PHYS-4420 THERMODYNAMICS & STATISTICAL MECHANICS
SPRING 2006
Homework Solutions
Assignment 2. Due Friday 2/03/06 : 4-9, 4-15, 5-4, 5-5, 5-11, 5-13
 –1
4-9. a) Tv
 1
= constant, so T1v1
 1
 T2v2
T v 
and 2   1 
T1  v2 
 1
T 
v 
. Then, ln  2   (  1) ln  1 
 T1 
 v2 
b) Find the value of  that is needed for this process.
T 
ln  2  ln  2 
T
5
  1   1      1.32 , so  = 2.32. The highest value of  is 1.67, for a
v 
1
ln  1  ln  
2
 v2 
monatomic gas, so this process is impossible.
4-15. The relationship between temperature and volume is known, and the volume is
proportional to the radius cubed.
V 
T1V1 1  T2V2 1 , so  2 
 V1 
T 
r2   1 
 T2 
1
3( 1)
 1
 3  10 K 

r1  
3
3

10
K


5
1
3
1
 r   T   1
T
V  T   1
 1 , and 2   1  . Then,  2    1 
so,
T2
V1  T2 
 r1   T2 
1
3(1.4 1)
(15 m)  100
1 / 1.2
(15 m)
r2 = 696 m
5-4. First prepare an expression that can be used for parts a) and b). Consider h = h(T,P).
 h 
 h 
dh    dT    dP . This can be written in simpler form because,
 T  P
 P T
 h 
 h 

  cP , and the cyclical relation can be used on   .
 T  P
 P T
 h 
 h   T 
        cP  . Then, dh becomes, dh  cP dT  cP dP
 P T
 T  P  P h
With this expression part a) can be done, but part b) involves dv. To bring that in, consider
P = P(T,v). Then,
 P 
 P 
dP    dT    dv . Again, the partials can be replaced.
 T v
 v T
1  v 
1
 P 
     , so     . From the cyclical relation, as was shown is class
v  P T
v
 v T
 v 



1
T  P 
 P 

dv . When this is put in the expression for
 . Then, dP  dT 

 

v

 v 
 T v
 
 P T
1


dv  , or
dh, the result is dh  cP dT  cP   dT 
v 

c 
  
dh  cP 1 
dT  P dv
 
v

 h 
  
a) Then, for constant v,
   cP 1 

 
 T v

c 
 h 
b) For constant T,
   P
 v T v
 T   v   h 
c) Again, from the cyclical relation 
      1 , so
 v h  h T  T v
 h 
cP 
 



v
 T 
 T  
v



 
 h 
  
   v   
 v h
cP 1 
v 1 




 
 
 T v


5-5.
5-11. |W| = |Q1| + |Q2| – |Q3|,
and
Q1 Q2 Q3


0
T1
T2
T3
Q2
Q3
1200 J


0
400 K 200 K 300 K
Q3
Q2
|Q3| – |Q2| = 1000 J
and

 3 J/K
300 K 200 K
These two equations must be solved for |Q2| and |Q3| . Multiply the second by 300 K.
|Q3| – |Q2| = 1000 J
and |Q3| – 1.5 |Q2| = 900 J.
Subtract these two equations, and get 0.5 |Q2| = 100 J, so
|Q2| = 200 J
Then, |Q3| = |Q2| + 1000 J = 200 J + 1000 J
|Q3| = 1200 J
200 J = 1200 J + |Q2| – |Q3| and
5-13. a) c 
Q2
Q2
T2


W
Q2  Q1 T2  T1
b) The temperatures must be converted from ºF to K.
32ºF = 0ºC = 273 K, and 68ºF = 20ºC = 293 K
T2
293 K
c

T2  T1 293 K  273 K
c = 14.6