IMTS Civil Eng. (Applied thermodynamics)

I ns t i t ut eofManage me nt & Te c hni c alSt udi e s

APPLI ED THERMODYNAMI CS 500

Ba c he l orofEngi ne e r i ng www. i mt s i ns t i t ut e . c om

IMTS (ISO 9001-2008 Internationally Certified) APPLIED THERMODYNAMICS

APPLIED THERMODYNAMICS

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UNIT I

01-05

DEFINITION AND EXPLANATION OF TERMS Learning

,

Objectives,

Introduction,

Terms

thermodynamics,

System

Surroundings, Isolated system, Closed system, Open system, Homogenous and Heterogenous system, Macroscopic properities, state of system, state variables, Thermodynamics Equilibrim, extensive and intensive properties.

UNIT II

06-11

THERMODYNAMICS PROCESS THERMODYNAMIC PROCESS,Cyclic Process,Reversible Process,Irreversible Process,Adiabatic

Process,Isentropic

Process,Polytropic

Process,Throttling

Process,SPONTANEOUS PROCESS,Key words,Summary,Hints and Answers to self assessment questions

UNIT-III

12-17

THERMODYNAMIC FUNCTIONS Learning Objectives,Introduction,WORK, HEAT, AND ENERGY,Modes of Transference of Energy,Zeroth law of thermodynamics

UNIT IV

18-34

FIRST LAW OF THERMODYNAMICS Internal Energy and Change in Internal Energy,Law of Conservation of Energy (First law of thermodynamics), Thermodynamic

Enthalpy (H) and Enthalpy Processes

for

an

Ideal

Gas

,Adiabatic

process,Relation between T & V, and T & p :,Heat capacity ,Relation between Cp & Cv :,Key words,Summary ,Hints and Answers to self assessment questions

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UNIT V

35-95

JOULE`S LAW, THERMOCHEMISTRY BOND ENERGY & SOLUTIONS

LEARNING

Experiment,The

OBJECTIVES,INTRODUCTION,Joule’s

Joule-

Thomson Experiment,THERMOCHEMISTRY,Factors Affecting Heat or Enthalpy Change,Standard

Enthalpy

of

Reaction,Types

of

Heat

(enthalpy)

of

Reactions,Heat of Formation or Enthalpy of Formation,Measurement of Enthalpy of Reactions,Determination of Enthalpy Change ,Enthalpy of Combustion,Heat or Enthalpy of Neutralization,Neutralization of weak acids and weak bases,Enthalpy of solution (ΔHsol),Enthalpy of fusion (ΔHfus),Laws of Thermochemistry,Hess's Law

of

Constant

reactions,Bond

Heat

Energy

Summation,Determination or

Bond

of

heats

Enthalpy,SOLUTIONS

of

various

,Types

of

Solutions,Solubility of Gases,Nature of the gas and the liquid,VAPOUR PRESSURE OF A SOLUTION,Vapor Pressure and Raoult's Solution,Solutions

with

Negative

Deviation,Solutions

with

Law,Ideal Positive

Deviation,AZEOTROPIC MIXTURES,COLLIGATIVE PROPERTIES,Elevation of Boiling

Point,Colligative

Properties

-

Osmotic

Pressure,Key

words,Summary,Hints and Answers to self assessment questions.

UNIT QUESTIONS-

96-97

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APPLIED THERMODYNAMICS

1

UNIT I DEFINITION AND EXPLANATION OF TERMS Structure 1.0

LEARNING OBJECTIVES

1.1

INTRODUCTION

1.2

Terms in Thermodynamics

1.2.1 SYSTEM 1.2.2 SURROUNDINGS 1.2.3 ISOLATED SYSTEM 1.2.4 CLOSED SYSTEM 1.2.5 OPEN SYSTEM 1.3

Homogeneous and Heterogeneous System

1.3.1 MACROSCOPIC PROPERTIES 1.3.1.1

State of a System

1.3.1.2

State Variables

1.4

Thermodynamics Equilibrium

1.5

Extensive and Intensive Properties

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APPLIED THERMODYNAMICS 1.0

LEARNING OBJECTIVES

At the conclusion of this course, you will be able to: 

Define the terms in Thermodynamics;



Define the thermodynamics properties;



Describe the type of thermodynamic systems, isolated, open and closed systems;



Distinguish between extensive and intensive properties;



Describe the thermodynamic processes, cyclic, reversible, non-reversible.

1.1

INTRODUCTION

Thermodynamics: The term thermodynamics implies flow of heat and it deals with energy Changes accompanying all types of physical and chemical processes. The Concepts of heat and work and the laws of thermodynamics have been Developed based on human experience. Many of the original formulations of Thermodynamics were based on observations of matter in macroscopic Aggregates and as such were not concerned at all with the atomic or molecular Structure of matter.

Thermodynamics is based on four generalizations that go under the names, Zeroth, First, Second and Third laws of thermodynamics. The branch of Thermodynamics that is concerned with the macroscopic properties of matter comes under classical thermodynamics while the one that concerns with the Atomistic nature of matter is considered under statistical thermodynamics.

The classical thermodynamics is also called equilibrium or reversible thermodynamics.

1.2

Terms in Thermodynamics

1.2.1

SYSTEM

A system refers to that portion of the matter under investigation. It may be simple or complex and may be made up of sub-systems. A system may consist of one or more sub-systems.

1.2.2

SURROUNDINGS

All the matter which is capable of interacting with the system is referred to as surroundings. The choice of system and surroundings is very arbitrary in the sense that the surroundings for one system by one individual may not be viewed as surroundings by another individual. In simple cases, air or a water-bath in which a system is investigated can be termed as its surroundings. A system may be open, closed or isolated.

1.2.3

ISOLATED SYSTEM

If the system is such that it does not interact with the surroundings, then it is considered to be isolated. (Or) A system, which can neither exchange mass nor energy with the surroundings is

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APPLIED THERMODYNAMICS called an isolated system. For e.g., a reaction carried out in a closed, well insulated container (such as thermos flask).

1.2.4

CLOSED SYSTEM

If matter cannot be added to or removed from a system, it is said to be a closed one. (Or) A system, which can exchange energy with the surroundings but not mass is called a closed system. For e.g., calcination of CaCO3 taken in sealed bulb. On heating, CaCO3 decomposes into CaO and CO2, however, CO2 cannot escape and remains trapped in the bulb.

1.2.5

OPEN SYSTEM

If matter can be added to or removed from a system, it is termed as an open system. (Or) A system, which can exchange matter as well as energy with the surroundings, is called an open system. For e.g., hot coffee in an open flask because it can gain or loose matter and energy with the surroundings S.A.Q - 1 Explain the terms a) System b) State variables _____________________________________________________________________________ _____________________________________________________________________________ ______________________________________________________________

1.3

Homogeneous and Heterogeneous System A system is homogeneous or Heterogeneous depending on whether it is a uniphase or a

multi-phase system.

Thus a system containing two or more immiscible liquids or a solid in

contact with a liquid in which it is not soluble, is a heterogeneous system. A liquid in contact with its vapour, being bi-phasic is also a heterogeneous system. On the other hand, if the whole system is uniform throughout (uni-phase), it is a homogeneous system.

A system is separated from its surroundings by a boundary or wall, real or imaginary, rigid or nonrigid, permeable or impermeable, a conductor of heat or an insulator. For instance, in the case of a glass tumbler containing a liquid immersed in a thermostat, the liquid is the system; the walls of the container serve as the boundary and the thermostat is the surroundings. If the boundary happens to conduct heat, it is called a diathermic boundary; if it is an insulator as in the case of a Dewar flask or a thermos flask, it is called an adiabatic boundary.

1.3.1

MACROSCOPIC PROPERTIES

These are the properties associated with a macroscopic system. Pressure, volume, temperature, mass, composition, density, viscosity, surface tension, refractive index, specific rotation, colour etc. fall under this category.

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1.3.1.1 State of a System The state of a system is decided by its macroscopic properties. i.e. it is said to be in a definite state if the macroscopic properties of the system have definite values; if there is a change in any one of the macroscopic properties, the system is said to pass into a different state.

In order to understand it, let H2O be considered as a chemical system. H 2O can exist in three physical states; ice, water and steam depending upon the conditions of temperature and o

pressure. For e.g., at 1 atmospheric pressure, H2O is: o

o

Liquid, if temperature is between 0 C 100 C.

Solid, if temperature is below 0 C. o

Gas, if temperature is above 100 C.

Thus, in

order to define the physical state of H2O we need the specification of its conditions of temperature o

o

and pressure. Consider H2O(l) at two temperatures, 25 C and 50 C respectively, both at one atmospheric pressure. Although physical state of the two samples is same, yet their macroscopic properties like, energy, viscosity, surface tension, refractive index, etc., are different. In other words, the thermodynamic state of two samples is different. Thus, state of the system may be defined as its conditions of existence of the system when its macroscopic properties have definite values. When the state of a system changes its macroscopic properties also change and acquire new definite values. The macroscopic properties are properties which arise out of collective behavior of large number of chemical entities.

The values of state functions depend only upon the state of the system but do not depend upon how that state is achieved. For e.g., consider one mole of CO2 at N.T.P., its volume will be 22.4 liters irrespective of the method through which the given sample of CO 2 is obtained.

Similarly, the values of other state variables are also fixed for this particular system at N.T.P. Some common state functions are volume, pressure, temperature, internal energy, enthalpy, entropy and free energy.

1.3.1.2 State Variables As the state of the system varies with change in any one of the macroscopic properties, these are called state variables. It, therefore, follows that when a system changes from one state (initial state) to another state (final state), there is a change in one or more of the macroscopic properties.

Pressure, volume, temperature, mass and composition are among the important

variables. In actual practice, it is not necessary to specify all these variables as some of them are inter-dependent. For a single gas, composition is not one of the variables as it reaming 100% pure. For an ideal gas obeying the relation PV = RT, if only two of the variables among P, V and T are known, the third one can be deducted.

The two variables generally specified viz. P and T, are called independent variables. The third variable, V is a dependent variable as its value depends on P and T. Mass is not a state variable in a closed system of one or more components.

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APPLIED THERMODYNAMICS 1.4

Thermodynamics Equilibrium

A system is said to be in thermodynamic equilibrium if the macroscopic properties of the system do not change with time. Thermodynamic equilibrium implies thermal, chemical, and mechanical equilibria. A system is said to be in thermal equilibrium if there is no flow of heat from one portion of the system to another. This is feasible if the temperature remains constant throughout the system. A system is said to be in chemical equilibrium if the composition of the various phases in the system remains the same throughout. A system is in mechanical equilibrium if no mechanical work is done by one part of the system on any other part of the same. This is possible only if the pressure remains the same throughout the system.

Control Volume A control volume is a fixed region in space chosen for the thermodynamic study of mass and energy balances for flowing systems. The boundary of the control volume may be a real or imaginary envelope. The control surface is the boundary of the control volume.

Steady State Steady state is that circumstance in which there is no accumulation of mass or energy within the control volume, and the properties at any point within the system are independent of time.

1.5

Extensive and Intensive Properties

Any property (mass, volume or energy) that depends upon the amount of substance(s) present in a system is known as extensive property. On the other hand, if a property is independent of the amount of substance(s) present in the system, then it is referred to as intensive property.

Pressure, temperature, viscosity, surface tension, specific heat refractive index, density, concentration etc. are intensive properties.

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APPLIED THERMODYNAMICS

6 UNIT II

THERMODYNAMICS PROCESS

2.0

THERMODYNAMIC PROCESS

2.1

Cyclic Process

2.1.1 Reversible Process 2.1.2 Irreversible Process 2.1.3 Adiabatic Process 2.1.4 Isentropic Process 2.1.5 Polytropic Process 2.1.6 Throttling Process 2.2

SPONTANEOUS PROCESS

2.3

Key words

2.4

Summary

2,5

Hints and Answers to self assessment questions

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APPLIED THERMODYNAMICS 2.0

THERMODYNAMIC PROCESS

Whenever one or more of the properties of a system change, a change in the state of the system occurs. The path of the succession of states through which the system passes is called the thermodynamic process. One example of a thermodynamic process is increasing the temperature of a fluid while maintaining a constant pressure. Another example is increasing the pressure of a confined gas while maintaining a constant temperature.

2.1

Cyclic Process

When a system in a given initial state goes through a number of different changes in state (going through various processes) and finally returns to its initial values, the system has undergone a cyclic process or cycle. Therefore, at the conclusion of a cycle, all the properties have the same value they had at the beginning. Steam (water) that circulates through a closed cooling loop undergoes a cycle.

2.1.1

Reversible Process

A reversible process for a system is defined as a process that, once having taken place, can be reversed, and in so doing leaves no change in either the system or surroundings. In other words the system and surroundings are returned to their original condition before the process took place. In reality, there are no truly reversible processes; however, for analysis purposes, one uses reversible to make the analysis simpler, and to determine maximum theoretical efficiencies.

Therefore, the reversible process is an appropriate starting point on which to base engineering study and calculation. Although the reversible process can be approximated, it can never be matched by real processes. One way to make real processes approximate reversible process is to carry out the process in a series of small or infinitesimal steps. For example, heat transfer may be considered reversible if it occurs due to a small temperature difference between the system and its surroundings. For example, transferring heat across a temperature difference of 0.00001 째F "appears" to be more reversible than for transferring heat across a temperature difference of 100 째F. Therefore, by cooling or heating the system in a number of infinitesamally small steps, we can approximate a reversible process. Although not practical for real processes, this method is beneficial for thermodynamic studies since the rate at which processes occur is not important.

2.1.2

Irreversible Process

An irreversible process is a process that cannot return both the system and the surroundings to their original conditions. That is, the system and the surroundings would not return to their original conditions if the process was reversed. For example, an automobile engine does not give back the fuel it took to drive up a hill as it coasts back down the hill.

There are many factors that make a process irreversible. Four of the most common causes of irreversibility are friction, unrestrained expansion of a fluid, heat transfer through a finite

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APPLIED THERMODYNAMICS temperature difference, and mixing of two different substances. These factors are present in real, irreversible processes and prevent these processes from being reversible.

2.1.3

Adiabatic Process

An adiabatic process is one in which there is no heat transfer into or out of the system. The system can be considered to be perfectly insulated.

2.1.4

Isentropic Process

An isentropic process is one in which the entropy of the fluid remains constant. This will be true if the process the system goes through is reversible and adiabatic. An isentropic process can also be called a constant entropy process.

