kittel thermal physcis
TRANSCRIPT
Physics 831: Statistical Mechanics
Russell Bloomer1
University of Virginia
Note: There is no guarantee that these are correct, and they should not be copied
1email: [email protected]
Contents
1 Problem Set 1 11.1 Kittel 8.1: Heat pump . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 Kittel 8.6: Room air conditioner . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.3 Kittel 8.9: Cooling of nonmetallic solid to T = 0 . . . . . . . . . . . . . . . . . . . . . . . . . 21.4 Sterling Heat Engine . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.5 Unavailability for work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.6 Gibbs Free Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
2 Problem Set 2 52.1 Spin model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52.2 Paramagnetism of a system of N localized spin-1/2 particles . . . . . . . . . . . . . . . . . . . 62.3 Kittel 2.3: Quantum harmonic oscillator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72.4 Review Thermal mechanics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72.5 Review Thermal mechanics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8
3 Problem Set 3 93.1 Kittel 3.2: Magnetic susceptibility . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93.2 Kittel 3.3: Free energy of a harmonic oscillator . . . . . . . . . . . . . . . . . . . . . . . . . . 103.3 Kittel 3.4: Energy fluctuations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103.4 Kittel 3.10: Elasticity of polymers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113.5 Ising spin chain . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12
4 Problem Set 4 134.1 Application of equal partition theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 134.2 Thermal Equilibrium of the Sun and Earth . . . . . . . . . . . . . . . . . . . . . . . . . . . . 144.3 Kittel 4.3: Average temperature of the interior of the Sun . . . . . . . . . . . . . . . . . . . . 144.4 Kittel 4.6: Pressure of thermal radiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 154.5 Kittel 4.7: Free energy of a photon gas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 154.6 Kittel 4.18: Isentropic expansion of photon gas . . . . . . . . . . . . . . . . . . . . . . . . . . 16
5 Problem Set 5 175.1 5.1: Kittel 4.14: Heat capacity of liquid 4He at low temperature . . . . . . . . . . . . . . . . . . . 175.2 5.2: Kittel 4.15: Angular distribution of radiant energy flux . . . . . . . . . . . . . . . . . . . . . . 175.3 5.3: Qualifying exam file problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 175.4 5.4: Helmholtz free energy for the Debye model . . . . . . . . . . . . . . . . . . . . . . . . . . . . 175.5 5.5: Kittel 5.3: Potential energy of gas in gravitational field . . . . . . . . . . . . . . . . . . . . . . 17
i
6 Problem Set 6 196.1 6.1: Kittel 5.4: Active transport . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 196.2 6.2: Kittel 5.7: States of positive and negative ionization . . . . . . . . . . . . . . . . . . . . . . . 196.3 6.3: Kittel 5.10: Concentration fluctuation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 206.4 6.4: Kittel 6.3: Distribution function for double occupancy statistics . . . . . . . . . . . . . . . . . 206.5 6.5: Kittel 6.7: Relation of pressure and energy density . . . . . . . . . . . . . . . . . . . . . . . . 216.6 6.6: Kittel 6.9: Gas of atoms with internal degree of freedom . . . . . . . . . . . . . . . . . . . . . 21
7 Problem Set 7 237.1 7.1: Pressure in types of gases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 237.2 7.2: Kittel 5.13: Isentropic expansion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 237.3 7.3: Kittel 6.8: Time for a large fluctuation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 237.4 7.4: Kittel 6.10: Isentropic relations of ideal gas . . . . . . . . . . . . . . . . . . . . . . . . . . . . 237.5 7.5: Kittel 6.12: Ideal gas in two dimensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 237.6 7.6: Kittel 7.4: Chemical potential versus temperature . . . . . . . . . . . . . . . . . . . . . . . . 237.7 7.7: The absorbtion of gas onto a surface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23
8 Problem Set 8 258.1 8.1: Mixing of two distinct atoms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 258.2 8.2: Kittel 7.3: Pressure and entropy of degenerate Fermi gas . . . . . . . . . . . . . . . . . . . . . 258.3 8.3: Kittel 7.6: Mass-radius relationship for white dwarfs . . . . . . . . . . . . . . . . . . . . . . . 268.4 8.4: Kittel 7.10: Relativistic white dwarfs stars . . . . . . . . . . . . . . . . . . . . . . . . . . . . 278.5 8.5: Electrons in the air off a conductor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27
9 Special Problem Set (9) 299.1 Problem 1: Properties of “Photon Gas” . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 299.2 Problem 2: Engine Cycle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 299.3 Problem 3: Vibrational Modes of a Molecule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 309.4 Problem 4: Relativistic Massless Bosons . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 319.5 Problem 5: Kittel & Kroemer 7.9 and more . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32
10 Problem Set 10 3310.1 10.1: Collisions with a wall for a Fermi Gas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3310.2 10.2: Free energy and pressure of a Boson gas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3310.3 10.3: Discontinuity in the slope of the heat capacity of a Bose gas . . . . . . . . . . . . . . . . . . 3410.4 10.4: Maximum work extracted from an ideal gas . . . . . . . . . . . . . . . . . . . . . . . . . . . 34
11 Problem Set 11 3711.1 11.1:A review problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3711.2 11.2: Dissociation of water . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3711.3 11.3: Practice with the Jacobian . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3811.4 11.4: More practice . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3911.5 11.5: van der Waals Gas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39
12 Problem Set 12 4112.1 12.1: Maxwell Relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4112.2 12.2: Equilibrium conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4112.3 12.3: Fluctuation in number of a Fermi gas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4212.4 12.4: Fluctuation in volume, pressure, entropy and temperature . . . . . . . . . . . . . . . . . . . 4212.5 12.5: Kittel 10.5: Gas-solid equilibrium . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42
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13 Problem Set 13 4513.1 13.1: Superconduction and Heat Capacity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4513.2 13.2: Kittel 10.8: First order crystal transformation . . . . . . . . . . . . . . . . . . . . . . . . . . 4513.3 13.3: Kittel 11.2: Mixing energy in 3He−4 He and Pb− Sn mixtures . . . . . . . . . . . . . . . . 4613.4 13.4: Kittel 11.4: Solidification range of a binary alloy . . . . . . . . . . . . . . . . . . . . . . . . 4613.5 13.5: Kittel 11.5: Alloying of gold into silicon . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47
iii
Chapter 1
Problem Set 1
1.1 Kittel 8.1: Heat pump
(a) Show that for a reversible heat pump the energy required per unit of heat delivered inside the building is givenby the Carnot efficiency
W
Qh= ηC =
τh − τl
τh(1.1)
What happens if the heat pump is not reversible?
For a reversible system: σh = σl. From the definition of entropy σl = Qlτl
; σh = Qhτh
. Then
W
Qh=
Qh −Ql
Qh=
σhτh − σlτl
σhτh
⇒σh = σl
τh − τl
τh= ηC X
(b) Assume that the electricity consumed by a reversible heat pump must itself be generated by a Carnot engineoperating between the temperatures τhh and τl. What is the ratio Qhh/Qh, of the beat consumed at τhh, tothe heat delivered at τh? Give numerical values for Thh = 600 K; Th = 300 K; Tl = 270 K.
For a Carnot engine
W
Qhh=
τhh − τl
τhh⇒ W =
τhh − τl
τhhQhh
The ratio is then
Qhh
Qh=
τhhW/(τhh − τl)
τhW/(τh − τl)⇒ Qhh
Qh=
(τh − τl)/τh
(τhh − τl)/τhh⇒ Qhh
Qh=
1− τlτh
1− τlτhh
For τhh = 600 K, τh = 300 K and τl = 270 K,
1− 270300
1− 270600
= .18 X
(c) Draw an energy-entropy flow diagram for the combination heat engine-heat pump, similar to Figures 8.1, 8.2and 8.4, but involving no external work at all, only energy and entropy flows at three temperatures.
1.2 Kittel 8.6: Room air conditioner
A room air conditioner operates as a Carnot cycle refrigerator between an outside temperature Th and a room at alower temperature Tl. The room gains heat from the outdoors at a rate A(Th − Tl); this heat is removed by the airconditioner. The power supplied to the cooling unit is P .
(a) Show that the steady state temperature of the room is
1
Tl = (Th + P/2A)−[(Th + P/2A)2 − T 2
h
]1/2(1.2)
The rate is dQldt
= A(Th − Tl), so the power is
P =dW
dt=
1
γ
dQ
dt= A
(Th − Tl)2
Tl
Solving for Tl
T 2l − (2Th − P/A) Tl + T 2
h = 0 ⇒ Tl = Th +P
2A−√
(Th + P/2A)2 − T 2h X
(b) If the outdoors is at 37oC and the room is maintained at 17oC by a cooling power 2 kW, find the heat losscoefficient A of the room in WK−1.
290 = 310 +1000
A−
√(310 +
1000
A
)2
− 3102 ⇒ 400−40000
A+
(1000)2
A2= 310
2+
620000
A+
(1000)2
A2− 310
2 ⇒ A = 1450 W/K X
1.3 Kittel 8.9: Cooling of nonmetallic solid to T = 0
Wa saw in Chapter 4 that the heat capacity of nonmetallic solids at sufficiently low temperatures is proportional toT 3, as C = aT 3. Assume it were possible to cool a piece of such a solid to T = 0 by means of a reversible refrigeratorthat uses the solid specimen as its (varying!) low-temperature reservoir, and for which the high-temperature reservoirhas a fixed temperature Th equal to the initial temperature Ti of the solid. Find an expression for the electrical energyrequired.Rate of change in work is dW = d(Qh −Ql) = γdQl = Th−Tl
TldQl. We are given C = aT 3
dQl = −dTl ⇒ dQl = −aT 3l dTl
Therefore the work is
dW = − a
Tl(Th − Tl) T 3
l dTl
For cooling Th → 0∫W = −a
∫ 0
Th
(ThT 2
l − T 3l
)dTl ⇒ W = −a
3ThT
∣∣0Th
+a
4T 4∣∣0Th⇒ W =
a
12T 4
h X
1.4 Sterling Heat Engine
The operation of a certain type of engine involves applying two isothermal steps and two isovolumetric steps per cycleto one mole of diatomic gas. The largest and smallest volumes shown are VL and VS , respectively. You do not needto consider the vibrational degree of freedom of the molecules.
(a) Find the efficiency of the engine in terms of Th, Tc, VL and VS .
From 1 → 2 the process is isothermal so pV = constant, so the work is
Q12 =
∫pdV ⇒ Q12 =
∫ VL
VS
cdV
V= c ln V
∣∣VL
VS⇒ Q12 = nRTh ln
(VL
VS
)
The heat along 2 → 3: Q23 = 52R (Tc − Th). For 3 → 4: Q34 = nRTc ln
(VLVS
). Finally, for 4 → 1: Q41 =
52R (Th − Tc). Then
W
Qh=
nRTh ln(
VLVS
)+ 5
2R (Th − Tc) + 5
2R (Tc − Th)− nRTc ln
(VLVS
)nRTh ln
(VLVS
)+ 5
2R (Th − TL)
⇒ η =Th ln
(VLVS
)− Tc ln
(VLVS
)Th ln
(VLVS
)+ 5
2(Th − Tc)
X
2
(b) Prove that this engine has a lower efficiency than that of Carnot engine operating at the highest and lowesttemperatures in the cycle.
