18.1. Solve: We can use the ideal-gas law in the form pV = NkBT to determine the Loschmidt number (N/V):
(
)
1.013 × 10 5 Pa N p = 2.69 × 10 25 m −3 = = V kB T (1.38 × 10 −23 J K )(273 K )
18.2. Solve: Nitrogen is a diatomic molecule, so r ≈ 1.0 × 10−10 m. We can use the ideal-gas law in the form pV = NkBT and Equation 18.3 for the mean free path to obtain p:
λ=
(1.38 × 10 −23 J k )(293 K) = 0.0228 Pa kB T kB T 1 = ⇒ p = = 2 4 2π ( N V )r 2 4 2πpr 2 4 2πλr 2 4 2π (1.0 m )(1.0 × 10 −10 m )
Assess: In Example 18.1 λ = 225 nm at STP for nitrogen. λ = 1.0 m must therefore require a very small pressure.
18.3. Solve: (a) Air is primarily comprised of diatomic molecules, so r ≈ 1.0 × 10−10 m. Using the ideal-gas law in the form pV = NkBT, we get N p = = V kB T
1.013 × 10 5 Pa 760 mm of Hg = 3.30 × 1012 m −3 J K )(293 K )
1.0 × 10 −10 mm of Hg ×
(1.38 × 10
−23
(b) The mean free path is
λ=
1 1 6 = 2 = 1.71 × 10 m 4 2π ( N V )(r 2 ) 4 2π (3.30 × 1012 m −3 )(1.0 × 10 −10 m )
Assess: The pressure p in the vacuum chamber is 1.33 × 10−8 Pa = 1.32 × 10−13 atm. A mean free path of 1.71 × 106 m is large but not unreasonable.
18.4. Solve: (a) The mean free path of a molecule in a gas at temperature T1, volume V1, and pressure p1 is λ1 = 300 nm. We also know that
λ=
1 ⇒ λ ∝V 4 2π ( N V )r 2
Although T2 = 2T1, constant volume (V2 = V1) means that λ2 = λ1 = 300 nm. (b) For T2 = 2 T1 and p2 = p1, the ideal gas equation gives p1V1 pV p1V2 = 2 2 = ⇒ V2 = 2V1 NkB T1 NkB T2 NkB (2T1 ) Because λ ∝ V , λ2 = 2λ1 = 2(300 nm) = 600 nm.
18.5. Solve: Neon is a monatomic gas and has a radius r ≈ 5.0 × 10−11 m. Using the ideal-gas equation,
(
)
(150) 1.013 × 10 5 Pa N p = 3.695 × 10 27 m −3 = = V kB T (1.38 × 10 −23 J / K )(298 K ) Thus, the mean free path of a neon atom is
λ=
1 1 −9 = m 2 = 6.09 × 10 2 −3 −11 27 4 2π ( N V )r 4 2π (3.695 × 10 m )(5.0 × 10 m )
Since the atomic diameter of neon is 2 × (5.0 × 10−11 m) = 1.0 × 10–10 m,
λ=
6.09 × 10 −9 m = 61 atomic diameters 1.0 × 10 −10 m
18.6. Solve: The number density of the Ping-Pong balls inside the box is N 2000 = = 2000 m −3 V 1.0 m 3 With r = (3.0 cm)/2 = 1.5 cm, the mean free path of the balls is 1 λ= = 0.125 m = 12.5 cm 4 2π ( N V ) r 2
( )
18.7. Solve: (a) In tabular form we have Particle
vx (m/s)
vy (m/s)
1 2 3 4 5 6 Average
20 30 −40 70 −80 −10 60 −20 0 −50 40 −20 0 0 r r r The average velocity is vavg = 0 iˆ + 0 jˆ . (b) The average speed is vavg = 59.2 m / s . (c) The root-mean-square speed is vrms =
(v ) 2
avg
v x2 (m/s)2
v y2 (m/s)2
v2 (m/s)2
v (m/s)
400 1600 6400 3600 0 1600
900 4900 100 400 2500 400
1300 6500 6500 4000 2500 2000 3800
36.06 80.62 80.62 63.25 50.00 44.72 59.20
= 3800 m 2 / s 2 = 61.6 m / s .
18.8. Solve: (a) The average speed is vavg =
220 m / s 1 n=25 = 20.0 m / s ∑ n m / s = 11 11 n=15
(b) The root-mean-square speed is 1
vrms =
(v ) 2
avg
1
1 n=25 2 4510 2 = ∑ n2 m / s = = 20.2 m / s 11 11 n=15
18.9. Solve: (a) The atomic mass number of argon is 40. This means the mass of an argon atom is m = 40 u = 40(1.661 × 10 −27 kg) = 6.64 × 10 −26 kg The pressure of the gas is
N 2 = p = 13 mvrms V
1 3
(2.0 × 10
25
)
m −3 (6.64 × 10 −26 kg)( 455 m / s) = 9.16 × 104 Pa 2
(b) The temperature of the gas in the container can be obtained from the ideal-gas equation in the form pV = NkBT:
T=
9.16 × 10 4 Pa pV = = 332 K NkB 2.0 × 10 25 m −3 (1.38 × 10 −23 J / K )
(
)
18.10. Solve: The pressure on the wall with area A = 10 cm2 = 10 × 10−4 m2 is p=
F ∆( mv) N = A A∆t
where N ∆t is the number of N 2 molecules colliding with the wall every second and ∆(mv) is the change in momentum for one collision. The mass of the nitrogen moleucle is m = 28 u = 28 (1.66 × 10−27 kg) = 4.648 × 10−26 kg and ∆v = 400 m / s − ( −400 m / s) = 800 m / s . Thus,
p=
(4.648 × 10
−26
kg)(800 m / s)(5.0 × 10 23 s −1 ) 1.0 × 10 −3 m 2
= 1.86 × 10 4 Pa
18.11. Solve: The pressure on the wall with area A = 10 cm2 = 10 × 10−4 m2 is p=
F ∆( mv) N = A A∆t
where N ∆t is the number of N2 molecules colliding with the wall every second and ∆(mv) is the change in momentum for one collision. Using the atomic mass number of oxygen, the number of collisions per second is
(
)
1.013 × 10 5 Pa (10 × 10 −4 m 2 ) N pA = = 1.91 × 10 24 s −1 = ∆t ∆( mv) (32 × 1.66 × 10 −27 kg)(500 m / s − ( −500 m / s))
18.12. Model: Pressure is due to random collisions of gas molecules with the walls. Solve:
According to Eq. 18.8, the collision rate with one wall is
rate of collisions =
N coll F pA = net = ∆t coll 2 mv x 2 mv x
where Fnet = pA is the force exerted on area A by the gas pressure. However, this equation assumed that all molecules are moving in the x-direction with constant speed. The rms speed vrms is for motion in three dimensions at varying speeds. Consequently, we need to replace vx not with (vx)avg, which is zero, but with
v x → (v x2 ) avg =
2 vrms v = rms 3 3
With this change,
rate of collisions =
3 pA = 2 mvrms
3 (2 × 101, 300 Pa)(0.10 m × 0.10 m) = 6.55 × 10 25 s −1 2(28 × 1.661 × 10 −27 kg)(576 m / s)
This collision rate can also be found by using the expression in Eq. 18.10, making the same change in vx, and using the ideal gas law to determine N/V.
