconsidersthedetailsofthemolecular collisionandtheintermediatecomplexesthatcanformduringtheevent.For example,kineticscanprovideameansofcomputingthecharacteristictimeof thedecayofthereactantsintheatmosphere.Oneshouldkeepinmindthatnegative Gibbsenergychangeforareaction(thermodynamicallyfavorableconditions)does notalwaysmeanthatthereactionwillproceedfastenoughtobeobserved.

8.4.1Reactionrate

Areactionratecanbedeﬁnedintuitivelyastherateatwhichtheproductsof thereactionareformed,whichisthesameastherateatwhichthereactantsare consumed.Asanexample,considerabimolecularreactionwithmoleculesCand DasproductsandAandBasreactants:

Therateofthisreaction(therateoflossofAorBandtherateofincreaseofC andD)is

where [X] denotestheconcentrationofspeciesXexpressedinmoleculescm 3 and k isthe reactionratecoefﬁcient.Theunitsof k dependontheorderofthereaction: forthebimolecularreaction(8.30) k isincm3 s 1 .Thereactionratecoefﬁcient k is uniqueforagivenreactionateachgiventemperature.Thetemperaturedependence k (T ) isdescribedbythe Arrheniusequation:

k (T ) = A exp Eact R∗ T [ratecoefﬁcientwithactivationenergy].(8.31)

Eact iscalledthe activationenergy.Alargevalueof Eact usuallyimpliesastrong temperaturedependenceofthereactionratecoefﬁcient.Theconstant A (nottobe confusedwiththeidentityofthespeciesAin(8.29))beforetheexponentialfunction isrelatedtothefrequencyofmolecularcollisionsandtheprobabilityformolecules tohaveanorientationinspacefavorableforareaction.Thedependenceof A on temperatureisusuallyweakcomparedtothatoftheexponentialfactor.

TheideaofactivationenergyisshownschematicallyinFigure 8.5.Thehorizontal axisrepresentsthereactioncoordinateforthereactants.Thereactioncoordinate canbethoughtofasthedistancebetweenthemoleculesAandBinthereaction (8.29).Theverticalaxisisthepotentialenergyofthereaction. H ◦ isthestandard enthalpyofformationforthisreaction.Notethat H ◦ isnegative,sothereaction isexothermic.

Figure8.5Schematicgraphofenergychangeforanexothermicreaction. Reactantshavetobeenergeticenoughtoovercomethebarrier Eact

Figure8.6Velocitydistributionofmoleculesattwodifferenttemperatures.The velocity v isexpressedinunitsof kB T /m0 .Thevelocity v ∗ correspondstothe kineticenergyequalto Eact

FortheproductsCandDtobeformedbythisreaction,thereactantsAandB musthaveenoughkineticenergytoovercometheenergybarrier Eact .Itshould benotedthatmanyreactionsintherealatmospheredonotproceedbecausethe activationbarrieristoolarge.Forexample,C + O2 → CO2 doesnottakeplacein theatmospherebecauseofthelargebarrier.

Equation(8.59)impliesthatreactionsproceedfasterathighertemperatures.4 Thiscanbeexplainedwiththehelpofkinetictheory.Itfollowsfrom(2.26)thatthe higherthetemperatureofthegas,thegreaterthefractionofmoleculesthathave kineticenergiesthatexceedacertaingivenenergy.Figure 8.6 showsthevelocity distributionfortwotemperatures.Athighertemperaturesmoremoleculeshave velocitieshigherthanthethresholdvelocity v ∗ correspondingtothekineticenergy 1 2 m0 v ∗2 whichisequalto Eact .Thismeansthatincreasingthetemperatureofthe gasincreasestheprobabilitythatmoleculeswillovercomethebarrier Eact andthat theproductswillbeformedatahigherrate.

4 Forsomereactionstheactivationenergyisactuallynegative(nobarrier).Therateofthesereactionsdecreases withincreasingtemperature.

8.4.2Conceptofchemicallifetime

Considerthecaseofaﬁrst-orderreaction,whenoneelement,A,decomposesinto twoelements,CandD:

TherateofdecreaseoftheconcentrationofelementAisproportionaltoits concentration:

where k isthereactionratecoefﬁcient.Afterrearrangingthetermsin(8.33)we have

and

Ifattheinitialtime t = t0 theconcentrationofAisequalto[A]0 ,then

ThisequationshowsthattheconcentrationofAdecaysexponentiallywith characteristictime tc = 1/k .Thetime tc requiredfortheconcentrationofAto decreasebyafactorofefromitsinitialvalueiscalledthechemicallifetime.The largerthereactionratecoefﬁcient k ,theshorterthelifetime tc .

Example8.7: halflife Thecharacteristictimeforaunimoleculardecayis tc .What isthehalflife,i.e.,whatisthetimeafterwhichhalftheconcentrationremains? Wehave

Then 1 2

Considernextthebimolecularreaction:

TherateoflossofAisgivenby

where k istheconstantforthisbimolecularreaction(8.39).Animportantcaseis whentheconcentration[B]ismuchlargerthan[A],then[B]canbeconsidereda constant,say[B0 ]inthelastequation.Forexample,gasAmightbeatracegassuch asatomicoxygenO,andgasBmightbeabackgroundgassuchasO2 orN2 (see e.g.,(8.17)).Thisleadsto

andthelifetimeofAis:

Foraphotochemicalreaction

thedecayoftheconcentrationofmoleculeAisgivenby

where J isthephotodissociationcoefficientexpressedins 1 .Thephotodissociation coefficient J intheinterval λ atheight z isdeterminedbythefluxofphotonswith wavelength λ atheight z , Fλ(z ),andtheabsorptioncross-section 5 ofmolecules absorbingnear λ, σ(λ).Notethat Fλ(z ) isthenumberofphotonsperunitarea,per unittime,perunitwavelength(unitsofphotonsm 3 s 1 ): Fλ(z ) = Fλ(top) exp( τ(z )/ cos )

(adiscussionof Fλ(z ) canbefoundinSection 8.2).Integratingoverabandof wavelengths λ (weassumeeachphotonstrikingamoleculedissociatesit),the photodissociationcoefﬁcientforthatwavelengthbandis

(z ) =

Onecanseefrom(8.44)thatthephotochemicallifetimeistheinverse J : tc = 1/J .(8.46)

5 Atypicalvalueof σ(λ) forthephotoabsorptioninthevisiblewavelengthrangebyNO2 is5×10 5 nm2 (Seinfeld andPandis, 1998).

Theconceptoflifetimeisusefulinmanyatmosphericproblems.Myriadsof atmosphericconstituentsundergochemicalreactionsandphotochemicalprocesses causedbysolarradiation.Inaddition,chemicalspeciesareadvectedbytransport processesintheatmosphere.Separationofprocesseswithdifferenttimescalescan simplifytheproblemsigniﬁcantly.Insomecasesthisistheonlywaytoanalyzethe variabilityintheverycomplicatedworldofatmosphericconstituents.Supposewe knowthatthephotochemicallifetimeofacertainconstituentismuchsmallerthan thecharacteristictimeofatmospherictransportatagivenheight.Thisisthecase, forexample,forstratosphericozone:ataltitudeshigherthan30kmthechemical lifetimeofozoneisseveralordersofmagnitudesmallerthanthetransporttimescale. Thisallowsustoneglecttheeffectoftransportintheﬁrstorderofapproximation whenanalyzingozonevariability.Nowconsidermethaneinthestratosphere.In thiscasethephotochemicallifetimeisseveralordersofmagnitudelargerthan thecharacteristictimefortransportprocesses.Thenwecantreatmethaneasin photochemicalequilibrium,whichmeansthatwecanneglectthechangeofmethane concentrationduetophotochemicalprocesses.Thefactthatmethanevariabilityis mainlydeterminedbytransportmakesitagoodtracerofatmosphericmassesin thestratosphere.

Example8.8 Considerthereactionofnitricoxideandozone,

AssumingthatthisreactionisthelonemechanismofNOdepletion,ﬁndthelifetime ofNOattemperature250K(typicalof z = 30kmintheatmosphere).

Answer:ThechangeofNOconcentrationduetothisreactioncanbedescribedby theequation:

d t [NO] =−k1 [NO][O3 ](8.48)

where k1 = 1.8 × 10 12 exp( 1370/T ) cm3 s 1 .TheconcentrationofO3 canbe consideredasaconstantsinceitismuchlargerthanthatofNO.If[NO]0 isthe concentrationofNOattheinitialtime,thenwehave [NO] = [NO]0 e k1 [O3 ]t .(8.49)

Thelifetime tc = 1/k1 [O3 ].With k1 ≈ 7.5 × 10 15 cm3 s 1 at T = 250Kand concentrationofO3 equalto3 × 1012 cm 3 at30km,weobtain tc ≈ 40s.

8.5Equilibriumconstant

Consideragenerictwo-bodies-to-two-bodiesreaction:

Asthereactionproceedstheproductsofthisreaction(CandD)becomesufﬁciently denseinnumberthattheywillbegintoreactandformAandBthroughthereverse reaction:

Wheneventuallytherateoftheforwardreaction(8.50)isthesameastherateof thereversereaction(8.51),thesystemisinchemicalequilibrium.Wecancombine (8.50)and(8.51)andwrite

Ifweweretryingtoﬁndtherateforwhichthereaction(8.52)proceedsfromthe pointofviewoftheindividualgasmolecules,wewouldsaytherateofincreaseof theconcentrationofCisgivenby

where kab and kcd arethereactionratecoefﬁcientsforforwardandreversereactions respectively.TheequationsimplystatesthattherateofbuildupofCisthesumof theratesofreactivecollisionsofAandBminusthereverseprocessinwhich CandDreact.Theﬁrsttermmustbeproportionaltotherespectivenumber densitiesandsimilarlyforthesecond(loss)term.Sinceforeverycreationofa CmoleculetheremustbeaBmolecule,theseratesofformationmustbeequal toeachotherandequaltothenegativeoftheratesofformationoftheAandB molecules.

Inequilibriumtheratesofchangeofthespeciesarezero.Thismeans

= 0equilibrium(8.54) or

Theconstant K iscalledthe equilibriumconstant forthereaction.If K isknown, wecandeterminetheratiosofconcentrationsoftheproductandreactantgasesin equilibrium.

Forthecaseofageneralreaction

theequilibriumconstantis(asderivedinphysicalchemistrytexts):

Therulecanbegeneralizedtocaseswheremorethantwospeciesareoneachside oftheequation.

Itisseenfrom(8.57)thatwhentheinversereactionrateisverysmall,which meansthatproductsdominateoverreactants, K islarge.Small K meanstherewill bearelativelylargeconcentrationofthereactantspecies.

Theequilibriumconstantdependsoncollisiondynamicsandinprinciple shouldhaveastrongtemperaturedependence,sincetheintermolecularrelative velocitywillbeanimportantfactorintherearrangements.Tofindthetemperature dependenceoftheequilibriumconstant,letuswritethereactionratecoefficients fortheforwardandreversereactions(8.52).UsingtheArrheniusequation(8.59) thereactionratecoefficientfortheforwardreaction, k1 ,is

Thereactionratecoefﬁcientforthereversereaction, k2 ,is

Theequilibriumconstant

Hence,knowingthestandardenthalpyassociatedwithareactionprovides informationabouttheequilibriumconcentrationsofthespecies.

Example8.9 ConsidertherecombinationofOHandO:

Writetheexpressionfortheequilibriumconstant.Byusingformula(8.60)ﬁndout whethertheproductswilldominateoverreactantsathighorlowtemperatures.

Answer:Inorderforthereactiontoproceed,anOHmoleculehastobumpintoan Oatominthesametinyvolume.Therateofgrowthtermmustbeproportionalto [OH][O].Theequilibriumconstantis

Toﬁndthetemperatureatwhichtheforwardreactionwillbefavoredweneedto ﬁndthestandardenthalpy:

Since H ◦ isnegative,from(8.60)wehavethattheequilibriumconstant K is largerwithlowertemperature.Thisexplainswhytheproductsofreaction(8.61) willdominateoverreactantsatthelowertemperatures.

Thenextstepistoﬁndtheconnectionbetweentheequilibriumconstantand theGibbsenergy.Theequilibriumofthechemicalreactionimpliesthatatagiven temperaturethereexistpartialpressuresofthegasesinvolvedinthereactionwith whichtherateoftheforwardreactionisequaltotherateofthereversereaction.Let usexpresstheequilibriumconstantintermsofpartialpressuresofeachconstituent. UsingtheIdealGasLawforthemolarconcentrationsweobtain:

,(8.64)

wherethe pi arepartialpressuresand = (a + b) (c + d ).Wecanrewrite(8.64) intheform: K = Kp (R∗ T ) ,(8.65) where Kp = pcC pdD paA pbB [equilibriumconstantforidealgases].(8.66)

Kp isusedasanequilibriumconstantforchemicalreactionsinvolvingspeciesin thegaseousstate.

