Table Of ContentIntroduction to Statistical Physics
Solution Manual
Kerson Huang
ii
Chapter 1
1.1
Mass of water =106g, temperature raised by 20 C.
◦
Heat needed Q=2 107cal = 8.37 107J.=23.2 kwh.
× ×
Work needed = mgh= 14 150 29000= 6.09 107 ft-lb =22.9 kwh.
× × ×
1.2
Work done along various paths are as follows
ab:
b b dV V
PdV =Nk T =Nk T ln b
B 1 V B 1 V
Za Za a
cd:
V
P (V V )=Nk T 1 b
d d− b B 3 − V
µ d¶
de:
e dV V
Nk T =Nk T ln a
B 3 V B 3 V
Zd d
No work is done along bc and ea. The total work done is the sum of the
above. Heatabsorbedequalstotalworkdone,sinceinternalenergyisunchanged
in a closed cycle.
1.3
(a)
1 ∂V bV Tb 1
α= = 0 −
V ∂T TbV
0
(b)
bV Tb 1
∆V = 0 − ∆T
Tb
0
Nk T Nk Tb
P = B = B 0T1 b
−
V V
0
Work done = P∆V =bNk ∆T
B
1
2 CHAPTER 1.
1.4
Consider an element of the column of gas, of unit cross section, and height
betweenzandz+dz. Theweightoftheelementis gdM,wheredM isthemass
−
of the element: dM = mndz, where m is the molecular mass, and n = P/k T
B
is the local density, with P the pressure. For equilibrium, the weight must
equal the pressure differential: dP = gdM .Thus, dP/P = (mg/k T)dz. At
B
− −
constant T , we have dp/P =dn/n.Therefore
n(z)=n(0)e−mgz/kBT
1.5
No change in internal energy, and no work is done. Therefore total heat
absorbed ∆Q=∆Q +∆Q =0. That is, heat just pass from one body to the
1 2
other. Suppose the final temperature is T. Then
∆Q =C (T T ), ∆Q =C (T T ). Therefore
1 1 1 2 2 2
− −
C T +C T
T = 1 1 2 2
C +C
1 2
1.6
Work done by the system is HdM. Thus the work on the system is
−
R
κ κH2
HdM = HdH =
T 2T
Z Z
1.7
Consider the hysteresis cycle in the sense indicated in Fig.1.6. Solve for the
magnetic field:
H = H +tanh 1(M/M )
0 − 0
±
(+forlowerbranch, forupperbranch.). UsingW = HdM,weobtain
− −
R
W = M0 dM[H +tanh 1(M/M )] −M0dM[ H +tanh 1(M/M )]
0 − 0 0 − 0
− − −
Z−M0 ZM0
= 4M H
0 0
−
1.8
3
A log log plot of mass vs. A is shown in the following graph. The dashed
line is a straightline for reference.
M
g
o
l
1 10 100 1000 10000
A
.
4 CHAPTER 1.
Chapter 2
2.1
Use the dQ equation with P,T as independent variables:
dQ=C dT +[(∂U/∂P) +P(∂V/∂P) ]dP
P T T
For an ideal gas (∂U/∂P) =0, P(∂V/∂P) = V. Thus
T T
−
dQ=C dT VdP.
P
−
The heat capacity is given by
C =C V(∂P/∂T) .
P path
−
The path is P =aVb, or equivalently Pb+1 =a(Nk T)b by the equation of
B
state. Hence
V(∂P/∂T) =[ab/(b+1)]V(Nk T)bT 1 =bNk /(b+1). Therefore
path B − B
b
C =C Nk
P − b+1 B
This correctly reduces to C for b=0.
P
2.2
Use a Carnot engine to extracted energy from 1 gram of water between 300
K and 290 K.
Max efficiency η =1 (290/300)=1/30.
−
1
W =ηC∆T = (4.164 J g 1K 1 1 g 10 K)=1.39 J
− −
30 × ×
Gravitational potential energy=1 g 9.8 kg s 2 110 m=1.08 J
−
× ×
2.3
The highest and lowest available temperatures are, 600 F = 588.7 K and 70
F = 294.3 K.
The efficiency of the power plant is W/Q =0.6[1 (294.3/588.7)]=0.3.
1
−
In one second: W =106 J.
So Q =2.33 106 J =C ∆T. Use C =4.184 J g 1K 1,
2 V V − −
×
5
6 CHAPTER 2.
Flow rate = 6000 (0.305m)3
×
2.33 106 J
∆T = (4.184 J g 1K 2)6000×(0.305 m)3 106 cm2/m3 =3.27×10−3K
− −
2.4
(a)
Since water is incompressible, a unit mass input gives a unit mass output.