18.13. Solve: The average translational kinetic energy per molecule in a gas is ε avg =
1 2 3 mvrms = kB T ⇒ vrms = 2 2
3kB T m
The mass of a helium molecule is m = 4 u = 4(1.66 × 10−27 kg) = 6.64 × 10−27 kg
⇒ vrms =
3(1.38 × 10 −23 J / K )(1273 K ) 6.64 × 10 −27 kg
= 2820 m / s
For argon m = 40 u, so the root-means-square speed is
vrms =
3(1.38 × 10 −23 J / K )(1273 K ) 40 × 1.66 × 10 −27 kg
= 891 m / s
18.14. Solve: The average translational kinetic energy per molecule is ε avg =
2 mvrms 1 2 3 mvrms = kB T ⇒ T = 3kB 2 2
For nitrogen molecules m = 28 u, so
(28 × 1.66 × 10 kg)(30 m / s) 3(1.38 × 10 kg) −27
T=
−23
2
= 1.0 K
18.15. Solve: Because the neon and argon atoms in the mixture are in thermal equilibrium, the temperature of each gas in the mixture must be the same. That is, using Equation 18.26, 2 2 mAr vrms Ar = mNe vrms Ne
vrms Ar = vrms Ne
mNe 20 u = 283 m / s = ( 400 m / s) mAr 40 u
18.16. Solve: The average translational kinetic energy per molecule is ε avg =
1 2 3 mvrms = kB T ⇒ vrms = 2 2
3kB T m
Since we want the vrms for H2 and N2 to be equal,
3kB TH2 mH 2
=
mH 2 2u 3kB T ⇒ TH2 = TN = (373 K) = 26.6 K mN 2 mN 2 2 28 u
18.17. Solve: The formula for the root-means-square speed as a function of temperature is
(vrms )T = (a) For (vrms ) T =
1 2
3kB T m
(vrms )STP , 3kB T = m
1 2
3kB (273 K ) 1 ⇒ T = (273 K) = 68.3 K m 4
(b) For (vrms ) T = 2(vrms )STP ,
3kB (273 K ) 3kB T =2 ⇒ T = 4(273 K ) = 1090 K m m
18.18. Solve: The formula for the root-means-square speed as a function of temperature is
(vrms )T =
3kB T m
The ratio at 20째C and at 100째C is
(vrms )100 (vrms )20
=
373 K = 1.13 293 K
18.19. Solve: Assuming ideal-gas behavior and ignoring relativistic effects, the root-mean-square speed of a molecule is
3kB T m is the speed of light (c) for a hydrogen molecule vrms =
The temperature where vrms
−27 8 2 mvrms (2 u)c 2 2(1.66 × 10 kg)(3.0 × 10 m / s) T= = = 7.22 × 1012 K = 3kB 3kB 3(1.38 × 10 −23 J / K ) 2
18.20. Solve: (a) The average translational kinetic energy per molecule is 2 ε avg = 12 mvrms = 23 kB T
This means εavg doubles if the temperature T doubles. (b) The root-mean-square speed vrms increases by a factor of 2 as the temperature doubles. (c) The mean free path is 1 λ= 4 2π ( N V )r 2 Because N/V and r do not depend on T, doubling temperature has no effect on λ.
18.21. Solve: (a) The total translational kinetic energy of a gas is K micro = 23 N A kBT = 23 nRT . For H2 gas at STP,
K micro =
3 2
(1.0 mol)(8.31 J / mol K)(273 K) = 3400 J
K micro =
3 2
(1.0 mol)(8.31 J / mol K)(273 K) = 3400 J
(b) For He gas at STP, (c) For O2 gas at STP, K micro = 3400 J. Assess: The translational kinetic energy of a gas depends on the temperature and the number of molecules but not on the molecule’s mass.
18.22. Solve: (a) The mean free path is λ=
1 4 2π ( N V )r 2
where r ≈ 0.5 × 10−10 m is the atomic radius for helium and N/V is the gas number density. From the ideal gas law,
N p 0.10 atm × 101,300 Pa / atm = 7.34 × 10 25 m −3 = = V kT (1.38 × 10 −23 J / K)(10 K) ⇒λ=
(
1
4 2π 7.34 × 10
25
m
−3
)(0.5 × 10
−10
m)
2
= 3.07 × 10 −7 m = 307 nm
(b) The root-mean-square speed is
vrms =
3kB T = m
3(1.38 × 10 −23 J / K )(10 K ) 4 × (1.661 × 10 −27 kg)
= 250 m / s
where we used A = 4 u as the atomic mass of helium. (c) The average energy per atom is ε avg = 23 kB T = 23 1.38 × 10 −23 J / K (10 K ) = 2.07 × 10 −22 J .