2.1.5

Polytropic Process

When a gas undergoes a reversible process in which there is heat transfer, the process frequently takes place in such a manner that a plot of the Log P (pressure) vs. Log V (volume) is a straight line. Or stated in equation form PVn = a constant. This type of process is called a polytropic process. An example of a polytropic process is the expansion of the combustion gasses in the cylinder of a water-cooled reciprocating engine.

2.1.6

Throttling Process

A throttling process is defined as a process in which there is no change in enthalpy from state one to state two, h1 = h2; no work is done, W = 0; and the process is adiabatic, Q = 0. To better understand the theory of the ideal throttling process let’s compare what we can observe with the above theoretical assumptions.

An example of a throttling process is an ideal gas flowing through a valve in mid position. From experience we can observe that: Pin > Pout, velin < velout (where P = pressure and vel = velocity).

These observations confirm the theory that hin = hout. Remember h = u + Pv (v = specific volume), so if pressure decreases then specific volume must increase if enthalpy is to remain constant (assuming u is constant). Because mass flow is constant, the change in specific volume is observed as an increase in gas velocity, and this is verified by our observations.

The theory also states W = 0. Our observations again confirm this to be true as clearly no "work" has been done by the throttling process. Finally, the theory states that an ideal throttling process is adiabatic. This cannot clearly be proven by observation since a "real" throttling process is not ideal and will have some heat transfer.

S.A.Q.- 2 .Which of the following is not a state function? Why? a) T b) H c) q d) S

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APPLIED THERMODYNAMICS

9

_____________________________________________________________________________ _____________________________________________________________________________ ______________________________________________________________ 2.2

SPONTANEOUS PROCESS

In our daily life we come across many processes which take place of their own under particular set of conditions. Some common examples are : Flow of heat from a hot body to a cold body Melting of ice at room temperature Evaporation of water Intermixing of gases All these processes take place in one direction only. They cannot be reversed automatically. For example, heat cannot flow from a cold body to a hot body of its own. Similarly, separation of gases from a mixture or formation of ice from water at room temperature cannot occur of their own. Such processes are known as spontaneous processes.

Thus, a spontaneous process may be defined as a process which has a natural tendency to occur in a particular direction either of its own or after proper initiation under a given set of conditions.

Spontaneous process does not mean that it takes place instantaneously. Spontaneous process may be extremely slow or very fast. Spontaneous processes are also known as feasible or probable processes.

Some of the familiar examples of spontaneous processes are listed below:

(a) Spontaneous processes that need no initiation. (i) Dissolution of sugar in water to form a solution. (ii) Reaction between nitric oxide (NO) gas with oxygen (O 2) to form nitrogen dioxide.

2NO (g) + O2 (g)

2NO2 (g).

(b) spontaneous processes that need initiation. (i) Hydrogen reacts with oxygen gas to produce water. This reaction is initiated by striking electric spark.

2H2 (g) O2(g)

2H2O (l)

(ii) Coal burns to give CO2 when ignited. Once burning of coal is `initiated, it continues of its own.

C (s) + O2 (g)

CO2 (g)

A process which has no natural tendency or urge to occur is said to be a non-spontaneous process.

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Non-spontaneous processes or changes may be carried out when energy from some external source is supplied continuously throughout the change. For example, electrolysis of water. The electrolysis of water takes place only so as long as electrical energy is supplied to water. Once the electric current is stopped, the eletrolysis of water also stops.

H2O (l)

H2 (g) + ½ O2 (g)

(non-spontaneous)

S.A.Q.-3 What do you understand by the term - Internal Energy? _____________________________________________________________________________ _____________________________________________________________________________ ______________________________________________________________ 2.3

Key words

Cyclic Process - a series of processes that results in the system returning to its original state Reversible process - a process that can be reversed resulting in no change in the system or surroundings.

Irreversible process - a process that, if reversed, would result in a change to the system or surroundings.

Adiabatic process - a process in which there is no heat transfer across the system boundaries.

Isentropic process - a process in which the entropy of the system remains unchanged.

Polytropic process - the plot of Log P vs. Log V is a straight line, PVn = constant.

Throttling process - a process in which enthalpy is constant h1 = h2, work = 0, and which is adiabatic, Q=0.

2.4

Summary

Defining an appropriate system can greatly simplify a thermodynamic analysis. A thermodynamic system is any three-dimensional region of space that is bounded by one or more surfaces. The bounding surfaces may be real or imaginary and may be at rest or in motion. The boundary may change its size or shape. The region of physical space that lies outside the selected boundaries of the system is called the surroundings or the environment.

The important information from this chapter is summarized below. 

A thermodynamic system is a collection of matter and space with its boundaries defined in such a way that the energy transfer across the boundaries can be best understood.



Surroundings are everything not in the system being studied.



Systems are classified into one of three groups:

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Isolated system - neither mass nor energy can cross the boundaries Closed system - only energy can cross the boundaries Open system - both mass and energy can cross the boundaries 

A control volume is a fixed region of space that is studied as a thermodynamic system.



Steady state refers to a condition where the properties at any given point within the system are constant over time. Neither mass nor energy is accumulating within the system.



A thermodynamic process is the succession of states that a system passes through.

Processes can be described by any of the following terms:

Cyclic Process, Irreversible process, Adiabatic process, Isentropic process , Polytropic process , Throttling process.

2,5

Hints and Answers to self assessment questions

1)

a). A system is defined as the part of the universe which is under

investigation.

b). The macroscopic properties of the system which change with the change in the state of a system are called state variable or state functions.

2)

(c) q.

3)

A fixed quantity of any substance is associated with a definite amount of energy

depends upon chemical nature of the substance and its state

which

of existence.

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12

UNIT III THERMODYNAMIC FUNCTIONS Structure

3.0

Learning Objectives

3.1

Introduction

3.2

WORK, HEAT, AND ENERGY

3.3

Modes of Transference of Energy

3.4

Zeroth law of thermodynamics

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APPLIED THERMODYNAMICS 3.0

LEARNING OBJECTIVES

At the conclusion of this course, you will be able to: 

Define the terms in Thermodynamics;



Define the thermodynamics properties;



Describe the type of thermodynamic systems, isolated, open and closed systems;



Distinguish between extensive and intensive properties;



Describe the thermodynamic processes, cyclic, reversible, non-reversible.

3.1

INTRODUCTION

For the purposes of physical chemistry, the universe is divided into two parts, the system and its surroundings. The system is the part of the world in which we have a special interest. It may be a reaction vessel, an engine, an electrochemical cell, a biological cell, and so on. The surroundings comprise the region outside the system and are where we make our measurements. The type of system depends on the characteristics of the boundary that divides it from the surroundings. If matter can be transferred through the boundary between the system and its surroundings the system is classified as open. If matter cannot pass through the boundary the system is classified as dosed. Both open and closed systems can exchange energy with their surroundings. For example, a closed system can expand and thereby raise a weight in the surroundings; it may also transfer energy to them if they are at a lower temperature.

An isolated system is a closed system that has neither mechanical nor thermal contact with its surroundings.

3.2

WORK, HEAT, AND ENERGY

The fundamental physical property in thermodynamics is work: work is motion against an opposing force. Doing work is equivalent to raising a weight somewhere in the surroundings. An example of doing work is the expansion of a gas that pushes out a piston and raises a weight. A chemical reaction that drives an electric current through a resistance also does work, because the same current could be driven through a motor and used to raise a weight.

The energy of a system is its capacity to do work. When work is done on an otherwise isolated system (for instance, by compressing a gas or winding a spring), the capacity of the system to do work is increased; in other words, the energy of the system is increased. When the system does work (when the piston moves out or the spring unwinds), the energy of the system is reduced and it can do less work than before.

Experiments have shown that the energy of a system may be changed by means other than work itself. When the energy of a system changes as a result of a temperature difference between the system and its surroundings we say that energy has been transferred as heat. When a heater is immersed in a beaker of water (the system), the capacity of the system to do work increases

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APPLIED THERMODYNAMICS because hot water can be used to do more work than the same amount of cold water. Not all boundaries permit the transfer of energy even though there is a temperature difference between the system and its surroundings.

An exothermic process is a process that releases energy as heat into its surroundings. All combustion reactions are exothermic. An endothermic process is a process in which energy is acquired from its surroundings as heat. An example of an endothermic process is the vaporization of water. To avoid a lot of awkward circumlocution, we say that in an exothermic process energy is transferred 'as heat' to the surroundings and in an endothermic process energy is transferred 'as heat' from the surroundings into the system. However, it must never be forgotten that heat is a process (the transfer of energy as a result of a temperature difference), not an entity. An endothermic process in a diathermic container results in energy flowing into the system as heat. An exothermic process in a similar diathermic container results in a release of energy as heat into the surroundings. When an endothermic process takes place in an adiabatic container, it results in a lowering of temperature of the system; an exothermic process results in a rise of temperature.

3.3

Modes of Transference of Energy

Chemical reactions are accompanied by the energy changes. The exchange of energy between the system and surroundings can occur in several ways. Two of the important modes of transference of energy are heat and work which have been described briefly as follows

Heat The transference of energy takes place as heat if the system (reaction mixture) and surroundings are at different temperatures. If the system is at higher temperature, the energy is lost to the surrounding as heat, causing a fall in the temperature of the system and a rise in the temperature of the surroundings. The energy transfer goes on until the system and the surroundings attain the same temperature.

If the system is at a lower temperature than the surroundings, the energy is gained by the system from the surroundings causing a rise in the temperature of the system. The amount of heat gained or lost by the system is represented by "q". According to the international conventions, a) heat absorbed by the system is positive, b) heat given out by the system is negative.

Work The transference of energy may take place as work if the system and surroundings have different pressures. For e.g., suppose a gaseous system is enclosed in a cylinder fitted with a movable piston. If pressure of the system is higher, the piston will be pushed up until the pressure of the

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APPLIED THERMODYNAMICS system becomes equal to the pressure of the surroundings. During expansion, work is done by the system on the surroundings. In this way energy is transferred to the surroundings as work. If the system is at lower pressure, piston will be pushed down until the pressure of the system becomes equal to that of surroundings. In this case work is done by the surroundings on the system and energy transfer takes place from the surroundings to the system as work. The amount of work done, by the system or on the system is denoted by "w". According to international conventions, i) work done on the system is positive, ii) work done by the system is negative.

Besides heat and work electrical energy and radiation energy are the other modes of energy exchange between system and surroundings. For e.g., (i) Use of electrical energy can cause chemical decomposition of water.

(ii) Burning of magnesium ribbon causes liberation of energy in the form of heat as well as light.

S.A.Q 1: What will be the sign of a system if a) heat is absorbed by the system b) heat given out by the system? _____________________________________________________________________________ ___________________________________________________________________ 3.4

Zeroth law of thermodynamics

The zeroth law of thermodynamics is a generalization about the thermal equilibrium among bodies, or thermodynamic systems, in contact. It results from the definition and properties of temperature. It can be stated as: "If A and C are each in thermal equilibrium with B, A is also in thermal equilibrium with C."

Zeroth law as equivalence relation A system is said to be in thermal equilibrium when its temperature does not change over time. Let A, B, and C be distinct thermodynamic systems or bodies. The zeroth law of thermodynamics can then be expressed as: "If A and C are each in thermal equilibrium with B, A is also in thermal equilibrium with C." The preceding sentence asserts that thermal equilibrium is a Euclidean relation between thermodynamic systems. If we also grant that all thermodynamic systems are (trivially) in thermal equilibrium with themselves, then thermal equilibrium is also a reflexive relation. Relations that are both reflexive and Euclidean are equivalence relations. One consequence of this reasoning is that thermal equilibrium is a transitive relation between the temperature T of A, B, and C: if T(A) = T(B)

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APPLIED THERMODYNAMICS

16

and T(B) = T(C) then T(A) = T(C).

S.A.Q 2: What will be the sign of a system if i) work done on the system and ii) work done by the system is negative. _____________________________________________________________________________ ___________________________________________________________________ Thermal equilibrium between many systems Many systems are said to be in equilibrium if the small exchanges (due to Brownian motion, for example) between them do not lead to a net change in the total energy summed over all systems.

A simple example illustrates why the zeroth law is necessary to complete the equilibrium description.

Consider N systems in adiabatic isolation from the rest of the universe (i.e no heat exchange is possible outside of these N systems), all of which have a constant volume and composition, and which can only exchange heat with one another.

The combined First and Second Laws relate the fluctuations in total energy, δU, to the temperature of the i th system,

and the entropy fluctuation in the ith system,

as follows:

.

The adiabatic isolation of the system from the remaining universe requires that the total sum of the entropy fluctuations vanishes, or:

That is, entropy can only be exchanged between the N systems. This constraint can be used to rearrange the expression for the total energy fluctuation and obtain:

where

is the temperature of any system j we may choose to single out among the N systems.

Finally, equilibrium requires the total fluctuation in energy to vanish, in which case:

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17

which can be thought of as the vanishing of the product of an antisymmetric matrix a vector of entropy fluctuations

and

. In order for a non-trivial solution to exist,

That is, the determinant of the matrix formed by

must vanish for all choices of N.

However, according to Jacobi's theorem, the determinant of a NxN antisymmetric matrix is always zero if N is odd, although for N even we find that all of the entries must vanish, , in order to obtain a vanishing determinant. Hence

at equilibrium. This

non-intuitive result means that an odd number of systems are always in equilibrium regardless of their temperatures and entropy fluctuations, while equality of temperatures is only required between an even number of systems to achieve equilibrium in the presence of entropy fluctuations.

The zeroth law solves this odd vs. even paradox, because it can readily be used to reduce an odd-numbered system to an even number by considering any three of the N systems and eliminating one by application of its principle, and hence reduce the problem to even N which subsequently leads to the same equilibrium condition that we expect in every case, i.e., . The same result applies to fluctuations in any extensive quantity, such as volume (yielding the equal pressure condition), or fluctuations in mass (leading to equality of chemical potentials). Hence the zeroth law has implications for a great deal more than temperature alone.

Temperature and the zeroth law In the space of thermodynamic parameters, zones of constant temperature will form a surface, which provides a natural order of nearby surfaces. It is then simple to construct a global temperature function that provides a continuous ordering of states. Note that the dimensionality of a surface of constant temperature is one less than the number of thermodynamic parameters (thus, for an ideal gas described with 3 thermodynamic parameters P, V and n, it is a 2 dimensional surface). The temperature so defined may indeed not look like the Celsius temperature scale, but it is a temperature function nonetheless.

For example, if two systems of ideal gas are in equilibrium, then P1V1/N1 = P2V2/N2 where Pi is the pressure in the ith system, Vi is the volume, and Ni is the "amount" (in moles, or simply the number of atoms) of gas.