η =(Th − Tc) ln
(VLVS
)52
(Th − Tc) + Th ln(
VLVS
) =1− Tc
Th
1 + 5/2ln(VL/VS)
(1− Tc
Th
) =ηC
1 + 52
ηCln(VL/VS)
where ηC is the Carnot’s efficiency. Now
5
2
ηC
ln (VL/VS)> 0
because ηC > 0 and VL > VS ⇒ ln (VL/VS) > 0. ∴ η < ηC X
1.5 Unavailability for work
A certain engine works in a cycle between reservoirs at Th and Tc. Find the difference in the work dome by anotherwise ideal engine (Carnot engine) and that done by the above engine in terms of ∆SU , where ∆SU is the changein entropy in the universe due to the operation of the engine.The change in the entropy on the Th is dUh = dQh = τhdσh. The low side, dUc = dQc = τcdσc. By conservationdQh + dQc = 0. Then dσirr = dσh + dσc = dQh/τh + dQc/τc. Then
dσirr =
(1
τh− 1
τc
)dQh =
τc − τh
τhτcdQh
The heat flow is independent of the temperature, so solving for Qh
∆σirr =τc − τh
τhτc∆Qh ⇒ ∆Qh =
τhτc
τc − τh∆σirr
An ideal engine has W = ηCQh where ηC = τh−τc
τh. Substituting the change in entropy to find the change in work
∆W =τh − τc
τh
τhτc
τc − τh∆σirr ⇒ ∆W = −τc∆σirr
For the universe the sign is change, so in conventional units
∆W = Tc∆SU X
1.6 Gibbs Free Energy
The Gibbs free energy of a certain system is given by the follow formula:
G = −nτ ln
(aτ5/2
P
)(1.3)
where a is a constant. Compute
(a) the entropy
G = U + pV − τσ ⇒(
∂G
∂τ
)pV
= −σ ⇒ σ = N ln
(aτ5/2
p
)+ Nτ
5
2
1
τ⇒ σ = N ln
(aτ5/2
p
)+
5
2N X
(b) the heat capacity at constant P , ie Cp
Cp = τ
(∂σ
∂τ
)p
= τ∂
∂τ
(σ = N ln
(aτ5/2
p
)+
5
2N
)= τ
(5N
2τ
)⇒ Cp =
5
2N X
3
(c) the equation of state relating P , V , and τ
V =
(∂G
∂p
)= −Nτ
(−1
p
)⇒ pV = Nτ X
(d) the energy
G = U − τσ + pV ⇒ U = G + τσ − pV = Nτ ln
(aτ5/2
p
)+
5
2Nτ −Nτ −Nτ ln
(aτ5/2
p
)⇒ U =
3
2Nτ X
(e) the chemical potential
µ =
(∂G
∂N
)pτ
⇒ µ = −τ ln
(aτ5/2
p
)X
4
Chapter 2
Problem Set 2
2.1 Spin model
In an isolated system of a large number N of localized particles of spin 1/2, each particle has a magnetic moment ofµ which can point either parallel or antiparallel to an applied magnetic field H. The energy of the system when n ↑and n ↓ is given by U = −(n ↑ −n ↓)µH. Consider the energy range between U and U + δU where δU is very muchsmaller than U but may be microscopically large so that δU µH.
(a) What is the total number of states Ω(U) lying in this energy range?
This problem is much like the random walk problem in that there are only two options for the spins to align alongand that they are equally as likely. With that, the multiplicity is a binomial distribution. Defining m ≡ n ↑ −n ↓,the energy becomes U = −mµH. From the random walk
Ω(m) =N !(
N+m2
)!(
N−m2
)!
Solving m in terms of the energy m = − UµH
, the distribution becomes
Ω(U) =N !(
N− UµH
2
)!
(N+ U
µH
2
)!
Because δU is small compared to the number of available state and that is much greater than µH, δU is the width ofthe distribution. This has to be unit of the number, so it becomes δU → δU
2µH. The final solution for U + δU becomes
Ω(U) =N !(
N− UµH
2
)!
(N+ U
µH
2
)!
δU
2µHX
(b) Assume that the energy U is in a region where Ω(U) is appreciable. Apply the Gaussian approximation to part(a) to obtain a simple expression for Ω(U)dU as a function of U .
To find the Gaussian approximation the logarithm of Ω needs to be expanded. The expansion of a binomial is givenby Equation 2.341.
log g ∼=1
2log 2/πN + N log 2− 2s2/N (2.1)
For this problem s = m2
= U2µH
. This approximation becomes
log (Ω(U) + δU) ∼=1
2log 2/πN + N log 2
− 2
(U
2µH
)2
/N + logδU
2µH
δU goes to dU as the function becomes continuous. This should become a Gaussian once the removal of logarithm.
Ω(U)dU = (2/πN)1/2 2Ne−2(
U2µH
)2/N dU
2µHX
1Kittel and Kroemer pg 20
5
2.2 Paramagnetism of a system of N localized spin-1/2 particles
(a) Using expression for Ω(U) calculated in problem 2.1(a) and applying Stirling approximation, find the relationbetween the absolute “temperature” τ = kBT and the total energy U of this system.
The temperature is defined by
1
kBT=
(∂ (log(g))
∂U
)N
(2.2)
The logarithm of 2.1(a) is
log Ω = log N !− log
(N − U
µH
2
)!− log
(N + U
µH
2
)!
Using Stirling’s approximation log N ! ' N log N −N
log Ω = N log N −N
−[(
µHN − U
2µH
)log
(µHN − U
2µH
)−(
µHN − U
2µH
)+
(µHN + U
2µH
)log
(µHN + U
2µH
)−(
µHN + U
2µH
)]= N log N −
[(µHN − U
2µH
)log
(µHN − U
2µH
)+
(µHN + U
2µH
)log
(µHN + U
2µH
)]Now differentiating with respect to U(
∂ (log(Ω))
∂U
)N
=1
2µH
[log
(µHN − U
µH
)− log
(µHN + U
µH
)]This means with respect to temperature c
(b) Under what circumstance is τ negative?
For the temperature to be negative the logarithm has to be negative. The only time that happens, when the terminside is less than one. This occurs whenever there are more spins against the field than pointing with it. In thatcase, U > 0, so
µHN − U
µHN + U< 1 → µHN − U < µHN + U X
(c) Find the magnetization as a function of H and T .
kBT = 2µH
(log
(µHN − U
µHN + U
))−1
e2µHkBT =
µHN − U
µHN + U
U
(1 + e
2µHkBT
)= µHN
(e
2µHkBT − 1
)
U = µHNe
2µHkBT − 1
1 + e2µHkBT
→ U = µHN tanh
(µH
kBT
)The magnetization is defined as the number of magnetic moments per unit volume. Define n = U/HV
M = (U/HV ) µ tanh
(µH
kBT
)→ nµ tanh
(µH
kBT
)X
6
2.3 Kittel 2.3: Quantum harmonic oscillator
(a) Find the entropy of a set of N oscillators of frequency ω as a function of the total quantum number n. Usemultiplicity function (1.55)2 and make the Stirling approximation log N ! ' N log N −N . Replace N − 1 by N .
Equation 1.55 is
g(N, n) =(N + n− 1)!
(N − 1)!(n)!(2.3)
Replacing N − 1 with N in equation 3
g(N, n) =(N + n)!
(N)!(n)!
The entropy is
σ = log g (2.4)
Then by Stirling approximation
σ = (N + n) log(N + n)− (N + n)−N log N + N − n log n + n
= N log
(N + n
N
)+ n log
(N + n
n
)X
(b) Let U denote the total energy n~ω of the oscillators. Express the entropy as σ(U, N). Show that the total energyat temperature τ is
U =N~ω
e~ω/τ − 1(2.5)
The energy is U = n~ω → U~ω
, so the entropy becomes
σ = N log
(N + U
~ω
N
)+
U
~ωlog
(N + U
~ωU~ω
)
Then the inverse temperature is
1
τ=
(∂ (σ)
∂U
)N
=1
~ωlog (N~ω/U + 1)
~ω
τ= log (N~ω/U + 1)
e~ω/τ − 1 =N~ω
U→ U =
N~ω
e~ω/τ − 1X
2.4 Review Thermal mechanics
Focus on the two stages in the engine cycle of problem 1.4: The isothermal expansion stage followed by the isovolu-metric stage. The working substance in the engine consist of N helium atoms. What is the change in the multiplicityof the gas as it traverses the two stages, with the volume increasing from VS(@τh to VL(@τc)?The entropy for an ideal gas (which helium closely approximates) is given by Equation 3.76, which is
σ = N
[log (nQ/n) +
5
2
](2.6)
where n ≡ N/V and n ≡ (mτ/2π~2)3/2. So the multiplicity is
g = eσ = e5N/2(nQ
n
)N
= e5N/2
(V
N
)N ( mτ
2π~2
)3N/2
2Kittel and Kroemer pg. 25
7
The multiplicity at VS(@τh
g1 = e5N/2
(VS
N
)N (mτh
2π~2
)3N/2
The multiplicity at VL(@τc
g2 = e5N/2
(VL
N
)N ( mτc
2π~2
)3N/2
So the difference is then
g1 − g2 = e5N/2
(VS
N
)N (mτh
2π~2
)3N/2
− e5N/2
(VL
N
)N ( mτc
2π~2
)3N/2
=e5N/2
NN
( m
2π~2
)3N/2 (V N
S τ3N/2h − V N
L τ3N/2c
)X
2.5 Review Thermal mechanics
Suppose a body at temperature τ and pressure P is immersed in a medium that is at the temperature τo and pressurePo, where τ and pressure P are not necessary equal to τo and pressure Po. The body and medium form a closedsystem. Now an external source does work to change the state of the body. The source of work is assumed to beisolated from both the body and the system.
(a) Prove that the minimum work Wmin needed to change the state of the body is given by:
Wmin = ∆(U + PoV + τoσ) (2.7)
The medium can be assume to be large enough that τo and Po are constants and then Uo is also constant. The heattransfer is then
dQo = −τo − dσoσodτo = dU + dUo + Wmin + PodVo
Because τo is constant and Uo is constant dτo = dUo = 0. Because the change in the volume of the medium mustequal to change in the volume in the body due to the constant pressure and temperature of the medium, dVo = dV .So the heat transfer is
−τodσo = dU + Wmin + PodV
According to the second law dσall ≥ 0. The change in all entropy is dσall = dσ + dσo. Therefore dσ ≥ −dσo. Whenit is at its minimum the two must be equal. With this final substitution
τodσ = dU + PodV + Wmin → Wmin = dU + PodV − τodσ
Now to pull the change out front on the right
Wmin = ∆(U + PoV − τoσ) X (2.8)
(b) Now consider the two special cases
(i) P = Po and τ = τo. What is Wmin equal to in this case.
