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University of Illinois at Chicago Department of Physics

SOLUTIONS

Thermodynamics and Statistical Mechanics Qualifying Examination

January 7, 2011 9:00 AM to 12:00 Noon

Full credit can be achieved from completely correct answers to 4 questions. If the student attempts all 5 questions, all the answers will be graded, and the top 4 scores will be counted towards the exam’s total score.

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Equation Sheet

exp[!b

!"

"

# x2] dx =$b

x2 exp[!b

0

"

# x2] dx =14

$b3

x2 exp[!

0

"

# x] dx = 2

dx x

ex !10

"

# =$ 2

6

dx x

ex +10

!

" =# 2

12

dx x2

ex !10

"

# = 2$ (3), where $ (3) can be considered to be just a number.

dx x2

ex +10

!

" = 32# (3)

n =

1

e(!"µ)/kT ±1

sinh(x) = 1

2(ex ! e!x )

cosh(x) = 1

2(ex + e!x )

ddx

sinh(x)!" #$ = cosh(x)

ddx

cosh(x)!" #$ = sinh(x)

xm

m=0

!" =

11# x

xm

m=0

n! =

1" xn+1

1" x

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Problem 1

A gas of N identical classical non-interacting atoms is held in a potential V (r) = ar , where

r = x2 + y2 + z2( )1/2

. The gas is in thermal equilibrium at temperature T.

(a) Find the single particle partition function Z1 of an atom in the gas. Express your answer

in the form Z1 = AT!a"# and provide an expression for the prefactor A and the exponents ! and ! . [Hint: convert the integral in r to spherical coordinates.]

Solution:

Z1 = exp[!E / kT ]" where E = p2 / 2m +V (r).

Z1 =1h3

exp[! px2 / 2mkT ]dpx

!#

#

"$

%&&

'

())

exp[! py2 / 2mkT ]dpy

!#

#

"$

%&&

'

())

exp[! pz2 / 2mkT ]dpz

!#

#

"$

%&&

'

())

* d+0

,

" d-0

2,

" dr0

#

" exp[!ar / kT ]r2 sin+

Using equation sheet values for integrals leads to

Z1 =2,mkT

h2

$%&

'()

3/2

4,( )2 kTa

$%&

'()

3

= 8,k3 2,mkh2

$%&

'()

3/2

T 9/2a!3 = AT.a!/

(b) Find an expression for the entropy S of this classical gas.

Solution:

Use F = !kT ln Z with Z =1N !

Z1N , then S = !

"F"T

#$%

&'( N ,V

F = !kT ! ln N !+ N ln Z1)* +, - !kT !N ln N + N + N ln Z1)* +,

S = k !N ln N + N + N ln Z1)* +, + kTN" ln Z1"T

#

$%&

'( N ,V

, where " ln Z1"T

#

$%&

'( N ,V

=9

2T

S = kN 112+ ln

Z1N

)

*.

+

,/

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Problem 2

A classical ideal gas is taken from state a to state b in the figure using three different paths: acb, adb, and ab. The pressure p2 = 2p1 and the volume V2 = 2 V1. (a) The heat capacity CV = 5

2 Nk . Starting from the First Law of Thermodynamics derive a value for Cp . No credit will be given for this part if you just state the answer.

Solution:

Cp =!Q!T

"#$

%&' p

. From the first Law dU = dQ ( pdV , let dQ = dU + pdV .

Write dU in terms of V and T as dU =!U!T

"#$

%&'VdT +

!U!V

"#$

%&'TdV .

The internal energy of an ideal gas depends only on temperature, so !U!V

"#$

%&'T

= 0.

Combining these results leads to dQ =!U!T

"#$

%&'VdT + pdV = CVdT + pdV .

Therefore, Cp =!Q!T

"#$

%&' p

= CV + p !V!T

"#$

%&' p

= CV + Nk, the latter is obtained from pV = NkT .

So, Cp = 72 Nk.

(b) Compute the heat supplied to the gas along each of the three paths, acb, adb, and ab, in

terms of N, k, and T1. Solution:

Paths acb, and adb consist of constant pressure and constant volume processes.

Q(acb) = CV dTa

c

! + Cp dTc

b

! = 52 Nk

V1Nk

dpp1

p2

! + 72 Nk

p2Nk

dVV1

V2

!

= 52 Nk

V1Nk

p2 " p1( ) + 72 Nk

p2Nk

V2 "V1( ) = 52 Nk

V1Nk

p1( ) + 72 Nk

2p1Nk

V1( ) = 192 NkT1

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Q(adb) = Cp dTa

d

! + CV dTd

b

! + = 72 Nk

p1Nk

dVV1

V2

! + 52 Nk

V2Nk

dpp1

p2

!