The surface PV/N = const defines surfaces of equal temperature, and the obvious (but not only) way to label them is to define T so that PV/N = RT, where R is some constant. These systems can now be used as a thermometer to calibrate other systems.

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UNIT IV FIRST LAW OF THERMODYNAMICS 4.0

Internal Energy and Change in Internal Energy

4.1

Law of Conservation of Energy (First law of thermodynamics)

4.2

Significance of

4.3 4.4

Thermodynamic Processes for an Ideal Gas

4.5

Adiabatic process

4.6

Relation between T & V, and T & p :

4.7

Heat capacity

4.8

Relation between Cp & Cv :

4.9

Key words

4.10

Summary

4.11

Hints and Answers to self assessment questions

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19

Internal Energy and Change in Internal Energy

Since chemical changes are always accompanied by energy changes, it indicates that reactants and products must be having certain amounts of energy. A fixed quantity of any substance is associated with a definite amount of energy, which depends upon chemical nature of the substance and its state of existence. This energy is called internal energy (E) or intrinsic energy of the substance. Internal energy is a state function, which means its value depends upon state of the substance but does not depend upon how that state is achieved.

For e.g., CO2 can be obtained by various methods such as by heating calcium carbonate or by burning coal. However, one mole of CO2 at N.T.P. is associated with a definite amount of internal energy which does not depend upon the source from which it is obtained.

Besides the state of

the system, the value of internal energy of the system also depends upon: (i) the chemical nature of substance, (ii) amount of the substance and (iii) the condition of temperature and pressure.

The various energies which contribute towards internal energy are: (a) Translational energy of the molecules (Et) (b) Rotational energy of the molecules (Er) (c) Vibrational energy of the molecules (Ev) (d) Electronic energy (Ed) (e) Nuclear energy (En) (f) Interaction energy of molecules (Ej).

It may be noted that the absolute value of internal energy cannot be determined because it is not possible to determine the exact values for the constituent energies such as translational, vibrational, rotational energies etc. However, changes in internal energy, which occur during

difference between the internal energies of the product and the reactants. (products)

- E(reactants) = Ep - ER

where Ep is the internal energy of the products, E R is the internal energy of the reactants. If the internal energy of the products is less than the internal en negative. On the other hand, when the internal energy of the products is more than the internal

such a way that the volume and temperature of the reacting system do not change, the internal

me.

the heat change occurring during the reaction at constant volume and temperature.

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20

Law of Conservation of Energy (First law of thermodynamics)

It is a common experience that energy cannot be generated without consuming energy of some other kind. If a certain amount of one kind of energy is produced, an equal amount of some other kind of energy disappears so that the total energy of the Universe remains constant. This observation forms the basis of law of conservation of energy (also known as First law of thermodynamics), which states that Energy can neither be created nor destroyed although it may be changed from one form to another.

From this statement it follows that total energy of the universe, For e.g., system and surroundings taken together as always constant during any physical or chemical process. In order to derive mathematical expression for first law of thermodynamics let us assume that a system having internal energy El absorbs a certain amount of heat energy (q).

1

Let w amount of work be done on it, so that its internal energy changes to E2 E2 = E1 + q + w E2 - E1 = q + w (Change in internal energy) = (Heat absorbed) + (Work done on the system) Let the change in volu referred to as pressure-volume work and its expression is given as Thus w = Now if there is only pressure-volume work the expression can be written as This is the mathematical form of first law of thermodynamics.

4.2 From first law of thermodynamics we have equation as -

Hence, qv occurring at constant volume and constant temperature.

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APPLIED THERMODYNAMICS 4.3 It has been seen that energy change occurring during the reaction at constant temperature and constant volume is given by internal energy change. However, most of the reactions in the laboratory are carried out in open beakers or test tubes, etc. In such cases, the reacting system is open to atmosphere. Since atmospheric pressure is almost constant, therefore, such reactions may involve the changes in volume. The energy change occurring during such reactions may not be equal to the internal energy change. In order to understand this, let us assume a chemical reaction involving gaseous substances which proceeds with the evolution of heat. When the reaction is carried out at constant pressure, two possibilities arise. (a) If the reaction proceeds with the increase in volume the system has to expand against the atmospheric pressure and energy is required for this purpose. The heat evolved in this case would be little less than the heat evolved at constant volume because a part of the energy has to be utilized. (b) If the reaction proceeds with decrease in volume at constant pressure, the work is done on the system and heat evolved will be greater than the heat evolved at constant volume.

Thus, it can be concluded from the above discussion that heat changes occurring at constant pressure and constant temperature are not simply due to the changes in internal energy alone but also include energy changes due to expansion or contraction against the atmospheric pressure.

In order to study the heat changes of chemical reactions at constant temperature and pressure a new function, enthalpy is introduced. Enthalpy is the total energy associated with any system which includes its internal energy and also energy due to environmental factors such as pressurevolume conditions. This can be understood as follows: A substance has to occupy a space in its surroundings depending upon its volume (V). It does so against the compressing influence of the pressure (P). Due to this, the substance possesses an additional energy called PV energy. . The sum of internal energy and PV energy of any system, under a particular set of conditions, is called enthalpy. It is denoted by H. Mathematically, it may be put as H = E + PV Some important features of enthalpy are: (i) It is a state function. (ii) It is also called heat content of the system (iii) Its value depends upon amount of the substance, chemical nature of the substance and conditions of temperature and pressure.

It is not possible to determine the absolute value of enthalpy of a system because absolute value

process can be determined. Change in enthalpy is equal to difference between the enthalpies of products and reactants. The change in enthalpy may be expressed as

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22

The enthalpy change of a reaction is equal to the heat absorbed or evolved during a reaction at constant temperature and constant pressure.

o follows from the first law of thermodynamics. = (E2 - E1) + P(V2 - V1) = E2 + PV2 - (E1 + PV1) = H2 - H1 The change in e constant pressure in a well-insulated calorimeter. The heat evolved or absorbed changes the temperature of the system. From the change in temperature, the heat evolved or absorbed during

In short:

The first law of thermodynamics is the application of the conservation of energy principle to heat and thermodynamic processes: -w The change in internal energy of a system is equal to the heat added to the system minus the work done by the system.

The first law makes use of the key concepts of internal energy, heat, and system work. It is used extensively in the discussion of heat engines. The standard unit for all these quantities would be the joule, although they are sometimes expressed in calories or BTU. It is typical for chemistry texts to write the first law as Î&#x201D;U=Q+W. It is the same law, of course - the thermodynamic expression of the conservation of energy principle. It is just that W is defined as the work done on the system instead of work done by the system. In the context of physics, the common scenario is one of adding heat to a volume of gas and using the expansion of that gas to do work, as in the pushing down of a piston in an internal combustion engine. In the context of chemical reactions and process, it may be more common to deal with situations where work is done on the system rather than by it.

System Work When work is done by a thermodynamic system, it is usually a gas that is doing the work. The work done by a gas at constant pressure is:

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For non-constant pressure, the work can be visualized as the area under the pressure-volume curve which represents the process taking place. The more general expression for work done is:

Work done by a system decreases the internal energy of the system, as indicated in the First Law of Thermodynamics. System work is a major focus in the discussion of heat engines.

4.4

Thermodynamic Processes for an Ideal Gas

Isothermal Process: Let us first consider the expansion and compression of an ideal gas from an initial volume V 1 to a final volume V2 under constant-temperature (isothermal) conditions. From the definition of pressure-volume work, we have:

If we assume the process occurs reversibly, then Pext = P where P is the pressure of the gas; therefore

Now let us use the equation of state for an ideal gas, which is PV = nRT where n is the number of moles of gas, R is the ideal gas constant, and T is the absolute (Kelvin) temperature. Substituting for P, we get

If we have a closed system, then n is a constant; moreover, since we have stipulated that this process occurs under isothermal conditions, T is also a constant. Therefore, the integral simplifies to:

This integral may be solved to yield

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Next we consider the change in the internal energy of the ideal gas upon isothermal expansion/compression. It is helpful to review the two key assumptions that underlie the ideal gas model: 1. Ideal gas particles are point particles.

2. Ideal gas particles have no interparticular forces.

From these two statements, we now contemplate what would happen to the internal energy (U) of an ideal gas upon isothermal expansion or compression. Because ideal gas particles are point particles, if the gas sample is compressed, the particles will not experience any change in energy because they have no internal volume to get "squeezed". Similarly, because they have no interparticular forces, they do not experience any change in attraction or repulsion between different particles if they are squeezed closer together. Hence, we may (correctly) reason that for an isothermal process on an ideal gas, the change in internal energy is zero. (isothermal process, ideal gas) (This same result may be rigorously proved using the thermodynamic master equations, or by using statistical thermodynamics.) Note that because U is a state function, Î&#x201D;U will always be equal to zero for any isothermal process on an ideal gas, whether or not it occurs reversibly.

Because of the First Law of Thermodynamics,

and because Î&#x201D;U = 0, the energy change associated with heat upon isothermal reversible expansion or compression is just the negative of the work, or

S.A.Q 3: Explain First Law of Thermodynamics _____________________________________________________________________________ _____________________________________________________________________________ ______________________________________________________________

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25

Adiabatic process

An adiabatic process is the thermodynamic process in which there is no heat interaction during the process, i.e. during the process, Q = 0. In these processes the work interaction is there at the Îł

expense of internal energy. The adiabatic process follows the law pV = constant where Îł is called adiabatic index and is given by the ratio of two specific heats (Cp/Cv). Thus, it is obvious that adiabatic expansion shall be accompanied by the fall in temperature while temperature will rise during adiabatic compression. The adiabatic expansion process is shown on fig given below.

Adiabatic Process

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In an adiabatic expansion, the work done W 1-2 by the fluid is at the expense of a reduction in the internal energy of the fluid. Similarly in an adiabatic composition process, all the work done on the fluid goes to increase the internal energy of the fluid. Îł

To derive pV = C: For a reversible adiabatic process To obtain the law relating P & V for a reversible adiabatic process, let us consider the non flow energy equation in the differential from.

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4.6

Relation between T & V, and T & p :

In a similar fashion, other results may be derived for processes on ideal gases. A summary of results is given below.

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30 m

Equation Table (PV = constant) Constant Pressure Constant Volume

Isothermal

Adiabatic

or

or

Variable

Work

Heat Capacity, Internal Energy,

Enthalpy, Entropy

S.A.Q. 4: Define Enthalpy _____________________________________________________________________________ _____________________________________________________________________________ ______________________________________________________________ 4.7 Heat capacity The internal energy of a substance increases when its temperature is raised. The increase depends on the conditions under which the heating takes place and for the present we suppose that the sample is confined to a constant volume. For example, the sample may be a gas in a container of fixed volume. If the internal energy is plotted against temperature, then a curve like that in Fig. shown below may be obtained. The slope of the tangent to the curve at any temperature is called the heat capacity of the system at that temperature. The heat capacity at constant volume is denoted Cv and is defined formally as

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The internal energy of a system increases as the temperature is raised; this graph shows its variation as the system is heated at constant volume. The slope of the tangent to the curve at any temperature is the heat capacity at constant volume at that temperature. Note that, for the system illustrated, the heat capacity is greater at B than at A. Heat capacity C describes the amount of heat required to change the temperature of a substance:



C= -1

-1

-1

-1

By definition, the heat capacity of water at 15°C is 1 cal K g or 18 cal K mol (i.e., the heat required to heat 1 gram of water from 14.5 to 15.5°C is 1 calorie). Heat capacities of solids approach zero as absolute zero is approached:



C=0

The heat capacity is written with a subscript P or V depending on whether it obtains for constant pressure CP or constant volume CV.

In this case, the internal energy varies with the temperature and the volume of the sample, but we are interested only in its variation with the temperature, the volume being held constant.

Heat capacities are extensive properties: 100 g of water, for instance, has 100 times the heat capacity of 1 g of water (and therefore requires 100 times the energy as heat to bring about the same rise in temperature). The molar heat capacity at constant volume, C V,m = CV/n, is the heat capacity per mole of material, and is an intensive property (all molar quantities are intensive).

-1

-1

Typical values of CV,m for polyatomic gases are dose to 25 J K mol . For certain applications it is useful to know the specific heat capacity (more informally, the 'specific heat') of a substance, which is the heat capacity of the sample divided by the mass, usually in grams: C V,s = CV/m. The specific heat capacity of water at room temperature is close to 4 J K

-1

-1

g . In general, heat

capacities depend on the temperature and decrease at low temperatures. However, over small ranges of temperature at and above room temperature, the variation is quite small and for approximate calculations heat capacities can be treated as almost independent of temperature.

The heat capacity is used to relate a change in internal energy to a change in temperature of a constant-volume system.

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Relation between Cp & Cv :

Consider a perfect gas being heated at constant pressure from T1 to T2.

According to non flow equation Q1-2 = (U2 – U1) + W 1-2 ------ (1) Also for perfect gas, U2 – U1 = m Cv (T2 – T1) In constant pressure process, the work done by the fluid, W 1-2 = p (V2 – V1) = m R (T2 – T1) [ pV = mRT, and p1 = p2 = p in this case] Heat transferred in the constant pressure process is given as Q1-2 = m Cp (T2 – T1) By equating the three expressions in (1) m Cp (T2 – T1) = m Cv (T2 – T1) + m R (T2 – T1) m Cp (T2 – T1) = (T2 – T1) (m Cv + m R) m Cp = m(Cv + R) Therefore Cp = Cv + R or Cp – Cv = R Dividing both sides by Cv, we get

Similarly, dividing both sides by Cp, we get

S.A.Q.5: How the enthalpy changes of a chemical reaction is calculated? _____________________________________________________________________________ _____________________________________________________________________________ ______________________________________________________________ Specific Heat of a Pure Substance Specific heat at constant volume Specific heat at constant volume, cv of a pure substance is defined as the ratio of change of the specific internal energy of the substance with temperature, when the specific volume is held constant.

Symbolically it is written as,

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Specific heat at constant pressure Specific heat at constant pressure, Cp of a pure substance is defined as the ratio of change of the specific enthalpy of the substance with temperature, when the pressure of the substance is held constant.

Symbolically it is written as,

4.9

Key words

Heat: Heat is one way of transferring energy.

Work: Work is another way of transferring energy.

Mechanical work: The amount of work done on an object equals the force exerted on it times the distance it is moved in the direction of the force.

Reversible process: A reversible process is defined to be one that can be reversed in direction by an infinitesimal change in the surroundings Zeroth law of thermodynamics: "If A and C are each in thermal equilibrium with B, A is also in thermal equilibrium with C." Exact Differentials: The differential of a function of two or more independent variables is called an exact differential.

