PHYS-4420 THERMODYNAMICS & STATISTICAL MECHANICS
FINAL EXAMINATION
SPRING 2005
Tuesday, May 10, 2005
NAME: ______ANSWERS__________________________
There are seven pages to this examination. Check to see that you have them all.
CREDIT GRADE
Problem 1
30%
Problem 2
20%
Problem 3
30%
Problem 4
20%
TOTAL
100%
To receive credit for a problem, you must show your work, or explain how you arrived at your
answer.
1. (30%) The Gibbs function is defined as G = U – TS + PV, where U is internal energy, T is
temperature, S is entropy, P is pressure, and V is volume.
a) (10%) Show that: dG = – SdT + VdP
dG = dU – TdS – SdT + PdV +VdP, but the First Law states, dU = TdS – PdV.
Then dG becomes,
dG = TdS – PdV – TdS – SdT + PdV +VdP = – SdT + VdP
G 

b) (10%) Use the equation given in part a) to show that: S  

T

P
G 
G 
 dT  
 dP
G is a function of T, and P. Therefore, dG  

T

P

P

T
G 

A term by term comparison with the equation given in part a) shows that S  
  T P
1
 S 
 V 
  

c) (10%) Use the equation given in part a) to show that: 

P

T

T

P
(This is one of the Maxwell relations.)
G 

Using the method of part b), we also see that V  
  p T
 S 
 V 
 V 
 2G
 2G
 2G
 2G
  
  
 
Then, 
, and 
, so  

TP
PT
 P T
  T P
  P T
  T  P TP
 S 
 V 
  

Therefore, 
  P T
  T P
2
2. (20%) In the vicinity of the triple point, the vapor pressure of liquid ammonia (in Pa, i.e.
N/m2) is represented by,
ln P = 24.38 – (3063/T)
where T is in Kelvins. This is the equation of the liquid-vapor boundary curve on a P-T
diagram. Similarly, the vapor pressure of solid ammonia is,
ln P = 27.92 – (3754/T).
a) (10%) What are the temperature and pressure of the triple point of ammonia?
Find the temperature and pressure at which the two curves cross. To get the temperature.
Set the pressures equal: 24.38 – (3063/T) = 27.92 – (3754/T). Then,
3754  3063
691
691
 27.92  24.38 , so
 3.54 , and T 
 195.1977
T
T
3.54
Then, since ln P = 24.38 – (3063/T), P  e
3063 

 24.38

T 

e
3063 

 24.38 

195.1977 

= 5933 Pa
TTP = ____195 K__________
PTP = __5933 Pa________
units
units
b) (10%) What are the latent heats of sublimation and vaporization near the triple point?
(Hint: Use the Clausius-Clapeyron equation with v  v and v  v . Then treat the
vapor as an ideal gas.)
dP
 13


 13 , since v  v .
dT T (v  v) Tv
Take the derivative with respect to temperature of the equation for the vapor-solid curve.
1 dP 3754
dP 3754 P 3754 P
d
d 
3754 
 2 , or


ln P 
 27.92 
 , which gives
P dT
T
dT
T2
T T
dT
dT 
T 
P R
dP 3754 R

From the ideal gas law,  , so
. Compare to the Clausius-Clapeyron
T v
dT
Tv
equation and we see that 13 = 3754 R = 3.12 ×107 J/kmole
In the same way,  23 = 3063 R = 2.54 ×107 J/kmole
13 = _3.12 ×107 J/kmole __
units
 23 = _2.54 ×107 J/kmole _
units
3
3. (30%) A container is divided into two equal chambers, each of the same volume V. One
chamber contains N0 molecules of an ideal gas at temperature T, and the other chamber is
completely empty, a perfect vacuum. A small hole, of area A, is punched in the wall
separating the two chambers, and gas begins to leak into the empty chamber. The
temperature of the gas is kept constant.
a) (10%) Find an expression for the rate at which molecules leave the filled chamber at the
instant the hole is punched. Express your answer in terms of A, V, N0, and v , the average
speed of an air molecule. (Hint: the flux of molecules moving in the + x direction can be
1
written,   n v , where n is the number of molecules per volume.)
4
dN
N 
 A  14 n v A  14  0 v A
dt
V 
dN N 0 v A