(
)
18.23. Solve: (a) The average kinetic energy of a proton at the center of the sun is ε avg = 23 kB T ≈
3 2
(1.38 × 10
−23
J / K )(2 × 10 7 K ) = 4 × 10 −16 J
(b) The root-mean-square speed of the proton is
vrms =
3kB T ≈ m
3(1.38 × 10 −23 J / K )(2 × 10 7 K ) 1.67 × 10 −27 kg
= 7 × 10 5 m / s
18.24. Solve: (a) Since the hydrogen in the sun’s atmosphere is monatomic, the average translational kinetic energy per atom is ε avg = 23 kB T =
3 2
(1.38 × 10
−23
J / K )(6000 K ) = 1.24 × 10 −19 J
(b) The root-mean-square speed is
vrms =
3kB T = mH
3(1.38 × 10 −23 J / K )(6000 K ) 1.67 × 10 −27 kg
= 1.22 × 10 4 m / s
18.25. Solve: The volume of the air is V = 6.0 m × 8.0 m × 3.0 m = 144.0 m3, the pressure p = 1 atm = 1.013 × 105 Pa, and the temperature T = 20°C = 293 K. The number of moles of the gas is
n=
pV = 5991 mols RT
This means the number of molecules is
N = nN A = (5991 mols)(6.022 × 10 23 mol −1 ) = 3.61 × 10 27 molecules. Since air is a diatomic gas, the room’s thermal energy is
Eth = Nε avg = N ( 25 kB T ) = 3.65 × 10 7 J Assess: The room’s thermal energy can also be obtained as follows:
Eth = nCV T = (5991 mols)(20.8 J/mol K)(293 K) = 3.65 × 107 J
18.26. Solve: The thermal energy of a solid is Eth = 3 NkB T = 3nRT The volume of lead V = 100 cm3 = 10−4 m3 which means the mass is M = ρV = (11,300 kg/m3)(10−4 m3) = 1.13 kg Because the atomic mass number of Pb is 207, the number of moles is
n=
M 1.13 kg = = 5.459 mol Mmol 0.207 kg / mol
⇒ Eth = 3(5.459 mols)(8.31 J / mol K )(293 K ) = 3.99 × 10 4 J
18.27. Solve: (a) For a monatomic gas, ∆Eth = nCV ∆T = 1.0 J = (1.0 mol)(12.5 J / mol K )∆T ⇒ ∆T = 0.0800°C or 0.0800 K (b) For a diatomic gas, 1.0 J = (1.0 mol)(20.8 J/mol K)∆T ⇒ ∆T = 0.0481°C or K (c) For a solid, 1.0 J = (1.0 mol)(25.0 J/mol K) ∆T ⇒ ∆T = 0.0400°C or K
18.28. Solve: The conservation of energy equation (∆Eth)gas + (∆Eth)solid = 0 J is ngas (CV ) gas (Tf − Ti ) gas + nsolid (CV )solid (Tf − Ti )solid = 0 J ⇒ (1.0 mol)(12.5 J/mol K)(−50 K) + (1.0 mol)(25.0 J/mol K)(∆T)solid = 0 ⇒ (∆T)solid = 25°C The temperature of the solid increases by 25°C.
Refer to Figure 18.13. At low temperatures, CV = 23 R = 12.5 J / mol K. At room temperature and modestly hot temperatures, CV = 25 R = 20.8 J / mol K. At very hot temperatures, CV = 27 R = 29.1 J / mol K. Solve: (a) The number of moles of diatomic hydrogen gas in the rigid container is
18.29. Visualize:
0.20 g = 0.1 mol 2 g / mol The heat needed to change the temperature of the gas from 50 K to 100 K at constant volume is Q = ∆Eth = nCV ∆T = (0.1 mol)(12.5 J/mol K)(100 K – 50 K) = 62.4 J (b) To raise the temperature from 250 K to 300 K, Q = ∆Eth = (0.1 mol)(20.8 J/mol K)(300 K – 250 K) = 104 J (c) To raise the temperature from 550 K to 600 K, Q = 104 J. (d) To raise the temperature from 2250 K to 2300 K, Q = ∆Eth = nCV∆T = (0.10 mol)(29.1 J/mol K)(50 K) = 145 J.
18.30. Solve: (a) Nitrogen is a diatomic gas. The thermal energy of the gas at T = 20°C = 293 K is Eth = =
pV 5 5 R T nRT = RT 2 2
(1.013 × 10
7
Pa )(15, 000 × 10 −6 m 3 )
(8.31 J / mol K)(293 K)
(20.8 J / mol K)(293 K) = 3.80 × 10 5 J
(b) The number density of the gas is
(1.013 × 10 7 Pa) = 2.51 × 1027 m−3 N p = = V kB T (1.38 × 10 −23 J / K )(293 K )
Since nitrogen is a diatomic molecule, the mean free path is
λ=
1 1 −9 m = 2 = 2.25 × 10 4 2π ( N V )r 2 4 2π (2.51 × 10 27 m −3 )(1.0 × 10 −10 m )
(c) For an isothermal process, ∆T = 0 K. Thus ∆Eth = 0 J.
18.31. Solve: (a) The thermal energy of a monatomic gas is E th = 23 NkB T = 23 nRT ⇒ T =
2 Eth 1 3 n R
2 5000 J 1 ⇒ TA = = 201 K 3 2.0 mol (8.31 J / mol K ) 2 8000 J 1 TB = = 214 K 3 3.0 mol (8.31 J / mol K ) Thus, gas B has the higher initial temperature. (b) The equilibrium condition is (ε A )avg = (ε B )avg = (ε tot )avg . This means
EAf E Etot = Bf = nA nB n A + nB ⇒ EAf =
nA 2 mols 5000 J + 8000 J = 5200 J Etot = ( ) 2 mols + 3 mols n A + nB EBf =
nB 3 mols Etot = (13,000 J) = 7800 J 5 mols n A + nB
18.32. Solve: The equilibrium condition is (εA)avg = (εB)avg = (εtot)avg ⇒
EAf E Etot = Bf = nA nB n A + nB
Thus the final thermal energies are
nA 4.0 mols 9000 J + 5000 J = 8000 J EAf = ( ) Etot = 4.0 mols + 3.0 mols n A + nB nB 3.0 mols 14,000 J = 6000 J EBf = ( ) Etot = 7.0 mols n A + nB Because EAi = 9000 J and EAf = 8000 J, 1000 J of heat energy is transferred from gas A to gas B.