4.10

Summary

Thermodynamics deals with the study of the laws that govern the conversion of energy from one form to another, the direction in which heat will flow, and the availability of energy to do work. It is based on the concept that in an isolated system anywhere in the universe, there is a measurable quantity of energy called the internal energy (E) of the system. This is the total kinetic and potential energy of the atoms and molecules of the system of all kinds that can be transferred directly as heat. The value of E can only be changed, if the system ceases to be isolated. In these circumstances U can change by the transfer of mass to or from the system, the transfer of heat

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APPLIED THERMODYNAMICS (Q) to or from the system, or by work (W) being done on or by the system. For an adiabatic (Q=0) system of constant mass, â&#x2C6;&#x2020;E=W. By convention, W is taken to be positive if work s done on the system and negative if work is done by the system. For non-adiabatic systems of constant mass, â&#x2C6;&#x2020;E = Q + W. This statement, which is equivalent to the law of conservation of energy, is known as the first law of thermodynamics.

4.11

Hints and Answers to self assessment questions

S.A.Q 1: What will be the sign of a system if a) heat is absorbed by the system b) heat given out by the system? a)

Positive

b)

Negative

S.A.Q 2: What will be the sign of a system if i) work done on the system and ii) work done by the system is negative. a)

Positive

b)

Negative

S.A.Q 3: Explain First Law of Thermodynamics The first law of thermodynamics provides the means to calculate amounts of work and heat transferred in various processes, including adiabatic processes.

S.A.Q. 4: Define Enthalpy The enthalpy is a state function whose change in a constant-pressure process is equal to the amount of heat transferred to the system in the process.

S.A.Q.5: How the enthalpy changes of a chemical reaction be calculated?

The enthalpy change of a chemical reaction can be calculated from the enthalpy changes of formation of all products and reactants.

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UNIT V JOULE`S LAW, THERMOCHEMISTRY BOND ENERGY & SOLUTIONS Structure 5.0 5.1 5.1.1 5.1.2 6.0 6.1 6.2 7.0 7.1 7.2 8.0 8.1 8.2 8.3 8.4 8.5 9.0 9.1 9.2 10.0 11.0 11.1 11.2 11.3 12.0 12.1 12.2 12.3 12.4 12.5 12.6 12.7 12.8 13.0 14.0 15.0 16.0 17.0

LEARNING OBJECTIVES INTRODUCTION Joule’s Experiment The Joule-Thomson Experiment THERMOCHEMISTRY Factors Affecting Heat or Enthalpy Change Standard Enthalpy of Reaction Types of Heat (enthalpy) of Reactions Heat of Formation or Enthalpy of Formation Measurement of Enthalpy of Reactions Determination of Enthalpy Change Enthalpy of Combustion Heat or Enthalpy of Neutralization Neutralization of weak acids and weak bases Enthalpy of solution (ΔHsol) Enthalpy of fusion (ΔHfus) Laws of Thermochemistry Hess's Law of Constant Heat Summation Determination of heats of various reactions Bond Energy or Bond Enthalpy SOLUTIONS Types of Solutions Solubility of Gases Nature of the gas and the liquid VAPOUR PRESSURE OF A SOLUTION Vapor Pressure and Raoult's Law Ideal Solution Solutions with Negative Deviation Solutions with Positive Deviation AZEOTROPIC MIXTURES COLLIGATIVE PROPERTIES Elevation of Boiling Point Colligative Properties - Osmotic Pressure Key words Summary Hints and Answers to self assessment questions FAQ Further reading

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LEARNING OBJECTIVES

At the conclusion of this course, you will be able to: 

Define the terms in Thermodynamics;



Define the thermodynamics properties;



Describe the type of thermodynamic systems, isolated, open and closed systems;



Distinguish between extensive and intensive properties;



Describe the thermodynamic processes, cyclic, reversible, non-reversible.



Understand Thermochemistry and Bond energy



describe the formation of different types of solutions;



state and explain Henry’s law and Raoult’s law;



explain deviations of real solutions from Raoult’s law;



describe colligative properties of solutions and correlate these with molar masses of the solutes;



explain abnormal colligative properties exhibited by some solutes in solutions.

5.1

INTRODUCTION

The transfer of heat and the performance of work may both cause the same effect in a system.

Energy which enters a system as heat may leave the system as work, or energy which enters the system as work may leave as heat. Hence, by the law of conservation of energy, the net work done by the system is equal to the net heat supplied to the system. The first law of thermodynamics can therefore be stated as follows: “When a system undergoes a thermodynamic cyclic process, then the net heat supplied to the system from the surroundings is equal to the net work done by the system on its surrounding”.

Mathematically i.e.,

The first law of thermodynamics can not be proved analytically, but experimental evidence has repeatedly confirms its validity and since no phenomenon has been shown to contradict it, therefore the first law is accepted as a ‘law of nature’.

5.1.1

Joule’s Experiment

The English scientist J.P.Joule conducted a series of experiments in 1840’s, and these experiments led to the first law of thermodynamics. The objective of joules experiment was to establish a relation between amount of work spent to bring about the liberation of heat and the amount of heat liberated.

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37

Jouleâ&#x20AC;&#x2122;s Experiment The experimental set up is shown in Fig-(a). Work is done on the fluid (system) kept in an insulated vessel by stirring of the paddle wheel. This work input to the fluid causes a rise in the temperature of the fluid. The amount of work on the fluid is calculated by the product of the weight and the vertical height (z) through which the weight descends. The system is then immersed in a water bath, as shown in Fig-(b), after removing the insulation. The heat is transferred from the fluid to the water bath till the original temperature is reached, which will be indicated by the thermometer. Thus the system undergoes a complete cycle with two energy interactions i.e. definite amount of adiabatic work transfer(W1-2) to the system followed by an amount of heat transfer(Q1-2) from the system, shown in fig-(c). The amount of heat rejected by the fluid is equal to the increase of energy of the water bath.

S.A.Q. 1 What is thermochemistry? _____________________________________________________________________________ _____________________________________________________________________________ ______________________________________________________________

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APPLIED THERMODYNAMICS Fig-C: Cycle completed by a system with two energy interactions i.e., work transfer followed by heat transfer.

Joule carried out many such experiments with different type of work interactions in a variety of systems, he found that the net work input to the fluid system was always proportional to the net heat transferred from the system regardless of work interaction. Based on this experimental evidence Joule stated that, “When a system (closed system) is undergoing a cyclic process, the net heat transfer to the system is directly proportional to the net work done by the system”. This statement is referred to as the first law for a closed system undergoing a cyclic process.

Where J = Joule’s equivalent or Mechanical equivalent of heat = 1Nm/J If both heat transfer and work transfer are expressed in same units as in the S.I. units then the constant of proportionality in the above equation will be unity and hence the mathematical form of first law for a system undergoing a cyclic process can be written as

If the cycle involves many more heat and work quantities as shown in fig-(d), the same result will be found.

For this cyclic process the statement of first law can be written as

The cyclic integral in the above equation can be split into a series of non cyclic integral as

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This is the first law for a closed system undergoing a cyclic process. i.e., it is stated as “When a closed system is undergoing a cyclic process the algebraic sum of heat transfers is equal to the algebraic sum of the work transfers”.

Equivalence of Heat and Work Based on the results of a series of thoroughly conducted experiments J.P.Joule discovered a direct proportionality between the spent work ‘W’ and the quantity of heat obtained ‘Q’.

or Q = AW where ‘A’ is a Proportionality factor Joule found that the proportionality factor ‘A’ remains same irrespective of the method of heat production, type of work, temperature of the body involved etc.

In other words, Joule established that when one and same amount of work is spent, one and same amount of heat is liberated. Thus the amount of liberated heat was shown to be equivalent to the amount equivalent of work spent; it is clear that this relation ship is true for the case when work is accomplished with the expense of heat.

Using the results of these measurement, Joule calculated the magnitude of ‘A’ which is known as thermal equivalent of work and of ‘J’ referred to as the mechanical equivalent of heat.

A = 0.002342 kcal/kgm J = 426.99 kgm/kcal

5.1.2

It is obvious that J = 1/A

The Joule-Thomson Experiment

For ideal gas energy U depends only on T and n, so that H depends only on T and n: H = U + PV = U(T , n) + nRT = H(T , n) (ideal gas) Therefore,

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40

Joule and Thomson carried out experiments in the 1850s to determine the value of (∂H/∂P)T ,n for real gases.

The apparatus itself is quite simple. Imagine a tube with a porous plate separating it into two parts. The porous plate will allow a gas to go through it, but only slowly. It acts as a throttle. On each side of the plate there is a piston that fits the tube tightly. Each piston can (in principle) be pushed up against the porous plate. The tube itself is insulated so that no heat can enter or leave the tube.

The experiment is also quite simple. Imagine that some gas is placed between the porous plate and the piston on the left side of the tube. This is side 1. On the other side the piston is flush against the porous plate. This is side 2.

The initial volume of gas on side 1 is V1. The pressure is p1 and the temperature T1.

Now during the experiment gas is pushed through the porous plate by pushing on the piston on the left side. At the same time the piston on the right side is pulled in such a way that the pressure on the right side is always p2 At the end of the experiment all of the gas has been pushed through the porous plate. The volume on side 1 is zero. The final volume on the right side (side 2) is V 2, the pressure is p2 and the temperature T2.

The curious result of this experiment is that careful measurement shows that T2 is not equal to T1.

Under some conditions it is higher, under others it is lower.

Initial Analysis

The total work done is then: w = p1V1 − p2V2 (1) Since the process is adiabatic, the total change in energy U is just the work, or ΔU = U2 − U1 = p1V1 = p2V2 U2 + p2V2 = U1 + p1V1

(2) (3)

so that: H2 = H 1

(4)

and the process is isoenthalpic.

The Joule-Thomson Coefficient What is measured experimentally is

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(5) which becomes, taking into account that the process is isoenthalpic:

(6) All that we’ve done so far is reduced the experiment to mathematics. The experimental result is that μ JT is sometimes positive and sometimes negative. In fact it is found that there is a certain temperature called the inversion temperature such that if the initial temperature T 1 is above the inversion temperature, the final temperature is higher than the initial temperature. If the initial temperature is below the inversion temperature, the final temperature is lower than the initial temperature. The inversion temperature is found, experimentally, to depend on the pressure.

To understand this we need to relate μJT to quantities that are experimentally measureable.

To do this we can start with the total differential of H assuming that T and p are the independent variables:

(7) Now our process is isoenthalpic, so dH = 0. This gives:

(8) Now I divide through by dp at constant H to get:

(9) Note the subscript H at the appropriate point in eq. (9). Lastly, let us solve for (∂T/∂p)H to get:

(10) This result simplifies a bit right away. We know that (∂H/∂T)p is the heat capacity Cp, so:

(11)

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APPLIED THERMODYNAMICS but we can’t make it any simpler. We need to express (∂H/∂p)T in terms that can be measured.

Indeed, this can be done, but it requires one slight bit of magic.

(12) which, when used in eq. (11) gives:

a neat equation in which all of the quantities are measurable.

6.0 THERMOCHEMISTRY Thermochemical Equation Chemical reactions involve energy changes. These changes can be represented in a standard way called the thermochemical equation. A chemical equation, which gives the heat change (evolved or absorbed) during the reaction, is called the thermochemical equation.

This equation can be written either by writing the heat evolved or absorbed as a term in the equation i.e.

or, by using ΔH notation, i.e., writing ΔH = -ve for exothermic and ΔH = +ve for endothermic reactions.

Writing thermochemical equations The following are the important conventions for writing the thermochemical equations. 

For exothermic reaction, ΔH is -ve ; For endothermic reaction, ΔH is + ve.



Unless otherwise mentioned in a given reaction, the DH values correspond to the standard states of the substances. The standard state is taken as 298 K and 1 atmospheric pressure.



While writing the thermochemical equation for a reaction, the physical states of the reactants and the products must be mentioned. The physical states are represented by the symbol 'l', 'g' and 'aq' for solid, liquid, gaseous and aqueous states respectively.

It is essential to mention the state because the physical states of the reactants and the products cause appreciable difference in the value of ΔH. For example, when one mole of liquid water is formed from gaseous hydrogen and gaseous oxygen, the heat evolved is equal to 285.8 kJ.

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On the other hand, when one mole of water in the gaseous state is formed from the gaseous oxygen and hydrogen, the heat evolved is equal to 241.8 kJ.



The coefficients of the substances in a thermochemical equation represent the number of moles of each substance involved in the reaction. When the coefficients in the chemical equation are multiplied or divided by a factor, ΔH value must also be multiplied or divided by the same factor. For example, in the equation,

If coefficients are multiplied by 2, we write the equation,



When the chemical equation is reversed, the sign of ΔH value is also changed but the magnitude remains the same. For example,

Thus, a reaction, which is exothermic in one direction, will be endothermic in reverse direction.

The amount of heat evolved or absorbed in a reaction is known as heat of reaction. As the amount of heat evolved or absorbed at constant temperature and pressure is called enthalpy, the heat changes for various reactions may also be called enthalpy changes. The heat change during a chemical reaction depends upon the amount of substance (number of moles) that has reacted.

Thus, heat of reaction may be defined as 'the amount of heat evolved or absorbed in a reaction when the number of moles of reactants reacts completely to give the products as given by the balanced chemical equation'. ΔHr = ΔHproducts - ΔHreactants For example, the heat change for the reaction of one mole of methane with two moles of oxygen to form one mole of CO2, and two moles of water is -890.3 kJ.

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APPLIED THERMODYNAMICS 6.1

Factors Affecting Heat or Enthalpy Change

The enthalpy change of reaction (ΔH) depends upon the following factors: Physical state of the reactants and the products The ΔH of a reaction depends upon the physical states of reactants and products. For example, when hydrogen and oxygen gases combine to give liquid water, the heat of reaction is different than when they combine to form gaseous water.

Quantities of reactants The amount of heat evolved or absorbed depends upon the amount of reactants. For example, the heat of combustion of 2 moles of carbon is double than heat of combustion of 1mole of carbon.

S.A.Q.2 Why heat? Why constant pressure? _____________________________________________________________________________ ___________________________________________________________________ Allotropic modification The amount of heat evolved or absorbed for different forms of the same substance are different.

For example,

Temperature The heat of reaction depends upon the temperature of reactants and products.

Pressure or volume The heat of reaction depends upon the conditions of constant pressure or volume. As, ΔH = ΔE+ PΔV DH may be equal, greater than or less than ΔE. As the heat of a reaction varies with temperature it is desirable to fix up a standard or a reference state. A substance is said to be in standard state when it is present in its most stable state at 298 K under a pressure of one atmosphere.

