PHYS-4420 THERMODYNAMICS & STATISTICAL MECHANICS
SPRING 2006
Class Activity - Class 4 January 27, 2006
Name________SOLUTION__________________
Do problem 4-11 from the textbook. Since many of you do not have the textbook with you, here
it is:
An ideal diatomic gas, for which cv = 5R/2, occupies a volume of 2 m³ at a pressure of 4 atm, and
a temperature of 20ºC. The gas is compressed to a final pressure of 8 atm. Compute the final
volume, the final temperature, the work done, the heat released, and the change in internal
energy for:
(a) A reversible isothermal compression.
(b) A reversible adiabatic compression.
(a) For an isothermal process, T is constant, so PV is constant. P1V1 = P2V2, so V2 
P1
V1
P2
4 atm
2 m3
V2 = 1 m3
8 atm
Since the process is isothermal, T does not change.
T = 20ºC = 293 K
V
PV
W   PdV , and P  1 1 , so
V
V
V2 dV
 1 m3 
V 

W  P1V1 
 P1V1 ln  2   (4 atm)(1.01  105 Pa/atm)(2 m3 ) ln 
3
V1 V
V
2
m
 1


V2 
2
1
W = – 5.6 ×105 J
Q = – 5.6 ×105 J
U = 0
Since T is constant, U = 0, so Q = W
(b) For an adiabatic process, we need , where  
Then,  
cP

cV
7
5
2

2
cP
, and cP  cV  R  52 R  R  72 R
cV
7
 1.4 .
5
1/ 
P
P
 4 atm 
3
P1V1  P2V2 , so V2  1 V1 , and V2   1  V1  
 2m
P
8
atm
P2


 2
PV
T
PV
(8 atm)(1.22 m3 )
(293 K)
PV = nRT, so 1 1  1 , and T2  2 2 T1 
P2V2 T2
P1V1
(4 atm)(2 m3 )1

V2
V1
P1V1  1 
   1 V  1 V
V2
1
V2 = 1.22 m3
T2 = 357 K = 84ºC
V dV
P1V1
. Then, W  P1V1 

V V
V
PV   1
1 
1  P1V1 P2V2  P1V1  P2V2
 1 1   1   1  



   1 V2
V1    1  V1 1 V2 1 
 1
W   PdV , and PV   P1V1 , so P 
W
1 / 1.4


2
1
[( 4 atm)(2 m3 )  (8 atm)(1.22 m3 )](1.01  105 Pa/atm)
1.4  1
Since the process is adiabatic,
Since Q = 0, U = – W = – (– 4.4 ×105 J)
W
W = – 4.4 ×105 J
Q=0
U = 4.4 ×105 J