18.33. Solve: The mean free path for a monatomic gas is λ=
1 V ⇒ = 4π 2 λ r 2 2 N 4π 2 ( N V )r
For λ = 2r, meaning that the mean free path equals the atomic diameter,
V/N 4π 3 V = 4π 2 (2r )(r 2 ) = r 6 2 ⇒ 4π 3 = 6 2 = 8.48 3 N 3 r
18.34.
Visualize:
Solve:
The average energy of an oxygen molecule at 300 K is
ε avg =
Eth 5 = 2 kB T = N
5 2
(1.38 × 10
−23
J / K )(300 K ) = 1.035 × 10 −20 J
The energy conservation equation Ugf + Kf = Ugi + Ki with Kf = ε avg is
mgyf + ε avg = mgyi + 0 J With yi = h and yf = 0 m, we have
mgh = 1.035 × 10 −20 J ⇒ h =
1.035 × 10 −20 J = 1.99 × 10 4 m (32 × 1.66 × 10 −27 kg)(9.8 m / s2 )
18.35. Solve: (a) To identify the gas, we need to determine its atomic mass number A or, equivalently, the
mass m of each atom or molecule. The mass density ρ and the number density (N/V) are related by ρ = m(N/V), so the mass is m = ρ(V/N). From the ideal-gas law, the number density is
N p 50,000 Pa = 1.208 × 10 25 m −3 = = V kT (1.38 × 10 −23 J / K )(300 K ) Thus, the mass of an atom is
m=ρ
V 8.02 × 10 −2 kg / m 3 = = 6.64 × 10 −27 kg N 1.208 × 10 25 m −3
Converting to atomic mass units,
A = 6.64 × 10 −27 kg ×
1u = 4.00 u 1.661 × 10 −27 kg
This is the atomic mass of helium. (b) Knowing the mass, we find vrms to be
vrms =
3kB T = m
3(1.38 × 10 −23 J / K )(300 K ) 6.64 × 10 −27 kg
= 1367 m / s
(c) A typical atomic radius is r ≈ 0.5 × 10−10 m. The mean free path is thus
λ=
1 1 −6 m = 1.86 µ m = 2 = 1.86 × 10 2 25 4 2π ( N V )r 4 2π 1.208 × 10 m −3 (0.5 × 10 −10 m )
(
)
18.36. Solve: (a) The number density is N / V = 1 cm −3 = 10 6 m −3 . Using the ideal-gas equation, N kB T ≈ (1 × 10 6 m −3 )(1.38 × 10 −23 J / K )(3 K ) V 1 atm = 4 × 10 −17 Pa × = 4 × 10 −22 atm 1.013 × 10 5 Pa
p=
(b) For a monatomic gas,
3kB T = m
vrms =
3(1.38 × 10 −23 J / K )(3 K ) 1.67 × 10 −27 kg
(c) The thermal energy is Eth = 23 NkB T , where N = (10 6 m −3 )V . Thus
Eth = 1.0 J =
3 2
(10
6
= 270 m/s
m −3 )V (1.38 × 10 −23 J / K )(3 K ) ⇒ V = 1.6 × 1016 m 3 = L3 ⇒ L = 2.5 × 10 5 m
18.37. Solve: Mass m of a dust particle is
[
]
m = ρV = ρ( 43 πr 3 ) = (2500 kg / m 3 ) 43 π (5 × 10 −6 m ) = 1.3 × 10 −12 kg 3
The root-mean-square speed of the dust particles at 20°C is
vrms =
3kB T = m
3(1.38 × 10 −23 J / K )(293 K ) 1.3 × 10 −12 kg
= 9.6 × 10 −5 m / s
18.38. Solve: Fluorine has atomic mass number A = 19. Thus the root-mean-square speed of 238UF6 is vrms ( 238 UF6 ) =
3kB T = m
3kB T 238 u + 6 × 19 u
The ratio of the root-mean-square speed for the molecules of this isotope and the 235UF6 molecules is
vrms vrms
( (
235 238
UF6
)=
UF6 )
352 (238 + 6 × 19) u = = 1.0043 349 (235 + 6 × 19) u
18.39. Solve: (a) If the electron can be thought of as a point particle with zero radius, then it will collide with any gas particle that is within r of its path. Hence, the number of collisions Ncoll is equal to the number of gas particles in a cylindrical volume of length L. The volume of a cylinder is Vcyl = AL = πr 2 L . If the number density
( )
of the gas is (N/V) particles per m3, then the number of collisions along a trajectory of length L is
N coll = Introducing a factor of
(
)
N N L 1 Vcyl = π r 2 L ⇒ λ electron = = V V N coll π ( N / V )r 2
2 to account for the motion of all particles, L 1 λ electron = = N coll 2π ( N / V )r 2
(b) Assuming that most of the molecules in the accelerator are diatomic,
5.0 × 10 4 m =
1
2π ( N / V )(1.0 × 10
−10
m)
2
⇒ N / V = 4.50 × 1014 m −3
From the ideal-gas equation,
p=
N kB T = ( 4.50 × 1014 m −3 )(1.38 × 10 −23 J / K )(293 K ) = 1.82 × 10 −6 Pa = 1.80 × 10 −11 atm V
18.40. Visualize: Please refer to Figure P18.40 in the textbook. Solve: (a) The most probable speed is 4 m/s. (b) The average speed is
vavg =
2 × 2 m / s + 4 × 4 m / s + 3× 6 m / s +1× 8 m / s = 4.6 m / s 2 + 4 + 3+1
(c) The root-mean-square speed is
2 × (2 m / s) + 4 × ( 4 m / s) + 3 × (6 m / s) + 1 × (8 m / s) = 4.94 m / s 2 + 4 + 3 +1 2
vrms =
2
2
2
