PHYS-4420 THERMODYNAMICS & STATISTICAL MECHANICS
SPRING 2006
Homework Solutions
Assignment 4. Due Friday 2/17/06 : 7-16, 7-17, 8-1, 8-7, 8-12, 8-14
7-16. c 

1.48

3

(162 kg/m )(9.43  108 Pa -1 )
7-17. a) Use a Tds equation.
The first Tds equation is: Tds  cv dT 
c = 311 m/s
T
dv . Since v is constant, the dv term is zero.

dT
Tv 2
Tv 2
Then, ds  cv
, However, c p  cv 
, so cv  c p 
, and
T



Tv 2  dT
dT v 2

ds   c p 
 cp

dT . Now integrate to get s.
 T
T


T
s  cP 
T0
dT v 2

T

T

T0
 T  v 2
dT  cP ln   
(T  T0 ) .
 T0  
dT
 vdP .
T
 P 
 P 
 P 
P = P(T,v), so dP  
 dT    dv . Since v is constant, dv = 0, dP  
 dT
 T v
 v T
 T v
dT
 P 
Then, ds  cP
 v   dT .
T
 T v
1  v 
 v 





T  P
v  T  P
 P 

From the cyclical relation, 

 . Then,
 
1  v 

 v 
 T v
  
 
v  P T
 P T
dT
dT

dT v 2
 P 
ds  cP
 v 
 v dT  cP

dT , as above.
 dT  cP
T
T

T

 T v
If you prefer, the second Tds equation is: Tds  cP dT  TvdP , so ds  cP
 T  v 2
b) s  cP ln   
(T  T0 )
T

 0
With numbers this becomes (note that specific volume, v = 1/density):
(4.9 105 K -1 ) 2
 310 K 
s  (390 Jkg -1K -1 ) ln 
(310 K  300 K)

3
3
-12
-1
 300 K  (9.85 10 kg/m )(7.7 10 Pa )
s = 12.5 J·kg-1·K-1
1
8-1. Use the first Tds equation in its original form, before introducing  and .
dT  P 
 P 
Tds  cv dT  T   dv , so ds  cv
   dv . For an isothermal process, dT = 0.
T  T v
 T v
 P 
ds  
 dv
 T v
RT
dv
R
 P 
For an ideal gas, P 
, so 
  , and ds  R .
v
v
 T v v
s  R 
v2
v1
For a van der Waals gas P 
v 
dv
 R ln  2 
v
 v1 
dv
RT
a
R
 P 
 2 , so 
, and ds  R
.
 
vb v
vb
 T v v  b
s  R 
v2
v1
 v b
dv
 .
 R ln  2
v b
 v1  b 
 v b
v 
 or ln  2  ? Since this is a compression, v1 > v2.
Which is larger, ln  2
 v1  b 
 v1 
Consider, v1v2 = v1v2 so, v1v2 – v2b > v1v2 – v1b since v1 > v2. Then,
v2 v2  b

v2(v1 – b) > v1(v2 – b) , so
.
v1 v1  b
Based on this result, you might conclude, as the author did, that s is greater for the ideal
v
v b
gas, but that is incorrect. Both 2 and 2
are less than one, so their logarithms are
v1
v1  b
v 
 v b 
v2 v2  b
 . That means that

, ln  2   ln  2
v1 v1  b
v
v

b
 1
 1

 v b
 v b
 , but the magnitude of ln  2
 is greater than
is less negative than ln  2
 v1  b 
 v1  b 
negative. Entropy is decreasing. Since
v 
ln  2 
 v1 
v 
the magnitude of ln  2  .
 v1 
Therefore, the van der Waals gas experiences the greater change (decrease) in entropy.
2
P
8-7. g  RT ln    AP
 P0 
RT
RT
 g 
 A , or manipulated to give
a) v    
 A . This can be left as v 
P
P
 P T
Pv  RT  PA , and finally, P(v  A)  RT
 P  dA
 g 
b) s  
P
   R ln   
 T  P
 P0  dT
c) g = u – Ts + Pv = f + Pv, so f = g – Pv.
P
f  RT ln    AP  Pv . This result is fine, but the equation of state can be used to
 P0 
simplify it.
P
P
Pv  RT  PA , so f  RT ln    AP  RT  PA  RT ln    RT
 P0 
 P0 
 P 
f  RT ln    1
  P0  
T
.
T
W
W T

The efficiency is also equal to  
. These are equal, so
.
Q2
Q2
T
8-12. The efficiency of the Carnot cycle shown is equal to,  
Since there is a change of phase, Q2   23 , and W is equal to the area of the Carnot cycle.
W T
Pv T
P



W = Pv. Then,
becomes
, and
 23 .
Q2
T
 23
T
T T (v)
dP

If we let P and T go to infinitesimals, we get,
 23
dT T (v)
 12
3.34 105 J/kg
 dP 


 1.35 107 Pa/K
8- 14. a) 

5
3



 dT 12 T (v  v ) (273 K)(9.05 10 m /kg)
 dP 
7
7
b) P  
 T  (1.35 10 Pa/K)( 2 K)  2.70 10 Pa
 dT 12
2.70 107 Pa
 267 atm
1.01105 Pa/atm
The final pressure is, P = P0 + P = 1 atm + 267 atm = 268 atm
In atmospheres, P 
3