Homework #7

1. (Kittel 6.4) Energy of gas of extreme relativistic particles. Extreme relativistic

particles have momenta p such that pc >> Mc2 , where M is the rest mass of the particle. The de Broglie relation λ = h p for the quantum wavelength continues to apply. Show that the mean energy per particle of an extreme relativistic ideal gas is

3τ if ε ≅ pc in contrast to 32

τ for the nonrelativistic problem. (An interesting variety

of relativistic problems are discussed by E. Fermi in Notes on Thermodynamics and Statistics, University of Chicago Press, 1966, paperback.)

Solution.

ε =εD(ε)e−βε dε

0

∞

∫D(ε)e−βεdε

0

∞

∫

For ε ≅ pc , D(ε ) ∝ ε 2 . Hence

ε =εε 2e−βεdε

0

∞

∫ε2e−βεdε

0

∞

∫

Let x = βε = ε kBT . Then

ε = kBTx3e−xdx

0

∞

∫x2e−xdx

0

∞

∫= kBT

3 × 22 ×1

= 3kBT

2. (Kittel 6.5) Integration of the thermodynamic identity for an ideal gas. From the

thermodynamic identity at a constant number of particles we have

dσ =dUτ

+pdVτ

=1τ

∂U∂τ

Vdτ +

1τ

∂U∂V

τdV +

pdVτ

Show by integration that for an ideal gas the entropy is

σ = CV logτ + N logV + σ1

where σ1 is a constant independent of τ and V.

Solution.

dσ =1τ

∂U∂τ

Vdτ +

1τ

∂U∂V

τdV +

pdVτ

For an ideal gas, pV = NkBT = Nτ .

∂U∂V

τ= T

∂S∂V

τ− p = T

∂p∂T

V− p = T

NkB

V− p = 0

dσ =1τ

∂U∂τ

Vdτ +

pdVτ

=CV

τdτ +

NV

dV

σ =CV

τdτ +

NV

dV∫ = CV ln τ + N lnV + σ1

3. (Kittel 6.7) Relation of pressure and energy density.

(a) Show that the average pressure in a system in thermal contact with a heat reservoir is given by

p = −∂εs ∂V( )N

exp −εs τ( )s∑

Z

where the sum is over all states of the system. (b) Show for a gas of free particles that

∂ε s

∂V

N= −

23

εs

V

as a result of the boundary conditions of the problem. The result holds equally well whether εs refers to a state of N noninteracting particles or to an orbital. (c) Show that for a gas of free non-relativistic particles

p = 2U 3V

where U is the thermal average energy of the system. The result is not limited to the classical regime; it holds equally well for fermion and boson particles, as long as they are nonrelativistic.

Solution.

p = −1Z s∑

∂εs

∂V

Ne−ε s τ

88

εs =�

2

2mks

2 with ks ≈1L

=1

V1 3

εs ∝1

V 2 3

Let εs ∝As

V 2 3 . Then

∂εs

∂V= −

23

As

V 5 3 = −23

εs

V

p =23

1Z

εs

Ve−ε s τ

s∑ =

23V

1Z

εse−ε s τ

s∑ =

2U3V

4. (Kittel 6.9) Gas of atoms with internal degrees of freedom. Consider an ideal

monatomic gas, but one for which the atom has two internal energy states, one an energy ∆ above the other. There are N atoms in volume V at temperature τ . Find the (a) chemical potential; (b) free energy; (c) entropy; (d) pressure; (e) heat capacity at constant pressure.

Solution.

N = λ e−βε

ε∑

ε =p2

2m+ε int

ε int = 0 ⇒ e−βε

ε∑ = Z = VnQ

ε int = ∆⇒ e−βε

ε∑ = e−β∆ Z = VnQe−β∆

N = λ e−βε

ε∑ = λVnQ 1+ e−β∆( )

λ =nnQ

11 + e−β∆

λ = eµ τ

µ = τ lnnnQ

− τ ln 1+ e−β∆( )

(b)

Z1 = 1+ e−β∆( )Zideal = 1 + e−β∆( )VnQ

Z =Z1

N

N!=

(VnQ)N

N!1+ e−β∆( )N

F = −τ ln Z = −τ lnVnQ

N!