18.41. Solve: (a) The cylinder volume is V = πr2L = 1.571 × 10–3 m3. Thus the number density is N 2.0 × 10 22 = = 1.273 × 10 25 m −3 V 1.571 × 10 −3 m 3 (b) The mass of an argon atom is m = 40 u = 40(1.661 × 10−27 kg) = 6.64 × 10−26 kg
⇒ vrms =
3kB T = m
3(1.38 × 10 −23 J / K )(323 K ) 6.64 × 10 −26 kg
(c) vrms is the square root of the average of v2. That is,
( )
2 vrms = (v 2 )avg = (v x2 )avg + v y2
avg
= 449 m / s
+ (vz2 )avg
( )
An atom is equally likely to move in the x, y, or z direction, so on average v x2 2 vrms = 3(v x2 )avg ⇒ (v x ) rms =
(v )
avg
( )
= v y2
avg
= (vz2 )avg . Hence,
vrms = 259 m / s 3 (d) When we considered all the atoms to have the same velocity, we found the collision rate to be 12 ( N V ) Av x (see Equation 18.10). Because the atoms move with different speeds, we need to replace vx with (vx)rms. The end of the cylinder has area A = πr2 = 7.85 × 10–3 m2. Therefore, the number of collisions per second is 1 2
2 x avg
=
( N / V ) A(v x ) rms = 12 (1.273 × 10 25 m −3 )(7.85 × 10 −3 m 2 )(259 m / s) = 1.296 × 1025 s−1
(e) From kinetic theory, the pressure is N N 2 = p = 13 m v 2 avg = 13 mvrms V V
( )
(f) From the ideal gas law, the pressure is
1 3
(1.273 × 10
25
)
m −3 (6.64 × 10 −26 kg)( 449 m / s) = 56,800 Pa
22 −23 NkB T (2.0 × 10 )(1.38 × 10 J / K )(323 K ) = 56,700 Pa = 1.571 × 10 −3 m 3 V Assess: The very slight difference with part (e) is due to rounding errors.
p=
2
18.42. Model: Pressure is due to random collisions of gas molecules with the walls. Solve:
According to Eq. 18.10, the collision rate with one wall is N 1N rate of collisions = coll = Av x ∆t coll 2 V
However, this equation assumed that all molecules are moving in the x-direction with constant speed. The rms speed vrms is for motion in three dimensions at varying speeds. Consequently, we need to replace vx not with (vx)avg, which is zero, but with
v x → (v x2 ) avg =
2 vrms v = rms 3 3
rate of collisions =
1 N Avrms 2 3 V
With this change,
The molecular mass of nitrogen is A = 28 u, thus the rms speed of the molecules at 20°C is
vrms =
3kB T = m
3(1.38 × 10 −23 J / K) (293 K) = 510 m / s 28(1.661 × 10 –27 kg)
With N = 0.10NA = 6.02 × 1022 molecules, the number density is
N 6.02 × 10 22 = = 6.02 × 10 25 m −3 V 0.10 m × 0.10 m × 0.10 m Thus
rate of collisions =
1 (6.02 × 10 25 m −3 )(0.10 m × 0.10 m)(510 m / s) = 8.86 × 10 25 s −1 2 3
18.43. Solve: (a) The number of molecules of helium is N helium =
(
)
2.0 × 1.013 × 10 5 Pa (100 × 10 −6 m 3 ) pV = 3.936 × 10 21 = kB T (1.38 × 10 −23 J / K)(373 K)
⇒ nhelium =
3.936 × 10 21 = 6.536 × 10 −3 mol 6.022 × 10 23 mol −1
The initial internal energy of helium is
Ehelium i = 23 N helium kB T = 30.4 J The number of molecules of argon is
N argon =
(
)
4.0 × 1.013 × 10 5 Pa (200 × 10 −6 m 3 ) pV = 8.726 × 10 21 = kB T (1.38 × 10 −23 J / K)(673 K)
⇒ nargon =
8.726 × 10 21 = 1.449 × 10 −2 mol 6.022 × 10 23 mol −1
The initial thermal energy of argon is
Eargon i = 23 N argon kB T = 121.6 J (b) The equilibrium condition for monatomic gases is (εhelium f)avg = (εargon f)avg = (εtotal)avg
⇒
Ehelium f nhelium
=
Eargon f nargon
=
Etot (30.4 + 121.6) J = = 7228 J / mol ntot (6.54 × 10 −3 + 1.449 × 10 −2 ) mol
⇒ Ehelium f = (7228 J / mol)nhelium = (7228 J / mol)(6.54 × 10 −3 mol) = 47.3 J Eargon f = (7228 J / mol)nargon = (7228 J / mol)(1.449 × 10 −2 mol) = 104.7 J
(c) The amount of heat transferred is
Ehelium f − Ehelium i = 47.3 J − 30.4 J = 16.9 J Eargon f − Eargon i = 104.7 J − 121.6 J = −16.9 J The helium gains 16.9 J of heat energy and the argon loses 16.9 J. Thus 16.9 J are transferred from the argon to the helium. (d) The equilibrium condition for monatomic gases is
(ε helium )avg = (ε argon )avg ⇒
Ehelium f N helium
=
Eargon f N argon
= 23 kB Tf
Substituting the above values,
47.3 J 104.7 J = = 3.936 × 10 21 8.726 × 10 21
3 2
(1.38 × 10
−23
J / K )Tf ⇒ TF = 580 K
(e) The final pressure of the helium and argon are
phelium f =
pargon f =
21 −23 N helium kB T (3.936 × 10 )(1.38 × 10 J / K )(580 K ) = 3.15 × 105 Pa = 3.11 atm = 100 × 10 −6 m 3 Vhelium