In this case, ∆U = τo∆σ − Po∆V + σ∆τ − V ∆P Making the substitution in equation 8
Wmin = τo∆σ − Po∆V + σ∆τ − V ∆P + Po∆V − τo∆σ) = σ∆τ − V ∆P = −∆GX (2.9)
(ii) P = Po and the process is carried out adiabatically. What is Wmin equal to in this case.
In this case there is no change in the entropy and temperature, ∆U = −Po∆V −V ∆P Making the substitutionin equation 8
Wmin = −Po∆V − V ∆P + Po∆V − τo∆σ = τo∆σ − V ∆P = −∆H X (2.10)
8
Chapter 3
Problem Set 3
3.1 Kittel 3.2: Magnetic susceptibility
(a) Use the partition function to find an exact expression for the magnetization M and the susceptibility χ = dM/dBas a function of temperature and magnetic field for the model system of magnetic moments in a magnetic field.The result of the magnetization is M = nm tanh(mB/τ), as derived in (46) by another method. Here n is theparticle concentration.
In this problem there are two different energies, with and against the field. For one particle the partition function is
Z1 = emB/τ + e−mB/τ = 2 cosh
(mB
τ
)For N particles the partition function becomes
Z = (Z1)N = 2N coshN
(mB
τ
)= 2N coshN (mBβ)
where β = 1/τ . The energy is then
U =1
Z
∂Z
∂β⇒ U =
1
2N coshN (mBβ)
∂
∂β
(2N coshN (mBβ)
)=
1
2N coshN (mBβ)2N coshN−1 (mBβ) sinh (mBβ) mB
U = mB tanh (mBβ) ⇒ M = nm tanh
(mB
τ
)Now for χ
χ =dM
dB=
d
dB
(nm tanh
(mB
τ
))⇒ χ =
nm2
τsech2
(mB
τ
)X
(b) Find the free energy and express the result as a function only of τ and parameter x ≡ M/nm.
The free energy is
F = −τ log Z = −τ log (2 cosh (mB/τ)) = −τ log
(2 sinh (mB/τ)
x
)= τ log x− τ log (2 sinh (mB/τ))
(c) Show that the susceptibility is χ = nm2/τ in the limit mB τ
In this case, mB/τ → 0 ∴
χ =nm2
τsech2
(mB
τ
)→ nm2
τsech20 ⇒ χ =
nm2
τX
9
3.2 Kittel 3.3: Free energy of a harmonic oscillator
A one-dimensional harmonic oscillator has an infinite series of equally spaced energy states, with εs = s~ω, where sis a positive integer or zero, and omega is the classical frequency of the oscillator. We have chosen the zero energyat the state s = 0.
(a) Show that for a harmonic oscillator the free energy is
F = τ log[1− e−~ω/τ
](3.1)
The partition function is
Z =
∞∑s=0
e−s~ω/τ = 1 + e−~ω/τ + e−2~ω/τ + . . .
This is a binomial expansion, as long as e−~ω/τ 1. Then
Z =1
1− e−~ω/τ=(1− e−~ω/τ
)−1
From Eq. 3.55
F = −τ log Z = −τ log(1− e−~ω/τ
)−1
⇒ F = τ log(1− e−~ω/τ
)In the case, τ ~ω: e−~ω/τ ≈ 1− ~ω/τ . Then
F = τ log(1− e−~ω/τ
)≈ τ log (1− 1 + ~ω/τ) ⇒ F = τ log (~ω/τ) X
(b) From (87) show that the entropy is
σ =~ω/τ
e~ω/τ − 1− log
[1− e−~ω/τ
](3.2)
From Eq. 3.49
σ = −∂F
∂τ= − ∂
∂τ
[τ log
(1− e−~ω/τ
)]=
τ~ωe−~ω/τ
τ2 (1− e−~ω/τ )− log
(1− e−~ω/τ
)⇒ σ =
~ω/τ
e~ω/τ − 1− log
(1− e−~ω/τ
)X
3.3 Kittel 3.4: Energy fluctuations
Consider a system of fixed volume in thermal contact with a reservoir. Show that the mean square fluctuation in theenergy of the system is
〈(ε− 〈ε〉)2〉 = τ2
(∂U
∂τ
)V
(3.3)
Here U is the conventional symbol for 〈ε〉.
(∂U
∂τ
)V
=1
Zτ2
∑s
ε2se−εs/τ − 1
Z2τ2
[∑s
εse−εs/τ
]2
τ2
(∂U
∂τ
)V
=1
Z
∑s
ε2se−varepsilons/τ − 1
Z2
[∑s
εse−εs/τ
]2
τ2
(∂U
∂τ
)V
= 〈ε2〉 − 〈ε〉2 X
10
3.4 Kittel 3.10: Elasticity of polymers
The thermodynamic identity for a one-dimensional system is
τdσ = dU − fdl (3.4)
when f is the external force exerted on the line and dl is the extension of the line. By analogy with (32) we form thederivative to find
−f
τ=
(∂σ
∂l
)U
(3.5)
The direction of the force is opposite to the conventional direction of the pressure.We consider a polymeric chain of N links each of length ρ, with each link equally likely to be directed to the right
and to the left.
(a) Show that the number of arrangements that give a head-to-tail length of l = 2|s|ρ is
g(N,−s) + g(N, s) =2N !(
12N + s
)!(
12N − s
)!
(3.6)
The length is given by l = 2|s|ρ → |s| = l/2ρ. |s| is the “steps” of the “random walk.” So
g(N, s) =N !(
N+2s2
)!(
N−2s2
)!
and g(N,−s) =N !(
N−2s2
)!(
N+2s2
)!
The total is the “steps” to the “left and right”
g(N,−s) + g(N, s) =2N !(
N+2s2
)!(
N−2s2
)!
X
(b) For |s| N show that
σ(l) = log [2g(N, 0)]− l2/2Nρ2 (3.7)
Using the Gaussian approximation
gtotal =2N !(
N+2s2
)!(
N−2s2
)!≈ 2g(N, 0)e−2s2/N
The entropy is σ = log g = log (2g(N, 0))− 2s2/N , because |s|2 = l2/4ρ2. Then
σ = log (2g(N, 0))− l2/2Nρ2 X
(c) Show that the force at extension l is
f = lτ/Nρ2 (3.8)
From Eq. 3.96
f = −τ
(∂σ
∂l
)V
= −τ∂
∂l
[log (2g(N, 0))− l2/2Nρ2]⇒ f =
τ l
Nρ2X
11
3.5 Ising spin chain
Based on the prescription of one-dimensional N spin system with only nearest-neighbor interactions J and -J for twoparallel spins and two antiparallel spins, respectively, you are asked to
(a) compute the partition function Z of the spin chain
The partition function
Z =∑
s
e−εs/τ =∑E
Ω(E)e−E/τ
If there are N particles then there are N−1 pairs. The energy is E = JNp−JNa = J(Np−Na), where Na = N−1−Np
∴ E = J(2Np − (N − 1)). The multiplicity is
Ω =(N − 1)!
Np!(N − 1−Np)!⇒ (N − 1)!
(N − 1− 1/2(E/J + N − 1))!(1/2(N − 1 + E/J))!
The partition function becomes
Z =
J(N−1)∑E=−J(N−1)
(N − 1)!
(N − 1− 1/2(E/J + N − 1))!(1/2(N − 1 + E/J))!e−E/τ X
(b) determine the entropy of the system at very high and very low temperatures.
The partition function for 1 pair of particles is
Z1 =(e−Jβ + eJβ
)where β = 1/τ . So for N − 1 pairs
Z =(e−Jβ + eJβ
)N−1
The entropy is given by
σ = ln Z − β
Z
∂Z
∂β
When at high τ , β → 0, so
σ = ln Z − 0
Z
∂Z
∂β= ln
(e−Jβ + eJβ
)N−1
⇒ σ ≈ (N − 1) ln 2
For low τ , let’s consider a shift in energy from -J to J → 0 to 2J . Then
Z =(e−Jβ + eJβ
)N−1
→ Z =(1 + e−2Jβ
)N−1
At low τ , β →∞, so
ln Z = ln(1 + e−2Jβ
)N−1
⇒ ln Z = ln (1 + 0)N−1 ⇒ ln Z = ln 1 = 0
So
σ = −τ ln Z ⇒ σ = 0 X
12
Chapter 4
Problem Set 4
4.1 Application of equal partition theorem
Each particle in a system of N particles has a mass m and is free to preform one-dimensional oscillations about itsequilibrium position. Assume the temperature is sufficiently high so that classical statistical mechanics is relevant.Calculate the heat capacity of this system of particles in each case of the following restoring forces:
(a) The force is proportional to the square of the particle’s displacement x from its equilibrium position.
The force is F ∝ x2i , so the energy from this force is Ui = hx3
i . The total energy is E = ε(xi)+E′, where ε(xi) = hx3i .
The average energy is
εi =
∫∞−∞ e−βEεidx1dx2 . . . dxfdp1dp2 . . . dpf∫∞−∞ e−βEdx1dx2 . . . dxfdp1dp2 . . . dpf
=
∫∞−∞ e−βεiεidxi∫−∞∞e−βεidxi
εi = − ∂
∂βlog
(∫ ∞−∞
e−βεidxi
)(4.1)
Now εi = hx3i , then
εi = − ∂
∂βlog
(∫ ∞−∞
e−βhx3i dxi
)Let y = β1/3xi, then
εi = − ∂
∂βlog
(∫ ∞−∞
e−βhx3i dxi
)⇒ εi = − ∂
∂βlog
(β−1/3
∫ ∞−∞
e−hy3dy
)= − ∂
∂β
(log β−1/3 + log
(∫ ∞−∞
e−hy3dy
))Because this derivative is with respect to β only
εi =1
3β⇒ εi =
1
3τ
The total energy is then
E = N(ε(pi) + ε(xi)) ⇒ E = N
(1
2τ +
1
3τ
)⇒ E =
5
6Nτ
So the heat capacity is then
CV =∂E
∂τ⇒ CV =
5
6N X
(b) The force is proportional to |x|3
13
The force is F ∝ x3, so the potential energy is U = hx4i , then ε(xi) = hx4
i . From equation 4.1
εi = − ∂
∂βlog
(∫ ∞−∞
e−βhx4i dxi
)Now, letting y = β1/4xi, then
εi = − ∂
∂βlog
(β−1/4
∫ ∞−∞
e−hy4dy
)= − ∂
∂β
(log β−1/4 + log
∫ ∞−∞
e−hy4dy
)⇒ εi =
1
4β=
1
4τ
For the total energy of the system
E = N(ε(pi) + ε(xi)) ⇒ E = N
(1
2τ +
1
4τ
)⇒ E =
3
4Nτ
So the heat capacity is
CV =∂E
∂τ⇒ CV =
3
4N X
4.2 Thermal Equilibrium of the Sun and Earth
The surface temperature of the sun is To; its radius is R (= 7×108m) while the radius of the earth is r (= 6.37×106m).The mean distance between the sun and the earth is L (= 1.5×1011m). Assume the earth has reached a steady stateof absorbing and emitting radiation so that its temperature does not change with time.