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45

The heat change that occurs when a process is carried out at 298K and one atmospheric pressure is called standard heat change. It is represented as ΔH° where the superscript (°) indicates the standard state. If the reactants and products are in their standard states, then the enthalpy of a reaction is termed as the standard enthalpy of reaction.

Problem What is the standard enthalpy change for the reaction:

o

-1

o

-1

Hf of SO2 = -296.9 kJmol Hf of H2S = -20.15 kJmol o

-1

Hf of H2O = -285.9 kJmol .

Solution o (products)

o

SHf

o

= 3 x DHf of (S) + 2 x DHf of [H2O(l)]

= 3 x 0 + 2 x (-285.9) kJ o

= -571.8 kJ (DHf of S = 0) o (reactants)

SHf

o

o

= DHf of [SO2(g)] + 2 x DHf of [H2S(g)]

= -296.9 +2 x (-20.15) kJ = -337.20 kJ o (reaction)

SHf

o (products)

= SHf

o (reactants)

- SHf

= -571.8 (-337.20) = -234.6 kJ 7.0

Types of Heat (enthalpy) of Reactions

The heat or enthalpy changes taking place during the chemical reactions are expressed in different ways depending upon the nature of the reaction. The various types of enthalpies of reactions are: 7.1

Heat of Formation or Enthalpy of Formation

The heat evolved or absorbed when 1 mole of a substance is formed from its elements is called heat of formation. It is denoted by ΔHf. For example, heat of formation of carbon dioxide and methane may be expressed as:

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46

These equations should always be written for one mole as per the definition of the substance to be formed. If for balancing, we require the coefficient 2, 3.... etc., in the equation then ΔHf values should also be multiplied by the same number as discussed earlier.

Standard heat of formation The heat of formation ΔHf depends upon the condition of temperature, pressure and physical state (gas, liquid or solid) of the reactants and the products. Therefore the heat change accompanying the formation of one mole of a compound from its elements when all the substances are in their standard states (1 atm pressure and 298 K), is called the standard heat of o

formation. It is expressed as ΔHf .

Since no heat changes are involved in the formation of elements form themselves in their standard states, the standard enthalpy of formation of all elements is zero. o

-1

For example, the standard enthalpy of formation (ΔHf ) for H2O(l) is - 286 kJ mol i.e., when one mole of liquid water is formed from its elements H2(g) and O2(g) at 298 K and 1 atm pressure, then -1

o

286 kJ mol of heat is released. The negative value of ΔHf indicates the formation of a stable compound.

Standard heat of reaction from standard heats of formation The knowledge of standard heats of formation of various substances can be used to calculate the heats of reactions under standard conditions. The standard heat of any reaction (ΔH°) is equal to o

the difference between the ΔHf of all the reactants i.e., ΔH° = Sum of the standard heats - Sum of the standard heats of formation of products of formation of reactants o

o

i.e., ΔH = SΔHf

o

(products)

- SΔHf

(reactants)

For a reaction, o

DH° = SΔHf o

o

(products) o

- SΔHf

(reactants) o

o

= [cΔHf (C) + dΔHf (Δ)] - [aΔHf (A) + bΔHf (B)] The heat of formation of all elements in their standard state is zero. On this basis, it is evident that the heat of formation of a compound is the heat of the compound.

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Heat of formation = Hf (compound) - Hf (elements) Problems The heat change for the reaction,

is -92.2 kJ. Calculate the heat of formation of ammonia.

Solution The heat of formation of ammonia is the heat change for the formation of 1 mole of ammonia from its elements, i.e.,

The heat change for the reaction

is ΔH = -92.2 kJ. This equation corresponds to formation of two moles of ammonia. Thus,

Calculate the heat change for the reaction

-1

The heat of formation of CH4(g), CO2(g) and H2O(l) are -74.8 kJmol

-393.5 kJmol

-1

and -

-1

285.8kJmol respectively.

Solution ΔH° for the reaction,

o (products)

is: ΔH° = ΔHf o

o

-ΔHf

(reactants)

o

o

o

{ΔHf [CO2(g)] + 2ΔHf [H2O(l)]} - {ΔHf [CH4(g)] + 2ΔHf [O2(g)]} o

-1

o

ΔHf [CO2(g)] = -393.5kJmol , ΔHf [H2O(l)] = - 285.8kJmol

-1

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-1

48

o

ΔHf [CH4(g)] = - 74.8kJmol , ΔHf [O2(g)] = 0 (by convention) ΔH° = {-393.5 + 2(-285.8)} - {-74.8 + 2 x 0)} = 965.1 + 74.8 = - 890.3kJmol-1 o

Calculate DHf for chloride ion from the data given below:

ΔH

o HCl

-1

= - 92.30 kJ mol

o

ΔH =- 75.14 kJ o

+

ΔH f of H ion = 0.00 kJ. Solution Consider the solution process of HCl(g)

o

o

Now, ΔHsol = SH f (Products) - SH f (reactants) o

-

o

+

o

= ΔH f (Cl )+ ΔH f (H )] - [ΔH f HCl(g)] o

-

or -75.14 kJ = ΔH f (Cl )+ 0 - (-92.30) o

-

or Δ H f (Cl ) = -75.14 - 92.30 = - 167.44 kJ. S.A.Q.3. What is heat capacity? _____________________________________________________________________________ _____________________________________________________________________________ ______________________________________________________________ 7.2

Measurement of Enthalpy of Reactions

The heat changes in chemical reactions are measured with the help of calorimeters. Depending upon the requirements of experiments different types of calorimeters are used. In general, reactions taking place at constant volume and involving gases are carried out in a closed container with rigid walls that can withstand high pressures such as a bomb calorimeter. The body of the bomb calorimeter is made of heavy steel. The steel vessel is coated inside with gold or platinum to avoid oxidation of steel during the chemical reactions. The vessel is fitted with a

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APPLIED THERMODYNAMICS tight screw cap. There are two electrodes R1 and R2, which are connected to each other through a platinum wire S, which remains dipped in a platinum cup just below it.

The substance under investigation is taken in platinum cup (few grams). The vessel is then filled with excess of oxygen at a pressure of about 15200 to 19000 mm of Hg and sealed. It is now dipped in an insulated water bath provided with a mechanical stirrer and a thermometer, sensitive enough to read upto 0.01°C (Beckmann's thermometer).

Bomb calorimeter The initial temperature of water is noted and the chemical reaction (combustion) is initiated by passing electric current through the platinum wire. The heat evolved during the chemical reactions raises the temperature of water, which is recorded from the thermometer. By knowing the heat capacity of the calorimeter and also the rise in temperature, the heat of chemical reaction or heat of combustion at constant volume can be calculated by using the expression.

where, Z = Heat capacity of calorimeter system Î&#x201D;T = Rise in temperature M = Molecular mass of substance w = Mass of substance taken.

8.0

Determination of Enthalpy Change

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50

The calorimeter is kept open to the atmosphere in the determination of change in enthalpy of a reaction. The calorimeter is immersed in an insulated water bath fitted with stirrer and thermometer. The temperature of the bath is recorded in the beginning and after the end of the reaction and the change in temperature is calculated. Knowing the heat capacity of water bath and calorimeter and also the change in temperature the heat absorbed or evolved in the reaction is calculated. This gives the enthalpy change (ΔH) of the reaction.

8.1

Enthalpy of Combustion

The enthalpy of combustion of a compound is the enthalpy change at normal pressure and at constant temperature accompanying complete combustion of one mole of the compound. It is denoted by ΔHc. Combustion here means the burning of the given compound to the highest oxides of the constituent elements in the presence of excess of oxygen.

For example, the enthalpy of combustion of benzene at 298 K is the enthalpy change of the reaction.

Thus, ΔHcomb C6H6(l) = -3268kJmol

-1 .

Enthalpy of combustion is generally obtained experimentally. For cases, where it is not possible to measure it experimentally, it is estimated from the enthalpies of formation of the various compounds involved in the process.

Problems -1

The heats of combustion of CH4 and C2H6 are -890.3 and -1559.7kJ mol respectively. Which of the two has greater efficiency of fuel per gram?

Solution The fuel efficiency can be predicted from the amount of heat evolved for every gram of fuel consumed. (i) The combustion of methane is as follows: ΔHc = 890.3 kJmol-1 Molar mass of CH4 = 16

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51

(ii) The combustion of ethane is as follows:

-1

Î&#x201D;Hc = 1559.7kJmol

Molar mass of C2H6 = 30

Thus, methane has greater fuel efficiency than ethane. (a) A cylinder of gas supplied by a company is assumed to contain 14 kg of Butane. If a normal family requires 20,000 kJ of energy per day for cooking, how long will the cylinder last? (b) If the air supplied to the burner is insufficient, a portion of gas escapes without combustion.

Assuming that 25% of the gas is wasted due to this inefficiency, how long will the cylinder last?

(Heat of combustion of butane = 2658 kJ/mol).

Solution (a) Molecular formula of butane = C4H10 Molecular mass of butane = 4 x 12 + 10 x 1 = 58 Heat of combustion of butane = 2658kJmol

-1

1 mole of 58 g of butane on complete combustion give heat = 2658 kJ 3

14 x 10 g of butane on complete combustion gives heat =

The family needs 20,000 kJ of heat for cooking per day.

641586 kJ of heat will be used for cooking by a family in

The cylinder will last for 82 days (b) 25 per cent of the gas is wasted due to inefficiency. This means that only 75% of butane is combusted. Therefore,

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52

the energy produced by 75% combustion of butane

-1

The enthalpy change involved in the oxidation of glucose is - 2880 kJ mol . Twenty five percent of this energy is available for muscular work. If 100 kJ of muscular work is needed to walk one kilometer, what is the maximum distance that a person will be able to walk after eating 120 g of glucose.

Solution -1

ΔHcomb of Glucose (C6H12O6) = - 2880 kJ mol

Calculate the enthalpy change of combustion of cyclopropane at 298 K. The enthalpy of formation -1

of CO2(g) , H2O(l) and propane(g) are -393.5, -285.8 and 20.42 kJmol respectively. The enthalpy of -1

isomerisation of cyclopropane to propene is -33.0 kJ mol .

Solution The required ΔH is

The given equations are:

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53

Multiply equation (i) and (ii) by 3 and add them. Now subtract equation (iii) and subsequently add equation (iv) from the resulting expression. ΔH = 3ΔH1 + 3ΔH2 - ΔH3 - ΔH4 = 3(- 393.5) + 3(- 285.8) - (20.42) + (-33.0) = - 2091.32 kJ. 8.2

Heat or Enthalpy of Neutralization

The reaction in which an acid and a base react to give a salt and water is called neutralization reaction. Neutralization reactions are exothermic in nature. The heat change when one gram equivalent of an acid is completely neutralized by a base or vice versa in dilute solution, is called heat of neutralization.

Examples of heat of neutralization are: Neutralization of HCl with NaOH

Neutralization of CH3COOH with NaOH

ΔH =-55.9 kJ It is important to note that the term gram equivalent is used in the definition of heat of +

-

neutralization. This is because neutralization involves 1 mole of H ions and 1 mole of OH ions to form 1mole of water and 57.1 kJ of heat is liberated.

Now, one gram equivalent of various acids on complete dissociation liberates one mole of H

+

+

ions. But one mole of the acid may produce more than one mole of H ions in solution depending +

upon its basicity; for example 1mol of H2SO4 gives 2 mol of H ions and 1mol of H3PO4 gives 3 +

mol of H ions on complete dissociation. But 1gram equivalent of both (H 2SO4 or H3PO4) +

produces only 1 mol of H ions.

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54

Thus, it is more appropriate to use the term gram equivalent in the definition of enthalpy of neutralization.

The average enthalpy of neutralization of any strong acid by a strong base is found to be - 57.7 kJ (- 13.7 kcal) irrespective of the nature of acid or the base. This suggests that the net chemical reaction in all neutralization reactions is the same, viz.,

This is because strong acids and strong bases are completely ionized in aqueous solutions. The +

aqueous solution of one gram equivalent of all strong acids contains the same number of H ions.

Similarly, aqueous solution of one gram equivalent of all strong bases also contains same -

number of OH . The neutralization reactions between strong acids and strong bases in aqueous +

-

solutions involve simply the combination of H ions (from an acid) and OH ions (from a base) to form unionized water molecules.

For example, in the reaction between hydrochloric acid and sodium hydroxide ion. The neutralization can be represented as:

Î&#x201D;H = -57.1kJ Cancelling common ions:

8.3

Neutralization of weak acids and weak bases

The heat of neutralization of a weak acid or a weak base is less than -57.1 kJ and is also different for different weak acids or bases.

For example for acetic acid the enthalpy of neutralization is -54.9 kJ. This can be explained as follows: Acetic acid (CH3COOH) is a weak acid. Weak acids (or weak bases) are ionised to a small extent in solutions. So acetic acid is only partially ionised in solution. Since the neutralization involves a +

-

reaction between H (from the acid) and OH (from the base), hence acetic acid must be fully ionized as per the reaction,

Î&#x201D;H = +1.2 kJ

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55

The ionization reaction is endothermic reaction. So, during ionization of acetic acid a small amount of heat (1.2 kJ) is absorbed. As a result, the enthalpy of neutralization of acetic acid is 1.2 kJ less than that for a strong acid-strong base pair.

Therefore, the aqueous solutions containing one gram equivalent of different weak acids do not +

contain 1 gm equivalent of H ions. Similarly, the aqueous solutions containing 1 gram equivalent -

of different weak bases do not contain 1 gram equivalent of OH ions. The weak acids, weak +

-

bases therefore, have to be dissociated to give 1 gram equivalent of H or OH ions but, neutralization of weak acid and strong base (or a weak base and strong acid) not only involves +

-

the combination of H and OH ions but also the dissociation of a weak acid (or a weak base).

The dissociation process is accompanied by the absorption of energy. This energy is called the heat of dissociation. Therefore, the overall liberated energy is less than 57.1 kJ (i.e., 57.1 of dissociation of acid or base).

The neutralization of acetic acid with sodium hydroxide can be explained as follows:

Î&#x201D;H = -55.9kJ Thus, heat of neutralization of acetic acid and sodium hydroxide is -55.9 kJ, because 1.2 kJ of heat energy is used up in dissociating acetic acid. Similarly, heat of neutralization of ammonium hydroxide and hydrochloric acid is -51.5 kJ as 5.6 kJ is the heat of dissociation of NH4OH. Problems 15. 100 ml of 1N of an acid and 100 ml of 1N of a base are mixed at 298K. During the experiment, the temperature arose to 298.0067 K. Calculate the heat of neutralization.

