Homework 11 Solution Physics 404 - Spring 2011

Homework 11 Solutions

Question 1

Kittel+Kromer, Chpater 10, Problem 1

a) We know the Helmholtz free energy of the vdW gas:

FvdW = −Nτ{

log

(nQ(V −Nb)

N

)+ 1

}− N2a

V(1)

The entropy can be found from

σ = −∂F∂τ

∣∣∣∣V

(2)

= N

{log

(nQ(V −Nb)

N

)+ 1

}+Nτ

{N

nQ(V −Nb)(V −Nb)

N

dnQdτ

}(3)

Using nQ =(mτ

2πh̄2

)3/2we obtain

σ = N

{log

(nQ(V −Nb)

N

)+

5

2

}(4)

b) The energy U can be obtained from F by differentiation w.r.t. τ or simply using

F = U − τσ:

U = F + τσ (5)

= −Nτ{

log

(NQ(V −Nb)

N

)+ 1

}− N2a

V+ τ ×N

{log

(nQ(V −Nb)

N

)+

5

2

}(6)

=3

2Nτ − N2a

V(7)

which is the ideal gas energy plus the vdW correction for attractive interactions.

Question 2

Kittel+Kromer, Chpater 10, Problem 2 We will use the C-C equation

dp

dT=

L

T∆V(8)

together with the ideal gas law ∆V ≈ τ/p. Here we have an approximate equality since the

volume of the liquid phase is not exactly zero, but it is negligible compared to the volume

of the gas at normal pressures.

dp

dT=

L

T 2kBp (9)

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Homework 11 Solution Physics 404 - Spring 2011

Also note that in the C-C equation L is latent heat per particle, whereas we are given the

latent heat per gram. To do the conversion it is convenient to multiply the numerator and

denominator by Avogadro’s number NA:

dP

dT=

LNA

NAkBT 2p (10)

where LNA is the latent heat per mole of H2O, which is 2260J/g×18g/mole = 40, 680J/mole

and NAkB is the universal gas constant R = 8.314J/moleK. Hence

dp

dT= 0.035

atm

K(11)

dT

dp= 28.4

K

atm(12)

Question 3

Kittel+Kromer, Chpater 10, Problem 3 The data given is summarized in a plot below.

FIG. 1: Summary of data given in the problem

The coexistence line is described by the C-C equation

dp

dT=

L

τ∆V≈ L

kBT 2p =

LNA

RT 2p (13)

where we have proceeded as in question 2, above. The latent heat of vaporization is:

LNA =RT 2

p

dp

dT(14)

At T = −1oC we estimate the pressure as P ≈ 4.58+3.882

mmHg = 4.23mmHg. The slope of

the coexistence line is

dp

dT≈ ∆p

∆T=

4.58mmHg − 3.88mmHg

2K= 0.35

mmHg

K(15)

Thus the latent heat of vaporization is

LNA =8.31 J

moleK(272K)2

4, 23mmHg

0.35mmHg

K= 5.09× 104 J

mole(16)

Question 4

Kittel+Kromer, Chpater 10, Problem 4

2

Homework 11 Solution Physics 404 - Spring 2011

a) The atoms in the solid are 3D harmonic oscillators with a ground state energy of

−ε0

εs = (nx + ny + nz)h̄ω − ε0 (17)

with nx, ny, nz = 1, 2, · · · . The partition function for a single atom in the solid is

Zs =∑nx

∑ny

∑nz

e−[(nx+ny+nz)h̄ω−ε0]/τ (18)

= eε0/τ

(∑n

e−nh̄ω/τ

)3

(19)

The summation is the geometric sum which we have already seen before and gives 11−e−h̄ω/τ .

Hence the partition function becomes:

Zs =eε0/τ

(1− e−h̄ω/τ )3 (20)

Now proceed to calcualte the activity of the solid as on page 286 of K+K:

λs = e− logZs =1

Zs= e−ε0/τ

(1− e−h̄ω/τ

)3(21)

We use the ideal gas approximation to describe the gas phase and find

λg =n

nQ=

p

τnQ=p

τ

(2πh̄2

Mτ

)3/2

(22)

which is Eq. (10.30). If the solid and vapor are in equilibrium, then they share the same

chemical potential, temperature and activity:

λg = λs (23)

eµg/τ = eµs/τ (24)

p

τ

(2πh̄2

Mτ

)3/2

= e−ε0/τ(1− e−h̄ω/τ

)3(25)

so

p = τnQe−ε0/τ

(1− e−h̄ω/τ

)3(26)

This is the vapor pressure over the solid. Now consider the high temperature limit τ � h̄ω

and do an expansion is terms of the small parameter h̄ω/τ :

1− e−h̄ω/τ ≈ 1−(

1− h̄ω

τ· · ·)

=h̄ω

τ(27)

3

Homework 11 Solution Physics 404 - Spring 2011

So the pressure becomes in this limit:

p ≈ τnQe−ε0/τ

(h̄ω

τ

)3

(28)

=ω3

τ 2

(Mτ

2π

)3/2

e−ε0/τ (29)

=ω3

√τ

(M

2π

)3/2

e−ε0/τ (30)

b) The latent heat can be found from the C-C equation with the ideal gas approxima-

tion:

d

dτlog p =

1

p

dp

dτ=L

τ 2(31)

First form log p from the result for part a).

log p = −1

2log τ − ε0

τ+ const. (32)

Take the τ derivative

d

dτlog p = − 1

2τ+ε0

τ 2(33)

=1

τ 2

(ε0 −

τ

2

)(34)

=L

τ 2by Eq. (10.18) (35)

Hence the latent heat per atom is

L = ε0 −τ

2(36)

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