(a) Find an approximate expression for the temperature T of the earth in terms of the parameters given.
The power of the Sun at its surface is
P = A⊙σT 4⊙where A⊙ is the surface area of the Sun. The power is constant, so the power at L is P = ALσT 4
L. More importantis the percent that the Earth receives, which is
P =AeA⊙
ALσT 4⊙
Now at equilibrium this must equal the amount emitted by the Earth
AseσT 4e =
AeA⊙AL
σT 4⊙ ⇒ 4πR2eσT 4
e =πR2
e4πR2⊙4πL2
σT 4⊙ ⇒ T =
(R⊙2L
)1/2
T⊙ X
(b) Calculate the temperature of the sun given T ≈ 290K.
The temperature of the Sun is
T⊙ =
(2L
R⊙)1/2
T ⇒√
2(1.5× 1011)
7× 108290 ⇒ T⊙ = 6000 K X
4.3 Kittel 4.3: Average temperature of the interior of the Sun
(a) Estimate by a dimensional argument or otherwise the order of magnitude of the gravitational self-energy of theSun, M⊙ = 2× 1033 g and R⊙ = 7× 1010 cm. The gravitational constant G is 6.6× 10−8 dyne cm2 g−2.
The self energy is negative. In cgs units energy is ergs=dyne·cm. The constant has units of dyne cm2
g2 . To remove g2,there needs to be M⊙ squared. For the distance, only the first power of R⊙ is needed. So the self-energy is
U = −GM2⊙R⊙ = −3.77× 1048 ergs X
(b) Assume that the total thermal kinetic energy of the atoms in the Sun is equal to -1/2 the gravitational self-energy.Estimate the average temperature of the Sun, assuming that there are 1× 1057 particles.
K = −U/2 ⇒ 3
2NkBT = −U/2 ⇒ T = − U
3NkB⇒ T =
GM2⊙3NkBR⊙ = 9.1× 106 K X
14
4.4 Kittel 4.6: Pressure of thermal radiation
Show for a photon gas that:
(a)
p = −(
∂U
∂V
)σ
= −∑
j
sj~(
dωj
dV
)(4.2)
From Eq. 4.39, U =∑j
〈sj〉~ωj . Then
p = −
(∂
∂V
(∑j
sj~ωj
))σ
= −∑
j
~ωj∂sj
∂V
∣∣∣∣σ
−∑
j
sj∂(~ωj)
∂V
∣∣∣∣σ
Because sj is the thermal average, it will not change if there is no change in entropy, so∂sj
∂V= 0. Then
p = −∑
j
sj~∂ωj
∂VX
(b)
dωj
dV= − ωj
3V(4.3)
dωj
dV=
(Nπc
L
)dV
= nπcd(V −1/3)
dV= −1
3
nπc
V 4/3= −1
3
nπc
LV= −1
3
ωj
VX
(c)
p =U
3V(4.4)
From parts a and b
p = −∑
j
sj~dωj
dV= −
∑j
sj
(−1
3
~ωj
V
)⇒ p =
1
3
U
VX
(d) Compare the pressure of the thermal radiation with the kinetic pressure of gas of H atoms at a concentrationof 1 mole cm−3. At what temperature (roughly) are the two pressure equal?
For the hydrogen gas p = nRTV
, for photons p =π2k4
B45~3c3
T 4. Then
nRT
V=
π2k4B
45~3c3T 4 ⇒ T =
(45~3c3nR
π2k4BV
)1/3
⇒ T = 3.21× 107 K X
4.5 Kittel 4.7: Free energy of a photon gas
(a) Show that the partition function of a photon gas given by
Z = τΠn
[1− e−~ωn/τ
]−1
(4.5)
where the product is over the modes n.
For one mode Z1 = 1
1−e−~ω1/τ . For the ith mode, Zi = 1
1−e−~ωi/τ . The total partition function is given by
Z = Z1 . . . Zi . . . =
(1
1− e−~ω1/τ
). . .
(1
1− e−~ωi/τ
). . . = Πn
[1
1− e−~ωn/τ
]⇒ Z = Πn
[1− e−~ωn/τ
]X
15
(b) The Helmholtz free energy is found directly from above as
F = τ∑
n
log[1− e−~ωn/τ
](4.6)
Transform the sum to an integral to find
T = −π2V τ4
45~3c3(4.7)
F = −τ log Z = −τ log Πn
[1− e−~ωn/τ
]−1
= τ log Πn
[1− e−~ωn/τ
]⇒ F = τ
∑n
log[1− e−~ωn/τ
]From Eq. 4.17
F = τ∑
n
log[1− e−~ωn/τ
]=
1
8
∫ ∞0
4πn2τ log[1− e−ncπ~/Lτ
]=
2τ
8
∫ ∞0
4πn2dnncπ~Lτ
e−nc~π/Lτ
1− e−nc~π/Lτ
Letting x = nc~πLτ
and Eq. 4.18 and 4.19
F = −π
3τ
(Lτ
~πc
)3 ∫ ∞0
x3dx
ex − 1= −π
3τ
(Lτ
~πc
)3π4
15= −π2L3τ4
45~3c3⇒ F = −π2V τ4
45~3c3X
4.6 Kittel 4.18: Isentropic expansion of photon gas
Consider the gas of photons of the thermal equilibrium radiation in a cube of volume V at temperature τ . Letthe cavity volume increase; the radiation pressure performs work during the expansion, and the temperature of theradiation will drop. From the results for the entropy we know that τV 1/3 is constant in such an expansion.
(a) Assume that the temperature of the cosmic black-body radiation was decoupled from the temperature of thematter when both were at 3000 K. What was the radius of the universe at that time, compared to now? Ifthe radius has increased linearly with time, at what time, at fraction of the present age of the universe did thedecoupling take place?
For the universe R = ct and τV 1/3 = constant, where R is the radius t is the time, and c is some constant. ThenτV 1/3 → τR = constant. Then
τiRi = τfRf ⇒Ri
Rf=
τf
τi
So RiRf
= 2.9/3000 ≈ 1/1000 so the radius was about 1/1000 the present radius. Because the growth is linear c = Riti
.
Then
Ri
titf = Rf ⇒
Ri
Rf=
tf
ti=
1
1000present time X
(b) Show that the work done by the photons during the expansion is
W =
(π2
15~3c3
)Viτ
3i (τi − τf ) (4.8)
where the subscripts i and f refer to the initial and final states.
dW = pdV ⇒ dW =1
3
U
VdV ⇒ dW =
π2τ4
45~3c3dV
By τV 1/3 = constant ⇒ τ3V = constant. Now
τ3dV = −3τ2V dτ ⇒ dV = −3τ−1V dτ
Remembering τ3V is a constant, so τ3V = τ3i Vi. The work is then
W = −∫ τf
τi
π2τ3i Vi
15~3c3dτ ⇒ W =
π2τ3i Vi
15~3c3(τi − τf ) X
16
Chapter 5
Problem Set 5
5.1 5.1: Kittel 4.14: Heat capacity of liquid 4He at low temperature
5.2 5.2: Kittel 4.15: Angular distribution of radiant energy flux
5.3 5.3: Qualifying exam file problem
5.4 5.4: Helmholtz free energy for the Debye model
5.5 5.5: Kittel 5.3: Potential energy of gas in gravitational field
17
18
Chapter 6
Problem Set 6
6.1 6.1: Kittel 5.4: Active transport
Consider the concentration of K+ is 104 inside the plant cell compared to the pond. Find the chemical potentialdifference.
Treating the ions as an ideal gas,the chemical potential difference is
∆µ = µin − µout = τ log
(nin
nQ
)− τ log
(nout
nQ
)= τ log
(nin
nout
)Inserting the values
∆µ =300 K
11605 K/eVlog(1000) ≈ 0.24 eV X
6.2 6.2: Kittel 5.7: States of positive and negative ionization
For the following system,
State # of electrons EnergyGround 1 −∆/2
Positive ions 0 −δ/2Negative ions 2 δ/2
Excited 1 ∆/2
Find the condition for⟨N⟩
= 1.
The grand partition function is
Z = λe∆/2τ + eδ/2τ + λ2e−δ/2τ + λe−∆/2τ (6.1)
So the average number is ⟨N⟩
=1
Z∑ASN
NλNe−εs/τ
⟨N⟩
=λ(e∆/2τ + e−∆/2τ
)+ 2λ2e−δ/2τ
λ (e∆/2τ + e−∆/2τ ) + λ2e−δ/2τ + eδ/2τ(6.2)
For⟨N⟩
= 1, then λ = eδ/2τ X
19
6.3 6.3: Kittel 5.10: Concentration fluctuation
(a) Show that
⟨N2⟩ =
τ2
Z∂2Z∂µ2
(6.3)
Now ⟨N2⟩ =
∑ASN N2e(Nµ−εs)/τ
Z =τ2
Z∑ASN
∂2
∂µ2e(Nµ−εs)/τ
=τ2
Z∂2
∂µ2
∑ASN
e(Nµ−εs)/τ ⇒⟨N2⟩ =
τ2
Z∂2Z∂µ2
X
(b) Show that
⟨(∆N)2
⟩=
τ∂⟨N⟩
∂µ(6.4)
Beginning with
τ∂⟨N⟩
∂µ= τ
∂
∂µ
(τ
Z
(∂Z∂µ
))= τ2
(1
Z∂2Z∂µ
− 1
Z2
∂Z∂µ
∂Z∂µ
)which reduces to
τ2
(1
Z∂2Z∂µ2
− 1
Z2
(∂Z∂µ
)2)
=⟨ (
N −⟨N⟩)2 ⟩
=⟨(∆N)2
⟩X
6.4 6.4: Kittel 6.3: Distribution function for double occupancy statistics
For a new system, which has orbitals 0,1,2 and energies 0,ε,2ε, respectively
(a) Derive⟨N⟩, so
Z = 1 + λe−ε/τ + λ2e−2ε/τ (6.5)
So⟨N⟩
⟨N⟩
=1
Z
(0 · 0 + 1 · λe−ε/τ + 2λ2e−2ε/τ
)⟨N⟩
=λe−ε/τ + 2λ2e−2ε/τ
1 + λe−ε/τ + λ2e−2ε/τ(6.6)
(b) Now if ε is double generate
The partition function is now
Z = 1 + 2λe−ε/τ + λ2e−2ε/τ =(1 + λe−ε/τ
)2
(6.7)
Then ⟨N⟩
=1
Z
(0 · 0 + 2 · λe−ε/τ + 2λ2e−2ε/τ
)=
2λe−ε/τ(1 + λe−ε/τ
)(1 + λe−ε/τ )
2 =2λe−ε/τ
1 + λe−ε/τ
Now ⟨N⟩
=2
λ−1eε/τ + 1⇒
⟨N⟩
=2
e(ε−µ)/τ + 1X (6.8)
20
6.5 6.5: Kittel 6.7: Relation of pressure and energy density
(a) Show that
p = −∑
s
(∂εs∂V
)N
e−εs/τ
Z(6.9)
So beginning with the partition function, Z =∑
s e−εs/τ , then the Helmholtz free energy is
F = −τ log Z = −τ log
(∑s
e−εs/τ
)so the pressure is then
p = −∂F
∂V
∣∣∣∣τ,N
=τ ∂Z
∂V
Z
p = −∑
s
(∂εs∂V
)N
e−εs/τ
ZX
(b) Show that for the free particles (∂εs
∂V
)N
= −2
3
εs
V(6.10)
for the boundary conditions of the system
The energy levels for these boundary conditions(∂εs
∂V
)N
=∂
∂V
(~2π2n2
2mV −2/3
)= −2
3
(~2π2n2
2mV −5/3
)= −2
3
εs
VX
(c) Show that
p =2U
3V(6.11)
From part 1 and 2
p = −∑
s
(∂εs∂V
)N
e−εs/τ
Z=
∑s
(2
3Vεs
)e−εs/τ
Z
Then using the definition of the partition function and energies
p =2
3V
(∑s εse
−εs/τ
Z
)⇒ p =
2U
3VX
6.6 6.6: Kittel 6.9: Gas of atoms with internal degree of freedom
For an ideal gas with an internal degree of freedom, ∆. The internal partition is for one particle is Zint =
e−ε/τ(1 + e−∆/τ
)(a) Find the chemical potential
Because the logarithm is being taken of the partition function the internal and ideal gas can be separated. This is infact true for all of this problem. From Equation 6.48 µ = τ [log (n/nQ)− log Zint]
µ = τ [log(n/nQ)− log Zint]
= τ[log(n/nQ) +
ε
τ− log(1 + e−∆/τ )
]X
21
(b) Free energy
From Equation 6.49, Fint = −Nτ log Zint
F = Nτ (log (n/nQ)− 1)−Nτ log(e−ε/τ
(1 + e−∆/τ
))= Nτ (log (n/nQ)− 1) + Nε−Nτ log
(1 + e−∆/τ
)X
(c) Entropy
From Equation 6.50, σint = − (∂F/∂τ)V
σ = σideal + σint
= N [log(n/nQ) + 5/2]− ∂
∂τ
(Nε−Nτ log
(1 + e−∆/τ
))= N [log(n/nQ) + 5/2] + N log
(1 + e−∆/τ
)+
N∆
τ (e∆/τ + 1)X
(d) Pressure
Notice that the internal partition function is independent of the volume. From Equation 3.49 −p =(
∂F∂V
)τ, then
−pint = 0, so the pressure is just the ideal gas, which is p = NτV
X
(d) Heat capacity at constant pressure.