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56 Solution

Heat capacity of solution = Mass of solution x Specific heat capacity Total mass of solution = 100 + 100 = 200 ml Heat capacity of solution = 200 x 4.2 = 840 JK

-1

Heat change in the reaction = Heat capacity x Rise in temperature -1

= (840 JK ) (298.0067 - 298)K = 840 x 0.0067 J = 5.63 J Now, one gram equivalent of acid = 1N HCl in 1000 ml 100 ml of 1N acid on neutralization gives heat = 5.63 J

= 56.3J Heat of neutralization = -56.3 J 16. Whenever an acid is neutralized by a base, the net reaction is:

Calculate the heat evolved for the following experiments: (a) 0.60 mol of HNO3 solution is mixed with 0.30 mol of KOH solution.

3

3

(b) 400 cm of 0.2 M H2SO4 is mixed with 600 cm of 0.1 M NaOH solution.

Solution According to the reaction,

+

-

When 1 mole of H ions and 1 mol of OH ions are neutralized to form 1 mol of water, 57.1 kJ of energy is released. (a) Heat evolved on mixing 0.60 mol of HNO 3 with 0.30 mol of KOH solution. Since HNO3 and KOH are strong acids and bases, O.60 mol of HNO3 -

+

0.60 mol of H ions 0.30 mol of KOH +

0.30

mol of OH ions In this case, out of 0.60 mol of H ions (from HNO3) only 0.30 mol will be

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57

-

+

neutralised (equal to OH ions present) by the base. 0.3 mol of H ions of HNO3 will remain unreacted. The net reaction is:

Now, heat evolved during the formation of 1 mol of H2O = 57.1kJ Heat evolved in the formation of 0.3 mol of H2O=57.1 x 0.3=17.13 kJ. 8.4

Enthalpy of solution (Î&#x201D;Hsol)

When a solute is dissolved in a solvent a solution is formed. During dissolution of a solute in any solvent, a certain amount of heat is either absorbed or evolved. Such heat changes under constant pressure conditions are known as the enthalpy of solution. 'The change in enthalpy when one mole of a solute is dissolved in a specified quantity of a solvent at a given temperature is called enthalpy of solution'.

To avoid the amount of solvent, heat of solution is usually defined for an infinite dilute solution.

Thus, heat of solution at infinite dilution is the heat change when one mole of a substance is dissolved in such a large quantity of solvent so that further dilution does not give any further heat change.

For example, dissolution of sodium chloride

Here 'aq' represents aqueous meaning a large excess of water.

For substances, which dissolve with the absorption of heat (endothermic), the enthalpy of solution is positive while for the substances which dissolve by liberating heat (exothermic), the enthalpy of solution is negative.

For example, when KCl is dissolved in water, heat is absorbed. Thus, the enthalpy of solution of KCl is positive. For a 200 times dilution (water : KCl = 200 : 1), the enthalpy change during the process,

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58

-1

So, the enthalpy of solution of KCl at a dilution of 200 is 18.6 kJ mol . The dissolution of CaCl2(s) in water is an exothermic process. So, the enthalpy of solution of calcium chloride (CaCl2) is negative. At a dilution of 400, the enthalpy change for the reaction,

So, the enthalpy of solution of CaCl2(s) at a dilution of 400 is -1

-75.3 kJ mol .

8.5

Enthalpy of fusion (Î&#x201D;Hfus)

The enthalpy of fusion of a substance is defined as 'the change in enthalpy when one mole of a solid substance is melted at its melting temperature'. For example, the enthalpy change of the reaction,

is the enthalpy of fusion of ice. Enthalpy of fusion for some common substances are given below: Substance: Ethanol Oxygen Hydrogen sulphide Sodium chloride (C2H5OH) (O2) (H2S) (NaCl) -1

Hfus kJmol : +4.8 +0.45 +2.0 +29 Melting point: 156 K 55 K 188 K 1074 K From the data given above, we see that the enthalpies of fusion for ionic substances are very high. This is due to strong Coulombic forces between the constituent ions in ionic solids. The solids such as O2 and H2S, which are molecular solids, have low heats of fusion because the forces of attraction between their molecules are weak forces. Thus, the heats of fusion of substances give an idea of nature of solid and the magnitude of forces acting between the particles constituting the solids.

Enthalpy of Vaporization (vap) 'The change in enthalpy when one mole of a liquid is converted into vapours at its boiling temperature is called enthalpy of vaporization' (Î&#x201D;Hvap).

Thus, the enthalpy change of the reaction

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is the enthalpy of vaporization of water.

The enthalpies of vaporization for certain common liquids are: Substance Boiling point ΔHvap kJmol

-1

He(l)

H2(l)

O2(l)

HCl(l)

H2O(l)

NaCl(l)

4K

20K

90K

188K

373K

1738 K

0.1

0.9

6.7

16.2

40.6

170

From the data given above, we can say ΔHvap (ionic liquids) > ΔHvap (polar liquids) > ΔHvap (nonpolar liquids) Thus, the ΔHvap depends upon the strength of intermolecular forces in any liquid.

Problem Determine the value of DH and DE for the reversible isothermal evaporation of 90.0 g of water at 100°C. Assume that water vapours behave as ideal gas and heat of evaporation of water is 540 -1

-1

-1

cal g . (R = 2 cal mol K ) Solution

Heat of evaporation of 1 g of water = 540 cal Heat of evaporation of 90 g of water = 540 x 90 = 48600 Cal. D H = 48600 Cal The evaporation of 5 mol of water is represented as

Dn = (5 - 0) = 5 ΔH = ΔE + ΔnRT or ΔE = ΔH - ΔnRT = 48600 - (5) (2.0) (373) = 44870 cal

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Enthalpy of Sublimation (ΔHsub) Sublimation is a process in which a solid substance directly changes into its vapours at any temperature below its melting point. Enthalpy of sublimation is defined as follows: The change in enthalpy when one mole of a solid substance is converted into its vapours without melting at a temperature below its melting point is called the enthalpy of sublimation.

For example, when one mole of solid iodine is converted into its vapours at room temperature, -1

heat equal to 62.4 kJ is absorbed. So, the enthalpy of sublimation of iodine is + 62.4 kJ mol , i.e.,

Compounds, which sublime on heating, are camphor, dry ice, ammonium chloride etc.

The heat of sublimation is related to heat of fusion and heat of vaporization as: ΔHsublimation = ΔHfusion + ΔHvaporization Problem 18. When 1 g of liquid naphthalene (C10H8) solidifies, 149 J of heat is evolved. Calculate the heat of fusion of naphthalene.

Solution The molecular mass of naphthalene is C10H8, = 10 x 12 + 8 x 1 =128 Heat evolved when 1 g of naphthalene solidifies = 149 J Heat evolved when 128 g of naphthalene solidifies = 149 x 128 = 19072 J

For the fusion reaction,

This reaction is the reverse of the above solidification reaction so that ΔHfusion = - ΔHsolidification ΔHfusion = 19072 J or = 19.072 kJ

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APPLIED THERMODYNAMICS Enthalpy of hydration This is defined as the heat change (evolved or absorbed) when one mole of the anhydrous salt combines with the required number of moles of water to form the specific hydrated salt.

9.0

Laws of Thermochemistry

All thermochemical reactions are governed by two laws.

Lavoisier and Laplace Law A.L.Lavoisier and P.S.Laplace gave this law in 1780 which states that 'the enthalpy of a reaction is exactly equal but opposite in sign for the reverse reaction'.

For example, if ΔH is the enthalpy change in going from A to B then the enthalpy change for the process B to A would be -ΔH. Thus, the enthalpy of formation of a compound is numerically equal but opposite in sign to the enthalpy of decomposition of the compound.

Whenever a thermochemical equation is reversed the sign of DH also gets reversed.

9.1

Hess's Law of Constant Heat Summation

G.H.Hess proposed a law regarding the heats or enthalpies of reaction in 1840 called the Hess's law. This law states that 'the heat change in a particular reaction is the same whether it takes place in one step or several steps'.

For example, a reactant 'A' changes to a product 'B' in one step and the heat change during this process is ΔH. If the reaction is carried out in two steps where 'A' first changes to 'C' an intermediate stage and then 'C' changes to 'B' in the following step then let the heat change during the formation of 'A' to 'C' be ΔH1 and that from 'C' to 'B' be ΔH 2. From Hess's law the heat change for the reaction is given as: ΔH = ΔH1 + ΔH2 S.A.Q.4 Define Homogeneous Mixture _____________________________________________________________________________ _____________________________________________________________________________ ______________________________________________________________

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Illustration of Hess's law This means that the amount of heat evolved or absorbed in a chemical reaction depends only upon the energy of initial reactants and the final products. The heat change is independent of the path or the manner in which the change has taken place.

The formation of carbon dioxide from carbon and oxygen can be illustrated as follows. Carbon can be converted into carbon dioxide in two ways. Firstly solid carbon combines with sufficient amount of oxygen to form CO2. The same reaction when carried in presence of lesser amount of oxygen gives carbon monoxide which then gets converted to CO 2 in step two in the presence of oxygen.

ΔH = ΔH1 + ΔH2 Thus, one can conclude that thermochemical equations can be added, subtracted or multiplied like algebraic equations to obtain the desired equation.

Application of Hess's Law Hess's law has been useful in determining the heat changes of reactions, which cannot be measured directly with calorimeter. Some of its applications are:

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Applications and problems Determination of heat of formation Compounds whose heats of formation cannot be measured directly using calorimetric methods because they cannot be synthesised from their elements easily e.g. methane, carbon monoxide, benzene etc are determined using Hess's Law. For example, the heat of formation of carbon monoxide can be calculated from the heat of combustion data for carbon and carbon monoxide as shown above.

Determination of heat of transition The heats of transition of allotropic modification of compounds such as diamond to graphite, rhombic sulphur to monoclinic sulphur, yellow phosphorous to red phosphorous etc. can be determined using Hess's Law. For example, the heat of transition of diamond to graphite can be calculated from the heat of combustion data for diamond and graphite, which is -395.4 kJ and -393.5 kJ respectively.

The thermochemical equations showing the combustion reaction of diamond and graphite are:

The conversion that is required is:

This can be obtained by subtracting the second equation from the first one.

Determination of heat of hydration The heats of hydration of substances is calculated using Hess's law.

For example the heats of hydration of copper sulphate can be calculated from the heats of solution of anhydrous and hydrated salts of copper. The heats of solution of CuSO 4 and -1

CuSO4.5H2O are -66.5 and -11.7 kJ mol . The corresponding thermochemical equations are:

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The process of hydration can be expressed as:

According to Hesss law, DH1 = DH + DH2 DH = DH1 - DH2 = -66.5 11.7 = -78.2 kJ/mol 9.2

Determination of heats of various reactions

Hess's law is useful in calculating the enthalpies of many reactions whose direct measurement is difficult or impossible.

Problems Calculate the standard heat of formation of carbon disulphide (l). Given that the standard heats of combustion of carbon (s) sulphur (s) and carbon disulphide (l) are 393.3, -293.72 and -1108.76kJ -1

mol respectively.

Solution The given data can be written in thermochemical equation form as:

The required equation is:

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65

Multiplying equation (ii) by 2 and adding to equation (i) we get,

Subtracting equation (iii) from the above equation we have,

Calculate lattice energy for the change,

-1

-1

Given that ΔHsubl. of Li = 160.67 kJ mol , ΔHDissociation of Cl2 = 244.34 kJ mol , ΔHionisation of Li(g) = -1

-1

o

-1

520.07 kJ mol , ΔHE.A of Cl(g) = - 365.26 kJ mol , ΔH f of LiCl(s) = - 401.66 kJ mol .

Solution Considering the different changes that occur in the formation of solid lithium chloride based on the data given the lattice energy of the above can be constituted as:

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or

-1

= - 839.31 kJ mol 10.0

Bond Energy or Bond Enthalpy

In chemical reactions the formation of a chemical bond is accompanied by the release of energy.

Conversely energy has to be supplied for the breaking of a bond.

Bond strengths are commonly described by their bond dissociation energy which is the energy required to break one mole of a bond of particular type. This is a definite quantity and is -1

expressed in kJ mol . For diatomic molecules the bond dissociation energy is same as bond energy, whereas in polyatomic molecules the bond energy is taken as the mean average of the various bond dissociation energies of the bonds of a given type.

The thermochemical data is useful in determining the bond energies of different bonds.

For example the bond energy of C-H bond in methane can be calculated form its heat of formation. The heat for formation of methane from carbon and hydrogen has been found to be -1

1663 kJ mol .

Methane has four C-H bonds and the energy required to break all the four C-H bonds is 1663 kJ. Therefore the average C-H bond energy is

21. Compute the average S-F bond energy in SF6. The values of standard enthalpy of formation -1

of SF6(g), S(g), F(g) are, 100, 275 and 80 kJ mol respectively.

Solution Consider the equation

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APPLIED THERMODYNAMICS D Hreaction = 6D Hf (F) + D Hf (S) - D Hf (SF6) = 6 x 80 + 275 - (-1100) = 1855 kJ

11.0

SOLUTIONS

A Solution is a homogeneous mixture of two or more chemically non-reacting substances called components. The composition of the components can be varied with in certain limits.

When two components are present - Binary solution.

When three components are present - Ternary solution.

When four components are present - Quaternary solution.

A component whose physical state is same as that of the resulting solution and that is present in excess is called solvent.

The component, which is in lesser quantity, is called solute.

Therefore a solution is made up of a solvent and one or more solute.

11.1

Types of Solutions

A solution can be solid, liquid or a gas depending upon the physical state. The various types of solutions are:

In this unit, we will consider only binary solutions and are in liquid state.

We will begin with the different ways in which the amount of the solute dissolved in a known amount of the solvent or solution is expressed, i.e., Units of concentration of a solution.

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S.A.Q.4 Define Homogeneous Mixture _____________________________________________________________________________ ___________________________________________________________________ Concentration of a Solution Mass Percentage or Volume Percentage The mass percentage of a component in a given solution is the mass of the component per 100g of the solution. For e.g., if W A is the mass of the component A, W B is the mass of the component B in a solution. Then,

Example: A 10% solution of sodium chloride in water (by mass) means that 10g of sodium chloride are present in 100g of the solution.

Volume percentage This unit is used in case of a liquid dissolved in another liquid. The volume percentage is defined as the volume of the solute per 100 parts by volume of solution.

For e.g., If VA is the volume of component A present is Vsol volume of the solution.

Then,

3

For e.g., a 10% solution of ethanol C2H5OH, in water (by volume) means that 10cm of ethanol is 3

present in 100cm of the solution.