= 72 Nk

p1Nk

V2 "V1( ) + 52 Nk

V2Nk

p2 " p1( ) = 72 Nk

p1Nk

V1( ) + 52 Nk

2V1Nk

p1( ) = 172 NkT1

The heat along path ab can be calculated by taking the difference between !U and W .!U between states a and b can be calculated along any path because it is a state function.W along path ab is also easy to calculate. First, calculate !U =W +Q for any path. We already have Q(adb), so

W (adb) = " pdVa

d

# " pdVd

b

# = " p1 dVV1

V2

# = " p1V1 = "NkT1

!U =W (adb) +Q(adb) = 172 NkT1 " NkT1 = 15

2 NkT1

Now, for W (ab) = " pdVa

b

# = " p1VV1dV

V1

V2

# = "p1

2V1V2

2 "V12( ) = "

3p1V12

= " 32 NkT1

Q(ab) = !U "W (ab) = 152 NkT1 " " 3

2 NkT1( ) = 9NkT1

(c) What is the heat capacity Cab of the gas for the process ab?

Solution:

Cab =dQdT

!"#

$%& ab

, so consider dQ = dU + pdV ='U'T

!"#

$%&VdT +

'U'V

!"#

$%&TdV + pdV = CVdT + pdV

(Cab = CV + p dVdT

!"#

$%& ab

Derive an expression for dVdT

!"#

$%& ab

using p = p1V1V and p = NkT

V)V 2 =

NkTV1p1

differentiate: 2VdV =NkV1p1

dT or dVdT

!"#

$%& ab

=NkV12Vp1

(Cab = CV + p NkV12Vp1

= CV +Nk2

= 3Nk

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Problem 3

Consider a one-dimensional stretched elastic string that is fixed at its two ends and vibrates only in a direction perpendicular to its length. The string consists of a very large number N of atoms arranged in a single row. Let the energies of vibration be quantized in units of hf, where f is the vibration frequency. This string is in thermal equilibrium with a heat bath at temperature T. (a) Determine an expression for the thermal energy of this string in terms of an integral over

the variable x = ! / kT . Solution:

The Planck distribution n =

1

e! /kT "1

(note that equation sheet has general expression n =

1

e(!"µ)/kT ±1).

U = 2

!

e! /kT "1n# , where the 2 is from the two polarizations of vibrations and the energy levels

are given by ! = hf =

hcs"

=nhcs2L

, L is the length of the string and cs is the speed of sound of

the vibrations.

For N atoms in this string, we can write U = 2 dn !

e! /kT "10

N

# .

Let’s rewrite this integral in terms of x = ! / kT with the limit xmax =

hcsN2LkT

.

U = 2

2Lhcs

!

"#$

%&kT( )2 dx x

ex '10

xmax

(

(b) Identify a characteristic temperature that separates low T and high T behavior.

Determine an expression for the thermal energy of this string in the limit of low and high T. Comment on these results in the context of the equipartition theorem.

Solution:

Rewrite xmax in terms of a characteristic temperature as xmax =

hcsN2LkT

=TDT

, where TD is a

characteristic (Debye) temperature for 1-d vibrations. Now, we can rewrite the energy integral

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as U =

2NkT 2

TDdx x

ex !10

TD /T

" .

At high T, we approximate ex ! 1+ x , so that

U !

2NkT 2

TDdx

0

TD /T

" =2NkT 2

TD

TDT

#

$%&

'(= 2NkT . This agrees with the equipartition result that

assigns an average thermal energy of 12

kT to each quadratic degree of freedom. The number of

vibrational modes for each polarization of a 1-dimensional string of atoms approaches N for large N. Each mode has 2 degrees of freedom (one for the kinetic energy and one for the potential energy), plus there are two polarizations.

At low T, we let TD / T !" . The integral

dx x

ex !10

"

# =$ 2

6 (from equation sheet), so that

U !

" 2NkT 2

3TD. Equipartition theorem does not say anything about this low T result.

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Problem 4

Consider a spherical drop of liquid water containing Nl molecules surrounded by N - Nl molecules of water vapor. The drop and its vapor may be out of equilibrium. (a) Neglecting surface effects write an expression for the Gibbs free energy of this system if

the chemical potential of liquid water in the drop is µl and the chemical potential of water in the vapor is µv. Rewrite Nl in terms of the (constant) volume per molecule in the liquid, vl, and the radius r of the drop.

Solution: G = µlNl + µv(N ! Nl ) = Nµv + Nl (µl ! µv ) = Nµv +4"r3

3vl(µl ! µv )

(b) The effect of the surface of the drop can be included by adding a piece Gsurface = !A to the

free energy, where ! is the surface tension (! > 0) and A is the surface area of the drop. Write Gtotal with the surface piece expressed in terms of r. Make two qualitative hand-sketches of Gtotal: one sketch with (µl ! µv ) > 0 and one sketch with (µl ! µv ) < 0. Describe the behavior of the drop in these two cases.