For an ideal gas, Cp = 52N . Now from Equation 6.37, but the volume is constant, so
Cp =
(∂U
∂τ
)p
+ p
(∂V
∂τ
)p
→ Cp =
(∂U
∂τ
)p
Now the energy needs to be found. So the relation Uint = Fint + σintτ yields the internal energy.
Uint = Nε−Nτ log(1 + e−∆/τ
)+ Nτ log
(1 + e−∆/τ
)+
N∆
e∆/τ + 1
=N∆
e∆/τ + 1
So the heat capacity for the internal energy is
Cp =∂
∂τ
N∆
e∆/τ + 1=
N∆2
τ2
e∆/τ
(e∆/τ + 1)2
The total heat capacity at constant pressure
Cp =5
2N +
N∆2
τ2
e∆/τ
(e∆/τ + 1)2 X
22
Chapter 7
Problem Set 7
7.1 7.1: Pressure in types of gases
7.2 7.2: Kittel 5.13: Isentropic expansion
7.3 7.3: Kittel 6.8: Time for a large fluctuation
7.4 7.4: Kittel 6.10: Isentropic relations of ideal gas
7.5 7.5: Kittel 6.12: Ideal gas in two dimensions
7.6 7.6: Kittel 7.4: Chemical potential versus temperature
7.7 7.7: The absorbtion of gas onto a surface
23
24
Chapter 8
Problem Set 8
8.1 8.1: Mixing of two distinct atoms
Calculate the entropy change when an impermeable partition separating two compartments each of volume V andeach containing an ideal monatomic gas of N identical atoms at temperature τ is removed. However, the atoms inone compartment are distinguishable from those in the other compartment.
To identify each set of particles one side will be A and the other B. The change in entropy for A particles is given
by σ2 − σ1 = N log(
V2V1
), but V2 = 2V1, so the change in entropy of A is then ∆σA = N log 2. This is identical for
B because there is no description of A in finding of the change in the entropy. So the change in the entropy for B is
∆σB = N log 2. The total change in entropy is then ∆σA + ∆σB = ∆σ = 2N log 2 X
8.2 8.2: Kittel 7.3: Pressure and entropy of degenerate Fermi gas
(a) Show that a Fermi electron gas in the ground state exerts a pressure
p =
(3π2)2/3
5
~2
m
(N
V
)5/3
(8.1)
From earlier in Kittel, the pressure is
p = −(
∂U
∂V
)σ
⇒ p = − ∂
∂V
(3
5NεF
)→ p = −3
5N
∂εF
∂V
Now expanding the Fermi energy
p = −3
5N
∂εF
∂V→ p = − ~2
2m
3
5N
∂
∂V
[(3π2N
V
)2/3]
=3
5
~2N5/3
2m
(3π2)2/3 ∂
∂V
(1
V 2/3
)
p =~2N5/3
5m
(3π2)2/3
V 5/3→ p =
(3π2)2/3
5
~2
m
(N
V
)5/3
X
(b) Find an expression for the entropy of a Fermi electron gas in the region τ εF
The heat capacity of a Fermi gas is CV = 12π2Nτ/τF . The heat capacity is related to the entropy by
CV = τ
(∂σ
∂τ
)V
→ ∂σ =CV
τ∂τ
Using the heat capacity of Fermi gas
dσ =
∫ τ
0
1
2π2Nτ/ (ττF ) dτ ⇒ σ =
1
2π2Nτ/τF X
25
8.3 8.3: Kittel 7.6: Mass-radius relationship for white dwarfs
Consider a white dwarf of mass M and radius R. Let the electrons be degenerate but nonrelativistic; the protons arenondegenerate.
(a) Show that to the order of magnitude of the gravitational self-energy is −GM2/R, where G is the gravitationalconstant.
The density for an uniform sphere is ρ = 3M4πR3 . So the potential self-energy is
∆V = −G∆mm(r)
r= −G4πr2∆rρm(r)
r
= −G4πr∆rρ24πr3
3= −16Gπ2ρ2r4∆r
3
= −3GM2
R6r4∆r
Then the total potential self-energy is
V = −∫ R
0
3GM2
R6r4dr → V = −3GM2
5R≈ −GM2
RX
(b) Show that the order of magnitude of the kinetic energy of the electrons in the state with mass of an electron,m, and mass of a proton, MH , is
~2N5/3
mR2≈ ~2M5/3
mM5/3H R2
(8.2)
The energy of one electron is ε = ~2
2m
(3π2N
V
)2/3
. There are N electrons in the Sun. The total energy would be then
K = ~2N2m
(3π2N
V
)2/3
. The volume of the Sun is V = 43πR3. The energy becomes
K =~2
2m
(9πN
4R3
)2/3
N =~2N5/3
mR2
[1
2
(9π
4
)2/3]
Because the mass of the proton is so much more than that of the electrons, so then all that has to be considered ismass of the proton. So the number of particles are N ≈ M/MH . The terms in [. . .] are on the order of two, so to theorder of magnitude, the kinetic energy is
K =~2N5/3
mR2
[1
2
(9π
4
)2/3]≈ ~2
mR2
(M
MH
)5/3
X
(c) Show that if the self-energy and kinetic energy are equal, that M1/3R ≈ 1020g1/3cm.
So immediately, then
GM2
R=
~2
mR2
(M
MH
)5/3
⇒ M1/3R =~2
mGM5/3H
M1/3R ≈(10−27
)210−810−27 (10−24)5/3
= 1020g1/3cm X
(d) What is the density of a white dwarf with its mass equal to the mass of the Sun (2× 1033g)
The density would be
M1/3R = 1020 ⇒(MR3)−1
= 10−60 ⇒ ρ =3M2
4π1060
ρ =3(2× 1033
)24π1060
→ ρ = 955000 g/cm3 X
26
(e) Show for a neutron star M1/3R ≈ 1017g1/3cm. What is the radius of the neutron star whose mass is equal tothat of the Sun?
Looking back at part c, M1/3R = ~2
mGM5/3H
, but the mass of a neutron is Mn ≈ 1000m and that MH ≈ Mn. Then
M1/3R =~2
MnGM5/3H
=~2
1000mGM5/3H
≈ 1017g1/3cm
The radius of a neutron star with mass of the Sun would be
M1/3R = 1017g1/3cm(2× 1033)1/3
R = 1017g1/3cm
R = 794000cm → R = 7.94km X
8.4 8.4: Kittel 7.10: Relativistic white dwarfs stars
Consider a Fermi gas of N electrons each of rest mass m in a sphere of radius R. Treating the electrons as relativisticfor a white dwarf, find the following
(a) Using the viral theorem find the value of N , neglecting the kinetic energy of the protons.
From problem 7.2, the kinetic energy for the electrons is K + 34NεF and the Fermi energy is εF = ~πc (3N/πV )1/3.
Using the volume of the sphere, the kinetic energy becomes K = 3π~c4R
(9
4π2
)1/3N4/3. The potential energy is from
Problem 7.6, V = 3GM2
5R≈ 3GN2M2
H5R
. Applying the viral theorem
3GN2M2H
5R=
3
4
π~c
R
(9
4π2
)N4/3
N2/3 =5π~c
4GM2p
(9
4π2
)1/3
N =
[5π~c
4GM2p
(9
4π2
)1/3]3/2
X
(b) Estimate the value of N
Inserting the values
N =
[5
4
π3× 108(1.05× 10−34
)6.6× 10−11 (1.67× 10−27)2
(9
4π2
)1/3]3/2
= 8.14× 1057 X
8.5 8.5: Electrons in the air off a conductor
The lowest possible energy of a conduction electron in a metal is −Vo below the energy of a free electron at infinity.The conduction electrons have a chemical potential energy µ. The minimum energy needed to remove an electronfrom the metal is Φ = Vo−µ and is called the work function of the metal. Consider an electron gas outside the metalin thermal equilibrium with the electrons in the metal at the temperature τ . Assume τ Φ, find the mean numberof electrons per unit volume outside the metal.
The mean number of electrons per unit volume outside the metal can be found by relating the chemical potentials.The chemical potential inside the conductor is µ. Outside the conductor, the chemical potential is µideal + Φ + µ.The chemical potential for an ideal gas is µ = τ log (n/nQ). Therefore
µideal + Φ + µ = µ → τ log (n/nQ) + Φ + µ = µ → n = nQe−Φ/τ
27
Now n = N/V , and nQ =(
mτ2π~2
)3/2. Then
N/V =( mτ
2π~2
)3/2
e−Φ/τ
Lastly, there is a spin 2 degeneracy, so
N
V= 2
( mτ
2π~2
)3/2
e−Φ/τ X
28
Chapter 9
Special Problem Set (9)
9.1 Problem 1: Properties of “Photon Gas”
The Helmholtz free energy of a certain system is given by F = bV τ4, where b is a constant and all other symbolshave their usual meaning.