Strength of a Solution Strength of a solution is defined as the amount of the solute in gms, present in one litre of the -1

solution. It is expressed as gL . Mathematically,

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Molarity Molarity of a solution is defined as the number of moles of solute dissolved per litre of solution.

Mathematically,

For e.g., If 'a' is the weight of the solute (in gms) present in VCC volume of the solution.

Then,

Molarity is expressed by the symbol M. It can also be expressed as,

Normality Normality of a solution is defined as the number of gram equivalents (gm.e) of a solute dissolved per litre of the given solution.

Mathematically it is,

For e.g., If a is the weight of the solute (in gms) present in VCC volume of the solution. Then,

Normality is expressed by the symbol N. It can also be expressed as,

Relationship between molarity and normality The molarity and normality of a solution is related to each other as follows:

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Molality Molality of a solution is defined as the number of moles of solute dissolved in 1000g of a solvent.

Mathematically, it is expressed as

Molality is expressed by the symbol m. Molality does not change with temperature.

Formality In case of ionic compounds like KCl, CaCO3 etc. Formality is used in place of molarity.

It is the number of gram formula masses of solute dissolved per liter of the solution. It is denoted by the symbol F. Mathematically it is given as,

Mole Fraction It is the ratio of number of moles of one component (solute or solvent) to the total number of moles of all the components (solute and solvent) present in the solution. It is denoted by the symbol X. Let us suppose that a solution contains two components A and B and suppose that n A moles of A and nB moles of B are present in the solution then,

Adding eq (i) and (ii) we get xA + xB = 1 Parts per million (ppm) When a solute is present in very small amounts, its concentration is expressed in parts per million. It is defined as the amount of the solute present in one million parts of the solution.

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It may be noted that the concentration units like molarity, mole fraction etc. are preferred as they involve the weight of the solute and solvent, which is independent of temperature. But units like, molarity, Normality etc., involve volume of the solution, hence changes with temperature.

11.2

Solubility of Gases

Gases are completely miscible with each other. Gases also dissolve in liquids and solids.

Gasses are soluble in liquid to different extents e.g., solubility of hydrochloric acid gas in water is much more than the solubility of oxygen in water.

Solubility of gas in a liquid depends on the following factors: 1) Nature of the gas and the liquid 2) Temperature 3) Pressure.

S.A.Q.5 When the unit Parts per million is used to express concentration term _____________________________________________________________________________ _____________________________________________________________________________ ______________________________________________________________ 11.3

Nature of the gas and the liquid

Solubility of nonpolar bromine is much more in nonpolar solvents like carbondisulphide or carbontetrachloride than in water.

Temperature Solubility of a gas in a liquid generally decreases with increase in temperature.

Pressure Solubility of a gas in a liquid increases with the increase in partial pressure of the gas above the solution. Quantitatively, this is governed by Henrys law, which states that at constant temperature.

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APPLIED THERMODYNAMICS The mass of a gas dissolved in a given volume of the liquid is directly proportional to the pressure of the gas present in equilibrium with the liquid.

where 'm' is the mass of the gas dissolved in a unit volume of the solvent. p is the pressure of the gas in equilibrium with the solvent. k is the proportionality constant called Henry's law constant characteristic of the nature of the solvent and the natures of the gas, also the temperature.

Dalton also concluded independently that, solubility of a gas in liquid solution is a function of the partial pressure of the gas.

If mole fraction is used to represent the solubility of the gas in the solution, then

where, p is the partial pressure of the gas in solution, x is the mole fraction of the gas in solution and KH is Henry's law constant.

Here are the values of Henry's Law Constant (KH) for some selected gases in water �Gas

�Temp/K

�KH/K bar

�He

�293

�144.97

�H2

�293

�69.16

�N2

�293

�76.48

�N2

�303

�88.84

�O2

�293

�34.86

�O2

�393

�46.82

S.A.Q.6 Define Molarity _____________________________________________________________________________ _____________________________________________________________________________ ______________________________________________________________

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APPLIED THERMODYNAMICS It is obvious from the above table that the value of KH increases with temperature, indicating that solubility of gases decreases with increase in temperature.

Henry's Law explains many biological phenomena and also has several applications in industries, like: 1) To increase the solubility of CO2 in soft drinks and soda water, the bottle is sealed under high pressure.

2) In lungs the partial pressure of oxygen is high so haemoglobin combines with. oxygen to form oxohaemoglobin. But in the tissues the partial pressure of oxygen is low therefore oxohaemoglobin releases oxygen for utilization in cellular activities.

3) The breathing gas for deep sea divers consists of oxygen diluted with less soluble helium gas.

This is to minimize the painful effects of deep sea divers due to decompression.

For e.g., concentration of CO2 which is dissolved in a carbonated beverage.

Example: Soda water is dependant directly on the partial pressure of CO2 in the gas phase.

4) Hydrogen bonding or dipole-dipole interaction between molecules of gas and molecules of the liquid. For e.g., dissolution of acetylene in acetone and dissolution of HCl in ether are due to hydrogen bonding.

Solid Solution Solid solutions are formed by mixing two solid components. Alloys (mixtures of metals) are the examples of solid solutions.

Solid solutions can be classified under two types: a) Substitutional solid solutions b) Interstitial solid solutions.

In substitutional solid solutions atoms, molecules or ions of one substance take place of particles of another substance in a crystalline lattice as shown in the figure.

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Example: Sulphides of zinc and cadmium form such mixtures in which cadmium ions randomly replace zinc ions in the ZnS crystal lattice. Another example is brass where copper and zinc atoms are randomly arranged in the lattice structure.

In interstitial solid solutions the atoms of one substance is present in the voids or interstices that exists in the lattice structure of the other substance, which is called the host lattice.

Example: Tungsten carbide, WC, the tungsten atoms are arranged in a face - centered cubic pattern with carbon atoms in octahedral holes. It is an extremely hard substance, which has many industrial use.

Example: Cutting tools Solubility of Solids in Liquids The extent to which a solute is soluble in a solvent is expressed in terms of solubility. Solubility of a substance may be defined as the amount of solute dissolved in 100gms of a solvent to form a saturated solution at a given temperature.

Solubility of a solid in a liquid depends on: a) Nature of solvent and solute b) Temperature.

Nature of solvent and solute Polar solvent dissolves polar solutes and nonpolar solvents dissolve nonpolar solutes. For e.g., polar solvents like water, liquid ammonia, liquid hydrofluoric acid etc. dissolve polar and ionic compounds. Whereas nonpolar solvents like benzene, carbon disulphide, carbontetrachloride etc. dissolve nonpolar compounds.

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75 Temperature

The substance, which dissolve in a solvent with absorption of heat (e.g., NH 4Cl, KNO3, NaNO3 etc.) show an increase in solubility with increase in temperature, whereas the solubility of substances which dissolve with evolution of heat (e.g. NaOH, (CH 3COO)2Ca etc.,) show a decrease in their solubility with rise in temperature.

Vapour Pressure of Liquids: Vapour pressure is defined as the partial pressure exerted by the vapors above the liquid surface in equilibrium with liquid at a given temperature.

In illustrations 1a and 1b, the liquid and vapor states of a given substance are represented. The individual molecules or particles are represented at a particular instant of their random motion.

When equilibrium is reached (illustration 1c), the number of molecules evaporating is equal to the number of molecules condensing. The pressure read on the manometer M is the vapor pressure of the liquid at that temperature.

Vapor pressure of a liquid depends on the following factors: a) Nature of the liquid b) Temperature of the liquid.

Nature of the liquid Liquids that have weak intermolecular forces are more volatile and have a higher vapor pressure.

For e.g., vapor pressure of ethyl alcohol is greater than that of the water.

Temperature of the liquid Vapor pressure increases with increase in temperature. This is because with an increase in temperature, rate of evaporation also increases.

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VAPOUR PRESSURE OF A SOLUTION

Let us suppose a small amount of a non-volatile solute (sugar) is dissolved in a solvent (water).

When evaporation of this solution takes place, only water evaporates as the other component (i.e., sugar) is non-volatile. The vapor pressure of the solution is found to be less than that of the pure solvent. The lowering of vapor pressure in this case can be explained on the basis of reduced tendency of the solvent molecules to go into vapor phase.

This is because some sugar molecules occupy position of water molecules at the surface of the liquid and they do not have the tendency to go into the vapor state.

There are lesser number of solvent molecules in the solution and in the vapor as compared to the pure solvent.

12.1

Vapor Pressure and Raoult's Law

In 1886 the French chemist Francois Marie Raoult carried out a series of experiments to study the vapor pressure of a number of binary solutions containing non-volatile as well as volatile non electrolyte solutes. On the basis of the results of the experiment he stated a law known as Raoult's law.

S.A.Q.7 Define Saturated Solution. _____________________________________________________________________________ _____________________________________________________________________________ ______________________________________________________________ It states that, "In a solution of volatile components, the partial vapor pressure of a component at a given temperature is equal to the mole fraction of that component in the solution multiplied by the vapor pressure of that component in the pure state".

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APPLIED THERMODYNAMICS Now, let us consider a mixture of two completely miscible volatile liquids A and B, having the mole fraction xA and xB. Suppose at a given temperature their partial vapor pressures are p A and pB and the vapor pressure in pure state are:

The total vapor pressure, P exerted by the solution is the sum of PA and PB as required by Dalton's law of partial pressures.

P = PA + PB or P = P

o A

o B

cA + P

=P

o A

(1-cB) + P

o B

=P

o A

-P

o

o B

P = (P

A

cB [because cA + cB = 1]

o B

cB + P

cB

cB

o

o A

- P A)cB + P

Similarly by putting cB = 1-cA the vapor pressure of the solution o A

P = (P

o

o B

- P B)cA + P

As the values of P

o A

and P

o B

are constants at a particular temperature, it reveals that total

pressure is a linear function of cB or cA i.e. the plot of P versus cA or P versus cB should be a straight line. The variation of P with mole fraction is given by the solid line III in the graph. The

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APPLIED THERMODYNAMICS solutions which obey Raoult's law are called ideal solutions. For such solutions, the vapor pressure of the solution always lies between the vapor pressure of the pure components.

The vapor pressure of a solution containing non-volatile solute is directly proportional to the mole fraction of the solvent. That is because, there is no contribution towards the vapor pressure of the solution from non volatile component (PB = 0).

where K is proportionality constant,

o A

Thus, PA = P

cA or, Psolution = PPure solvent x mole fraction of solvent.

In terms of symbols,

But xA + xB = 1 or xA = 1-xB â&#x20AC;Śâ&#x20AC;Ś(2) From equations (1) and (2)

Rearranging

o

In the above equation pA - ps represents lowering of vapor pressure and

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79

the mole fraction of the non-volatile solute in the solution.

12.2

Ideal Solution

An Ideal solution is the one which obeys Roault's Law at all concentrations and temperature and during the formation of which no change in enthalpy and volume takes place i.e., o

PA = P

A

XA

o

PB = P BXB If the solution is made up of two components which are liquids then,

In the ideal solutions, solvent- solute interactions are equal to solute-solute and solvent-solvent interactions.

Few examples are: (i) n-hexane and n-heptane (ii) Chloro benzene and bromo benzene (iii) Ethylene bromide and ethylene chloride (iv) Carbon tetrachloride and silicon tetrachloride.

More examples of ideal solutions: 1) Ethyl bromide and ethyl iodide 2) Benzene and toluene.

Non-ideal Solutions In general, most of the solutions deviate from ideal behavior. There are two types of deviations, positive deviation and negative deviations. Such solutions are called non ideal solutions.

12.3

Solutions with Negative Deviation

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APPLIED THERMODYNAMICS In this type of deviation, the partial vapor pressure of each component of solution is less than the vapor pressure as expected according to Raoult's law.

Solvent-solute interaction is greater than solute-solute and solvent- solvent interaction. Therefore the escaping tendency decreases which results in the decrease in their partial vapor pressures. In such solutions, total vapor pressure of the solution is also less than the vapor pressure expected according to Raoult's law. An e.g., of solution exhibiting negative deviation is chloroform and acetone, the molecular interaction of which can be shown as below.

Since a new force of attraction exists between the molecules, the partial vapor pressure decreases.

Examples of negative (-ve) deviation, Acetic acid and pyridine 1) CH3COOH + C5H5NChloroform and acetone 2) CHCl3 + (CH3)2COCloroform and benzene

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APPLIED THERMODYNAMICS 3) CHCl3 + C6H6Chloroform and Diethylether 4) CHCl3 + (C2H5)2OWater and hydrochloric acid gas 5) H2O + HClAcetone and aniline 6) (CH3)2CO + C6H5 NH2 12.4

Solutions with Positive Deviation

These types of solutions where Î&#x201D;H of mixing is >0 and Î&#x201D;V of mixing >0 are found to exhibit positive deviation from Raoult's law. Here solute-solute interactions and solvent-solvent interactions are greater than solute - solvent interactions.

Consider a solution of liquid mixtures A and B.

As expected from Raoult's law.

An e.g., of such liquid mixture is ethanol n hexane mixture. The H-bond in ethanol molecules are broken by the n hexane molecules coming in between the ethanol molecules. As a result the escaping tendency increases there by increases the vapor pressure of individual component therefore a positive deviation is observed.

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APPLIED THERMODYNAMICS Other examples are: 1) (CH3)2CO + CS2Acetone

Carbon-di-sulphide

2) C6H6 + (CH3)2COBenzene acetone 3) CH3OH + H2OMethanol water 4) CCl4 + C6H6 5) CCl4 + CHCl3 6) CCl4 + C6H5CH3 7) H2O + CH3OH 8) H2O + C2H5OH 12.5

AZEOTROPIC MIXTURES

In case of solutions showing positive deviation from Raoult's law, at one of the intermediate compositions, the total vapor pressure is the highest and the boiling point is the lowest. Similarly in case of solutions showing negative deviations from Raoult's law, at one of the intermediate compositions, the vapor pressure is the lowest and boiling point is the highest. For such solutions the composition in the liquid and the vapor phase is the same. In other words such solutions vaporizes without any change in the composition i.e., they behave as pure liquid.

Such mixture of liquids that boil at constant temperature behaving like a pure liquid, and its composition in liquid phase is same as in vapor phase, is called Azeotropic mixture.

Solutions that show positive deviation form minimum boiling azeotropy.

Solutions of ethanol and water show a large negative deviation from Raoult's law and there exists a minimum boiling mixture which consists of 95% of ethanol by volume. To separate the components of this mixture distillation cannot be used.