Solution: Gtotal = Nµv +4!r3

3vl(µl " µv ) + 4!#r

2

For (µl ! µv ) > 0, the drop radius is always zero, so the drop just evaporates. For (µl ! µv ) < 0, the drop radius is either zero or the drop increases in size depending upon whether or not the radius is larger than a critical radius.

0 1 2 3 4 5 6r

100

200

300

400G

1 2 3 4 5 6r

- 20- 10

1020

G

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(c) Under appropriate conditions, there is a critical radius, rc, that separates drops which

grow in size from those that shrink. Determine this critical radius. Solution: We saw in part (b) that for (µl ! µv ) < 0 the drop either evaporates completely or grows. The critical radius separates these two types of behaviors and can be determined from the maximum in G. dGdr rc

= !4"rc

2

vl(µv ! µl ) + 8"#rc = 0

rc =2#vlµv ! µl

(d) Assume that the vapor behaves as an ideal gas and recall that the chemical potential of an

ideal gas is given by µv = µvo + kT ln(p / po ) . Write the chemical potential difference

(µv ! µl ) in terms of the vapor pressure and a reference pressure po , where po is taken to be the pressure of a vapor in equilibrium with a large flat surface of water. Then, derive and comment on the dependence of the relative humidity p / po on rc .

Solution: If po is the pressure of a vapor in equilibrium with a large flat surface of water, then the reference chemical potential is just the chemical potential of the bulk liquid, µv

o = µl .

Therefore, µv ! µl = kT ln(p / po ) and rc =

2!vlkT ln(p / po )

.

So, p / po = exp(2!vl / kTrc ) . Smaller critical radii require higher relative humidity, so drops grow more easily in a more humid environment.

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Problem 5

Consider a paramagnetic material whose magnetic particles have angular momentum J, which is a multiple of !. The projections of the angular momentum along the z-axis can take 2J – 1 values ( Jz = !J ,!J +1,!J + 2,..., J ), which leads to 2J – 1 allowed values of the z – component of a particle’s magnetic moment (

µz = !J"µ ,!(J +1)"µ ,..., J"µ ). The energy of

the magnetic moment in a magnetic field pointing in the +z direction is !µz B . (a) Derive an expression for the partition function Z1 of a single magnetic particle in a

magnetic field B pointing in the +z direction. Write your answer in terms of hyperbolic

sin functions, where sinh(x) = 1

2(ex ! e!x ) . You may find it convenient to use the

variable b = !µB" , where

! = 1

kT .

Solution: The allowed energies are

E = J!µB,(J +1)!µB,...,"J!µB .

The partition function is

Z1 = e!J"µB#

+ e!(J+1)"µB#

+ e!(J+2)"µB#

+ ...+ eJ"µB#

, where # = 1kT

Let b = "µB#,

Z1 = e!Jb + e!(J+1)b + e!(J+2)b + ...+ eJb

= e!Jb 1+ eb + e2b + ...+ e2Jb$%&

'()

= e!Jb 1+ eb + eb( )2 + ...+ eb( )2J$

%&&

'

())

= e!Jb1! eb( )2J+1

1! eb

$

%

&&&&

'

(

))))

=e!Jb ! e(J+1)b

1! eb

To write this in terms of sinh functions,

Z1 =

e!Jb ! e(J+1)b

1! ebe!b/2

e!b/2

"

#$$

%

&''=

e!(J+1/2)b ! e(J+1/2)b

e!b/2 ! eb/2=

sinh (J +1 / 2)b() *+sinh(b / 2)

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(b) Derive an expression for the average energy of the particle in part (a). Write your

answer in terms of the hyperbolic cotangent function coth(x) =

cosh(x)sinh(x)

.

Solution:

E = !1Z

dZd"

= !1Z

dbd"

dZdb

= !sinh(b / 2)

sinh (J +1 / 2)b#$ %&'µB( )

J + 12( )sinh b

2cosh b J + 1

2( )#$(

%&) !

12

sinh b J + 12( )#

$(%&)cosh b

2

sinh b2( )2

= !'µB J + 12( )coth b J + 1

2( )#$(

%&)! 1

2coth b

2#$(

%&)

(c) Use the expression for the average energy in part (b) to determine the magnetization M

(the average z-component of the total magnetic moment) of a system of N identical, independent magnetic particles. Comment on its behavior as T ! 0 .

Solution: The average z-component of the magnetic moment of a single particle is just the average energy times -1/B. The total M will be the average energy times –N/B.

M = N!µ J + 1

2( )coth b J + 12( )"

#$%&'( 1

2coth b

2"#$

%&'

As T ! 0 , b!" , and the coth functions go to 1, so

M = NJ!µ . This is equivalent to

having all N particles in their lowest energy magnetic state with µz = J!µ .