(a) Compute the chemical potential of the system.
The chemical potential is related to the free energy by
µ =
(∂F
∂N
)τ,V
=∂
∂NbV τ4 ⇒ µ = 0 X
(b) What is the energy of the system?
The energy of the system is related to the free energy by
U = −τ2 ∂(F/τ)
∂τ= −τ2 ∂
∂τbV τ3 ⇒ U = −3bV τ4 X
(c) Find the equation of state that describes the system.
The equation of state is related to the pressure, so
p = −(
∂F
∂V
)τ
⇒ p = −bτ4 or p =U
3VX
(d) What is the work done during an isentropic expansion from τ1 to τ2?
So first the entropy has to be found σ = −(∂F/∂τ)V = −4bV τ3.
dW = pdV1.c⇒ dW =
1
3
U
VdV
1.b⇒ dW = −bτ4dV
An isentropic expansion σ =constant, so τ3V =constant. Then dV = −3τ−1V dτ . The work is then dW = 3bV τ3dτ ,but because the expansion is isentropic V τ3 =constant. With that choose it to be the initial state 1. Finally
W =
τ2∫τ1
3bV1τ31 dτ ⇒ W = 3bV1τ
31 (τ2 − τ1) X
9.2 Problem 2: Engine Cycle
For an ideal-gas engine cycle the processes AB and CD are isentropic, process BC is isobaric, and process DA isisovolumetric.
(a) Calculate the engine efficiency η in terms of the temperatures τA, τB , τC , τD and γ (= Cp/CV ).
29
The efficiency is η = Qh−QlQh
. So the heat comes and leaves the system in only two places, because AB and CD are
isentropic. The heat in is Qh = Cp (τC − τB). The heat out is Ql = CV (τD − τA). So the efficiency is
η =Qh −Ql
Qh=
Cp (τC − τB)− CV (τD − τA)
Cp (τC − τB)
η = 1− τD − τA
γ (τC − τB)X
(b) Find η/ηC , where ηC is the efficiency of a Carnot engine operating at the highest and lowest temperatures ofthe cycle.
First find the highest and lowest temperatures. From the graphs and defined process: VCτB = VBτC and VC > VB →τC > τB ; pDτA = pAτD and pD > pA → τD > τA; and τAV γ−1
A = τBV γ−1B and VA > VB → τB > τA. The Carnot
efficiency is ηC = 1− τAτC
. Also from the ideal gas law, τB =VB
VCτC . τD =
(VCVD
)γ−1
τC . V γ−1D =
VγB
VC
τCτA
. Combining
the statements τD =
(VC
VB
)γ
τA The efficiency is then
η = 1−τA
((VCVB
)γ
− 1)
γτC (1− VB/VC)
Then the ratio is
η
ηC=
1−τA
((VCVB
)γ−1)
γτC(1−VB/VC)
1− τAτC
X
(c) Is η/ηC> 1 or < 1.
From previous knowledge, it should be η/ηC < 1. Now the check
1−τA
((VCVB
)γ−1)
γτC(1−VB/VC)
1− τAτC
< 1 ⇒τA
((VCVB
)γ
− 1)
γτC (1− VB/VC)>
τA
τC
(VC/VB)γ − 1 > γ (1− VB/VC)
This statement is true as long as γ > 1, which is true by the fact that γ = Cp/CV = (CV + 1)/CV > 1. X
9.3 Problem 3: Vibrational Modes of a Molecule
Suppose that in a certain polyatomic ideal gas, the vibrational motions of each molecule can be described in termsof three independent simple harmonic oscillators having the same frequency ω/2π. The vibrational energy of eachmolecule is then given by εn =
(na + nb + nc + 3
2
)~ω, where n = na + nb + nc with all the n’s being 0, 1, 2, . . .
(a) What is the degeneracy of the vibrational states of each molecule as a function of n?
The degeneracy for S independent harmonic oscillators is (S+n−1)!/(S−1)!(n)!. Here S = 3 then g(n) =(n + 2)!
n!2X
(b) Write down an expression to represent the vibrational part of the molecular partition function (call it Zvib).
The partition function is the sum of the Boltzmann factor. So Zvib =
∞∑n=0
1
2
(n2 + 3n + 2
)e−3~ω/2τe−n~ω/τ X
(c) Calculate Zvib at high temperature, i.e. assume ~ωτ 1.
30
The partition function at high temperature
Zvib =e−3~ω/2τ
2
∫ ∞0
(n2 + 3n + 2
)e−n~ω/τdn
=e−3~ω/2τ
2
[Γ(3)
(~ω/τ)3+
3Γ(2)
(~ω/τ)2+
2Γ(1)
~ω/τ
]Now applying ~ω τ
Zvib =
(1− 3
2
~ω
τ
)[3τ3
(~ω)3+
3τ2
(~ω)2+
τ
~ω
]
(d) Evaluate the heat capacity and energy per molecule at high temperature.
The energy is Uvib = τ2 ∂ log Zvib∂τ
, which is
Uvib = τ2 ∂
∂τlog
[(1− 3
2
~ω
τ
)[3τ3
(~ω)3+
3τ2
(~ω)2+
τ
~ω
]]= τ2
(18τ2 − 6τ − z
(2τ − 3) (3τ2 + 3τ + 1)
)~ωτ⇒ Uvib = τ2 18τ2
6τ3
The total energy per molecule is Utot = Uideal + Uvib ⇒ Utot = 32τ + 3τ → Utot =
9
2τ . The heat capacity is
CV = ∂U∂τ⇒ CV = ∂
∂τ92τ ⇒ CV =
9
2X
9.4 Problem 4: Relativistic Massless Bosons
Consider an ideal gas of massless bosons in thermal equilibrium. The number of such bosons in the system is constant.
(a) Derive an expression for the number of thermally excited bosons above the ground state if the energy of thebosons happens to exhibit the dispersion relation ε = ~ck.
First k is the wavenumber or k = nπ/L. The partition function would be Z =∞∑
n=0
e−n~π/Lτ . Then
〈n〉 = Z−1∑
ne−n~π/Lτ ⇒ 〈n〉 =1
e~π/Lτ − 1
Then
Ne =∑
〈n〉 =1
8
∫ ∞0
4πn2dn
en~π/Lτ − 1=
V τ3
2π2~3c3
∫ ∞0
x2dx
ex − 1
Ne = 2ζ(3)V τ3
π2~3c3⇒ Ne ≈ 1.202
V τ3
π2~3c3X
(b) Find the critical temperature if the number density of 1020 cm−3 below which Bose-Einstein condensation (BEC)occur.
For a BEC, N = Ne. Then
1020V = 1.202V τ3
π2~3c3→ τ = ~c
3
√1020π2
1.202⇒ T = 21458 K X
31
9.5 Problem 5: Kittel & Kroemer 7.9 and more
(a) Calculate the integral for Ne(τ) for a one-dimensional gas of noninteracting bosons, and show that the integraldoes not form in one dimension. Take λ = 1 for the calculation.
From Equation 7.86, D(ε) = Lπ
(2m~2ε
)1/2. Then
Ne =
∫dεD(ε)f(ε, τ) =
∫ ∞0
dεL
π
(2m
~2ε
)1/21
eε/τ − 1
=L
π
(2m
~2
)1/2
τ1/2
∫ ∞0
x−1/2dx
ex − 1
Taking the first order term on the exponent∫∞0
x−1/2dx/(1 + x− 1) = −1/2x−1/2∣∣∞0
, which does not converge. Anyhigher order terms will diverge also. X
(b) What functional form does the density of state (DOS) need to have for an abrupt transition to occur in a 1-Dand 2-D Boson gases? What is the range of α in 1-D? In 2-D?
32
Chapter 10
Problem Set 10
10.1 10.1: Collisions with a wall for a Fermi Gas
Calculate the average rate of collision per unit area with a wall in a Fermi gas at T = 0, which will be called Φ.
Let’s think of this problem in momentum space. With T = 0, f = 1 of the fermi-dirac distribution. For an isotropicsphere in momentum space: sin θdθdφp2dp/(2π~)3. Then the velocity is v · n = v cos θ. Remembering that v = p/m.So
Φ =
pF∫0
π/2∫0
2π∫0
(2s + 1)
(2π~)3p3
mcos θ sin θdφdθdp
=π(2s + 1)
m(2π~)3
pF∫0
p3dp =(2s + 1)π
4(2π~)3p4
F
m
Now, pF = 2mεF , and εF = ~2
2m
(6π2N
(2s+1)V
)2/3
Φ =π(2s + 1)
4(2π~)34m2
m
(~2
2m
)2(6π2N
(2s + 1)V
)4/3
Φ =3(6π2)1/3
16(2s + 1)1/3
~m
(N
V
)4/3
X
10.2 10.2: Free energy and pressure of a Boson gas
Determine the free energy F and pressure P on a Boson gas at τ τE . Include the degeneracy factor (2s + 1) inyour answers.
From the class notes U = NeτI, where Ne = 1.306V4
(2Mπ~2
)3/2τ5/2(2s + 1) and I = .7702. Now U = −τ2∂(F/τ)/∂τ .
Then
F/τ =
∫− 1
τ2Udτ
F/τ = − .7702(1.306)V
4
(2M
π~2
)3/2
(2S + 1)
τ∫0
τ1/2dτ
F = − .7702(1.306)V
4
(2M
π~2
)3/22
3(2s + 1)τ5/2
F = −.4742V (2s + 1)
(M
π~2
)3/2
τ5/2 = −2
3U X
33
Now for the pressure
p = −(
∂F
∂V
)τ
⇒ p =2(.7702)(1.306)
(3)4
(2M
π~2
)3/2
(2s + 1)τ5/2
p = .4742(2s + 1)
(M
π~2
)τ5/2 =
2
3
U
VX
10.3 10.3: Discontinuity in the slope of the heat capacity of a Bose gas
Provide the missing steps on the derivation of the heat capacity of a Bose gas near τE . That is, use ∆U expression
to show that ∆CV = 0 and ∆(
∂CV∂τ
)= −3.66N
τEat τ = τE
From the lecture notes
Uex − Uo = ∆U = −3π2~6
m3Nµ=0
(Nµ=0 −N
V τ
)2
Now Nµ=0 = N(
ττE
)3/2
. So at τ = τE → Nµ=0 = N . It is obvious that the squared term is 0 from Nµ − N = 0,
then ∆U = 0. Now CV = ∂U∂τ
, and ∆CV = ∂(Uo+∆U)∂τ
− ∂Uo∂τ
= ∂∆U∂τ
= ∆CV . Which is
∆CV = −3π2~6
m3V 2
[∂Nµ=0
∂τ
(Nµ=0 −N
τ
)2
+ 2Nµ=0
(Nµ=0 −N
τ
)(1
τ
∂Nµ=0
∂τ− Nµ=0 −N
τ2
)]Every term is multiplied by Nµ=0 −N , which is 0 at τ = τE . Now for ∆ ∂CV
∂τ, which is
∆∂CV
∂τ=
∂2(Uo + ∆U)
∂τ2− ∂2Uo
∂τ2=
∂∆CV
∂τ
The only nonzero term is the second term by the same argument as before.
