PHYS-4420 THERMODYNAMICS & STATISTICAL MECHANICS
SPRING 2006
Homework Solutions
Assignment 5. Due Friday 2/24/06 : 10-1, 10-2, 10-7
 S 
 S 
 dQ 

10-1. CV  
 . For a reversible process, đQ = TdS, so CV  T    
 T V  (T /T ) V
 dT V
 S 
CV  

  ln T V
 S 
10-2. In the notes from the last class, and in the textbook, it is pointed out that   must go to
 P T
zero as T goes to zero. That is a consequence of the third law which states that all entropy
changes go to zero as T goes to zero. However, one of the Maxwell relations shows that
 V 
 S 
 V 

    , so 
 must go to zero as T goes to zero. That is easy to check for
 T  P
 T  P
 P T
an ideal gas.
nRT
nR
 V 
V
, so 
which does not go to zero as T goes to zero.
 
P
 T  P P
We could try the same for the van der Waals gas, but that would get complicated. It is
 S   P 
better to look at the Maxwell relations again, and note that one is 
    , which
 V T  T V
 P 
 P 
tells us that 
 , or 
 must go to zero as T goes to zero. This is easier to calculate
 T V
 T v
for a van der Waals gas.
RT
a
R
 P 
P
 2 , so 
which does not go to zero as T goes to zero.
 
vb v
 T v v  b
Therefore, both the ideal gas equation and the van der Waals gas equation violate the third
law at T = 0, and cannot hold there.
T
T
T
T
dT
dT
dT
 S 0  A T 1 2
 B T 3
 S0  A T 1 2 dT  B  T 2 dT
0 T
0
0
0
0
T
T
B
S  S0  2 AT 1 2  T 3 . The entropy change, S = S – S0, goes to zero as T goes to zero, so
3
this could be legitimate. There is no reason to reject the paper on the basis of the third law.
10-7. S  S0  CV 
T
1