PHYS-4420 THERMODYNAMICS & STATISTICAL MECHANICS
SPRING 2006
Homework Solutions
Assignment 1. Due Friday 1/27/06: 1-8, 1-11, 2-3, 2-8, 2-13, 3-3, 3-8, 3-11
1-8. E = aT + bT2
For the numbers given,
60 mv = a 200ºC + b (200ºC)2
40 mv = a 400ºC + b (400ºC)2
Multiply the top equation by 2.
120 mv = a 400ºC + b (400ºC)(200ºC)
Now subtract the last two to eliminate a.
– 80 mv = b (400ºC)(200ºC) = b 80000 (ºC) 2
b = – 1.0 ×10-3 mv/(ºC)2
From the first equation,
60 mv  b(200C) 2
a
 0.3 mv/ C  b(200C)  0.3 mv/ C  [1.0  103 mv/( C)2 ]( 200C)
200C
a = 0.50 mv/(ºC)
Then,
E = [0.50 mv/(ºC)] T – [1.0 ×10-3 mv/(ºC)2] T2
For E = 30 mv, this becomes,
30 mv = [0.50 mv/(ºC)] T – [1.0 ×10-3 mv/(ºC)2] T2, or
T2 – (500ºC)T + 30000 (ºC)2 = 0
This has solutions T  250C  (250C)2  30000(C)2  250C  180.3C .
The data shows that emf decreases as temperature increases, so only the positive sign
makes sense.
T = 430.3ºC
1-11. a) T(ºC) = T(K) – 273.15 K = 77.35 K – 273.15 K
b) T(ºF) = (9/5)(T(ºC) + 32ºF = (9/5)(– 195.80ºC) + 32ºF
c) T(R) = T(ºF) + 459.67 R = – 320.44ºF + 459.67 R
T(ºC) = – 195.80ºC
T(ºF) = – 320.44ºF
T(R) = 139.23 R
2-3. a) P = Av, and from the ideal gas law, v 
RT
RT
P2
, so P  A
. Then, A 
.
P
P
RT
A
P12
RT1
b)
Pv
A
A
, but P = Av, so T  v 2 . Then, T1  v12 .
R
R
R
A
A
A 
Now, T2  v22  (2v1 ) 2  4 v12   4T1  4(200 K)
T2 = 800 K
R
R
R 
c) From the ideal gas law, T 
1  v 
2-8. a)   

v  T  P
 P 


a
RT  RTv
T v
 v 

e
From the cyclical relation , 
, and P 
.
 
vb
 P 
 T  P
 
 v T
R  RTv RT  RTv  a 
 P 
e

e

 

2 
vb
 T v v  b
 RT v 
a
a
 RT  RTv RT  RTv  a 
 P 
e

e
  

2
2 
vb
 v T (v  b)
 RTv 
a
a
R  RTv RT  RTv  a 
 P 
a
e

e



1
2 
vb
vb
 v 
 T  v
 RT v  
RTv


 
a
a
T
a
 P 
 RT  RTv RT  RTv  a 
 T  P
 
e

e

 v  b  Rv 2
2
2
 v T
(v  b )
vb
 RTv 
a
a

1
1  v 
1
RTv
    
v  T  P v  T  a

 v  b Rv 2
a
1
a
a

 1




2
2
1
v RTv

  v RTv
 T 1  a  T  1  a



2 

 v  b RTv 2
 v  b RTv 






b) For an ideal gas, v 
1  v 
1    RT
RT
, so   
  

v  T  P v  T  P
P
1R
R
1


 .
  
  P v P RT T
2-13. The volume that will spill equals the increase in volume of the water.
V = V0[1 + (T – T0)], so V= V – V0 = V0(T – T0).
V= (250 cm³)(0.21 ×10-3 K-1)(50ºC – 20ºC) = (250 cm³)(0.21 ×10-3 K-1)(30 K)
V= 1.575 cm³ = 1.58 cm³
3-3. a)
nRT
nRT
 AV , and
, and P = AV. Then,
V
V
AV22 A(V1 / 2) 2 1 AV12 1
AV12
, and T2 


 T1 .
T1 
nR
nR
4 nR
4
nR
b) PV = nRT, so P 
T
AV 2
.
nR
T2 = ¼ T1
V2
V 2 
A
c) W   PdV  A VdV  A   [V22  V12 ]
V
V
2
 2 V
nRT nRT
From part b), A  2  2 1 . Since P = AV, V2 = ½ V1 . Then,
V
V1
V2
V2
1
1
1
2
 nRT1 2  1  nRT1  3 
A 2
nRT1  V1 
2
2
W  [V2  V1 ] 
V
1 

  V1  
2 
2 1 
2
2V1  2 
2  4 
 4 
 2V1
3
W   nRT1
8
3-8. a) Ub – Ua = U = Qacb – Wacb
U = 80 J – 30 J = 50 J
Qadb = U + Wadb = 50 J + 10 J
b) For this path, U = Ua – Ub = – 50 J
Q = U + W = – 50 J – 20 J
Qadb = 60 J
Q = – 70 J
c) Qad = Uad + Wad ; Uad = Ud – Ua = 40 J;
Qad = Uad + Wad = 40 J + 10 J
Wad = Wadb = 10 J, since Wdb = 0.
Qad = 50 J
Qdb = Udb + Wdb = Udb;
Udb = (Ub – Ua) – (Ud – Ua) = 50 J – 40 J = 10 J
Qdb = 10 J
3-11. a) V = AP + B, so V1 = AP1 + B and V2 = AP2 + B. Subtracting the two equations,
V  V 1 m3  1.8 m3
V1 – V2 = A(P1 – P2), so A  1 2 
P1  P2 1 bar  12 bar
A = – 1.6 m3/bar = – 1.6 ×10-5 m3/Pa
V1 = AP1 + B, so B = V1 – AP1 = 1 m3 – (– 1.6 m3/bar)(1 bar)
b) PV = nRT, so
B = 2.6 m3
1 bar (1.8 m3 ) T  0.9 T  0.9(300 K)
P1V1 T1
PV
 , so T2  2 2 T1  2
1
1
P2V2 T2
P1V
(1 bar)(1 m3 )
T2 = 270 K
V2
P2
c) W   PdV , and V = AP + B, so dV = AdP. Then, W  A PdP
V1
P1


A 2
 1.6 m /bar 1
P2  P12 
[( 2 bar) 2  (1 bar) 2 ]  0.6 bar  m3
2
2
3
5
W = (0.6 bar·m )(10 Pa/bar) = 6 ×104 N·m
W = 6 ×104 J
P2
W  A PdP 
P1
3
d) Q = W + U = 6 ×104 J + (800 J/K)(270 K – 300 K)
Q = 3.6 ×104 J