PHYS-4420 THERMODYNAMICS & STATISTICAL MECHANICS
Quiz 2
SPRING 2004
Friday, April 16, 2004
NAME: ________ SOLUTION _________________________
To receive credit for a problem, you must show your work, or explain how you arrived at your
answer.
1. (30%) The Gibbs free energy is defined as G = E – TS + pV, where E is internal energy, T is
temperature, S is entropy, p is pressure, and V is volume.
a) Show that: dG = – SdT + Vdp +dN
dG = dE – TdS – SdT + pdV +Vdp, but the First Law states, dE = TdS – pdV +dN.
Then dG becomes,
dG = TdS – pdV +dN – TdS – SdT + pdV +Vdp = – SdT + Vdp +dN
b) Use the equation given in part a) to show that: V 
G is a function of T, p, and N. Therefore, dG 
G

 p T , N
G
G
G 
 dT 
 dp 
 dN
 T  p, N
 p T , N
 N T , p
A term by term comparison with the equation given in part a) shows that V 
G

 p T , N
c) Use the equation given in part a) to derive any Maxwell relation.
Three are possible, one for each pair of terms in the expression for dG. One is derived here.
G 
 . Then,
Using the method of part b), we also see that S  
 T  p , N
S 
V 
V 
 2G
S 
 2G
 
 

  
, and
. Then, 
 p T , N  T  p , N
 T  p , N  T p
 p T , N
 p T
The others are: 
V 

S 

 

 

and
 N T , p  T  p , N
 N  T , p  p  T , N
Note: It is sufficient to state that dG is an exact differential, and then quote one of the
Maxwell relations.
1
2. (20%) An ideal Carnot engine is being used as a refrigerator to cool a room where some
students are working with computers. The students plus computers are generating 10,000
watts of heat (10,000 joules/second). The Carnot refrigerator removes heat from the room,
where the temperature is 20°C, and delivers it outside the building, where the temperature is
30°C. How much power must be put into the Carnot refrigerator to maintain the room at
20°C?
T  T2
W
W
Q1 T1
, or Q1  . We know
 , and   1
. Then,
T1
Q1

Q2 T2
T T
T
WT1
Q
W
W
becomes Q2 1 
, and 2 
, so W  Q2 1 2
Q1 
T2 T1  T2
T2 T1  T2
T2

dW
303 K  293 K
dW dQ2 T1  T2
 (10,000 W)

Finally,
dt
293 K
dt
dt
T2
For a Carnot cycle,  
P=
dW
= 340 W
dt
Alternatively, if you remember the coefficient of performance for a refrigerator:
Q
Q2
T2
T T
c 2 

, so W  Q2 1 2 etc.
W Q1  Q2 T1  T2
T2
Or, W  Q1 and Q1  Q2  W lead to W   (Q2  W ) . Solve for W, and W 
Then use  
T1  T2
.
T1
2
Q2
.
1 
3. (25%) Consider a system consisting of N distinguishable particles, each of which has only
three possible energy states. The energies of the states are, E0 = , E1 = , and E2 = 2. The
system is in equilibrium with a heat reservoir which is at temperature T.
a.) (5%) What is the partition function  for a single particle? (Note that it has only three
energy levels.)
   e  E  e   0  e    e   2  1  e    e 2 
n
n
  1  e   e2 
b.) (5%) What is the partition function for the entire system of N particles? (Note that the N
particles are distinguishable.) You can give your answer in terms of  from part (a).
Z   N . This answer is sufficient for this part. If it is completed,
Z  (1  e   e2  ) N
c.) (10%) What is the average energy of the N- particle system at temperature T?
E 






ln Z  
ln (1  e    e 2  ) N  
N ln( 1  e   e 2  )




 e   2e2 
e   2e 2 
E  N
 N
1  e   e 2 
1  e   e 2 
E  N
e   2e2 
1  e   e 2 
d) (5%)The temperature of the system is adjusted so that kT = . , and it is found that the
total energy of the N-particle system equals 1000 kT. How many particles are in the
system? (Hint: use your answer to part c.)
1
e   2e2 
,  = 1. Then E  N
becomes,
kT
1  e   e 2 
e 1  2e 2
1  e 1  e 2
1000 kT  NkT
N

1000
,
and
1  e 1  e  2
e 1  2e  2
Since  
N = 2354
3
4. (25%) In a classroom activity, you analyzed the frictional force on a disk that floated on a
layer of air. In this problem, you will examine it again. The diameter of the disc is 10 cm
(0.10 m) and its mass is 0.30 kg. It is gliding on an air table with a speed of 1.0 m/s. You can
assume that air is mainly nitrogen. The diameter of a nitrogen molecule is 4.0 ×10-10 m. and
its mass is 4.65 ×10-26 kg Assume the air temperature is 300K.
a) (10%) Calculate , the coefficient of viscosity for the air.

1
3
4kmT

1
3 (4.0  1010 m) 2


1
3 d 2
4kmT

4(1.38  1023 J/K)(4.65  1026 kg)(300 K)

 = 1.04 ×10-5 poise
units
b) (10%) The magnitude of the frictional force on the disk is measured to be 1.0 ×10-3 N.
How thick is the layer of air on which the disk is gliding?
 v A (1.04  105 poise)(1.0 m/s)  (0.05 m) 2
F
dv
v
  x   x . Then, y  x 
F
1.0  10 3 N
A
dy
y
y = 8.2 ×10-5 m = 0.082 mm
units
c) (5%) A careful examination of the equations reveals that the frictional force is not
constant. It depends on the speed of the disk. With that in mind, find the time needed for
the disk to slow to one half of its original speed. (Remember, the frictional force is
negative, which means it is directed opposite to the velocity.)
F
dv
v
A
dv
A
  x   x , so F  
vx .
vx , or m x  
A
dy
y
dt
y
y
Integrating gives, ln vx  
A
my
t = 0. If vx = ½ vx0 we have,
t
1
2
t  Constant , or vx  vx 0e
e

A
t
my
, and ln 2 
(0.30 kg)( 8.2  105 m)
ln 2
(1.04  105 poise)  (0.05 m) 2
A
my

Then,
A
t
my
t , or
dvx
A

dt
vx
m y
, where vx0 is the speed at
t
my
ln 2
A
t = 210 s = 3 min 30 sec
The calculation is easier if you note that the numerical value of
4
A
y
 1.0  103 .