N

− Nτ ln 1 + e−β∆( )= Nτ lnnnQ

−1

− Nτ ln 1+ e−β∆( )

(c)

σ = −∂F∂τ

V= N ln

nQ

n+

52

+ N ln 1 + e−β∆( )+ Nτ e−β∆

1 + e−β∆∆τ 2

ie. σ = N lnnQ

n+

52

+ N ln 1+ e−β∆( )+

N∆τ

1eβ∆ +1

(d)

p = −∂F∂V

τ=

NτV

(e)

Cp = τ ∂σ∂T

p= kBτ

∂σ∂τ

p= kBτ

∂∂τ

N lnnQ

n+

52

+ kBτ

∂∂τ

N ln(1 + e−β∆ ) +N∆τ

1eβ∆ +1

Hence,

Cp =52

NkB + NkBτe−β∆

1 + e−β∆∆τ 2 −

1eβ∆ +1

∆τ 2 +

∆τ

eβ∆

(eβ∆ +1)2∆τ 2

=

5NkB

2+ NkB

∆2

τ 2eβ∆

(eβ∆ +1)2

5. (Kittel 6.11) Convective isentropic equilibrium of the atmosphere. The lower 10-

15 km of the atmosphere - the troposphere -is often in a convective steady state at constant entropy, not constant temperature. In such equilibrium pV γ is independent of altitude, where γ = Cp CV . Use the condition of mechanical equilibrium in a uniform gravitational field to: (a) Show that dT dz = constant , where z is the altitude. This quantity, important in meteorology, is called the dry adiabatic lapse rate. (Do not use the barometric pressure relation that was derived in Chapter 5 for an isothermal atmosphere.) (b) Estimate dT dz , in degrees Celsius per km. Take γ = 7 5. (c) Show that p ∝ ργ , where ρ is the mass density. If the actual temperature gradient is greater than the isentropic gradient, the atmosphere may be unstable with respect to convection.

Solution. (a)

p(z) − p(z + dz)[ ]A = nmAdzg ⇒ −dpdz

= nmg

dTdz

=dTdp

dpdz

= −dTdp

nmg (Eqn 1)

pV γ = constant, with pV = NkBT gives pTp

γ

= constant and so T γ

pγ −1 = constant

lnTγ

pγ −1 = constant ⇒ γ ln T − (γ − 1)ln p = constant

γT

dT − (γ −1)dpp

= 0

2) (Eqn )1(p

TdpdT

γγ −=

Eqn 2 can also be obtained from

τdσ = dU + pdV = cV dT + pdV For isentropic processes dσ = 0 and so

CVdT + pdV = 0

pdV + Vdp = NkBdT

CVdT + NkBdT − Vdp = 0

CpdT −NkBT

pdp = 0

(since γ =Cp

CV

and NkB = Cp − CV ).

γdT − (γ −1)Tdp

p= 0 ⇒

dTdp

=γ −1

γTp

Substitute Eqn 2 into Eqn 1 to get

dTdz

= −γ −1

γTpp

mg

Now p = nkBT and so

dTdz

= −γ −1

γmgkB

= constant

(b) Take m the mass of N2

m = 28 ×1.67 ×10−27kg = 4.68 ×10−26kg

Now γ =75

and so

dTdz

= −7 5 −1

7 54.68 ×10−26

1.38 ×10−23 9.8K m = 9.5 ×10−3 K m

T drops about 10 degrees Celsius every km. (c)

pV γ = constant, and ρ ∝ V −1⇒ pρ−γ = constant, and so p ∝ ργ

6. (Kittel 6.12) Ideal gas in two dimensions. (a) Find the chemical potential of an ideal

monatomic gas in two dimensions, with N atoms confined to a square of area A = L2 . The spin is zero. (b) Find an expression for the energy U of the gas. (c) find an expression for the entropy σ . The temperature is τ .

Solution. (a) N = λZ1 with Z1 = ∑e−βε = D(ε)∫ e−βεdε .

D(ε )dε =A2πpdp

h 2 =2πAh2

12

dp2 =2πAm

h2 dε

Z1 =2πAm

h 2 e−βε dε0

∞

∫ =2πAm

h2 τ

N = eµ τ 2πAmh 2 τ ⇒ µ = τ ln

Nh2

2πAmτ

(b)

ε =12

kBT +12

kBT = kBT

U = Nε = NkBT

(c)

Z =Z1

N

N!⇒ F = −τ ln Z = −τ N ln

2πAmh2 τ

− N ln N + N

F = −τN ln2πAmτ

h2N+1

σ = −∂F∂τ

A= N ln

2πAmτh2 N

+1

+ Nτ 1

τ= N ln

2πAmτh2 N

+ 2

U = F + τσ = Nτ