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The boiling point diagram for ethanol-water (not-drawn to scale) Solution that show negative deviation form maximum boiling azeotropy.

A solution of HNO3 and H2O of approximately 68% nitric acid and 32% of water by mass is an example of maximum boiling azoetrope.

12.6

COLLIGATIVE PROPERTIES

Certain properties of solutions depend only on the number of particles of the solute (molecule or ions) present in definite volume of the solution and do not depend on the nature of solute. Such properties are called colligative properties.

A few colligative properties are: (i) Relative lowering of vapor pressure (ii) Elevation of boiling point (iii) Depression in freezing point (iv) Osmotic pressure.

Relative Lowering of Vapor Pressure The relative lowering of vapor pressure is a colligative property as it depends only on the concentration of the solute and it is independent of its nature.

Consider a binary solution consisting of non volatile solute in a volatile solvent. The vapor pressure of the solution (P) is equal to vapor pressure of solvent (P A) in the solution, which in turn is directly proportional to its mole fraction in solution.

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APPLIED THERMODYNAMICS =P

o A

84

cA ...(i)

where cA = mole fraction of solvent.

If mole fraction of solute is cB then cA + cB = 1 or cA = 1-cB ...(ii) From eq (i) and (ii) PA = P o A

P

o

o

A

o

(1-cB) = P A-P AcB

o

- PA = P AcB

is called the relative lowering of vapor pressure and is equal to the mole fraction of the solute.

Calculation of molecular mass from relative lowering of vapor pressure A known mass W B of the non-volatile solute is dissolved in a known mass (W A) of the volatile solvent to prepare a dilute solution and relative lowering of vapor pressure is determined experimentally. Knowing the molecular mass (MA) of the solvent, molecular mass (MB) of the solute can be determined as follows:

Relative lowering of vapor pressure is given by:

For dilute solutions W B/MB << W A/MA and hence in the above expression W B/MB may be neglected in the denominator as compared with W A/MA.

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APPLIED THERMODYNAMICS

In this expression all the parameters are known except MB and hence MB can be calculated.

12.7 Elevation of Boiling Point The boiling point of a liquid may be defined as the temperature at which the vapor pressure of the liquid becomes equal to the atmospheric pressure.

The vapor pressure curves for the pure solvent and solution consisting of a nonvolatile solute is represented diagrammatically. AB is the vapor pressure curve for pure solvent and CD is vapor pressure curve for the solution. T 0 represents the boiling point of pure solvent and T 1 represents the boiling point of solution. In the presence of a non-volatile solute the solution has to be heated to a slightly higher temperature so that the vapor of the solution becomes equal to atmospheric pressure and solution begins to boil. The differences in the boiling points are represented as Î&#x201D;Tb, which is the elevation in boiling point. The elevation in boiling point is proportional to the solute concentration.

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86

in the denominator.

If W A is the mass of the solvent in kg, then nB/W A is equal to the molality ('m' of the solution) Tb = kMAm where kMA is replaced by a constant k b where kb is called ebulloiscopic constant defined as elevation in boiling point caused by one mole of nonvolatile solute dissolved per kilogram (1000g) -1

of the solvent. The units of k b are K.kg mol . The value of kb depends upon the solvent and is independent of the nature of the solute and concentration of the solution.

Calculation of molecular mass of the solute from elivation in boiling point ΔTb = kb m ……(1) If W B gms of the solute is dissolved in W A gms of the solvent then

(where MB is the molar mass of the solute) From (1) and (2)

Colligative Properties - Depression in Freezing Point Freezing point of a substance is defined as the temperature at which the vapor pressure of the solid form is in equilibrium with liquid form of the substance. In other words, solid and liquid forms of the substance has same vapor pressure.

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87

The curves AB, BC, and A'B' are the vapor pressure curves of solid solvent (ice) liquid water and o

solution containing non volatile solute respectively. T K is the freezing point of pure solvent T 1K is the freezing point of the solution. The freezing point depression depends upon the molal concentration of the solute and does not depend upon the nature of solute. Therefore it is a colligative property.

denominator as compared to W A/MA. Therefore,

If W A is the mass of solvent in kg then nB/WA is the molality of the solution.

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88 Or,

For, a given solvent MA is fixed hence KMa is a constant (Kf). Î&#x201D;Tf = kf m Cryoscopic constant (kf) is defined as the depression in the freezing point when one mole of non volatile solute is dissolved per kilogram of solvent.

-1

Units of kf are K kg mol .

The molar mass of the unknown solute can be calculated by applying the formula Î&#x201D;Tf = kf m ......(1) But,

(where W B is the weight of the solute in gms and MB is molar mass of the solute and W A is the weight of the solvent in gms in which W B gms of solute is dissolved) From equations (1) and (2)

Knowing all the other values on the right hand side MB can be calculated.

Depression in freezing point is used to lower the freezing temperature of water and clear roads during severe winters. Depression in freezing point is also useful in antifreeze, which is used in the radiators of cars to prevent the water in the radiators from freezing during extreme winters.

Ethylene glycol is added as antifreeze.

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APPLIED THERMODYNAMICS 12.8

Colligative Properties - Osmotic Pressure

The excess pressure that must be applied to the solution/conc.solution to prevent osmosis.

Principle of measuring osmotic pressure

If two solutions of identical composition but different concentration are separated by a semi permeable membrane (one that permits only the solvent to pass through it), the direction of flow will be from the more dilute to the more concentrated solution (left to right in the illustration). This flow will continue until the hydrostatic pressure is developed at the concentrated solution front.

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90

This pressure prevents the further movement of solvent molecules. This pressure is called osmotic pressure.

Or Osmosis is defined as the passage of solvent from pure solvent or from solution of lower concentration into solution of higher concentration through a semi-permeable membrane.

The osmotic pressure of a solution at a particular temperature may be defined as the excess hydraulic pressure that builds up when the solution/conc.solution is separated from the solvent/dil solution by a semi permeable membrane.

Vant Hoff 's Laws of Osmotic Pressure Vant Hoff, analyzing the available data on osmotic pressure of solutions concluded that osmotic pressure of dilute solutions varied with their concentration and temperature in the same way as pressure of the gas.

He enunciated some laws for solutions which are parallel to those relating to gases.

Vant Hoff Boyles law Osmotic pressure (p) of a solution at a constant temperature is directly proportional to its concentration C (i.e., moles per liter)

where n is the number of moles of solute present in volume V liters of solution.

Vant Hoff Charless Law For a solution of fixed concentration the osmotic pressure (p) of a solution is directly proportional to its absolute temperature (T).

Combining both the laws,

The constant of proportionality also turns out to be the same as gas

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Measurement of osmotic pressure by the Berkeley and Hartley's apparatus

Berkley and Hartley's apparatus The apparatus consists of a strong steel vessel into which a porous pot is fitted. The walls of the porous pot are coated with copper ferrocyanide. Due to osmosis water moves into the steel vessel from the porous pot. The water level falls in the indicator tube. This can be stopped by applying pressure using the plunger. The pressure applied is equal to the osmotic pressure. This can be recorded using a pressure gauge.

Osmotic Pressure as a Colligative Property For a given solvent the osmotic pressure depends only upon the molar concentration of solute but does not depend on its nature.

The following relation relates osmotic pressure to number of moles of solute: pV=nRT Van't Hoff 's solution equation (p = Osmotic pressure) or

but,

Therefore, p = CRT where C = Concentration of solution in moles per liter

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APPLIED THERMODYNAMICS R = Gas constant T = Temperature n = Number of moles of solute v = Volume of solution in liters Determination of Molecular Mass from Osmotic Pressure In order to determine the molecular mass of unknown non volatile compound, a known mass (say w g) of the compound is dissolved to prepare a known volume (say v liters) of solution. The osmotic pressure of the solution is determined and the molar mass is calculated as follows:

where nB is the number of moles of the solute and is given by W B/MB. Hence W B is the mass of the solute in gms and MB is the molecular mass of the solute.

Thus,

This method is exceptionally suitable for the determination of molecular masses of macromolecules such as proteins and polymers. This is due to the reason that for these substances the values of other colligative properties such as elevation in boiling point or depression in freezing point are too small to be measured. On the other hand, osmotic pressure of such substances is measurable.

Fractional distillation It is the separation of a mixture into its component parts, or fractions, such as in separating chemical compounds by their boiling point by heating them to a temperature at which several fractions of the compound will evaporate. It is a special type of distillation. Generally the component parts boil at less than 25째C from each other under a pressure of one atmosphere (atm). If the difference in boiling points is greater than 25 째C, a simple distillation is used.

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Key words

Hessís Law of Constant Heat Summation: If a reaction takes place in several steps then its standard reaction enthalpy is the sum of the standard enthalpies of the intermediate reactions into which the overall reaction may be divided at the same temperature.

Thermochemical equations: A balanced chemical equation together with the value of its ΔrH is called a thermochemical equation.

Standard enthalpy of formation: The standard enthalpy change for the formation of one mole of a compound from its elements in their most stable states of aggregation (also known as reference states) is called Standard Molar Enthalpy of Formation.

Solubility: Solubility of a substance is its maximum amount that can be dissolved in a specified amount of solvent. It depends upon the nature of solute and solvent as well as temperature and pressure.

Henry’s law: The law states that at a constant temperature, the solubility of a gas in a liquid is directly proportional to the pressure of the gas.

Raoult’s law: The law states that for a solution of volatile liquids, the partial vapour pressure of each component in the solution is directly proportional to its mole fraction.

14.0

Summary

A solution is a homogeneous mixture of two or more substances. Solutions are classified as solid, liquid and gaseous solutions. The concentration of a solution is expressed in terms of mole fraction, molarity, molality and in percentages. The dissolution of a gas in a liquid is governed by Henry’s law, according to which, at a given temperature, the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas. The vapour pressure of the solvent is lowered by the presence of a non-volatile solute in the solution and this lowering of vapour pressure of the solvent is governed by Raoult’s law, according to which the relative lowering of vapour pressure of the solvent over a solution is equal to the mole fraction of a nonvolatile solute present in the solution. However, in a binary liquid solution, if both the components of the solution are volatile then another form of Raoult’s law is used. Two types of deviations from Raoult’s law, called positive and negative deviations are observed. Azeotropes arise due to very large deviations from Raoult’s law.

The properties of solutions which depend on the number of solute particles and are independent of their chemical identity are called colligative properties. These are lowering of vapour pressure, elevation of boiling point, depression of freezing point and osmotic pressure. The process of osmosis can be reversed if a pressure higher than the osmotic pressure is applied to the solution.

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APPLIED THERMODYNAMICS Colligative properties have been used to determine the molar mass of solutes. Solutes which dissociate in solution exhibit molar mass lower than the actual molar mass and those which associate show higher molar mass than their actual values.

Quantitatively, the extent to which a solute is dissociated or associated can be expressed by vanâ&#x20AC;&#x2122;t Hoff factor i. This factor has been defined as ratio of normal molar mass to experimentally determined molar mass or as the ratio of observed colligative property to the calculated colligative property.

15.0

Hints and Answers to self assessment questions

S.A.Q. 1 What is thermochemistry?

The study of the heat released or absorbed by a chemical reaction when it is carried out under conditions of constant pressure.

S.A.Q.2 Why heat? Why constant pressure?

Many chemical reactions either are initiated by heating under conditions of constant pressure, or release heat under constant pressure conditions. Because the amount of heat released under these conditions is so important we give it a special name, the Enthalpy, Î&#x201D;H (or more archaically, heat of reaction). Since this quantity is involved in so many reactions it is of interest to chemists to know how much heat a reaction requires or how much heat it releases.

S.A.Q.3. What is heat capacity?

The heat capacity just tells how much heat it takes to increase the temperature of some system by 1K or equivalently 1 C. The heat capacity has two parts, the amount of heat it takes to increase the temperature of the calorimeter itself by 1K, called the calorimeter constant, Ccal, and the amount of heat it takes to increase the contents of the calorimeter by 1K. S.A.Q.4 Define Homogeneous Mixture Solutions are homogeneous mixtures of two or more than two components.

S.A.Q.5 When the unit Parts per million is used to express concentration term When a solute is present in trace quantities, it is convenient to express concentration in parts per million (ppm)

S.A.Q.6 Define Molarity Molarity (M) is defined as number of moles of solute dissolved in one litre (or one cubic decimetre) of solution.

S.A.Q.7 Define Saturated Solution.

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APPLIED THERMODYNAMICS A solution in which no more solute can be dissolved at the same temperature and pressure is called a saturated solution.

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UNIT QUESTIONS UNIT-I QUESTIONS 1.

Distinguish Between (i) Open, closed and isolated system (ii) Isothermal and adiabatic changes (iii) Extensive and Intensive Properties (iv) Reversible and irreversible process.

2.

DESCRIBE the following types of thermodynamic systems: a. Isolated system b. Closed system c. Open system

3.

DEFINE the following terms concerning thermodynamic systems: a. Thermodynamic surroundings b. Thermodynamic equilibrium c. Control volume d. Steady-state

UNIT-II QUESTIONS DESCRIBE the following terms concerning thermodynamic processes: a. Thermodynamic process b. Cyclic process c. Reversible process d. Irreversible process e. Adiabatic process f. Isentropic process g. Throttling process h. Polytropic process UNIT-III AND IV QUESTIONS 1.

What do you mean by internal energy of molecules? Is it possible to determine its absolute value?

2.

Explain the law of conservation of energy. How do you justify its validity?

3.

What do you understand by the term a) Internal energy b) Enthalpy or Heat constant?

4.

Do you agree with the statement that energy of the universe is conserved?

5.

Derive the relationship between Cp & Cv

6.

Derive pV = C For a reversible adiabatic process

Îł

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96

APPLIED THERMODYNAMICS UNIT-V QUESTIONS 1.

Define the term solution. How many types of solutions are formed? Write briefly about each type with an example.

2.

What role does the molecular interaction play in a solution of alcohol and water?

3.

State Raoult’s law and mention some important applications?

4.

What is meant by positive and negative deviations from Raoult's law and how is the sign of ΔmixH related to positive and negative deviations from Raoult's law?

5.

Determine the amount of CaCl2 (i = 2.47) dissolved in 2.5 litre of water such that its osmotic pressure is 0.75 atm at 27° C.

6.

Enthalpy of combustion of carbon to CO2 is –393.5 kJ mol–1. Calculate the heat released upon formation of 35.2 g of CO2 from carbon and dioxygen gas.

7.

Calculate the number of kJ of heat necessary to raise the temperature of 60.0 g of aluminium from 35°C to 55°C. Molar heat capacity of Al is 24 J mol–1 K–1.

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APPLI EDTHERMODYNAMI CS

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