ForareversibletransformationatconstanttemperaturethechangeofGibbs energyis(see(4.97))

For1molofidealgas

(NotethatthemolarGibbsenergy G isalsocalledthechemicalpotential.)By integratingthisequationfromthestandardpressurelevelof1atmtosomearbitrary pressurelevel p holding T constantweobtain

where p isinatm.Inthisformula G istheGibbsenergyatpressure p andtemperature T and G ◦ T istheGibbsenergyat1atmandtemperature T . Using(8.69)wecanwritethechangeofGibbsenergyforthegeneralchemical reaction

intheform

Theargumentofthelogarithmicfunctioninthelastformulaistheequilibrium constant Kp (see(8.66)).Therefore,

Atequilibrium G = 0,andthechangeofGibbsenergyatapressureof1atmand arbitrarytemperature T , G ◦ T ,isrelatedtotheequilibriumconstantatpressure p andtemperature T , Kp ,bythesimplerelation:

Theequation(8.73)tellsusthatif G ◦ T ispositive, Kp shouldbelessthanunity, whichmeansthatatequilibriumtheconcentrationsofthereactantswillexceed thoseoftheproducts.If,ontheotherhand, G ◦ T isnegativeand,moreover,is large,then Kp islargeandtheproductswilldominateintheequilibrium.

Forstandardconditionsatapressureof1atmandatemperatureof25 ◦ Cwe obtain,injoules,

ThechangeinGibbsenergy G isespeciallyusefulbecauseitsimultaneously takesintoaccountboththeFirstandSecondLawsofThermodynamics.Itdoesso insuchawaythatifthetemperatureandpressureareheldconstant(andtheyoften areinatmosphericproblems)wehaveafunctionwhichcanbeappliedmuchmore broadly.

Example8.10 ConsiderthereactionofrecombinationofNO2 andO3 : NO2 + O3 NO3 + O2 .(8.75)

Dothereactantsorproductsdominatefortheforwardreactionat1atmand25 ◦ C? Answer:WehavetoﬁndthechangeoftheGibbsenergyforthisreaction.

From(8.74)weobtain Kp = 1.9 × 1017 .Withsuchalargevalueof Kp theproducts willdominatefortheforwardreactionatequilibrium.

8.6Solutions

Chemistryrefresher A solution isahomogeneousmixtureofseveralcomponents. Consideratwo-componentsolution.Onecomponenthasamolefraction ηw (theratio ofthenumberofmolesofacomponenttothetotalnumberofmoles),andtheother componenthasmolefraction η .Thecomponentwithagreatermolefraction,letitbe ηw ,iscalledthe solvent.Thesolventdeterminesthestateofmatterofthesolution (gas,liquidorsolid).Thecomponentwiththesmallermolefraction, η ,iscalledthe solute.InChapter5weconsideredaclouddropletasanexampleofasolutionwith waterasasolventandthesaltasasolute.Asolutionofasaltinasolventsuchas wateris saturated whentheratesofdissolvingandcrystallizationareequal.Inthis casetherecouldbesomesubstanceinthecrystallineformpresentinthecomposite system.Forexample,theremightbeasaltcrystalinsideaclouddroplet.Theamount ofdissolvedmaterial(solute)inthesaturatedsolutioniscalledits solubility,which mightdependstronglyontemperatureandweaklyonpressure.Weareinterestedin theeffectofthedissolvedsoluteonthevaporpressureofthesolvent.

Raoult’sLaw Considerasolutionthatisinchemicalequilibrium.Thevapor pressureofeachcomponentofthesolutionisapproximately

where ppure i isthevaporpressureofapure i thcomponentand ηi isthemolefractionof an i thcomponentofthesolution.Thesolutioniscalled ideal whenbothsolventand soluteobeyRaoult’sLaw.Raoult’sLawapplieswhenthecomponentsofthesolution arepresentinhighconcentrations.WeusedRaoult’sLawwhenweconsideredthe equilibriumvaporpressureoveradropletcontainingdissolvedelectrolytes.

Forsolutionsatlowconcentrationsthevaporpressureofthesoluteobeys Henry’s Law.AccordingtoHenry’sLaw,thevaporpressureofasolute, p,isaproductofthe molefractionofthesolute, η ,andanempiricaltabulatedconstant, KH ,expressedin unitsofpressure:

Generally,thevalueof KH increaseswithincreasingtemperature.Thus,atthesame pressurethemolefractionofasolutedecreaseswithincreasingtemperature.

Whentheatmosphericpressuredecreases,thepartialpressureofagasdecreases, andthemolarsolubilityofagasdecreases.Forexample,highinthemountainsthe atmosphericpressureislow;asaresultthesolubilityofoxygeninhumanblood decreases,whichcancauserespirationproblems.Attheoppositeend,thehigherthe pressure,thehigherthesolubilityofgases.Youmightsaythegasis“squeezed”into thesolution.

Example8.11 Calculatethemolarsolubilityofnitrogendissolvedin1lof water at25 ◦ Candatmosphericpressureof1atm.Henry’sLawconstantfornitrogenat 25 ◦ Cis8.68 × 109 Pa.ThepercentagebyvolumeofN2 indryairis78.1.

Answer:ThepartialpressureofN2 at1atmis pN2 = 0.781 × 1atm = 7.91 × 104 Pa.FromHenry’sLaw ηN2 = pN2 /KH = (7.91 × 104 Pa)/(8.68 × 109 Pa) = 9.1 × 10 6 .Themolefractionofnitrogen ηN2 = νN2 /(νN2 + νH2 O ) ≈ νN2 /νH2 O since νN2 νH2 O .ThenumberofmolesofH2 Oin1lis (1000/18) mol.Then, νN2 = (9.1 × 10 6 × 1000/18) mol = 5.05 × 10 4 mol.Themolarsolubilityof nitrogenis5.05 × 10 4 moll 1 .

Example8.12 CalculatethemolarsolubilityofCO2 inmolesperliterdissolved inwaterat25 ◦ CandCO2 pressureof2.4atm(pressureusedtocarbonatesoda). Henry’sLawconstantforCO2 at25 ◦ Cis1.67 × 108 Pa.

Answer:2.4atm=2.43 × 105 Pa.ThemolefractionofCO2 accordingtoHenry’s Lawis ηCO2 = pCO2 /KH = (2.43 × 105 Pa)/(1.67 × 108 Pa).Sincethereis (1000/18) molofH2 Oin1l,themolarsolubilityofCO2 is (2.43 × 105 Pa × 1000/18moll 1 )/(1.67 × 108 Pa) = 8.1 × 10 2 moll 1 .Whenoneopensabottle ofsoda,thepressuredecreases;asaresultthesolubilityofCO2 decreases,andthe

Figure8.7Notation.(a)Puresolventinequilibriumwithitsvaporatpressure es . (b)Solutioninequilibriumwiththesolvent’svaporatpressure e .

bubblesofCO2 emergefromthesolution.Athighertemperaturethesolubilityof CO2 decreasessinceHenry’sconstantincreases(sodafromafridgesparklesless thansodaheldatroomtemperature).

8.6.1MolarGibbsenergyofanidealsolution

InthissectionwewillﬁndouthowtheGibbsenergyofapuresolventchangeswhen asmallamountofsoluteisadded.Wewillconsiderliquidsthatareatequilibrium withtheirvapors.ThismeansthattheGibbsenergyofthevaporisequaltothe Gibbsenergyoftheliquid.Letusdenotetheequilibriumvaporpressureovera puresolventas es ,andtheequilibriumvaporpressureoverthesolutionas e (see Figure 8.7).From(8.69)themolarGibbsenergy(theGibbsenergypermole)of vaporatpressure es is

where es isinatm, G isthemolarGibbsenergy 6 atpressure p andtemperature T and G ◦ T isthemolarGibbsenergyat1atmandtemperature T .Sinceatequilibrium themolarGibbsenergyofaliquidisequaltothatofthevapor, G vapor = G liquid , theGibbsenergyofapuresolvent,denotedas G w ,is

Whenasoluteisadded,themolarGibbsenergyofthesolvent, G ,whichisat equilibriumwithitsvaporatpressure e ,is

6 ThemolarGibbsenergyisoftencalledthechemicalpotential.

Subtracting(8.80)from(8.81),weobtainthedifferencebetweenthemolarGibbs energyofthesolventinthesolutionandthepuresolvent:

FromRaoult’sLaw, e /es = ηw ,where ηw isthemolefractionofsolvent(e.g. water)inthesolution.Then,

Wewillconsideradilutesolution,whenthemolefractionofthesolute, η ,ismuch smallerthanthatofthesolvent, η ηw .Since η + ηw = 1,werewrite(8.83)as

Takingintoaccountthatfor η 1,thelogarithmicfunctioncanbewrittenas ln (1 η) ≈−η ,weobtain

Wewilluse(8.85)inthenextsectiontoﬁndthetemperatureatwhichthesolution boilsandfreezes.

8.6.2Theelevationoftheboilingpointandtheloweringofthefreezingpoint ofasolution

Whensomesoluteisdissolvedinapuresolvent,theboilingandfreezingpointsof thesolutionarenotthesameasforthepuresolvent.Wewillshowthatthechange isproportionaltotheamountofsolute.Wewillconsidernonvolatilesolutes(for example,asalt).Inthiscasethevaporofthesoluteisapuregas.

Example8.13 Showthattheadditionofadissolvedsoluteinasolutionwillelevate theboilingpointcomparedtotheboilingpointofpuresolvent.

Answer:Theboilingpointisthetemperatureatwhichthesaturatedvaporpressure ofaliquidisthesameastheatmosphericpressure.Weconsidertwocases.

Case1: equilibriumbetweenapuresolventanditsvapor(seeFigure 8.8a).The puresolventboilsattemperature T0 .Thevaporpressureis1atm.Atequilibrium, themolarGibbsenergiesofthepuresolvent,denotedbelowas G w ,anditsvapor, denotedas G v ,areequaltoeachother:

Gv(T0, p)

Gw(T0, p)

Figure8.8Notation.(a)MolarGibbsenergyforthevaporoverthepuresolvent, G v (T0 , p),andmolarGibbsenergyforthepuresolvent, G w (T0 , p).(b)MolarGibbs energyofsolventvaporoverthesolution, G v (T , p),andmolarGibbsenergyof thesolvent, G (T , p).

Case2: thereisanamount η ofnonvolatilesoluteinthesolvent.Thesolutionis atequilibriumwiththesolvent’svapor(seeFigure 8.8b).Thevaporpressureatthe boilingpointis1atm,thesameasincase1,buttheboilingpointdiffers.Assume thatthesolutionboilsatatemperature T = T0 + T where T T0 .From (8.85)themolarGibbsenergyofthesolventinthesolutionis G = G w R∗ T η . Atequilibrium,themolarGibbsenergyofthesolventinthesolution, G ,isequal tothemolarGibbsenergyofthevapor, G v :

Since T T0 wecanexpandbothsidesof(8.88)inTaylor’sseriesretaining onlythelinearterm:

Substitutingtheseexpansionsin(8.88)andtakingintoaccount(8.86),weobtain

Herewealsoreplaced T η by T0 η since η hasasmallvalueproportionalto T .

AsshowninSection4.8,thetemperaturederivativeoftheGibbsenergyata constantpressureisminustheentropy.So,

and

where sw istheentropyof1molofthesolventand sv istheentropyof1molof thesolvent’svapor.Then,from(8.91)weobtain

and

Multiplyinganddividingtheright-handsideofthelastequationby T0 andtaking intoaccountthat (sv sw )T0 istheamountofheatrequiredtoevaporate1molof thesolventattheboilingpoint,inotherwords (sv sw )T0 = vap H ◦ ,weobtain theﬁnalformulafortheelevationoftheboilingpoint T :

Since T ispositive,theboilingpointofthesolutionishigherthanthatofthepure solvent.Thechangeintheboilingpointisproportionaltotheamountofsolute η .

Example8.14 Whatisthechangeintheboilingtemperatureof1lofwaterwith 15gofNaCldissolvedinit?

Answer:ThemolecularweightofNaClis58.44gmol 1 .Substitutingin(8.96) η = (15/58.44)/(

,weobtain T

Example8.15 Calculatethechangeinthefreezingpointofthesolution. Answer:Weassumethatonlypuresolventisfrozen,whilethesoluteremainsin thesolution.Thenthecalculationofthefreezingpointofthesolutionissimilarto thecalculationoftheboilingpointexceptwehavefusioninsteadofvaporization. fus H ◦ istheheatreleasedwhen1molofthesolventisfrozenatatemperature T0 ;

thisheatisnegative.Substituting fus H ◦ in(8.96),weﬁndthatthefreezingpoint ofasolutionislowerthanthatofthepuresolvent.Thedifference T is:

Thedecreaseinthefreezingpointisproportionaltotheamountofsolute.

Example8.16 Whatisthechangeinthefreezingtemperatureof1lofwaterwith 15gofNaCldissolvedinit?

Answer: | fus H ◦ | forwateris6.008kJmol 1 , T0 = 273K, η = (15/58.44)/ (1000/18), R∗ = 8.31JK 1 mol 1 .Accordingto(8.97)thechangeinfreezing temperature T =−0.48K.

Notes

Thereareexcellentbooksonphysicalchemistry,forexampleAtkins(1994). ThebookbyHouston(2001)givesareadableaccountofthekinetictheoryof gasesandreactionkinetics.ThebookbyHobbs(2000)onphysicalchemistryfor theatmosphericsciencesisataboutthesamelevelasthisbookanditdelves muchmoreintothesubjectofreactionsintheatmosphere.Amorethorough discussionofchemicalequilibriumiscontainedinDenbigh(1981).Thebookon atmosphericphysicsandchemistrybySeinfeldandPandis(1998)isthemost comprehensive.Booksonthemoregeneralsubjectofatmosphericchemistry includethosebyBrimblecombe(1986)andWarneck(1999).Acomprehensive bookonthechemistryofthemiddleatmosphereisbyBrasseurandSolomon(2005). Finlayson-PittsandPitts(2000)coverboththeupperandloweratmosphereand includemanyusefultables.