N argon kB T Vargon
(8.726 × 10 )(1.38 × 10 21
=
200 × 10
−6
−23
m
J / K )(580 K ) 3
= 3.49 × 105 Pa = 3.45 atm
18.44. Solve: (a) The number of moles of helium and oxygen are nhelium =
2.0 g = 0.50 mol 4.0 g / mol
noxygen =
8.0 g = 0.25 mol 32.0 g / mol
Since helium is a monoatomic gas, the initial thermal energy is
Ehelium i = nhelium ( 23 RThelium ) = (0.50 mol)( 23 )(8.31 J / mol K )(300 K ) = 1870 J Since oxygen is a diatomic gas, the initial thermal energy is
Eoxygen i = noxygen
(
5 2
RToxygen
)
= (0.25 mol)( 25 )(8.31 J / mol K )(600 K ) = 3116 J
(b) The total initial thermal energy is Etot = Ehelium i + Eoxygen i = 4986 J As the gases interact, they come to equilibrium at a common temperature Tf. This means
4986 J = nhelium ( 23 RTf ) + noxygen ( 25 RTf ) ⇒ Tf =
4986 J
( R)(3nhelium + 5noxygen ) 1 2
=
1 2
4986 J = 436.4 K = 436 K (8.31 J / mol K)(3 × 0.50 mol + 5 × 0.25 mol)
The thermal energies at the final temperature Tf are
Ehelium f = nhelium ( 23 RTf ) = ( 23 )(0.50 mol)(8.31 J / mol K )( 436.4 K ) = 2720 J Eoxygen f = noxygen ( 25 RTf ) = ( 25 )(0.25 mol)(8.31 J / mol K )( 436.4 K ) = 2266 J (c) The change in the thermal energies are
Ehelium f − Ehelium i = 2720 J − 1870 J = 850 J
Eoxygen f − Eoxygen i = 2266 J − 3116 J = −850 J
The helium gains energy and the oxygen loses energy. (d) The final temperature can also be calculated as follows:
Ehelium f = (nhelium ) 23 RTf ⇒ 2720 J = (0.50 mol)(1.5)(8.31 J/mol K)Tf ⇒ Tf = 436.4 K = 436 K
18.45. Solve: The thermal energy of any ideal gas is related to the molar specific heat at constant volume by Eth = nCVT Since CP = CV + R ,
20.8 J / mol K = CV + R ⇒ CV = 12.5 J / mol K The number of moles of the gas is
n=
1.0 × 10 20 = 1.66 × 10 −4 mol 6.02 × 10 23 mol −1
Thus
T=
(1.0 J) = 482 K (1.66 × 10 mols)(12.5 J / mol K) −4
18.46. Solve: For a system with n degrees of freedom, the molar specific heat is CV = nR/2. The specific heat ratio is
CP CV + R R R R = = 1+ ⇒ = γ − 1 = 1.29 − 1 = 0.29 ⇒ CV = = 3.5 R = 27 R CV CV CV CV 0.29 Thus, the system has 7 degrees of freedom.
γ =
18.47. Solve: The molar specific heat at constant volume CV can in general be written as CV = 12 f R , where f is the number of degrees of freedom. Using CP = CV + R ,
γ =
CP CV + R R 2R f + 2 = = 1+ = 1+ = CV CV CV fR f
For a monatomic gas γ = (3 + 2) 3 = 1.67 and for a diatomic gas γ = (5 + 2) 5 = 1.40 .
18.48. Solve: As the volume V1 of a gas increases to V2 = 2V1 at a constant pressure p1 = p2, the temperature of the gas changes from T1 to T2 as follows: p2 V2 pV V p = 1 1 ⇒ T2 = T1 2 2 = 2T1 T2 T1 V1 p1 Since the process occurs at constant pressure the heat transferred is
Q = nCP ∆T = nCP (T2 − T1 ) = nCP (2T1 − T1 ) = nCP T1 For a monatomic gas,
CP = CV + R = 23 R + R = 25 R For the diatomic gas,
CP = CV + R = 25 R + R = 27 R Thus
n( 27 R)T1 Qdiatomic = 5 = 1.40 Qmonatomic n( 2 R)T1
18.49. Solve: Assuming that the systems are thermally isolated except from each other, the total energy for the two thermally interacting systems must remain the same. That is,
E1i + E2i = E1f + E2f ⇒ E1f − E1i = E2i − E2f = −( E2f − E1f ) ⇒ ∆E1 = − ∆E2 No work is done on either system, so from the first law ∆E = Q. Thus Q1 = −Q2 That is, the heat lost by one system is gained by the other system.
18.50. Solve: (a) From Equation 18.26 vrms = 3kBT m . For an adiabatic process γ −1 f
Tf V
= Ti Vi
γ −1
V ⇒ Tf = Ti i Vf
γ −1
⇒ Tf = Ti (8) 3 5
−1
= 4Ti
The root-mean-square speed increases by a factor of 2 with an increase in temperature.
[
(b) From Equation 18.3 λ = 4 2π ( N / V )r 2 (c) For an adiabatic process,
]
−1
. A decrease in volume decreases the mean free path by a factor of 1/8.
V Tf Vfγ −1 = Ti Viγ −1 ⇒ Tf = Ti i Vf
γ −1
= Ti (8) 3 5
−1
= 4Ti
Because the decrease in volume increases Tf , the thermal energy increases by a factor of 4. (d) The molar specific heat at constant volume is CV = 23 R, a constant. It does not change.