∂∆CV
∂τ= −
6π2~6
m3V 2
[(∂Nµ=0
∂τ
)2 (Nµ=0 −N
τ2
)+ Nµ=0
∂2Nµ=0
∂τ2
(Nµ=0 −N
τ2
)+ Nµ=0
1
τ2
(∂Nµ=0
∂τ
)− 2Nµ=0
∂Nµ=0
∂τ
(Nµ=0 −N
τ3
)]Again, by Nµ=0 −N = 0, only one term remains
∂∆CV
∂τ= −6π2~6
m3V 2Nµ=0
(1
τ
∂Nµ=0
∂τ
)2
Now using Nµ=0 = N(τ/τE)3/2. Then
∂∆CV
∂τ= −6π2~6
m3V 2N
(τ
τE
)3/2(
3
2
N
τ1/2τ3/2E
)2
⇒ ∂∆CV
∂τ= −6π2~6
m3V 2
9
4
N3
ττ3E
∣∣∣∣τ=τE
∂∆CV
∂τ=
6π2~6
m3V 2
(9
4
)N2( m
2π~2
)3(
2.612V
N
)2N
τE= −3.66
N
τEX
10.4 10.4: Maximum work extracted from an ideal gas
Determine the maximum work that can be extracted from an ideal gas system held at constant volume by cooling itfrom temperature τ1 to a temperature τo of the medium.
The amount of heat that the system gives off can be turned into work is
∆Qgain =
τ1∫τo
NCV dτ ⇒ ∆Q = NCV (τ1 − τo)
34
But there is some energy lost to the change in entropy
∆σ =
τ1∫τo
NCV
τdτ ⇒ ∆σ = NCV log
(τ1
τo
)
The heat is ∆Qlost = τ∆σ = NCV τo log(
τ1τo
). The difference of the two is the maximum work that can be done by
the ideal gas
W = NCV
(τ1 − τo − τo log
(τ1
τo
))X
35
36
Chapter 11
Problem Set 11
11.1 11.1:A review problem
The entropy of a monatomic ideal gas can be written as σ = (3N/2) ln τ+ terms that involve N , V , M and someconstants. By including the internal degrees of freedom in the entropy of a polyatomic ideal gas, prove that the factor3N/2 must be replaced by the factor N
γ−1.
From problem 11.3,
CV = τ
(∂σ
∂τ
)V
⇒ dσ =CV
τdτ ⇒ σ = CV ln τ
Now
CV = τ
(∂σ
∂τ
)V
= Cp + τ
(∂σ
∂p
)τ
(∂p
∂τ
)V
= Cp − τ
(∂V
∂τ
)p
(∂p
∂τ
)V
Now applying the ideal gas law pV = Nτ
Cp − τ
(∂V
∂τ
)p
(∂p
∂τ
)V
= Cp − τN
p
N
V= Cp −N
Then CV = Cp −N .
N
γ − 1=
CV N
Cp − CV=
CV N
N= CV → N
γ − 1= CV
Then from the first equation
σ = CV ln τ ⇒ σ =N
γ − 1ln τ X
11.2 11.2: Dissociation of water
A certain number of moles of H2O are introduced into a container of volume V . At some high temperature Tdissociation takes place according to 2H2O → 2H2 + O2. Let x denote the fraction of water vapor molecules whichare dissociated at T corresponding to a total pressure P .
(a) Derive an expression that relates x to p and a constant that depends only on temperature.
The partial pressure is pi = niτ , so the total pressure is
p =∑
i
niτ
Now for each water molecule that dissociates, there are 3/2 molecules. Then
p = (1− x)nτ +3
2xnτ =
(1 +
x
2
)nτ
37
Then from class
N2H2O
N2H2
HO2
= KN (τ)
Factoring the concentration n = N/V
(1− x)2n2
(x2)x2n3
= Kn(τ) ⇒ 2(1− x)2
x3Kn(τ)= n
Placing that into the pressure
p =2(1 + x/2)(1− x)2
x3
τ
Kn(τ)X
(b) Write down an equation that relates the chemical potentials of H2O, H2 and O2 in equilibrium.
In equilibrium, the Gibb’s free energy is 0.
G = µiNi = 0
Now, rewriting the chemical equation as
−2H2O + 2H2 + O2 = 0 ⇒ −2NH2O + 2NH2 + NO2
Combining the two equations
µiNi = −2µH2ONH2O + 2µH2NH2 + µO2NO2 = 0
Setting the number equal as part of the equilibrium conditions
− 2µH2ONH2O + 2µH2NH2 + µO2NO2 = 0
(−2µH2O + 2µH2 + µO2
)N = 0⇒ µH2 +
1
2µO2 = µH2O X
11.3 11.3: Practice with the Jacobian
Since any thermodynamical quantity of a system with a fixed number of particles can be written as a function oftwo variables that are not necessarily the same as those used in class, thus you may select the appropriate functionalforms to derive the following expressions: (
∂U
∂τ
)V
= τ
(∂σ
∂τ
)V(
∂U
∂V
)τ
= −p + τ
(∂σ
∂V
)τ
Express the Helmholtz free energy F as a function of T and V to prove that(
∂σ∂V
)τ
=(
∂p∂τ
)V
so that one of the above
expressions can be also be written as(
∂U∂V
)τ
= −p + τ(
∂p∂τ
)V
Writing the energy in terms of entropy and volume
U(σ, V ) ⇒ dU = τdσ − pdV
dU = τ
(∂σ
∂τ
)V
dτ +
[τ
(∂σ
∂V
)τ
− p
]dV
Now for the relationship between the energy and temperature holding the volume constant(∂U
∂τ
)V
= τ
(∂σ
∂τ
)V
X
38
Then the relationship between the energy and volume holding the temperature constant(∂U
∂V
)τ
= −p + τ
(∂σ
∂V
)τ
X
Writing the free energy in terms of temperature and volume
dU = d(τσ)− σdτ − pdV ⇒ dF = −σdτ + pdV
Then
dF =
(∂F
∂τ
)V
dτ +
(∂F
∂V
)τ
dV(∂F
∂τ
)V
= −σ;
(∂F
∂V
)τ
= −p
∂2F
∂τ∂V=
∂2F
∂V ∂τ⇒(
∂σ
∂V
)τ
=
(∂p
∂τ
)V
X
11.4 11.4: More practice
Prove that(
∂CV∂V
)τ
= τ(
∂2p∂τ2
)V
From problem 11.3 (∂CV
∂V
)τ
=∂
∂V
[τ
(∂σ
∂τ
)τ
]= τ
∂2σ
∂V ∂τ= τ
(∂
∂τ
)V
(∂σ
∂V
)τ
Again from problem 11.3
τ
(∂
∂τ
)V
(∂σ
∂V
)τ
= τ
(∂
∂τ
)V
(∂p
∂τ
)V
= τ
(∂2p
∂τ2
)V
X
11.5 11.5: van der Waals Gas
Use some of the results from problems 11.3 and 11.4 above to obtain a formula for the energy of a van der Waalsgas which obeys the equation of state
(p + a/V 2
)(V − b) = (R/kB) τ , where R is the gas constant. Note that your
answer will also involve the specific heat function CV .
The pressure is
p =R
kB
τ
V − b− a
V 2
From problem 11.3
dU = CV dτ +
[τ
(∂p
∂τ
)V
− p
]dV ⇒ U
′+ dU
′′= CV τ +
[τ
(∂p
∂τ
)V
− p
]dV
Then with the pressure from above
τ
(∂p
∂τ
)V
− p =R
kB
τ
V − b− R
kB
τ
V − b+
a
V 2=
a
V 2∫a
V 2dV = − a
V
Placing this in the second equation
U = CV τ − a
V+ constant X
39
40
Chapter 12
Problem Set 12
12.1 12.1: Maxwell Relations
The formula Cp − CV = −τ(
∂σ∂p
)τ
(∂p∂τ
)V
was derived in class
(a) Show that Cp − CV = τV Kτ
(∂p∂τ
)2V
, where Kτ = − 1V
(∂V∂p
)τ
is isothermal compressibility.
So what needs to be shown here is that(
∂σ∂p
)τ
=(
∂V∂p
)τ
(∂p∂τ
)V
. Now using the Jacobian and that(
∂p∂τ
)V
=(
∂σ∂V
)τ(
∂σ
∂p
)τ
=
(∂V
∂p
)τ
(∂σ
∂V
)V
=∂(V, τ)
∂(p, τ)
∂(σ, τ)
∂(V, τ)=
∂(σ, τ)
∂(p, τ)=
(∂σ
∂p
)τ
From this
Cp − CV = −τ
(∂σ
∂p
)τ
(∂p
∂τ
)V
= −τ
(∂V
∂p
)τ
(∂σ
∂V
)V
(∂p
∂τ
)V
= −τV
V
(∂V
∂p
)τ
(∂p
∂τ
)V
(∂p
∂τ
)V
Cp − CV = τV Kτ
(∂p
∂τ
)2
V
X
(b) If the entropy σ(p, T ) = a(p)τn at the low temperature, where a(p) is a function of pressure, show that Cp−CV =
1V Kτ
(dadp
)2
τ2n+1. This result implies that there is no distinction between Cp and CV at very low temperatures.
From the first part, 1Kτ
= −V(
∂p∂V
)τ. Matching the terms, it needs to be shown that
(∂p∂V
)τ
(∂σ∂p
)τ
=(
∂p∂τ
)V
. Then(∂p
∂τ
)V
=
(∂p
∂V
)τ
(∂σ
∂p
)τ
=∂(p, τ)
∂(V, τ)
∂(σ, τ)
∂(p, τ)=
∂(σ, τ)
∂(V, τ)=
(∂σ
∂V
)τ
From the first part(
∂σ∂V
)τ
=(
∂p∂τ
)V
. So(
∂p∂V
)τ
(∂σ∂p
)τ
=(
∂p∂τ
)V
, Placing this into the original equation with the
definition of Kτ .
Cp − CV = −τ
(∂σ
∂p
)τ
(∂p
∂τ
)V
=1
V Kτ
(∂σ
∂p
)2
τ
Now σ(p, T ) = a(p)τn, so
1
V Kτ
(∂(a(p)τn)
∂p
)2
τ ⇒ Cp − CV =1
V Kτ
(da
dp
)2
τ2n+1 X
12.2 12.2: Equilibrium conditions
A substance is in equilibrium in the presence of externally applied forces. A small number of particles or a smallamount of energy, or both, are allowed to transfer between any two infinitesimal volume elements of this substance.Prove that the temperature and chemical potential must each have a constant value.