The net heat supplied per unit mass is ∆Q=C(T T) C(T T ),
1 2
− − −
whereC isthespecificheatofwater(perunitmass.) Insteadystatev2/2=
∆Q. This gives
v = 2∆Q= 2C(T +T 2T)
1 2
−
(b) p p
The entropy depends on the temperature like lnT. A unit volume of water
from each of the input streams has total entropy lnT +lnT This makes two
1 2
unit volumes in the output stream, with entropy 2lnT. Therefore the change
in entropy is ln T2/T T 0. Thus T √T T , and
1 2 1 2
≥ ≥
¡ ¢
v =√2C T T
max 1 2
−
¯p p ¯
¯ ¯
¯ ¯
2.5
(a)
PVγ =2P Vγ, PVγ =2 P Vγ
(V 1/V )γ =020. (L¯+2 a)/(\L 0 a0) γ =2.
1 2
−
£ a¤ 21/γ 1
= −
L 21/γ +1
(b)
∆U =∆Q W, ∆Q=0.
−
C ∆T = W, ∆T = W/C .
V V
− −
T =2T +∆T =2T (W/C ), T =T ∆T =T +(W/C ).
1 0 0 V 2 0 0 V
− −
RT R[2T (W/C )]
P = 1 = 0− V
V A(L+a)
1
(c)
a
W =A dx(P P )
P =2P0Vγ/[A1(L−+2x)]γ, P =P Vγ/[A(L x)]γ.
1 R0 0 2 0 0 −
P V a 2a
W = 0 0 1 1
γ 1 − L γ − L
− ³ − ´µ ¶
7
2.6
(a)
PV =U/3, U =σVT4.
P =σT4/3.
dS =dQ/T =(dU +PdV)/T.
Integrate along paths with T=const, V=const.
4
S = σVT3.
3
(b)
S =Constant. ∴T3 V−1. Thus
∼
T R 1
−
∼
2.7
The heat absorbed by an ideal gas in an isothermal process is
∆Q=NkT ln(V /V )
f i
where V and V are respectively the final and initial volume.The temperature
f i
T in this formula is the ideal-gas temperature.
Draw a Carnot cycle on the PV diagram, and label the corners 1234 clock-
wise from the upper left.
The heat absorbed at the upper temperature T , and the heat rejected at
2
the lower temperature T , are
1
Q =NkT ln(V /V )
2 2 2 1
Q =NkT ln(V /V )
1 1 3 4
Because 23 and 12 lie on adiabatic lines, we have
V2T2γ−1 =V3T1γ−1
V1T2γ−1 =V4T1γ−1
Dividing one equation by the other yields V /V =V /V .
2 1 3 4
The efficiency of the cycle is therefore
Q T
η =1 1 =1 1
− Q − T
2 2
2.8
Diesel cycle:
Q =C (T T )
2 P 3 2
−
Q =C (T T )
1 V 4 1
−
η =1 (Q /Q )=1 γ 1[(T T )/(T T )]
1 2 − 4 1 3 2
− − − −
We have P =P , hence
3 2
T /T =V /V =r
3 2 3 2 c
The processes 12 and 34 are adiabatic, with TVγ 1 = constant. V =V .
− 4 1
Thus
8 CHAPTER 2.
T3V3γ−1 =T4V1γ−1
T2V2γ−1 =T1V1γ−1
Using the three relations derived, we obtain
1 rγ 1
η =1 c −
− γrγ 1(r 1)
− c−
2.9
Otto cycle:
Q =C (T T )
2 V 3 2
−
Q =C (T T )
1 V 4 1
−
η =1 (Q /Q )=1 [(T T )/(T T )]
1 2 4 1 3 2
− − − −
The processes 12 and 34 are adiabatic, with TVγ 1 = constant. We have
−
V =V , V =V Thus
4 1 3 2
T1V1γ−1 =T2V2γ−1.
T3V2γ−1 =T4V1γ−1.
Taking the ratio of these equations, we have
T /T =T /T =rγ 1.
2 1 3 4 −
Thus
η =1 r1 γ
−
−
2.10
First note T /T =V /V =2.
b a b a
Work done Heat absorbed
a b P (V V )=P V =NkT C ∆T =C T
a b a a a a P P a
→ −
b c 0 C T
V a
→ −
c a PdV = NkT ln2 NkT ln2
a a
→ − − −
R
W (Net work done) =NkT (1 ln2)
a
−
Q (Heat absorbed) =C T = 5NkT
2 P a 2 a
W 2
η = = (1 ln2)=0.12
Q 5 −
2
In comparison, η =1 (T /T )=0.5.
Carnot − b a
2.11
First note T =4T . The P,V,T for the points A,B,C,D are as follows:
2 1
P V T
A P V =NkT /P T
1 1 1 1 1
B 2P 2V 4T
1 1 1
C 2P V 2T
1 1 1
D P 2V 2T
1 1 1