NotationandabbreviationsforChapter8

c speedoflightinvacuum(ms 1 )

G ◦ changeinGibbsenergypermole,atstandardconditions (kJmol 1 )

H ◦ changeinenthalpypermole,atstandardconditions(kJmol 1 )

Eact activationenergy(J)

E energy(J)

η mixingratio

f frequencyofelectromagneticwave(Hz)

F energyﬂux(Jm 2 s 1 )

Fλ(z ) ﬂuxofelectromagneticenergyatwavelength λ,elevation z (Jm 2 s 1 m 1 )

g,l,aqgaseous,liquid,aqueousphase

h Planck’sconstant(Js)

H ◦ standardenthalpy(kJmol 1 )

J photodissociationcoefﬁcient(s 1 )

k reactioncoefﬁcient

K , Kp equilibriumconstant

KH Henry’sLawconstant

λ wavelengthofanelectromagneticwave(m)

n0 numberdensityofphotons(photonsm 3 )

N (z ) concentrationofabsorbersatlevel z (moleculesm 3 )

R∗ universalgasconstant(Jmol 1 K 1 )

σ(λ) absorptioncross-section(m2 )

tc lifetime(s)

t1/2 halflife(s)

τ opticaldepth(dimensionless) zenithangle

v ∗ thresholdvelocitytoexceedanactivationbarrier(ms 1 )

[X ] concentrationofX(moleculescm 3 )

[X ]0 concentrationofXat t = 0

Problems

8.1Computethestandardenthalpyofreactionforthefollowingreactions: CO + O2 → CO2 + O O + O3 → 2O2 .

Arethesereactionsexothermicorendothermic?

8.2CalculatethestandardGibbsenergyforthereaction: NO3 + H2 O → HNO3 + OH.

Canthisreactionproceedspontaneously?

8.3Considerthereaction 2NO → N2 + O2

Determinewhetherhighorlowtemperaturesarefavorablefortheforwardandreverse reactions.(Hint:Useformula(8.60).)

8.4ThereactionofHNO3 dissociationis

HNO3 + hf → OH + NO2 .

Byexaminationofthestandardenthalpyﬁndoutwhatphotonenergyisrequiredfor thisreactiontoproceed(useTable 8.1).

8.5Calculatetheequilibriumconstantat298Kforthereaction

NO + O3 → NO2 + O2 .

Arethereactantsorproductsfavoredforthisreactionatequilibriumat298K?Doyou needtoraiseorlowerthetemperaturetohavemoreproductsattheequilibrium?

8.6Findtheexpressionfortheequilibriumconstantforthereaction

A + B AB,

intermsofreactioncoefficients,ifthecoefficientfortheforwardreactionis k1 ,and thecoefficientforthereversereactionis k2

8.7Estimatethemolarsolubilityofoxygeninwaterat25 ◦ Cand1atm.Henry’sconstant foroxygenis4.4 × 109 Pa.ThepercentagebyvolumeofO2 indryairis20.95%.

8.8Estimatebyhowmuchtheamountofnitrogeninadiver’sbloodwillchangewhen thediverisrisingfromadepthof80m.Anadulthumanmalehasanaverageblood volumeofabout5l.UseHenry’sLawconstantfornitrogenat25 ◦ C, KH = 8.68 × 109 Pa.(Fastrising,i.e.,arapiddecreaseinpressureandthusnitrogensolubility,cancause theformationofnitrogenbubblesinthebloodstreamwhichoftenleadstodeath.)

8.9Showthatthedifferencebetweenthevaporpressureofasolutionatagiventemperature T andthevaporpressureofapuresolventatthesametemperatureis

where η isamolefractionofanonvolatilesolute, v isthevolumeof1molofasolvent, vv isthevolumeof1molofvapor. Hint:Suggestthatthesolutionisinequilibriumwith thevaporatpressure e = es + p where es istheequilibriumvaporpressureforthe puresolventand p issmall.Thefactthat p issmallallowsyoutoexpandtheGibbs energyinTaylor’sseriesandretainonlythelinearterm(similartothecalculationof theboilingpointofasolution).

Thethermodynamicequation

Inthischapterwederivetwoofthefundamentalequationsofatmosphericscience, the equationofcontinuity andthe thermodynamicequation.Theequationof continuityexpressestheconservationofmassintheformofapartialdifferential equation,theformneededtoimplementitinnumericalsimulationsorforecasts. ThethermodynamicequationexpressesthecombinedFirstandSecondLawsof Thermodynamicsintoasimilarform.Butbeforewecometotheseimportant formulasweneedsomeexperiencewithscalarandvectorﬁelds.Muchofthechapter isconcernedwithelementaryvectoranalysiswhichshouldhavebeencoveredin theprerequisitecalculuscourse.Hence,somestudentscanskipoverthereview sections,butweadviseallstudentstorefreshtheirmemories.Allthismachineryis toprepareforthenextstepinaneducationinatmosphericsciences: dynamics

Vectorrefresher Athree-dimensionalvector,denotedinboldface, a ,isa mathematicalobjectwhichhasbothlengthanddirection.Intwo-spaceitcanbe representedbyanarrowasinFigure 9.1.Ittakesthreenumberstorepresenta 3-vector,twoanglesandalength, (θ , φ ; |a |).Alternatively,itcanberepresentedbyits threecomponentsalongthethreeCartesiancoordinateaxes,(ax , ay , az ).Notethatthe vectorisanabstractobjectinspaceindependentofthechoiceofcoordinatesystem, butthethreenumbersneededtospecifyitmayindividuallydependonthecoordinate systemchosenbytheanalysttodescribethevector.Forexample,itisconventionalin meteorologytosetupaCartesiancoordinatesystemwiththeoriginatapointonthe Earth’ssurface,the x -axisincreasingintheeastwarddirectionandthe y -axis increasinginthenorthwarddirection.

Multiplicationbyascalar Let α beascalar,i.e.,anumberwhichisindependentof ourchoiceofcoordinatesystem, a beavector.Thenmultiplicationofavectorbya scalariswrittenas b = α a .Thedirectionof b isthesameasthatof a andthelengthis |α ||a |.Inotherwordsthetwoanglesdesignatingthedirectionof b and a arethe

Figure9.1Thevector a intwodimensions.

same,butthelengthisstretched.If α isnegative,thedirectionofthevectoris reversed.InthecaseofCartesiancomponentrepresentation,allthreecomponentsare multipliedby α : (α ax , α ay , α az ).

Addingvectors Thesumoftwovectorsisavector c = a + b whosecomponents aredeﬁnedbyaddingtheCartesiancomponentsoftheaddends:

Clearly a + b = b + a .Thesumofthetwovectorscanalsobeunderstood geometricallybyjoiningthetailofonevectortotheheadoftheotherandtheline joiningtheﬁrsttailtothesecondheadisthevectorsum.

Cartesianunitvectors Aunitvectorisavectorwhichhasunitlength.Itisvery convenientandcommontouseunitvectorswhichpointalongthethreeCartesian axes:

Thisnotationallowsusto expand avectorintoitsCartesiancomponents

Thelengthof a isgivenby |

Notethatthelengthofavectorisindependentofthecoordinatesystemchosen.

Wind inmeteorologyisusuallydenotedby v = u i + v j + w k .Thusawindblowing totheeastis u i ;if u > 0itiscalleda westerly.Awindblowingtothenorthis v j; if v > 0itiscalled southerly.Risingair(averticalwind)isdenoted w k .Notethat thecomponentsarereferredtoaCartesiancoordinatesystemwhoseoriginisatthe surfaceoftheEarthataﬁxedlocation.Wind speed isgivenby √u 2 + v 2 + w 2 .

Example9.1 Findthelengthofthevector a = 2i + 3j 4k .

Answer: |a |= 22 + 32 + ( 4)2 = √29.

Example9.2 Findthesumofthevectors a = 2i + 3j 4k , b =−4i + 3j + 2k . Answer: a + b =−2i + 6j 2k .

Itissometimesusefultouseacurvilinearcoordinatesystemsuchas cylindrical coordinates todescribephenomenaintheatmosphere.Weusetheunitvectors pointingalongthecylindricalcoordinatedirections(dependentonthepoint):

r , ˆ nθ , ˆ k .Avector a canbewrittenas:

Example9.3 Apurelycyclonicwindblowscounterclockwise(northernhemisphere) aboutacenterofaction.Atadistance r fromthecenterthewindcanbedenotedas v

Example9.4 Apointonarotatingdiskwithaxisatthedisk’scenterhasvelocity r ω ˆ nθ ,where ω isangularvelocityinrads 1 and r istheaxisofrotation.

Vectorrefresher:dotproduct The dot or scalarproduct oftwovectorsresultsina scalar.Eventhoughthecomponentsofthetwovectorsformingthescalarproduct dependonthechoiceofcoordinatesystem,thescalarproductdoesnot:

where θa,b istheanglebetweenthetwovectorsasshowninFigure 9.2.Notethat sincecos θa,b isanevenfunctionofitsargument,theorderofthevectorsinthe productmakesnodifference: a · b = b · a .

Analternativewaytodeﬁnethedotproduct a b is

Itcanbeshownthatthetwodeﬁnitionsareequivalent.Usingtheseconddeﬁnition,

Figure9.2Illustrationofthevectors a and b inadotproduct.

wecanshowthat c · (a + b) = c · a + c · b. Itisalsoclearthat i i = 1, i j = 0,etc.

Example9.5 Find a · b where a = 2i + 3j 4k , b =−4i + 3j + 2k

Answer: a · b = (2) × ( 4) + 3 × 3 + ( 4) × (2) =−7.

Example9.6 Whatistheanglebetween a and b?

Answer:cos θ = a · b/|a ||b|=−7/ 29;arccos ( 7/ 29) = 1.815.

Crossproduct The vector or crossproduct isindicatedbythenotation c = a × b.(9.10)

Thevector c isperpendiculartotheplanedefinedby a and b (seeFigure 9.3).The righthandruleisusedtodetermineitsdirection(pointtherightindexfingeralong thefirstvectorintheproduct,thensweepittowardthesecondvector’sdirection;the thumbpointsinthedirectionofthevectorproduct).Itslengthisgivenby

Bytherighthandrulewesee:

Ausefulformforthecrossproduct a × b is

Figure9.3Illustrationofthevectors c = a × b inacrossproduct. c is perpendiculartotheplaneformedby a and b.

Mathrefresher: 3 × 3 determinant Recallthatitcanbeexpandedalonganyrow orcolumn.Intherepresentationofthecrossproduct:

Recallthealternatingsignsoftheunitvectorsasoneexpandsalongthetoprow.Note thatiftworows(vectors)ofadeterminantareproportional,thedeterminantvanishes. Thisissimplythestatementthatsin (θa,b ) = 0.

Vectorrefresher:theboxproduct Thisisdeﬁnedby:

Ausefulformfortheboxproductis

.(9.15)

Thethreevectors a , b and c formarectangularparallelepiped.Theboxproductisthe volumeofthatgeometricalﬁgure.Therulesofdeterminantscanbeusefulhere.For example,cyclicpermutationoftherowsleadstoanequivalentdeterminant. Interchangingadjacentrowsﬂipsthesign.Itisusefultoknowthatthethreevectors

Figure9.4Sketchoftheparallelepipedformedbythevectors a , b and c.The magnitudeoftheboxproduct a b × c isthevolumeoftheparallelepiped.

ofaboxproductcanbeidentiﬁedasthethreeedgesofasolidparallelepiped.Thebox productisthevolumeoftheparallelepiped(seeFigure 9.4).Notethat

9.1Scalarandvectorﬁelds

Ascalarfieldisafunctiondefinedonthethree-dimensionalspacecoordinatesand possiblyalongthetimeaxis.Anexampleisthetemperaturefield T (x , y , z ; t ) ≡ T (r , t ),wherethepositionvector r isdefinedby

(9.17) and i , j, k areunitvectorspointingalongthe x , y and z axes(seeFigure 9.5).A smallincrementin r isdenotedas 1

Figure9.5Schematicdiagramofapositionvector r whosecomponentsare x , y and z .

1 Herewereplacethesmallvalues x , δ x ,etc.,withinfinitesimalsd x ,etc.,withtheapproximatesign ≈ replaced bytheequalitysign =.Thismeansthatinthisnotationsecond-orderquantitiessuchas (d x )2 areneglected(set tozero)whenadditivetofirst-orderterms.Whilethis operational shortcutmightcausesometocringe,itshould notdisturbtheflowofourstory.

Figure9.6Schematicdiagramofapositionvector r andanincrementofitd r

Considertheestimationofthetemperatureﬁeldatthepoint r + d r (Figure 9.6), giventhatweknowitsvalueatthepoint r ,namely T (r ):

WemayusetheﬁrsttwotermsoftheTaylorexpansion:

Wecanalsowritethisasadotproduct:

Aftersubstitutingd T = T (r + d r ) T (r ),weobtain d T = d r ·∇ T [differentialofascalarﬁeld].(9.21)

Thevector ∇ T (r ) iscalledthe gradientofT.Wewillusethemodernnotation ∇ T todenotethegradient(insomeoldertextsitisdenoted grad T ).

[gradientofascalarﬁeld].(9.22)

Thegradientisa vectorﬁeld.Ateachpointinspace r ithasanassociatedlength anddirection.

