Introduction To Statistical Physics
Solution Manual 2nd Edition
Author: Kerson Huang
,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: Z B Z B
Dv Vb
Cd: P Dv =A N Kbt1 a
= Nkbt1 Ln
V V
P (V — V ) = Nk T µ ¶
D D B
V
B 3 1− b
Vd
De: Z E
Dv Va
N k BT 3 = N kBT3 ln
d V Vd
No Work Is Done Along Bc And Ea. The Total Work Done Is The Sum Of
The Above. Heat Absorbed Equals Total Work Done, Since Internal Energy Is
Unchanged In A Closed Cycle.
1.3
(A)
1 ∂V Bv0t B−1
Α= =
V ∂T T0b v
(B)
Bv0t B−1
∆V = ∆T
T0B
Nkbt Nk b t b
0 T 1−B
P = =
V V0
Work Done = P ∆V = Bn Kb∆T
1
,
, 2 CHAPTER 20.
1.4
Consider An Element Of The Column Of Gas, Of Unit Cross Section, And
Height Between Z And Z+Dz. The Weight Of The Element
− Is Gdm , Where Dm
Is The Mass Of The Element: Dm = Mndz, Where M Is The Molecular
Mass, And N = P/Kbt Is The Local Density, With P The Pressure. For
Equilibrium, The Weight Must Equal
— The Pressure Differential:
− Dp = Gdm
.Thus, Dp/P = (Mg/Kbt )Dz. At 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 = ∆Q1 + ∆Q2 = 0. That Is, Heat Just Pass From One Body
To The Other. Suppose The Final Temperature Is T . Then
∆Q1 = C1(T − T1), ∆Q2 = C2(T − T2). Therefore
C1T 1 + C2T 2
T =
C1 + C2
1.6
R
Work Done By The System Is − Hdm . Thus The Work On The System Is
Z ΚZ Κh2
Hdm = Hdh =
T 2T
1.7
Consider The Hysteresis Cycle In The Sense Indicated In Fig.1.6. Solve For The
Magnetic Field:
H = ±H0 + Tanh−1(M/M0)
R
( + For Lower Branch, − For Upper Branch.). Using W = − Hdm , We Obtain
Z M0 Z −M0
W =− Dm [H0 + Tanh−1 (M/M0 )] − Dm [−H0 + Tanh−1 (M/M0 )]
−M0 M0
= −4M0H0
1.8
Solution Manual 2nd Edition
Author: Kerson Huang
,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: Z B Z B
Dv Vb
Cd: P Dv =A N Kbt1 a
= Nkbt1 Ln
V V
P (V — V ) = Nk T µ ¶
D D B
V
B 3 1− b
Vd
De: Z E
Dv Va
N k BT 3 = N kBT3 ln
d V Vd
No Work Is Done Along Bc And Ea. The Total Work Done Is The Sum Of
The Above. Heat Absorbed Equals Total Work Done, Since Internal Energy Is
Unchanged In A Closed Cycle.
1.3
(A)
1 ∂V Bv0t B−1
Α= =
V ∂T T0b v
(B)
Bv0t B−1
∆V = ∆T
T0B
Nkbt Nk b t b
0 T 1−B
P = =
V V0
Work Done = P ∆V = Bn Kb∆T
1
,
, 2 CHAPTER 20.
1.4
Consider An Element Of The Column Of Gas, Of Unit Cross Section, And
Height Between Z And Z+Dz. The Weight Of The Element
− Is Gdm , Where Dm
Is The Mass Of The Element: Dm = Mndz, Where M Is The Molecular
Mass, And N = P/Kbt Is The Local Density, With P The Pressure. For
Equilibrium, The Weight Must Equal
— The Pressure Differential:
− Dp = Gdm
.Thus, Dp/P = (Mg/Kbt )Dz. At 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 = ∆Q1 + ∆Q2 = 0. That Is, Heat Just Pass From One Body
To The Other. Suppose The Final Temperature Is T . Then
∆Q1 = C1(T − T1), ∆Q2 = C2(T − T2). Therefore
C1T 1 + C2T 2
T =
C1 + C2
1.6
R
Work Done By The System Is − Hdm . Thus The Work On The System Is
Z ΚZ Κh2
Hdm = Hdh =
T 2T
1.7
Consider The Hysteresis Cycle In The Sense Indicated In Fig.1.6. Solve For The
Magnetic Field:
H = ±H0 + Tanh−1(M/M0)
R
( + For Lower Branch, − For Upper Branch.). Using W = − Hdm , We Obtain
Z M0 Z −M0
W =− Dm [H0 + Tanh−1 (M/M0 )] − Dm [−H0 + Tanh−1 (M/M0 )]
−M0 M0
= −4M0H0
1.8