From the first law of thermodynamics, ∆W = ∆U −∆Q. The system is at equilibrium so ∆W = ∆U = 0. Now let’sconsider two subsystems 1 and 2, then ∆U1 = ∆U2 = 0, ∴ ∆Q1 = ∆Q2. Now from the definition of heat, δ(σ1τ1) =δ(σ2τ2). From equilibrium terms σtot is constant, so are any smaller sections. Then σ1δτ1 = σ2δτ2 → δτ1 = δτ2. Theonly way this can be true at equilibrium is if δτ1 = 0 = δτ2. Now the total number of particles is constant. Thismeans δ(µ1N1) = δ(µ2N2) → µ1δN1 = µ2δN2 ∴ µ1 = µ2 or that the chemical potential is constant. X
41
12.3 12.3: Fluctuation in number of a Fermi gas
Determine (∆N)2 for an electron gas at temperatures much lower than the Fermi temperature.
From lecture notes
(∆N)2 = τ
(∂N
∂µ
)τ,V
At temperatures below the Fermi temperature, the number is
N =V
3π2
(2mµF
~2
)3/2
Then
(∆N)2 = τ
(∂N
∂µ
)τ,V
= τ∂
∂µ
(V
3π2
(2mµF
~2
)3/2)
= τV
2π2
(2m
~2
)3/2
µ1/2F =
V
2π2
(2m
~2
)3/2(
~2
2m
)1/2(3π2N
V
)1/3
(∆N)2 =31/3mV τ
π4/3~2
(N
V
)1/3
X
12.4 12.4: Fluctuation in volume, pressure, entropy and temperature
Find
(a) ∆V ∆p
From class
∆σ∆τ − ∆V ∆p =CV
τ∆τ
2 +
(∂σ
∂V
)τ
∆V ∆τ −(
∂p
∂V
)V
∆τ∆V −(
∂p
∂V
)τ
∆V2 =
CV
τ∆τ
2 −(
∂p
∂V
)τ
∆V2
By the fact that we are looking for function of p and V means we need 〈∆V ∆p〉 =(
∂p∂V
)τ〈∆V 2〉. From the class
notes 〈∆V 2〉 = −τ(
∂V∂p
)τ. Substituting this into equation above
〈∆V ∆p〉 = −τ
(∂p
∂V
)τ
(∂V
∂p
)τ
= −τ X
(b) ∆σ∆τ
From the previous problem, 〈∆σ∆τ〉 = CVτ〈∆τ2〉. From class, 〈∆τ2〉 = τ2
CV. Combining these two
〈∆σ∆τ〉 =CV
τ〈∆τ2〉 =
CV
τ
τ2
CV= τ X
12.5 12.5: Kittel 10.5: Gas-solid equilibrium
Consider the gas-solid equilibrium under the extreme assumption that the entropy of the solid may be neglected overthe temperature range of interest. Let −εo be the cohesive energy of the solid, per atom. Treat the gas as idealand monatomic. Make the approximation that the volume accessible to the gas is the volume V of the container,independent of the much smaller volume occupied by the solid.
(a) Show that the total Helmholtz free energy of the system with the total number of atoms N = Ns+Ng is constant.
F = Fs + Fg = −Nsεo + Ngτ [log (Ng/V nQ)− 1] (12.1)
The entropy of the solid can be neglected, because it has no effect in the free energy. Fs = Us = −Nsεo The freeenergy of a gas is given in Kittel and Kroemer as Fg = Ngτ [log (n/nQ)− 1] here n = Ng/V . The total free energyis then
F = −Nsεo + Ngτ [log (Ng/V nQ)− 1]
42
(b) Find the minimum of the free energy with respect to Ng; show that in the equilibrium condition
Ng = nQV e−εo/τ (12.2)
At equilibrium(
∂F∂Ng
)N,τ,V
= 0. Then
∂
∂Ng
([N − Ng
]εo + Ngτ
[log
(Ng/V nQ
)− 1
])= εo + τ
[log
(Ng/V nQ
)− 1
]+ τ = εo + τ log
(Ng/V nQ
)= 0
Finally
εo + τ log (Ng/V nQ) = 0 ⇒ Ng = nQV e−εo/tau X
(c) Find the equilibrium vapor pressure.
From part b and the ideal gas law
p =Ngτ
V= τnQe−εo/τ ⇒ p = τ5/2
( m
2π~2
)3/2
e−εo/τ X
43
44
Chapter 13
Problem Set 13
13.1 13.1: Superconduction and Heat Capacity
A certain metal in zero magnetic field and at atmospheric pressure has a heat capacity Cn = ατ in the normal state,and a heat capacity Cs = γτ3 in the superconducting state. Here α and γ are constants.
(a) Express these constants in terms of each other and the critical temperature τc.
The heat capacity is given by Ci = τ(
∂σ∂τ
)i. Then
Cn = τ∂σn
∂τ⇒ ατ = τ
∂σn
∂τ⇒ σn = ατ
Cs = τ∂σs
∂τ⇒ γτ3 = τ
∂σs
∂τ⇒ σs =
1
3γτ3
At the critical temperature σn = σs. Then
ατc =1
3γτ3
c ⇒ α =γ
3τ2
c X
(b) What is the difference between the internal energy of the metal in the normal and superconducting states τ = 0?Express the answer in terms of γ and τc.
The internal energy difference comes from Un =∫ τc
0Cndτ = 1/2ατ2
c and Us =∫ τc
0Csdτ = 1/4γτ4
c . Then
∆U =1
2ατ2
c −1
4γτ4
c =1
6γτ4
c −1
4γτ4
c ⇒ ∆U = −γ
12τ4c X
13.2 13.2: Kittel 10.8: First order crystal transformation
Consider a crystal that can exist in either of two structures, denoted by α and β. We suppose that the α structureis the stable low temperature form and the β structure is the stable high temperature form of the substance. If thezero of the energy scale is taken as the state of separated atoms at infinity, then the energy density U(0) at τ = 0will be negative. The phase stable at τ = 0 will have the lower value of U(0); thus Uα(0) < Uβ(0). If the velocity ofsound vβ in the β phase is lower than vα in the α phase, corresponding to lower values of the elastic moduli for β,then the thermal excitations in the β phase will have larger amplitudes than in the α phase. The larger the thermalexcitation, the large the entropy and the lower the free energy. Soft systems ten to be stable at high temperatures,hard systems at low.
(a) Show from Chapter 4 that the free energy density contributed by the phonons in a solid at a temperature muchless than the Debye temperature is given by −π2τ4/30v3~3, in the Debye approximation with v taken as thevelocity of all phonos.
The energy density of the phonons at τ θ is
U =3π4Nτ4
5 (kBθ)3=
π2τ4
10 (~v)3
45
Now
∂(F/τ) = − U
τ2∂τ ⇒ F/τ = −
∫π2τ2
10 (~v)3dτ
F/τ = U(0)− π2τ3
30 (~v)3⇒ F = U(0)− π2τ4
30 (~v)3X
(b) Show that at the transformation temperature
τ4c =
(30~3/π2) [Uβ(0)− Uα(0)] /
(v−3
β − v−3α
)(13.1)
There will be a finite real solution if vβ < vα. This example is simplified model of a class of actual phasetransformations in solids.
At the phase transformation the Gibb’s free energy are equal. If there are no volume changes Fα(τc) = Fβ(τc). Fromabove
Fα(τc) = Fβ(τc)⇒ Uα(0)−π2τ4
c
30 (~vα)3= Uβ(0)−
π2τ4c
30 (~vβ)3
τ4c
(1
v3β
−1
v3α
)=
30~3
π2(Uβ(0)− Uα(0))⇒ τ
4c =
30~3
π2
Uβ(0)− Uα(0)
v−3β − v−3
α
X
(c) The latent heat of transformation is defined as the thermal energy that must be supplied to carry the systemthrough the transformation. Show that the latent heat of this model is
L = 4 [Uβ(0)− Uα(0)] (13.2)
The latent heat is the change in enthalpy. If the volume change is small, the change is just the change of internalenergy.
L = Hβ −Hα = Uβ(τc)− Uα(τc)
= Uβ(0)− Uα(0)−(
π2τ4c
10 (~vα)3− π2τ4
c
10 (~vβ)3
)= Uβ(0)− Uα(0)− (−Uβ(0) + Uα(0)) = 4 (Uβ(0)− Uα(0)) X
13.3 13.3: Kittel 11.2: Mixing energy in 3He−4 He and Pb− Sn mixtures
The phase diagram of liquid 3He−4 He mixtures in Figure 11.8 shows that the solubility of 3He in 4He remains finite(about 6 percent) as τ → 0. Similarly, the Pb−Sn phase diagram of Figure 11.14 shows a finite residual solubility ofPB in solid Sn with decreasing τ . What do such finite residual solubilities imply about the form of the function u(x)?
The finite residual solubilities imply that at straight line can be drawn from x = 0 that is tangential to the minimumof u(x), for Helium is x ≈ .94 and for Pb− Sn x ≈ .97. X
13.4 13.4: Kittel 11.4: Solidification range of a binary alloy
Consider the solidification of a binary alloy with the phase diagram of Figure 11.10. Show that, regardless of theinitial composition, the melt will always become fully depleted in component B by the time the last remnant of themelt solidifies. That is, the solidification will not be complete until the temperature has dropped to TA.
Starting as some concentration x, as it is cooled some will solidify, xS1. This amount is removed from x. From thediagram, it will consist of B, so the remaining liquid, xL1, will have a lower concentration of B. Again, cooling thesample more some more will solidify, xS2. Now the liquid, xL2, will have more of B removed, so xL2 < xL1. This canbe done continuously until all of B has been removed.X
46
13.5 13.5: Kittel 11.5: Alloying of gold into silicon
(a) Suppose a 1000 A layer of Au is evaporated onto a Si crystal, and subsequently heated to 400o C. From theAu − Si phase diagram, Figure 11.11, estimate how deep the gold will penetrate into the silicon crystal. Thedensities of Au and Si are 19.3 and 2.23 g cm−3.
The total number of moles per unite area is n, and nAu, nSi, which are the number of moles per unite area of goldand silicon, respectively. At a given temperature, the number of gold moles in a mixture is nAu = xAun. The samecan be said for silicon. The number of moles of gold per unit area is nAu = tAuρAu/MAu. Then same can be side forsilicon. From figure 11.11, at 400o C, xSi = .32 ∴ xAu = .62. Then
nAu
xAu=
nSi
xSi⇒ tAuρAu
MAuxAu=
tSiρSi
MSixSi⇒ tSi =
tAuρAuMSixSi
ρSiMSixAu=
1000A19.3g cm−328g mol−1.32
2.33g cm−3197g mol−1.68⇒ tSi = 554A X
(b) Redo the estimate for 800o C.
For 800o C, xSi = .45 then xAu = .55. Then
tSi =tAuρAuMSixSi
ρSiMSixAu=
1000A19.3g cm−328g mol−1.45
2.33g cm−3197g mol−1.55⇒ tSi = 963A X
47