Ifyouwanttoknowtherateofchangeofthefieldinaparticulardirection,say alongthedirectiondefinedbytheunitvector, n,itcanbefoundbydefining2 the vectorincrementd r tobe n d s whered s isaninfinitesimaldistanceand n defines thedirectionalongwhichtheincrementistobetaken.Using(9.21)wecanwrite

Thisderivativetakenalongthedirectionofthespeciﬁedunitvector n iscalledthe directionalderivative,andisoftengiventhenotation ∂ T /∂ n astherateofchangeof T alongacertaindirection,deﬁnedbytheunitvector n.Theconventionalnotation forthedirectionalderivativeis:

If n liesinthetangentplanetoanisothermal(stillthinkingofthescalarfieldas temperature)surface,thedirectionalderivativevanishessincethereisnochangein anydirectionlyinginthisplane.Thismeansthatthecomponent(projection)ofthe gradientvectortangenttotheisothermalsurfacevanishes.Thegradientvectoris perpendiculartoisothermalsurfaces(ingeneralso-called levelsurfaces).Thiscan beseenforafixedgradientvector ∇ T .Justvarytheunitvectorinalldirections. Thelengthsof n and ∇ T arefixed,sothemaximumoccurswhentheanglebetween n and ∇ T iszero(cos θn,∇ T = 1),inotherwordswhen n isparallelto ∇ T

Example9.7 Considertheﬁeld

Findthegradientasafunctionof x and y .

Answer:

SeethecontourmapinFigure 9.7

Example9.8 Findthedirectionalderivativeoftheﬁeldinthelastexampleinthe direction n = (1/√2)(i + j) (thisisaunitvectorinthe x –y planedirected45◦ abovethe x -axis).

Answer:Takethedotproductof n withthegradient:

2 Rememberthatthereaderhasthepowertochoosed r,itstinylengthanddirection.

Figure9.7Contourmapofaﬁeld T (r ) = cos2π x cos π y showingconstant T lines.Arrowsindicatedirectionofthegradientvectorevaluatedatthepointswhere thearrowsoriginate.

Figure9.8Asmallrectangularparallelepipedofsidewidthsd x ,d y ,d z ,indicating thepressureforcesonthesidesperpendiculartothe x -axis.

9.2Pressuregradientforce

Thegradientofthepressurefield ∇ p(r , t ) isveryimportantinmeteorology. Consideraparcelofaircontainedintherectangularparallelepipedd x d y d z ,and whosecenterislocatedatthepoint r .Thisvolumeelementisembeddedina surroundingpressurefield p(r ) = p(x , y , z ).Letuscomputethe x componentof thenetforceonthevolumeelement.AsindicatedinFigure 9.8,theleft-mostface experiencesaforceduetotheexternalfieldtotheright: F leftface x = p(x

theright-mostfaceexperiencesaforcetotheleft

Thenetforceonthevolumeelementis

whered V isthevolumeoftheinﬁnitesimalmaterialelement.Newton’sSecond Law(forceismasstimesacceleration)tellsusthat

where ax isthe x componentofaccelerationandd M isthemasscontainedinthe parcel.Wecandivideeachsidebyd V andobtain

where ρ isthedensityoftheairintheparcel.Putinmoreconventionalformwe have:

Ifweevaluatethe y and z componentsinasimilarfashionwecansummarizethe resultinvectorform

Theforceperunitmass a asgivenhereiscalledthe pressuregradientforce.Wehave encountereditsverticalcomponentearlierinestablishingthehydrostaticequation. Abovetheatmosphericboundarylayeritshorizontalcomponentsareverynearly balancedbytheCoriolisforceinmidlatitudes(called geostrophicbalance).

Example9.9:horizontalaccelerationofaparcelinthetropics Supposeaparcel whosedensityis0.7kgm 3 isembeddedinaﬁeldofpressurewithagradient 10hPaover1000km.Whatistheaccelerationoftheparcel(ignoringtheCoriolis force)andwhatisitsincreaseinspeedfromrestinpassingover1000km?

Answer:Theaccelerationistowardlowpressureandisgivenby a = 1.43 × 10 3 ms 2 v = √2ax = 53ms 1

9.3Surfaceintegralsandﬂux

Considerthevectorfieldofvelocityofafluid, v (r ),ataninstantoftime(disregard timedependencefornow).Thefluidismovinglocallyalonglinesegments tangentialtothelocalvelocityvector v (r ).Thetrajectoriesofindividualparcels followtheflowlinesinsteadyflow(∂ v /∂ t = 0)butdifferinunsteadyflow (∂ v /∂ t = 0).Animaginarysurfaceisplacedinthefluidflow(sayapenetrable screen)asindicatedinFigure 9.9.Letanelementofthesurfacebedenotedbyd S whered S isthemagnitudeofthe(vector)areaelementandthedirectionofd S isthe localperpendiculartothesurfaceareaelement.Inaclosedsurface,byconvention, thevectorpointsoutwards;otherwise,ithastobespecifiedaccordingtocontext. Firstimagineaparallelflowthatisuniformoveritscross-sectioninapipeinthe x direction.Then v (r ) = v0 i .Supposethesurface S isaperpendicularcross-section ofthepipe.Then S = S i .Howmuchmasspassesthrough S perunittime?First considerthe“front”offluidpassingthrough S attime t .Attime t + d t thefront advancesbyadistance v0 d t .Theamountofvolumesweptoutbythisfrontinthe timed t isjust v0 S d t = v · S d t asinFigure 9.9.Ifthedensityofthefluidis ρ ,we canconvertthistoamassflux

Figure9.9Advanceofamaterialsurfacethroughacylindricalpipeduringthe timed t .Theﬂowistakentobeofuniformvelocity v overthecross-section.

Figure9.10Illustrationofaﬂowthroughasurfaceatanangletothenormal.The volumeofﬂuidsweptoutinthetimed t is |S||v |d t cos θS,v = S v d t

Figure9.11Portionofasurface S depictinganareaelementonasurfaced S throughwhichﬂuidofdensity ρ ispassingatvelocity v .

Ifthesurfaceweretiltedwithrespecttothe y –z planetheamountofmassperunit timepassingthroughitwouldbethesame(seeFigure 9.10).Theonlythingthat mattersistheprojectionof S ontothevelocity v .Ifthesurfacewerecurvedwe wouldhavetogeneralizeto(seeFigure 9.11)

Theamountof massflux passingthroughtheareaelementd S is massflux = ρ v · d S [fluxofmassthroughanareaelement].(9.37)

Figure9.12Theannulusistheshadedarea.Itscircumferenceis2π r anditswidth isd r .Theareaoftheannulusis2π r d r .

Themassﬂuxdensityis

where n isaunitvectorparalleltotheinﬁnitesimalareavectord S (i.e.,

Example9.10:fluidflowinapipe Viscousflowinapipeisslowernearthewalls thanalongthecenterlineofthepipe.Asimplesteadyflowsolutionforan incompressiblefluidisgivenby

where r isthecylindrical-coordinatedistanceperpendiculartothecenterlineand a istheradiusofthepipe. u0 isthevelocityatthecenterofthecross-section. Whatistheﬂuxofmassthroughaplaneparalleltothecenterlineofthepipe?

Answer: Firstformanannulus(aring)(seeFigure 9.12)inthecross-section.The areaoftheringis2π r d r .Themassﬂuxthroughtheringis ρ u (r )2π r d r .Thetotal ﬂuxis

9.4Conductionofheat

TheFourierLawofheatconductionstatesthattheamountofheat(enthalpy)flowing acrossaunitperpendiculararea(thevectorenthalpyflux q)isproportionaltothe gradientofthetemperaturefield, ∇ T .TheFourierLawworkswellinsolidssince

thetransferisfromonemoleculetoitsneighborinamediumwherethereisno relativemacroscopicmotionfromonelocationinthesolidtoanother.

Inliquidsorgasesthestorycanbemuchmorecomplicatedbecausethese macroscopicmotionsarepermitted.Buoyancyforexamplemightcausedifferential forcesmovinglighter(usuallywarmer)materialupwardsleavingthemoredense fluidbehind.Thisresultsinanettransferofheatupwardsinthemediumat themacroscopiclevel.Whiletheactualtransferofheattakesplacefromone infinitesimalelementtoanotherviamolecularcollisions(arelativelyslowprocess whenconsideredatmacroscopicscales),themacroscopicmotionscanmoveheat aroundmuchmorerapidlythanpuremoleculartransfersatthesmallerscalefrom oneinfinitesimalelementtoanotherbuildinguptothemacroscopicscale.

Thetransferofheatbywindsorcurrentsiscalled thermaladvection.In atmosphericapplicationsthetransferisdominatedbyadvectionbylargeeddies (ﬂuctuatingorirregulardeparturestypicalinturbulencefromthelargerscaleﬂows). Forexample,inthemorningboundarylayerwhereturbulenceiscommon,theair atthesurfacewhichhasbeenheatedbytherisingsuncanbebuoyedinparcelsto heightsofakilometerortwo(whereitsrisemightbelimitedbyincreasedstability atthoselevels).Theeddiesnecessarilybringwarmairinaparcelintocontactwith coolerairatthesamelevelwithanensuinglargethermalgradientatboundaries separatingwarmerandcoolerparcelsandultimatelyenthalpyistransferredatthe molecularlevel:

where q isaheatfluxdensityand κH isacoefficientknownasthethermal conductivity(itvariesfromonesubstancetoanotherandcanbefoundintables). Heatflowsfromwarmtowardcoolregionsinthedirectionoppositethegradient vector.Theamountofheattransferredbymolecularprocessesperunittimeflowing throughasurface S (flux), FS ,is

Example9.11 Steadyheatﬂowsalongarodwithcircularcross-section(area A) andlength L withitsleftandrightendsattachedtoreservoirsoftemperatures TL and TR .Let x = 0attheleftendand x = L attherightendoftherod.Theﬂuxof heat F (x ) atthepoint x is

Buttheﬂuxofheatmustbeconstantatanypointalongtherodotherwiseheat energywouldaccumulateatsomepoint.Then F (x ) = F0 .Wecannowintegrate

Figure9.13Temperaturedependencealongarodoflength L,heldattemperature TL at x = 0and TR at x = L. thelastequationfrom0to L toﬁnd:

Thisallowsustoevaluate F

andweﬁndthe x dependenceof T (x ) byintegratingfrom0to x :

SeeFigure 9.13.Itisinterestingthatthecurvedoesnotdependon κH or A.

9.5Two-dimensionaldivergence

Webeginourstudyofdivergenceintwodimensions(inthe x –y plane).Weare examininganimportantpropertyofavectorfieldsuchasthetwo-dimensional velocity V (x , y ).Istheremorefluidflowingoutofasmallfixedareaintheplane thaniscomingin?Thetwo-dimensionaldivergenceisrelevantinmeteorological applications.Forexample,atabox(fixedinspace)surroundingalowpressure areaatthesurface,airspiralsincounterclockwiseinthenorthernhemisphere (cyclonically)towardsthecenterofthelow.Inthe x –y plane(thesurface)thereisa netflowofairintothebox.Whathappenstoit?(Afterall,massisconserved.)The answerisitgoesupinthe z direction.Whenairgoesupweknowwhathappens:it

Figure9.14Atwo-dimensionalrectangularboxcenteredatthepoint(x0 , y0 )ina constant z planeillustratingtheﬂuxissuingfromthebox.

rains(assumingmoistureisavailable,etc.).Hencewecareaboutthenetﬂowinto oroutofﬁxedhorizontalboxesatdifferentlevelsintheatmosphere.3

Thereisamathematicalwayofexpressingthenetflowintooroutofatwodimensional(2D)box.Letusfocusourquestiononasmallregioninspace consistingofarectangleofsidesdx anddy whosecenterisat(x0 , y0 ).Wecan evaluatethefluxpassingthrougheachofthefouredgesandaddthemuptofindout whetherthereisanetfluxissuing from thebox.Toobtainthefluxcomingthrough therighthandedge(seeFigure 9.14)weneedtotake

Theﬂuxgoingoutoftheleftverticaledgeissimilar:

Whenweaddtheseleftandrightedgeﬂuxcontributionstogetherweobtain

netﬂuxouttheleftandrightedges = ∂ Vx ∂ x d x

.(9.49)

3 Notethattheairintheboxmighthavesimplybecomemoredenseduringthenetinﬂow,butweimplicitlymade theapproximationthattheairisverynearlyincompressible.

Thethermodynamicequation

Itiseasytoseethatthenetﬂuxpassingthroughtheupperandloweredgesis just

netﬂuxoutfromtheupperandloweredges =

Ifweadduptheﬂuxesfromallfouredgesweobtain:

The divergence ofthe2Dvectorfield V (x , y ) isdefinedtobethelimitasthebox shrinkstozerooftheemergingfluxdividedbytheareaofthebox.Wecanexpress itinCartesiancoordinatesusingourresultsfromabove:

wherewehaveemployedthesubscript2tomakeitclearthatwearedealingwith twodimensionsonly.Butthedeﬁnitionholdsmoregenerallyforanyinﬁnitesimal loop(rectangular,parallelogram,circle,etc.)aroundthetinyshrinkingarea:

Inthelaststepwemadeuseoftherulefortriplevectorboxproducts: a · (b × c) = (a × b) · c.

Example9.12:easterlyflowrateincreasing Suppose v (x ) = λx i , λ> 0.The divergenceof v (x ) isdiv2 (v ) = λ.Thisisadivergentflowsincemorefluidis leavingatinybox(fixedinspaceanywhereintheflow)ontheeastsideperunit timethanisenteringonthewestside(perunitboxareaandtime).