18.51. Solve: (a) The thermal energy of a monatomic gas of N molecules is Eth = Nε avg , where ε avg = 23 k BT . A monatomic gas molecule has 3 degrees of freedom. However, a two-dimensional monatomic gas molecule has only 2 degrees of freedom. Thus,
Eth = N ( 22 kB T ) = NkB T = nRT If the temperature changes by ∆T, then the thermal energy changes by ∆Eth = nR∆T . Comparing this form with ∆Eth = nCV ∆T , we have CV = R = 8.31 J/mol K. (b) A two-dimensional solid has 2 degrees of freedom associated with the kinetic energy and 2 degrees of freedom associated with the potential energy (or 2 spring directions), giving a total of 4 degrees of freedom. Thus, Eth = Nεavg and ε avg = 24 kB T . Or Eth = 2NkBT = 2nRT. For a temperature change of ∆T, ∆Eth = 2nR∆T = nCV∆T ⇒ CV = 2 R = 16.6 J / mol K.
18.52. Solve: (a) A polyatomic molecule can have 3 degrees of freedom associated with the translational kinetic energy and 3 degrees of freedom associated with rotations about three mutually perpendicular axes, giving a total of 6 degrees of freedom. Thus, Eth = Nε avg = N ( 62 kB T ) = 3 NkB T = 3nRT ⇒ ∆Eth = 3nR∆T = nCV ∆T ⇒ CV = 3 R = 25.0 J/mol K (b) Since CP = CV + R, CP =3R + R = 4R. The specific heat ratio of a polyatomic molecule is predicted to be C 4R γ = P = = 1.33 CV 3 R The prediction of 1.33 is very close to water’s value of 1.31.
18.53. Solve: (a) The rms speed is vrms hydrogen 3kB T ⇒ = vrms oxygen m
vrms =
32 u =4 2u
(b) The average translational energy is ε = 23 kB T . Thus
ε avg hydrogen ε avg oxygen
=
Thydrogen Toxygen
=1
(c) The thermal energy is
Eth = 25 nRT
⇒
Eth hydrogen Eth oxygen
=
nhydrogen noxygen
=
mhydrogen 2.0 g / mol
32.0 g / mol = 16 moxygen
18.54. Solve: The thermal energy of the neon (monatomic) gas is Eneon = 23 nneon RT =
3 2
(1 mol) RT = 23 RT
The thermal energy of the nitrogen (diatomic) gas is
Enitrogen = 25 nnitrogen RT =
5 2
(2 mol) RT = 5 RT
Thus the total thermal energy is
Eth = 23 RT + 5 RT =
13 2
RT ⇒ ∆Eth =
13 2
R∆T = nCV ∆T
where n is the total number of moles equal to nneon + nnitrogen = 3 mol. Therefore, we have
CV =
13 6
R = 18.0 J / mol K
18.55. Solve: The thermal energy of a monatomic gas of n1 moles is E1 = 23 n1 RT . The thermal energy of a diatomic gas of n2 moles is E2 = 25 n2 RT . The total thermal energy of the mixture is
Eth = E1 + E2 =
1 2
(3n1 + 5n2 ) RT ⇒ ∆Eth = 12 (3n1 + 5n2 ) R∆T
Comparing this expression with
∆E th = ntotal CV ∆T = (n1 + n2 )CV ∆T we get
(n1 + n2 )CV =
(3n1 + 5n2 ) R 2
The requirement that the ratio of specific heats is 1.50 means
γ = 1.50 =
CP CV + R R = = 1+ ⇒ CV = 2 R CV CV CV
The above equation is then
(n1 + n2 )(2 R) =
(3n1 + 5n2 ) R ⇒ 4n 2
1
+ 4n2 = 3n1 + 5n2 ⇒ n1 = n2
Thus, monatomic and diatomic molecules need to be mixed in the ratio 1:1. Or the fraction of the molecules that are monatomic needs to be 12 .
18.56. Solve: (a) The thermal energy of a monatomic gas of n1 moles is E1 = 23 n1 RT . The thermal energy of a diatomic gas of n2 moles is E2 = 25 n2 RT . The total thermal energy of the mixture is
Eth =
1 2
(3n1 + 5n2 ) RT ⇒ ∆Eth = 12 (3n1 + 5n2 ) R∆T
Comparing this expression with
∆E th = ntotal CV ∆T = (n1 + n2 )CV ∆T we get
CV =
(3n1 + 5n2 ) R 2(n1 + n2 )
(b) For a diatomic gas, n1 → 0, and CV = 25 R . For a monotomic gas, n2 → 0, and CV = 23 R .
18.57. Solve: (a) The thermal energy is Eth = ( Eth ) N + ( Eth ) O = 25 N N 2 kB T + 25 N O2 kB T = 25 N total kB T 2
2
where Ntotal is the total number of molecules. The identity of the molecules makes no difference since both are diatomic. The number of molecules in the room is
N total =
pV (101,300 Pa )(2 m × 2 m × 2 m ) = = 2.15 × 10 26 −23 kB T J / K 273 K 1 . 38 × 10 ) ( )(
The thermal energy is
Eth =
5 2
(2.15 × 10 )(1.38 × 10 26
−23
J / K )(273 K ) = 2.03 × 10 6 J
(b) A 1 kg ball at height y = 1 m has a potential energy U = mgy = 9.8 J. The ball would need 9.8 J of initial kinetic energy to reach this height. The fraction of thermal energy that would have to be conveyed to the ball is 9.8 J = 4.83 × 10 −6 2.03 × 10 6 J (c) A temperature change ∆T corresponds to a thermal energy change ∆Eth = 25 N total kB ∆T . But 25 N total kB = Eth T . Using this, we can write
∆Eth =
Eth ∆Eth −9.8 J ∆T ⇒ ∆T = 273 K = −0.00132 K T= 2.03 × 10 6 J T Eth
The room temperature would decrease by 0.00132 K or 0.00132°C. (d) The situation with the ball at rest on the floor and in thermal equilibrium with the air is a very probable distribution of energy and thus a state with high entropy. Although energy would be conserved by removing energy from the air and transferring it to the ball, this would be a very improbable distribution of energy and thus a state of low entropy. The ball will not be spontaneously launched from the ground because this would require a decrease in entropy, in violation of the second law of thermodynamics. As another way of thinking about the situation, the ball and the air are initially at the same temperature. Once even the slightest amount of energy is transferred from the air to the ball, the air’s temperature will be less than that of the ball. Any further flow of energy from the air to the ball would be a situation in which heat energy is flowing from a colder object to a hotter object. This cannot happen because it would violate the second law of thermodynamics.