dt
4V
b) (10%) The number of molecules in the chamber that was initially filled will decrease
until the two chambers are equally populated (each with N0/2 molecules). Then, the rate
at which molecules leave the chamber will be equal to the rate at which they enter it from
the other side. Derive an expression for N, the number of molecules in the initially filled
chamber, as a function of time after the hole is punched. Again, the result can be in terms
of A, V, N0, and v .
If there are N molecules in the first chamber, there are N0 – N molecules in the other.
N vA
Then based on part a), the rate at which molecules are leaving is
, and the rate at
4V
( N 0  N )v A
which they are entering is
, Then
4V
dN
N vA ( N 0  N )v A
vA



(2 N  N 0 )
dt
4V
4V
4V
N 
dN
vA 

 N  0  . This differential equation can be separated and solved.
dt
2V 
2 
dN
vA
N 
vA


dt , and after integrating, ln  N  0   
t  const . Then,
N0
2
V
2
2
V


N
2
vA
N

t
N
N0
 Ce 2V . At t = 0, N = N0, so C  0 . Then,
2
2
N
4
vA
vA 

t
N 0 N 0  2V t N 0 

e

1  e 2V 

2
2
2 

c) (10%) As the gas redistributes itself between the two chambers, the entropy of the system
increases. Calculate the difference between the entropy of the final state with both
chambers holding N0/2 molecules, and the initial state when all the molecules were in one
chamber.
w 
S  S f  Si  k ln w f  k ln wi  k ln  f 
 wi 
wi 
N 0!
1
N 0!0!
wf 
N 0!
N 0!

( N 0 / 2)!( N 0 / 2)! ( N 0 / 2)!2

w 
N 0! 
  k[ln N 0!2 ln( N 0 / 2)!]
S  k ln  f   k ln 
2 


(
N
/
2
)!
 wi 
0


S  k[ N ln N 0  N 0  2{( N 0 / 2) ln( N 0 / 2)  N 0 / 2}]  k[ N 0 ln N 0  N 0ln( N 0 / 2)]
 N 
S  N 0 k ln  0 
 N0/ 2 
S =___ N0k ln 2________
5
4. (20%) A new satellite that uses a Carnot engine for power, is being designed. For a high
temperature reservoir, it will use a nuclear source at a fixed temperature, T2. Its low
temperature reservoir will be a set of cooling fins that radiate heat into space. The
temperature of the fins depends on the rate at which they radiate energy, and that is
proportional to the fourth power of their temperature, T1. They will maintain a constant
temperature when the rate at which the Carnot engine delivers heat to the fins is equal to the
rate at which the fins radiate heat to space:
d Q1
 AT14
dt
where A is a constant determined by the design of the fins. For this system, the Carnot engine
will have a maximum power output for a specific value of T1 that is between absolute zero
and T2 .
a) (5%) Explain briefly why the power output should be a maximum somewhere in this
temperature range.
(Hint: Power is the rate at which work is done, and
dW d Q2 d Q1


.)
dt
dt
dt
If T1 and T2 are close together, T1 will be high and able to radiate heat rapidly, so it is
possible to run the engine at a high rate, but the engine is not efficient. If T1 is lowered,
the engine becomes more efficient, but heat can not be radiated as rapidly, so the engine
has to be run at a slower rate. At some temperature a compromise is found, and a
maximum power output results.
b) (10%) Find the value of T1 for which the power output is a maximum. Express your
answer in terms of T2.
P
Q2 T2
dW d Q2 d Q1 d Q2
 , so



 AT14 . In addition, we know that
Q1 T1
dt
dt
dt
dt
Q2 
T2
Q1 , and
T1
d Q2 T2 d Q1 T2

 AT14  AT2T13
dt
T1 dt
T1
P  AT2T13  AT14  A(T2T13  T14 ) . To find the maximum, solve
dP
 A(3T2T12  4T13 )  0,
dT1
possible. Instead,
dP
 0.
dT1
(3T2  4T1 )T12  0 . This has two solutions, but T1 = 0 is not
3
T1  T2 .
4
6
c) (5%) What is the efficiency of the Carnot engine when operating at the value of T1 found
in part (b), i.e. when the power output is a maximum.
3
T2
T
3 1
  1  1  1  4  1    25%
T2
T2
4 4
7