Example9.13:divergenceexpressedinpolarcoordinates FollowingFigure 9.15 weusealooparoundanareaelementinpolarcoordinates.Thefoursidesare(1) r → r + d r , θ .(2)Attheouterradius r + d r ,let θ → θ + d θ .(3)Nowdecrease r : at θ + d θ , r + d r → r .(4)Nowbacktothestartingpoint,at r , θ + d θ → θ Thetwoangle-changingsidesyieldfortheemerging(radial)ﬂuxes(steps1 and3): netradialﬂuxes = Vr (r + d r , θ)(r + d r ) d θ Vr (r , θ)r d θ = ∂(

Figure9.15ViewinpolarcoordinatesoftheboxdescribedinExample9.13.

Nextconsidertheangularﬂuxes(steps2and4):

Theareaofthelittleelementis r d θ d r .Upondividingthroughbythisareaand takingthelimitasitshrinkstozerowehave:

(V ) = 1 r

Example9.14:rigidlyrotatingﬂuid Thisissimilartoarotatingdisk.Thevelocity ﬁeldis V = v

where

isaunitvectorperpendicularto ˆ r ﬁxedatthe point(r , θ )andpointinginthedirectionofincreasing θ (see(9.7)).Wecancompute div2 (V ) = 0.Notethat Vr = 0inthiscase.Thisisageneralresultinthecaseof V = f (r ) ˆ nθ .

9.6Three-dimensionaldivergence

Thegeneralizationtothethree-dimensionaldivergenceisstraightforward.This timethetinyboxintheplanebecomesa3Dboxfixedinthespace.Considera closedsurface S .Thevelocityfluxemanatingfromtheenclosedvolumeis emergingflux = S v (r ) · d S.(9.57)

Nowlettheenclosedvolumebecomeverysmall.FortheCartesiancoordinatecase takeittobearectangularparallelepipedwhosesidesareoflengthsd x ,d y ,d z .Its volumeisthereforetheproduct, (d V )S = d x d y d z wherethesubscript S isusedto

Thethermodynamicequation indicatethesurfacesurroundingtheinfinitesimalvolumeelement.The divergence ofthevelocityfieldisdefinedtobetheemergingfluxperunitvolume:

Thisdefinitionofthedivergenceisactuallyindependentoftheshapeofthevolume forreasonablywell-behavedfunctions v (r ).Wetakeitheretobearectangular parallelepipedforconvenience.Notethatthedivergenceisascalarfielddefined overthespacewhosepointsaredesignatedby r .

Whilethisappearstobeausefulconcept,sofaritseemstobearatherdifﬁcult thingtocompute.Nextwewillﬁndaconvenientwaytocomputethedivergence. Inrectangularcoordinateswetakethesurfacetobetherectangularparallelepiped mentionedbefore.

Theintegrationoverthesixfacesoftheboxissosimilartothetwo-dimensional casethatweneednotrepeatithere.Theresultis

Thedivergenceisthen:

Fromourearliernotationwiththe ∇ operator,wecanwrite

Example9.15:divergenceoftheproductofscalarandvector Wecanfindthe divergenceofaproductofascalarfieldandavectorfieldbyexpandingthe individualtermsintotheirCartesian[x , y , z ]components.Let G (r ) beascalar fieldand a (r ) beavectorfield:

9.7Divergencetheorem

Consideramacroscopicvolume V surroundedbyaclosedsurface S inwhicha vectorfield A (r ) isdefined.Letthefluxoutofthevolumebe

Nowsubdividethevolumeintoboxeseachoneofwhichissmallenoughthatwe canusetheapproximation

Notethatfluxflowingoutofthesidesofonelittleboxflowsintothesidesofits neighbor.Ifweformthesum

Butsincetheﬂuxleavingthewholevolumeisjustthealgebraicsumoftheﬂuxes leavingthetinyboxes

Wehaveatlastthe divergencetheorem:

Thedivergencetheoremisveryusefulinthatitsayswecanapplythemicrodefinitionessentiallytomacroscopicvolumes.Wesimplyintegratethedivergence formicro-volumesuptoobtainthefluxissuingfromthemacroscopicvolume. WhilethemathematicalexpressionforthedivergenceinCartesiancoordinates isveryusefulforcomputationfromanalyticalformulas,weoftenfindthatin meteorologicalapplicationstheintegralformsareeasiertoapply.Forexample inthe2Dcasewecanintegratearoundaboxonaweathermapanddividebyits areatoobtainagoodapproximationtothedivergence.

Example9.16 Acylinderofradius R ofairhasdensityproﬁle ρ(z ).Itisrigidly risingat w0 ms 1 .Whatistheﬂuxofmasspassingthroughalevelat z = z0 ?

Answer:Massﬂux = ρ(z ) · w0 · π R2 .Justforfuntake R = 1km; ρ = ρ0 e z /H ,

ρ0 = 1.2kgm 3 , H = 104 m;andﬁnally w0 = 0.01ms 1 .Thenat z = 0,we have3.77×104 kgs 1 .At z = H /2thisbecomes2.29 ×104 kgs 1 .

9.8Continuityequation

Considerthedivergenceoftheproduct ρ(r )v (r ).Accordingtothedefinitionofthe divergencethisisthefluxof mass perunittimeissuingfromaninfinitesimalvolume perunitvolume.Iftheboxfromwhichthemassisissuingis fixedinspace (and therearenosourcesofmassinside),themassinsidetheboxhastobechanging:

where M isthemassinsidetheﬁxedbox.Dividingbythevolumeoftheboxand lettingitshrinktozero,weobtain:

Theminussigntakesintoaccountthattheﬂuxoutoftheboxrepresentsanegative rateofchangeofmassinthebox.Notethatweusedthe partial derivativeinthe lastformulabecausewearereferringtoa ﬁxed positionforourbox.Rearranging wehavetheEulerianformofthe equationofcontinuity:

t +∇· (ρ v ) = 0 [Eulerformoftheequationofcontinuity].(9.70)

Byexpandingthedivergenceoftheproductwecanwriteitinanotherform:

Thislastequationhasaveryspecialmeaningifweregrouptheﬁrsttwoterms

Dρ Dt =−ρ ∇· v [Lagrangianformoftheequationofcontinuity](9.72) wherethedifferentialoperator

D Dt ≡ ∂ ∂ t + v ·∇ [materialderivative](9.73) iscalledthe materialderivative,andwewillreturntoitinthenextsection. Analternativeandperhapsamorephysicalderivationofthecontinuityequation (intwodimensionsforsimplicity)canbeconductedasfollows.Considerasmall

Figure9.16Schematicviewofaboxelementinmotionatvelocity v .Thebox mightextenditswidthorheightduringthemotion,butmassmustbeconserved.

rectangularboxwithsidesd x andd y ,whoseareaisd x d y .Themassdensityof materialintheboxis ρ givingthemassintheboxas M = ρ d x d y = constant sincemasswillbeconservedalongthepathofthebox.Theboxmightbedistorted duetothedifferentialmotionsoftheﬂuid(e.g.,shear).Wecanwrite:

where xR (t ) representsthelocationoftherighthandedgeoftheboxasitmoves, xL (t ) similarlyrepresentsthelocationofthelefthandedge(seeFigure 9.16).The samenotationgoesfortheupperandloweredgesofthebox.Notethat xR (t ) might bemovingatadifferentspeedfrom xL (t ) andthereforetheboxmightbestretched orsqueezedinthatdirection.Aswetakethederivativethroughtheexpressionwe obtain

wherebyd u wemean (d x /d t )(R ) (d x /d t )(L),andbyd v wemean (d y /d t )(U ) (d y /d t )(L).Inthelastequationwehavetorecognizethatthederivativeof ρ is along themotion,aswasthederivativeofthemass.Afterdividingthroughbythe aread x d y andtakingthelimit:

Nowwiththefancynotation:

Inthelastequationweusedthematerialderivativefortherateofchangeof ρ since itistakenalongthemotion.Notethatthisisthesamestatementasin(9.72).The generalizationtothreedimensionsisstraightforward.

9.9Materialderivative

Asaparcelmoves,itstemperatureorsomeotherpropertymightchangealongthe pathoftheparcel.Ontheotherhand,somepropertiesareconservedalongthe path,forexample,thepotentialtemperatureinadiabaticflows.Therateofchange ofafieldalongthefluid’smotionisanimportantpointofviewtotakebecause manyphysicallawsaremosteasilyexpressedinthisform.Forexample,therate ofchangeofthemomentuminreactiontoanimposedforceistobetakenalong thepathoftheparcel(Newton’sSecondLaw).

Takeasmallrectangularparallelepipedofdimensiond x ,d y ,d z .Letthecenter ofthefiguremovewiththelocalvelocityofthefluidinwhichitisembedded, v (r , t ). Thevelocityfield v (r , t ) mightbechanginginbothspaceandtime.Considerthe temperature T (r , t ) asanexample.Ifweknowthevalueofthetemperatureata certainpoint(r , t ),say T (r , t ),whatcanwesayaboutitsvalueatneighboringpoints inspace-time?Thetotaldifferentialcanbeusedtomakeanestimate.Wecanwrite thetotaldifferentialforthetemperaturefieldas

Thisisthechangeinthetemperatureduetoadisplacementfrompoint r to r + d r andfrom t to t + d t . Nowdividethroughbyd t

Ingenerald r canbeinanydirection,butinthisspecialcaseitshouldbethe displacementd r thatoccursduetothemotionofthefluidduringthetimed t .In otherwords,d r = v (r , t ) d t where v (r , t ) isthevelocityofthefluidmotionfield

evaluatedat(r , t ).Afterthiscrucialsubstitutionwecanwrite:

Statedagain,thisistheexplicitrecognitionthatthesmallincrementd r whichis associatedwiththetimeintervald t isidentifiedwiththemotionofthefluid,namely itisthespatialincrementthatisadvancedbythefluiditselfduringthetimed t . Weagainencountertheimportantcombinationofderivativeswhichwetermedthe materialderivative,

Thechangeofthetemperaturefieldalongthemotion,DT /Dt ,iscomposedoftwo terms:thefirstterm, ∂ T /∂ t ,representsthe local changeofthetemperatureina certainvolumefixedinspace(aweatherstationforexample);thesecondterm, v ·∇ T ,isduetotheadvectionofhotterorcolderairintothisfixedvolumebythe wind.

9.10Thermodynamicequation

Considertheheatingofamovingparcel.WeapplytheFirstLawof Thermodynamicstoanindividualparcel.Wecanwrite

wherewehaveusedd Q toindicatetheamountofdiabaticheating 4 oftheparcel intimed t .Therateofchangeoftheenthalpyoftheparcelasitmovesalongits pathisgivenbythematerialderivative:

whereDQ /Dt istherateofdiabaticheatingoftheparcel(Js 1 ).Sincetheaircan betreatedasanidealgaswemaywrited H = cp M d T :

ThisisanexpressionfortheFirstLawofThermodynamicsforaﬂuidinmotion. The thermodynamicequation followsfrom(9.84)bydividingthroughby M,the

4 Thediabaticheatingincludessolarandterrestrialinfraredabsorptionheatingaswellasheatingdueto condensation.

Thethermodynamicequation massoftheparcel(sincemassisconservedalongthepath):

[thermodynamicequation](9.85)

whereDQM /Dt indicatestheheatingrateperunitmassofair;itiscalledthe diabatic heatingrate(perunitmass).Theleft-handsideofthethermodynamic equationistheheatingrateofaparcelalongitspath.Theheatingrateisproportional totherateofchangeofthetemperature.Theright-handsideoftheequationtells uswhatcontributestothatheatingrate.Firstistheheatingperunitmassofthe parcelbysuchactionsasradiativeheatingorcondensationheating,collectively calledDQM /Dt .Thesecondtermcontributestotherateoftemperaturechange becauseofcompressionorexpansionoftheparcelasitmovesalongitspathfrom onepressuretoanother.

Example9.17 Supposea1kgparcelofairmoveshorizontallyalonganisobaric surfaceandisheatedbyradiationby4Wkg 1 .Whatistherateofchangeof temperatureoftheparcel?

Answer:NotethatthederivativeDp/Dt vanishesbecausetheparcelmovesalong anisobaricsurface.Thenwecanﬁndtherateofchangeoftemperaturefrom (4Wkg 1 )/(1004Jkg 1 K 1 ) ≈ 0.004Ks 1 .

Example9.18 Inthepreviousexamplesupposetheparcelismovingeastward alongthehorizontalisobaricsurfaceatavelocityof3ms 1 andthattheeastward componentofthegradientoftemperatureisgivenby1.5Kkm 1 .Whatisthelocal (ﬁxedposition)rateofchangeoftemperature?

Answer:Weneedtowritethematerialderivative:

t + v ·∇ T = 0.004Ks 1 .

Weseekthevalueof ∂ T /∂ t .Thehorizontalvelocityis v = 3 i (ms 1 )and ∇ T = 0.0015 i (Km 1 ).Hence, ∂ T /∂ t = (0.004 0.0045) Ks 1

Example9.19 Whatistherateofchange(alongitspath)oftemperatureofa parcelofdensity ρ = 1.0kgm 3 whichisrisingsuchthatDp/Dt =−0.3Pas 1 ? Assumethediabaticheatingrateiszero.