18.58. Solve: This is not a wise investment. Although an invention to move energy from the hot brakes to the forward motion of the car would not violate energy conservation, it would violate the second law of thermodynamics. The forward motion of the car is a very improbable distribution of energy. It happens only when a force accelerates the car and then sustains the motion against the retarding forces of friction and air resistance. The moving car is a state of low entropy. By contrast, the random motion of many atoms in the hot brakes is a state of high probability and high entropy. To convert the random motion of the atoms in the brakes back into the forward motion of the car would require a decrease of entropy and thus would violate the second law of thermodynamics. In other words, the increasing temperature of the brakes as the car stops is an irreversible process that cannot be undone.
18.59. Solve: (a) We are given that 3kB Tf 3kB Ti vrms f ) = = 2(vrms i ) ( m m This means that Tf = 4Ti. Using the ideal-gas law, it also means that pfVf = 4piVi. Since the pressure is directly proportional to the volume during the process, we have pi/Vi = pf/Vf. Combining these two equations gives pf = 2pi and Vf = 2Vi.
(vrms i ) =
(b) The change in thermal energy for any ideal gas process is related to the molar specific heat at constant volume by ∆Eth = nCV (Tf − Ti ) . The work done on the gas is
W = − ∫ pdV = − (area under the p-versus-V graph) = − 23 pi Vi
The first law of thermodynamics ∆Eth = Q + W can be written Q = ∆Eth − W = nCV (Tf − Ti ) + 23 pi Vi = 3nCV Ti + 23 pi Vi
= 3n( 25 R)Ti + 23 pi Vi =
15 2
pi Vi + 23 pi Vi = 9 pi Vi
18.60. Solve: (a) The escape speed is the speed with which a mass m can leave the earth’s surface and escape to infinity (rf = ∞) with no left over speed (vf = 0). The conservation of energy equation Kf + Uf = Ki + Ui is 2 0 + 0 = 12 mvesc −
GMe m ⇒ vesc = Re
2GMe Re
The rms speed of a gas molecule is vrms = (3kBT/m)1/2. Equating vesc and vrms, and squaring both sides, the temperature at which the rms speed equals the escape speed is
2GMe 3kB T 2GMe = ⇒ T = m m Re 3kB Re For a nitrogen molecule, with m = 28 u, the temperature is
2(6.67 × 10 −11 N m 2 / kg 2 ) (5.98 × 10 24 kg) T = (28 × 1.661 × 10 −27 kg) = 141, 000 K 3(1.38 × 10 −23 J / K) (6.37 × 10 6 m) (b) For a hydrogen molecule, with m = 2 u, the temperature is less by a factor of 14, or T = 10,100 K. (c) The average translational kinetic energy of a molecule is ε avg = 23 kB T = 6.1 × 10–21 J at a typical atmosphere 2 temperature of 20°C. The kinetic energy needed to escape is K esc = 12 mvesc . For nitrogen molecules, Kesc = –18 2.9 × 10 J. Thus εavg/Kesc = 0.002 = 0.2%. Earth will retain nitrogen in its atmosphere because the molecules are moving too slowly to escape. But for hydrogen molecules, with Kesc = 2.1 × 10–19 J, the ratio is εavg/Kesc = 0.03 = 3%. Thus a large enough fraction of hydrogen molecules are moving at escape speed, or faster, to allow hydrogen to leak out of the atmosphere into space. Consequently, earth’s atmosphere does not contain hydrogen.
18.61. Solve: (a) The thermal energy of a monatomic gas of n1 moles at an initial temperature T1i is E1i = n1 RT1i . The thermal energy of a diatomic gas of n2 moles at an initial temperature T2i is E2i = 25 n2 RT2i . Consequently, the total initial energy is 3 2
3n1 RT1i + 5n2 RT2i (3n1T1i + 5n2 T2i ) R = 2 2
Etot = E1i + E2 i =
After the gases interact, they come to equilibrium with T1f = T2f = Tf. Then their total energy is
Etot =
(3n1 + 5n2 ) RTf 2
No energy is lost, so these two expressions for Etot must be equal. Thus,
(3n1T1i + 5n2 T2i ) R = (3n1 + 5n2 ) RTf 2
2
⇒ Tf =
3n1T1i + 5n2 T2i 2 Etot 2 ( E1i + E2i ) = = 3n1 + 5n2 R(3n1 + 5n2 ) R (3n1 + 5n2 )
The thermal energies at the final temperature Tf are
E1f = 23 n1 RTf = 23 n1 R
3n1 2 E1i + E2i = ( E1i + E2i ) R (3n1 + 5n2 ) 3n1 + 5n2
E2f = 25 n2 RTf = 25 n2 R
2 E1i + E2i 5n2 = ( E1i + E2i ) R 3n1 + 5n2 3n1 + 5n2
(b) In part (a) we found that
Tf =
3n1T1i + 5n2 T2i 3n1 + 5n2
(c) 2 g of He at T1i = 300 K are n1 = (2 g)/(4 g/mol) = 0.50 mol. Oxygen has an atomic mass of 16, so the molecular mass of oxygen gas (O2) is A = 32 g/mol. 8 g of O2 at T2i = 600 K are n2 = (8 g)/(32 g/mol) = 0.25 mol. The final temperature is
Tf =
3(0.50 mol)(300 K ) + 5(0.25 mol)(600 K ) = 436 K 3(0.50 mol) + 5(0.25 mol)
The heat flows to the two gases are found from Q = nCV∆T. For helium,
Q = n1CV ∆T = (0.50 mol)(12.5 J / mol K )( 436 K − 300 K ) = 850 J For O2,
Q = n2 CV ∆T = (0.25 mol)(20.8 J / mol K )( 436 K − 600 K ) = −850 J So 850 J of heat is transferred from the oxygen to the helium.