Answer:WecancalculateDT /Dt =−0.3Pas 1 /(1.0kgm 3 × 1004JK 1 kg 1 ) ≈−0.0003Ks 1 .

Example9.20 Continuingthepreviousexample,whatisthelocal(ﬁxedposition) rateofchangeofthetemperature,assumingtheenvironmentallapserateis 10Kkm 1 ?

Answer:Firstweneedtoestimatetheverticalvelocity, w .Assuminghydrostatic balancewecancalculate

andd p/d z =−ρ g ;hence, w ≈ 0.03ms 1 .Nowproceedwith

Thus,

Finally,wecancombinethetwoandﬁndthat ∂ T /∂ t = 0.Thisisbecausewechose theadiabaticlapserateasourverticalgradient.Thereadershouldthinkofthe consequencesofalapseratethatisgreaterorlessthantheadiabaticlapserate.

9.11Potentialtemperatureform

Thethermodynamicequationcanalsobeexpressedintermsofthepotential temperaturebystartingwiththeSecondLawofThermodynamics:

θ ,orindifferentialform

Usingthematerialderivativeweﬁnd

[potentialtemperatureformofthethermodynamicequation]. (9.87)

Thisformisequivalenttotheform(9.85)anditshowsmoredirectlytheinﬂuence ofboththeFirstandSecondLawsofThermodynamics.Itisleftasanexercisefor thereadertoshowtheequivalenceofthetwo(Problem9.13).

9.12ContributionstoDQM /Dt

DQM /Dt isamaterialderivative,whichmeanstherateofchangeistakenalong themotionoftheinﬁnitesimalvolumeelement.Localheatingofaninﬁnitesimal volumeelementcanbeduetoseveralsources.Welistafewofthemhere.

Heatingbyconduction Atthemolecularleveltheheatexchangefromoneinﬁnitesimal volumeelementtoitsneighborswithdifferingtemperatureisgivenbythedivergence oftheheatﬂux ∇· h(r , t ).Thisgivesthecoolingrateperunitvolumeofthemoving element.Toobtainthecoolingrateperunitmass,onemustdividebydensity ρ(r , t )

Heatingbyphasechange Asamoistparcelmovesitmightexperienceatemperature changeandthiscouldleadtocondensation(orevaporation)ontodroplets.Therelease

Thethermodynamicequation

ofenthalpyisgivenby L d Mvap = Mair L d ws ,where ws (T ) isthesaturationmixing ratioofthevolumeelementcontaining Mair ofair.Hence,theheatingrateperunit massissimply L d ws (notethatd ws isnegativeforcondensation).

Heatingbyradiation Inthiscasewehaveacertainradiationﬂuxdensityofenergy F(r , t ).Theheatingrateperunitvolumeis −∇· F(r , t ).Andtheheatingrateperunit massis (1/ρ)∇· F(r , t ).

Tosummarizewehave

Sometimesafrictionalheatingtermisincludedaswell.Generallyinapplications suchasnumericalweatherforecastingandclimatemodeling,theﬁrsttermabove issmallcomparedtotheothersandisneglected.

Notes

Thischapterisreallyanintroductiontodynamics.Mostdynamicsbookscover thesesubjectsandmanydosoinmoredetail,seeforexample,Holton(1992).

NotationandabbreviationsforChapter9

a , b etc.arbitraryvectors

a , ax theseareusedforvectoraccelerationandits x component

ax , ay , az theCartesiancomponentsofvector a A avectorﬁeld

div2 V divergenceofvectorﬁeld V intwodimensions

D/Dt materialderivative

DQM /Dt therateofheatingofamovingparcelperunitmass (Js 1 kg 1 )

∇· F(r , t ) divergenceofthevectorﬁeld F(r , t )

∇ T = (∂ T /∂ x )i +··· gradientof T

i , j, k theCartesianunitvectors

κH thermalconductivity(JK 1 m 1 s 1 )

L enthalpyofevaporation(latentheat)(Jkg 1 )

n unitvector

nθ unitvectorinthethetadirection(polarcoordinates)

∂ T /∂ n directionalderivativeof T

r = x i + y j + z k positionvector

ˆ r unitvectorinthe r direction(polarcoordinates)

ρ density(kgm 3 )

S,d S surface,surfaceelementvector

v three-dimensionalvelocityvectorﬁeld(ms 1 )

V uppercaseusuallyindicatestwo-dimensionalvelocity

Problems

9.1Bywritingoutthecomponents,provethat

9.2Ahorizontalwindisblowingfromsouthwesttonortheastat30◦ northofeast.Express itintermsof i , j, k .

9.3Findaunitvector45◦ northofeast.Findaunitvectorperpendiculartothisoneinthe x –y plane.

9.4Thevectors2i + 3j,4i k determineaplane.Finda unit vectorperpendiculartothis plane.

9.5Aparcelmovesalongapath:

If γ = 0,whatisthecurvethatdescribesthemotion?Describethemotionwhen γ> 0.Whatisthepositionvectorasafunctionoftime?Whatisthevelocityvector?

9.6Acertainpressureﬁeldvariesspatiallyas

Howmanymaximaarethereatthesurface(z = 0)andatthecenterofthechannel (y = 0)asonegoesaroundthecircle(x goesfrom0to2π R)?Whatisthetotal differentialofthepressure?Whatisthepressuregradient?

9.7Thewindhastheform v = (v0 /r ) nθ , nθ =− sin θ i + cos θ j.Whatare u , v ?What isthedivergenceofthiswind,div2 (v )?

9.8Acertainsurfacepressureﬁeldisgivenby

Describethispressuresurfacenear (x0 , y0 ).Whatisthepressuregradientasafunction of x and y ?Whatisthedirectionalderivativeinthedirection45◦ northofeast?

9.9The500hPalevelisgivenby

Z500 = 5500 + 50sin (φ ω t ) cos π y

, L ≤ y ≤ L

wheretheverticalheightsareinmetersandthelongitudinaldistancesareinkilometers. L isabout30◦ oflatitude.Suppose ω = 2π radianspermonth.Describethis disturbance.

9.10Supposethediabaticheatingrateforaparcelis10Wkg 1 andthewindisblowing atanangleof60◦ northofeastat10ms 1 .Alsosupposethetemperaturegradientis 0.1Kkm 1 anddirecteddueeast.Youmayassumethehorizontalgradientofpressure iszero.Findtherateofchangeoftemperaturealongthemotionandﬁndthelocalrate ofchangeofthetemperatureataﬁxedpoint.

9.11Sinkingair.Supposetheatmosphericproﬁleisisothermal.Aparcelisdescendingat 2cms 1 andthediabaticheatingrateiszero.Whatistherateoftemperatureincrease alongthemotion;ataﬁxedaltitude?(Usereasonablevaluesforparametersnotgiven.)

9.12Dryairisblowingfromwesttoeastataspeedof u = 12ms 1 .Thetemperature gradientis ∇ T = 2.0 × 10 6 (Km 1 ) i .Theairisbeingheateddiabaticallyatarate 1.1 × 10 5 Kkg 1 s 1 .Theairmovesalonganisobaricsurface.

(a)Writeanexpressionfortherateofchangeofthetemperatureoftheairalongits pathintermsoftheheatingrates(thermodynamicequation).

(b)Basedupon(a)evaluatethematerialderivative.

(c)Theairispassingoveraﬁxedstation.Findtherateofchangeofthetemperature atthestation.

9.13Showthatthetwoforms(9.85, 9.87)ofthethermodynamicequationareequivalent byusing θ = T (p/p0 ) κ .

AppendixA

Unitsandnumericalvaluesofconstants

Theunitsusedinatmosphericsciencearethe StandardInternational (SI)units. TheseareessentiallytheMKSunits.Theunitsoflengthare meters,abbreviated m;thoseformassarethe kilogram,abbreviatedkg;andfortimetheunitisthe second,abbreviateds.Theunitsforvelocitythenarems 1 .Theunitofforceis the newton (1kgms 2 ,abbreviatedN).Tables A1–A3 showtheSIunitsforsome basicphysicalquantitiescommonlyusedinatmosphericscience.

Theunitofpressureisthe pascal (1Nm 2 = 1Pa).Thisisofspecialimportance inmeteorology.Inparticular,atmosphericscientistslikethe millibar (abbreviated mb),butinkeepingwithSIunitsmoreandmoremeteorologistsusethe hectopascal (abbreviatedhPa;1hPa = 100Pa = 1mb).The kilopascal (1kPa = 10hPa)isthe formalSIunitandsomeauthorspreferit. Oneatmosphere (abbreviated1atm)of pressureis

1atm = 1.013bar

= 1013.25mb

= 1013.25hPa

= 101.325kPa

= 101325Pa

= 1.01325 × 105 Pa(A1) and1mb = 1hectopascal = 100Pa.Sometimesoneencounterspressureininches ofmercury(inHg)ormillimetersofmercury(mmHg);1atm = 760.000mmHg = 29.9213inHg.

The dimensions ofaquantitysuchasdensity, ρ ,canbeconstructedfromthe fundamentaldimensionsoflength,mass,timeandtemperature,denotedbythe letters L, M , T , θ respectively.Thedimensionsofdensity,indicatedwithsquare brackets [ρ ],are ML 3 .IntheSIsystemits units arekgm 3 .Manyquantitiesare purenumbersandhavenodimension;examplesincludeargumentsoffunctions suchassineorlog.Theradianisreallynotaunitinthesenseusedhere.

TableA1 Usefulnumericalvalues

Universal

gravitationalconstant(G )6.673×10 11 Nm2 kg 2

universalgasconstant(R∗ )8.3145JK 1 mol 1

Avogadro’snumber(NA )[grammole]6.022 ×1023 moleculesmol 1

Boltzmann’sconstant(kB )1.381 ×10 23 JK 1 molecule 1

protonrestmass1.673 ×10 27 kg

electronrestmass9.109 ×10 31 kg

PlanetEarth

equatorialradius6378km

polarradius6357km

massofearth5.983×1024 kg

rotationperiod(24h)8.640×104 s

accelerationofgravity(atabout45 ◦ N)9.8067ms 2

solarconstant1370Wm 2

Dryair

gasconstant(Rd )287.0JK 1 kg 1

molecularweight(Md )28.97gmol 1

speedofsoundat0 ◦ C,1000hPa331.3ms 1 densityat0 ◦ Cand1000hPa1.276kgm 3

speciﬁcheatatconstantpressure(cp )1004JK 1 kg 1 speciﬁcheatatconstantvolume(cv )717JK 1 kg 1

Watersubstance

molecularweight(Mw )18.015gmol 1

gasconstantforwatervapor(Rw )461.5JK 1 kg 1 densityofliquidwaterat0 ◦ C1.000×103 kgm 3 latentheatofvaporizationat0 ◦ C2.500×106 Jkg 1

TableA2 Selectedphysicalquantitiesandtheirunits

QuantityUnitAbbreviation masskilogramkg lengthmeterm timeseconds forcenewtonN pressurepascalPa = Nm 2 = 0.01hPa energyjouleJ

temperaturedegreeCelsius ◦ C temperaturedegreeKelvinK speedms 1 densitykgm 3

speciﬁcheatJkg 1 K 1

TableA3 Selectedconversionsto SIunits

QuantityConversion energy/heat4.186J = 1cal 1kWh = 3.6×106 J

pressure1atm = 760mmHg

1atm = 29.9213inHg

distance1m = 3.281ft

temperature T (K) = T (◦ C) + 273.16

Eachsideofanequationmusthavethesamedimensions.Thisprinciplecan oftenbeusedtoﬁnderrorsinaproblemsolution.

AppendixB

Notationandabbreviations

atmpressureunit,oneatmosphere(Chapter 1)

a , b,etc.arbitraryvectors(Chapter 9)

a , ax vectoraccelerationandits x component(Chapter 9)

a Bohrradius(Chapter 2)

a dropletradius(m)(Chapter 5)

a , b empiricalcoefﬁcientsusedinvanderWaalsequationofstate (Chapter 5)

a ∗ criticaldropletradius(m)(Chapter 5)

ax , ay , az theCartesiancomponentsofvector a (Chapter 9)

A avectorﬁeld(Chapter 9)

A horizontalareaofaslab(m2 )(Chapter 6)

c speedoflightinvacuum(ms 1 )(Chapter 8)

cv , cp speciﬁcheats(heatcapacityperkg)atconstantvolume,pressure (Jkg 1 K 1 )(Chapter 3)

c v , c p molarspeciﬁcheats(Jmol 1 K 1 )(Chapter 3)

c v , c p , G overbarindicatesquantitiesexpressedpermole(Chapter 4)

CAPEconvectiveavailablepotentialenergy(Jkg 1 )(Chapter 7)

CINconvectiveinhibitionenergy(Jkg 1 )(Chapter 7)

Cv , Cp heatcapacitiesatconstantvolume,pressure(JK 1 )(Chapter 3) div2 V divergenceofvectorﬁeld V intwodimensions(Chapter 9)

d H /d t rateofheating(Chapter 9)

d H /d t ,d Q /d t timerateofchangeofenthalpy,heatingrate(Js 1 )(Chapter 3) (d H )p changeinenthalpyatconstantpressure(J)(Chapter 3)

d Qrev inﬁnitesimalabsorptionofheat,subscriptindicatingthatthe changeisreversible(J)(Chapter 4)

d QM /d t rateofheatingperunitmass(Jkg 1 s 1 )(Chapter 9)

d Q ,d W differentialsforheat,work,baremphasizespathdependence(J) (Chapter 3)

d SU ,V inﬁnitesimalchangeinentropyduringwhich U and V areheld constant(JK 1 )(Chapter 4)

D/Dt materialderivative(Chapter 9)

f H ◦ (X ) standardheatoffusionofsubstance X (Jmol 1 )(Chapter 3)

G ◦ changeinGibbsenergypermole,atstandardconditions (kJmol 1 )(Chapter 8)

V changeinvolume(m 3 )(Chapter 3)

vap H ◦ (X ) standardheatofvaporizationofsubstance X ,overbarindicates 1molofsubstance,superscriptoindicatesatstandard conditions(usually25◦ C)(Jmol 1 )(Chapter 3)

x displacementin x (Chapter 3)

H ◦ changeinenthalpypermole,atstandardconditions(kJmol 1 ) (Chapter 8)

t timeinterval(s)(Chapter 2)

e , es vaporpressure,saturationvaporpressure(Pa)(Chapter 5)

e vaporpressureoverasolution(Pa)(Chapter 5)

Eact activationenergy(J)(Chapter 8)

E energy(J)(Chapter 8) = Mw /Md = 0.622(dimensionless)(Chapter 5)

η mixingratio(Chapter 8)

f frequencyofelectromagneticwave(Hz)(Chapter 8)

f numberofdegreesoffreedomofamolecule(Chapter 3)

fx partialderivativewithrespectto x (Chapter 3)

F force(N)(Chapters 2, 3)

F energyﬂux(Js 1 )(Chapter 8)

Fλ(z ) ﬂuxofelectromagneticenergyatwavelength λ,elevation z (Js 1 m 1 )(Chapter 8)

F , F /M force,perunitmass(Nkg 1 )(Chapter 7)

F force(N)(Chapter 4)

g,l,aqgaseous,liquid,aqueousphase(Chapter 8)

g accelerationduetogravity(9.81ms 2 )(Chapter 6)

g Gibbsenergyperkilogram(Jkg 1 )(Chapter 4)

g , gz , g0 accelerationduetogravity,itsvalueasafunctionofaltitude,its valueatthesurface(ms 2 )(Chapter 1)

gl , gg speciﬁcGibbsenergyforliquid,gas(Jkg 1 )(Chapter 5)

gv , gw speciﬁcGibbsenergyforvapor,liquidwater(Jkg 1 )(Chapter 5)

G gravitationalconstant(Chapter 1,Table A1)

G Gibbsenergy(J)(Chapters 4, 8)

G Gibbsenergypermole(Jmol 1 )(Chapter 4)

γ ratioofspeciﬁcheats cp /cv (dimensionless)(Chapter 3)

d , m , e lapserate, d T /d z ofdryairascendingadiabatically,ofmoist adiabat,oftheenvironment(Km 1 )(Chapter 6)

∇ T ∇ T = (∂ T /∂ x )i +··· ,gradientof T (Chapter 9)

h height(Chapter 1)

h heightaboveareferencelevel(Chapter 6)

h speciﬁcenthalpy(Jkg 1 )(Chapter 5)

h Planck’sconstant(Js)(Chapters 2, 8)

h, u speciﬁcenthalpy,internalenergy(Jkg 1 )(Chapter 7)

h0 initialheight(Chapter 2)

H enthalpy(J)(Chapters 3, 4)

H scaleheight(Chapter 6)

H (r ) ﬂuxofheatcrossingthesurfaceofasphere(Js 1 )(Chapter 5)

Ha scaleheightoftheatmosphere(Chapter 3)

Hw ascaleheightforwatervapor(morkm)(Chapter 6)

H ◦ standardenthalpy(kJmol 1 )(Chapter 8)

i , j, k theCartesianunitvectors(Chapter 9)

J photodissociationcoefﬁcient(s 1 )(Chapter 8)

k reactioncoefﬁcient(Chapter 8)

kB Boltzmann’sconstant(Chapters 1, 2, 3,Table A1)

K , Kp equilibriumconstant(Chapter 8)

KH constantinHenry’sLaw(Chapter 8)

K kineticenergy(J)(Chapter 7)

κ = R/cp (dimensionless)(Chapters 3, 4, 6)(0.286fordryair)

κH thermalconductivitycoefﬁcient(Jm 1 s 1 K 1 )(Chapters 3, 5, 9)

LCLliftingcondensationlevel(Chapter 5)

L dimensionlength(m)(Chapter 1)

L enthalpyofevaporation(latentheat)(Chapter 9)

L lengthofboxedge(m)(Chapter 2)

λ meanfreepath(m)(Chapter 2)

λ wavelengthofanelectromagneticwave(m)(Chapter 8)

L = Hvap theenthalpy(latentheat)ofevaporation(Jkg 1 )(Chapters 5, 6)

Levap latentheatofevaporation(Jkg 1 )(Chapter 7)

mbpressureunit,onemillibar = 1hPa(Chapter 1)

me electronmass(kg)(Chapter 2)

m0 massofamolecule(kg)(Chapter 2)

M dimensionmass(kg)(Chapter 1)

Meff effectivemolecularweight(gmol 1 )(Chapter 2)

MG grammolecularweightofagas(gmol 1 )(Chapter 2)

Mv , Md , Me grammolecularweightofvapor,dryairandeffective(gmol 1 ) (Chapter 5)

M mass(kg)(Chapters 2, 3, 4)

M massofanobjectorsystem(Chapter 1)

ME massoftheEarth(Chapter 1)

Ml , Mg bulkmassofliquid,gas(kg)(Chapter 5)

n unitvector(Chapter 9)

nθ unitvectorinthethetadirection(polarcoordinates)(Chapter 9)

ns numberdensityofvapormoleculesatsaturation(moleculesm 3 ) (Chapter 5)

nsat numberdensityofvapormoleculesatsaturation(moleculesm 3 ) (Chapter 5)

nw numberdensityofwatermoleculesinvapor(moleculesm 3 ) (Chapter 5)

n0 (z ) moleculardensityasfunctionofheight(moleculesm 3 ) (Chapter 2)

n0 numberdensityofphotons(photons/m 3 )(Chapter 8)

n0 numberdensity(moleculesm 3 )(Chapters 3, 5)

Nnewtons(Chapter 2)

N totalnumberofmolecules(Chapter 2)

NA Avogadro’snumber(moleculesmol 1 )(Chapters 1, 2, 5,Table A1)

N (z ) concentrationofabsorbersatlevel z (absorbersm 3 )(Chapter 8)

ν , νA , νB numberofmoles,numberofmolesofspeciesA,B(Chapters 3, 4)

ω , f angularfrequency(rads 1 ),frequency(cycless 1 ≡ Hz) (Chapter 6)

p pressure(Pa)(allchapters)

p, pG pressure,partialpressureforgasG(Chapter 2)

p, p(z ), p0 pressure,asafunctionof z ,atareferencelevel(hPa)(Chapter 6)

p(V ) pressureasafunctionofvolume;expressionforacurveinthe p–V plane(Chapter 4)

Paunitofpressure,1Pa = Nm 2 (Chapter 1)

P (z ) probabilitydensityfunction(Chapter 2)

P (vx , vy , vz ) jointprobabilitydensityfunctionforvelocitycomponents (Chapter 2)

(z ), 1 , 2 geopotentialheightasafunctionofheight,attwolevels(meters, onchartsoftenindecameters,dm)(Chapter 6)

q speciﬁchumidity(kgwatervapor/kgofmoistair)(Chapter 5)

rr = x i + y j + z k positionvector(Chapter 9)

r unitvectorinthe r direction(polarcoordinates)(Chapter 9)

r relativehumidity(Chapter 5)

r0 effectivemolecularradius(m)(Chapter 2)

R gasconstantforaparticulargas(Jkg 1 K 1 )

Rd gasconstantfordryair(Jkg 1 K 1 )(Chapter 2,Table A1)

Reff effectivegasconstantforamixtureofspecies(Jkg 1 K 1 ) (Chapters 2, 5)

Rw thegasconstantforwatervapor(Chapters 2, 5,Table A1)

R∗ universalgasconstant(Jmol 1 K 1 )(Table 1.1)

ρ density(kgm 3 )(allchapters)

ρ , ρ0 , ρe density,atareferencelevel,oftheenvironment(kgm 3 ) (Chapter 6)

ρa , ρe densityforadiabatandenvironment(kgm 3 )(Chapter 7)

s entropyperunitmass(lowercaseindicatesperunitmass) (JK 1 kg 1 )(Chapter 4)

sl , sg speciﬁcentropyforliquid,gas(JK 1 kg 1 )(Chapter 5)

SIStandardInternationalsystemofunits(Chapter 1)

S,d S surface,surfaceelementvector(Chapter 9)

S entropy(JK 1 )(Chapter 5)

S , SA , SB entropy,entropyofstateA,stateB(JK 1 )(Chapter 4)

Ssys , Ssurr , Suniverse entropyforthesystem,surroundings,universe(sys+surr) (Chapter 4)

σ standarddeviation(Chapter 2)

σ surfacetension(Jm 2 )(Chapter 5)

σ 2 variance(Chapter 2)

σ(λ) absorptioncross-section(m2 )(Chapter 8)

σc collisioncross-section(m2 ,nm2 )(Chapter 2)

t0 , tc bothstandforlifetime(s)(Chapter 8)

t1/2 halflife(s)(Chapter 8)

T Kelvintemperature(K)(Chapters 2, 3, 5)

TC temperaturemeasuredindegreesCelsius(◦ C)(Chapter 5)

TD dewpointtemperature(K)(Chapter 5)

Tv virtualtemperature(K)(Chapter 5)

Tw wet-bulbtemperature(K)(Chapter 5)

T dimensiontime,alsoperiodofarepeatingprocess,and temperature(Chapter 1)

Temp temperaturedimension(Chapter 1)

T , T (z ), T0 temperature,asafunctionof z ,atareferencelevel(K) (Chapter 6)

Te (z ), Ta (z ) temperatureofenvironment,ofanadiabat(K)(Chapter 6)

T verticalaveragetemperatureinalayerofair(K)(Chapter 6)

θ potentialtemperature(K)(allchapters)

θe equivalentpotentialtemperature(K)(Chapter 5)

θw wet-bulbpotentialtemperature(K)(Chapter 5)

θs saturationequivalentpotentialtemperature(K)(Chapter 5) zenithangle(Chapter 8)

τ opticaldepth(dimensionless)(Chapter 8)

(u , v , w ) thevelocitycomponents (vx , vy , vz ) (Chapter 2)

U internalenergy(J)(allchapters)

UA , UB internalenergyatstatesA,B(J)(Chapter 4)

v three-dimensionalvelocityvectorﬁeld(ms 1 )(Chapter 9)

v speed(ms 1 ),sometimeswithasubscriptindicatingvelocity componentalongacoordinateaxis(e.g., vx )(Chapter 1)

vesc escapevelocity(kms 1 )(Chapter 2)

vl , vg speciﬁcvolumeforliquid,gas(m3 kg 1 )(Chapter 5)

v ∗ thresholdvelocitytoexceedanactivationbarrier(ms 1 ) (Chapter 8)

v meanspeed(ms 1 )(Chapter 2)

v meanmolecularspeed(ms 1 )(Chapter 5)

v 2 meansquarevelocity(m2 s 2 )(Chapter 2)

v 2 x meansquareof x componentofvelocity(m2 s 2 )(Chapter 2)

V uppercaseusuallyindicatestwo-dimensionalvelocity (Chapter 9)

V volumeofasystem(m3 )(Chapters 1, 3)

V volume(m3 )(Chapter 3)

Va , Ve volumesofaparcelalonganadiabatandoftheenvironment (m3 )(Chapter 7)

VA , VB initialandﬁnalvolumes(Chapter 3)

w , ws mixingratio,saturated(kgwatervaporperkgdryair) (Chapters 5, 6, 7)

WA →B , QA →B workdonebythesystem,heattakenintothesystemingoing fromstateAtostateB(J)(Chapter 4)

WV1 →V2 workingoingfrom V1 to V2 (Chapter 3)

W , Q workdonebythesystem,heattakenintothesystem(Chapter 3)

[X ] concentrationofX(Chapter 8)

[X ]0 concentrationofXat t = 0(Chapter 8)

z elevation(m)(Chapter 2)

z , z verticaldistance,incrementofit(m)(Chapter 6)

Zp altitudeofpressurelevel p (m)(Chapter 6)

AppendixC

Answersforselectedproblems

Chapter1

1.126428ftfor H = 8km.

1.20.0013.

1.3819hPa(using H = 8km,1mile =1.6km).

1.4(a)40200km,(b)111.7kmdeg 1 ,(c)96.7kmdeg 1

1.550000km2 .43000km2

1.811.2kms 1 .

1.92 2h/g .

Chapter2

2.11.29,1.25,1.2kgm 3 .

2.20.80kgm 3 .

2.3212hPa.

2.412.7N.

2.52.69×1025 moleculesm 3

2.6 Rd = 2.87hPaK 1 m3 kg 1

2.7Differentiate P (v ) withrespectto v andsetittozero.SeeTable 2.2

2.8462,493,413,1846ms 1 .

2.9630Pa.1kgm 1 .Pressurewithmoisture = 78646Pa.Densityofmoistmix = 1kgm 3 .Densityofdryairatsametemperatureandpressure:1.003kgm 3 .

2.101.6×108 J.

2.114.3×107 J.

2.12 n0 = 2.69 × 1025 moleculesm 3 , H = 8000m, N = 2.15×1029 molecules.

2.13 zλ=H = H ln (n0 σc H ).

2.14 (gp0 /RT0 ) ∞ 0 z e z /H d z = gp0 H 2 /RT0 .

2.15 1 2 Nm0 v 2 = 3 2 NkB T ,seeProblem2.12for N .

2.16 fyellow = c /λ = 6.0 × 1014 s 1 . fcoll = n0 σc v ≈ 6.5 × 109 s 1 .Anexcited atommightsuffertensofcollisionsbeforeitrelaxestothelowerenergy level.

Chapter3

3.1(d) p0 e 1/2 /2H .

3.2(a) ρ R.(b)and(c)Let1/T ≡ x ,thentakethepartialwithrespectto x

3.3 κT = 1/p;toget κθ use pV γ = p0 V γ 0 ;result1/γ p.

3.4 β = 1/T

3.6195.8kJ.

3.7 T = 205.1K. T =−67.9K.Therefore, U = 48.7J = W ,since Q = 0.

3.8(a)10kJ,(b)7.17kJ,(c)d U + V d p → d U .

3.9(a)59.7kJ,(b)59.7kJ.

3.101.202kgm 3 . Q =−7500J.Usethemassofthedryaironlyasan approximation. T = 8.7K.

3.11 T600 = 248.7K. U = Mcv T , H =−21.4kJ.

3.120.02Kkg 1 s 1 ,20Jkg 1 s 1 , U perunittime = 14.3Jkg 1 s 1

3.13(a)12,500kg.(b)Massofan“air”molecule = 0.029/NA = 4.82 × 10 26 kg. N = 2.6 × 1029 molecules.(c)1.3×1044 .

3.14 vsound = √RT forisothermal; √γ RT , γ = 1.40foradiabaticcompression waves.

3.15 θ = T0 (p0 /p)0.286 .

Chapter4

4.1Zeroifreversible.

4.223.5JK 1 .

4.3 θ = 65.8K.

4.4 H = 500J. T = 0.5K. S = 2JK 1

4.5(a) ( S )a = 23.5JK 1 ,(b) ( S )b = 120.6JK 1 ,(c) ( S )c = 8.6JK 1 .

4.6(b) V2 = 3.44m3 , T2 = 227K,(c) H =−146.7kJ, U =−104.7kJ, (d) W = 104.7kJ.

4.7(a) p2 = 37877Pa, W2 =−65.2kJ,(b) W1 = 104.7kJ, W2 =−65.2kJ, Wtot = 39.5kJ,(c) T3 = 113.6K, ( S )2 = 1.39kJK 1 .

4.8Usetheisothermjoiningthebeginningandendpoints. S = 199JK 1 . U = H = 0.

4.9Bothzeroforthecycle,ifitisreversible. Tmax isatthepointA. Tmin isatthe oppositepointonthecircle.

4.10Slidingupanddownanisotherminthe T S diagramenclosesnoarea.

4.14 19.2kJ.

Chapter5

5.1(a)2.5 × 106 J,(b)9.15JK 1 .

5.2283.6K.

5.3 es = 65hPa,z = 22km × ln (1000/66.75) = 21.7km.

5.41.4%.

5.53.8cm.

5.6(a)23.6hPa,(b)11.8hPa,(c)1.171kgm 1 ,(d) w = 0.0074kgkg 1 , ws = 0.0149kgkg 1 ,(e) w .

5.7(0.01,6.80hPa,112.8%),(0.10,6.10hPa,101.2%),(1.0,6.03hPa,100.1%), (10,6.025hPa,100.0%).

5.8eb/a = 1.002.

5.9101.4%.

5.10RadiusatthemaximumoftheKöhlercurve a = 3B/b;RH = 1 + 2b1.5 /31.5 B0.5 × 100%; a = 6 µm,RH = 100%.

Chapter6

6.1 p(z ) = p0 exp (gz0 /RT0 )e(z / z0 1) .

6.20.12ms 2 .

6.30.6ms 1 ,1.2ms 1 ,3.6ms 1 ,3min.

6.4Adiabatically0.03ms 2 ,isothermally0.02ms 2 .

6.5If p0 = 1000hPa, T0 = 300K, = 6Kkm 1 , z10 = 10km, T1 = T0 z10 = 240K,then p(z ) =

6.6 = T0 exp z / z0 + κ gz0 (ez /z0 1 )/RT0 .

6.7 ≈ 5Kkm 1 .

6.8 Z (p) =− Ra g ln p p0 Rb 2gp0 (ln p p0 )2 .

6.9926m.

6.100.516ms 2 .

6.11 ω = 5.7 × 10 3 rads 1 , f = 9.1 × 10 4 Hz, T = 18min.

6.13(a) A = 10m, B = 0,(b) A = 0, B = 175m.

Chapter7

7.1(a)16gkg 1 ,(b)22gkg 1 ,RH = 73%,(c)20 ◦C,(d)940hPa,(e)23 ◦ C, (f)yes,(g)yes,(h)8.5gkg 1 ,(i)12gkg 1 .

7.2(a)270hPa,(b)coldanddry,(c)verylowto400hPa,abovethat T TD ≈ 10 ◦ C,(d)between950and910hPa,750and700hPa,above200hPa,(e)no, yes.

7.3(a)150hPa,(b)unstable,(c)moderatebelow700hPa,dryabove,(d)at 650hPa,(e)largeCAPE,noCIN,unstable.

7.4(a) w = 3.6gkg 1 ,RH = 45%, θ = 284K,(b)840hPa,(c) θe = 294K, (d)thesameasin(a),(e)thesameasin(c).

7.5(a)3.5gkg 1 ,(b) T = 30 ◦ C, TD =−30 ◦ C.Note:thenumbershereare unrealisticbuteasytoworkwithonthecharts.

Chapter8

8.1 33.81kJmol 1 , 391.9kJmol 1 ;bothexothermic.

8.2 H ◦ = 72.14kJmol 1 ,no.

8.3Lowtemperaturesforforward,highforreverse(thisishowNOisformedin combustion).

8.43.4 × 10 19 J = 2.1eV.

8.56 × 1034 ,Lowerthetemperature.

8.6 k1 /k2 .

8.72.68 × 10 4 moll 1 .

8.82.5 × 10 2 mol.

Chapter9

9.1d (ax bx )/d t + d (ay by )/d t = bx d ax /d t + ax d bx /d t +··· .

9.2 |v |

9.3 √2 2 (i + j), √2 2 (i j).

9.4 0.239 i + 0.160 j 0.958 k .

9.5Circlewithradius α centeredat (x0 , y0 ).Pointontherimofarollingwheel.

9.6Threemaxima.

9.7 u =−y v0 /r +1 , v = x v0 /r + 1.div2 (v ) = 0,since vr = 0, vθ hasno θ dependence.

9.8Ellipticallyshapedbowl. ∇ p = 4a (x x0 )3 i + 2b(y y0 )j, ∂ p/∂ n = √2 2 (4a (x x0 )3 + 2b(y y0 ))

9.9Thisisawavethattravelswesttoeastandtakesonemonthtocircletheglobe.

9.100.01Ks 1 , ∂ T /∂ t = 0.009Ks 1 .

9.11Warming:0.00020Km 1 ,sameatﬁxedaltitude,sinced T /d z = 0.

9.12(b)1.1 × 10 5 Ks 1 ,(c) 1.3 × 10 5 Ks 1 .

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activationenergy, 202 additivityprinciple, 78 adiabat, 11 adiabatic, 57 air, 28

Archimedes’Principle, 144 Arrheniusequation, 202 atm, 2 atomicnumber, 36 Avogadro’snumber, 36

Boltzmann’sconstant, 21 boxproduct, 224 Brunt–Väisäläfrequency, 148 bulk, 7 buoyancyforce, 144

CAPE, 181

Carnotcycle, 84 Cartesiancomponents, 221 CentralLimitTheorem, 30 chemicalpotential, 93 CIN, 183

Clausius–Clapeyronequation, 110 closedsystem, 15 coefﬁcientofexpansion, 67 collisioncross-section, 22 collisionfrequency, 23 component, 8 compositesystem, 78 compressibility, 66 conditionalinstability, 179 conservativeforce, 73 conservativeproperty, 58 continuityequation, 240 convectiveavailablepotentialenergy(CAPE), 181 convectiveinhibitionenergy(CIN), 183 coordinates, 7 criticalisotherm, 102 crossproduct, 223 cyclonicwind, 222 cylindricalcoordinates, 222 cylindricalpipe, 230

Index

degreesoffreedom, 52 density, 2 determinant, 224 dewpointtemperature, 114 diabatic, 75, 244 diathermal, 9 differential, 63 dimensions, 2 directionalderivative, 227 divergence, 234, 236 divergencetheorem, 239 dotproduct, 222 dynamics, 1, 220

efﬁciency, 85 effusion, 35 endothermic, 192, 194 enthalpy, 44, 61, 64 entropy, 74 equationofcontinuity, 220, 240 equationofstate, 21 equilibration, 30 equilibriumconstant, 207 equilibriumstate, 7 equivalentpotentialtemperature, 156 escapevelocity, 19 exothermic, 192 exponentialfunction, 15 extensivevariable, 13

ﬂux, 230, 231 Fourier’sLaw, 232 freeatmosphere, 159 freeconvection,levelof(LFC), 178 freeexpansion, 77, 88 functionofstate, 54

geopotential, 141 geopotentialheight, 141 Gibbsenergy, 89 gradient, 226 gravitationalgeopotential, 141

harmonicoscillator, 148 haze, 129 heat, 48 heattransfer, 10 heating, 48 heatingbyradiation, 246 hectopascal, 2 Henry’sLaw, 212 heterogeneousnucleation, 122 homogeneousnucleation, 122 hydrologicalcycle, 99 hydrostaticequation, 135, 146 hygroscopic, 126

idealgas, 21, 70 inﬂectionpoint, 102 integratingfactor, 70, 72 integrationconstant, 110 intensivevariable, 13 internalenergy, 44, 51 internalenergyofanidealgas, 53 internallyuniformproperty, 7 inversion, 179 irreversiblechange, 47 irreversibleprocess, 12 isochoric, 50 isolatedsystem, 11 isotherm, 10 isothermalpath, 76 isotope, 36 isotropic, 32

jointdistribution, 32

Kelvin, 26 kilogram, 2 kilopascal, 2

Lagrangemultiplier, 160 latentheat, 63 LCL, 119 levelsurface, 227 LFC, 178 liftingcondensationlevel,LCL, 119 Loschmidtnumber, 22

macroscopic, 7 massﬂuxdensity, 232 materialderivative, 240, 242 maximumwork, 47 Maxwell–Boltzmanndistribution, 32 meanfreepath, 22 meansquaredvalue, 29 meanvalue, 29 millibar, 2 mixingratio, 113 mode, 30 moist, 156 moistadiabat, 155 moiststaticenergy, 158 mol, 7

molarGibbsenergy, 91 mole, 7 molecularweight, 36 mostprobablevelocity, 30

neutralbuoyancy,levelof(LNB), 179 Newton, 2, 5 Newton’sSecondLaw, 144 partialpressure, 39 pascal, 2 perfectdifferential, 70 photochemicalreactions, 194 photodissociation, 194 physicalchemistry, 191 Poisson’sequation, 58, 59 polarcoordinates, 236 positionvector, 225 potentialenergy, 6 potentialinstability, 180 potentialtemperature, 58, 75 pressure, 20 pressuregradientforce, 228, 229 probabilitydensityfunction,pdf, 30 products, 192 pseudo-adiabaticprocess, 154 quasi-staticprocess, 12 radiation,heatingby, 246 randomvariable, 29 Raoult’sLaw, 211 reactants, 9, 192 reactionratecoefficient, 202 realizations, 29 rectangularparallelepiped, 224 reducedvolume, 103 relativehumidity, 114 reservoir, 9 reversibleadiabaticprocess, 75 reversibleprocess, 12 reversiblework, 47 rootmeansquare, 28 rotatingfluid, 237 saturated, 9 saturationequivalentpotentialtemperature, 157 saturationvapormixingratio, 114 saturationvaporpressure, 102 scalarfield, 225 scalarproduct, 222 scaleheight, 19, 24, 137 SIunits, 2 slingpsychrometer, 118 slopeofdryadiabat, 138 sounding, 147 specificentropy, 75, 109 specificGibbsenergy, 91 specificheatcapacity, 50 specifichumidity, 115 specificvolume, 103 stablelayer, 145

stable,stability, 145 standardconditions, 38 standardenthalpyforfusion, 64 standardenthalpyforvaporization, 64 standardenthalpyofformation, 192 StandardInternationalunits, 2 stateofasystem, 44 STP, 3, 38 supercooled, 111 supersaturated, 104 surfaceintegral, 230 surfacetension, 122 surroundings, 9, 77 temperature, 26 thermaladvection, 233 thermalconductivity, 233 thermalenergy, 48 thermodynamiccoordinates, 7 thermodynamicequation, 220, 243 thermodynamicpotential, 81

thermodynamicsystem, 7 thickness, 142 tracer, 155 triplepoint, 106 unitvector, 221 universalconstant, 3, 21 universalgasconstant, 3, 38 universe, 77 unstable, 145 vaporization, 64 variance, 29 vectoraddition, 221 vectorﬁeld, 226 virtualtemperature, 113 viscousﬂow, 232 weight, 5 wet-bulbpotentialtemperature, 157 work, 13, 45