Assignment 8, solutions

Problem 1:

Inside a blackbody cavity, the energy density per unit frequency interval, ρ(ν), is given by Planck's formula
ρ(ν) = (8πν2/c3) hν/(exp(hν/kT) - 1).
The intensity per unit frequency interval, I(ν), of the radiation emitted by the blackbody is given by I(ν) = ¼ ρ(ν) c.
Commonly, Wien's displacement law is written as λmax (m) = (2.9*10-3 m K)/T.
Derive the Wien's displacement law for the frequency νmax (s-1), and show that νmax is NOT equal to c/ λmax.  Why?

Solution:

 Concepts: Blackbody radiation Reasoning: We are asked to derive the Wien law in terms of frequency. Details of the calculation: The Wien Law in terms of frequency gives the frequency of the peak of the radiation distribution. dI(ν)/dν = 0. I(ν) = (2πh/c2) (ν3/(exp(hν/kT) - 1)). d(ν3/(exp(hν/kT) - 1))/dν = 0. Let x =  hν/kT. d(x3/(exp(x) – 1))/dx = 0. xex/(ex – 1) = 3. Use a calculator to find x ~ 2.82.  νmax = (5.88*1010 Hz/K) T(K),  νmax is proportional to the absolute temperature T. c/λmax = (3*108 m/s)*T/(2.898* 10-3 m K).  c/λmax = ~(1*1011 Hz/K) T(K),  νmax is NOT equal to c/λmax. This is because  I(λ) is not equal to I(ν). |I(ν)dν| = |I(λ)dλ|,  I(λ) = I(ν)|dν/dλ| = (c/λ2)I(ν).

Problem 2:

Consider a system of one mole of an ideal gas A and three moles of an ideal gas B at the same pressure P and temperature T, in volumes of VA and VB respectively.  The two gases are separated by a partition so they are each sequestered in their respective volumes.  If the partition is removed, calculate the change in entropy of the system.

Solution:

 Concepts:Entropy, the ideal gas law Reasoning:Change in entropy: ΔS = ∫if dS = ∫if dQr/T, where the subscript r denotes a reversible path. The gases will mix.  To calculate the entropy change, we treat the mixing as two separate gas expansions, one for gas A and another for gas B. Here we have and expansion at constant temperature.  For a reversible path leading from the initial to the final state for each gas dU = 0, dQ = PdV = nRTdV/V,  dS = nRdV/V. Details of the calculation: For each gas ∆S = nRln(Vf/Vi).  So for the two separate gas expansions we have ∆SA = nARln((VA + VB)/VA),  ∆SB = nBRln((VA + VB)/VB). The total entropy change is  ∆S = ∆SA + ∆SB = nARln((VA + VB)/VA) +  nBRln((VA + VB)/VB). PV = nRT implies that the initial volume is directly proportional to the number of moles.  Therefore  ∆S = nARln((nA + nB)/nA) +  nBRln((nA + nB)/nB). ∆Smix = Rln(4) + 3Rln(4/3) = 2.25 R = 18.7 J/K.

Problem 3:

A pipe of length 180 m, open on one end and closed on the other, lies at the bottom of a 200 m deep lake.  A light movable piston is placed inside the pipe.  The space between the closed end of the pipe and the piston is filled with air.  The piston is in equilibrium 20 m away from the closed end of the pipe.
The open end of the pipe is very slowly raised until the pipe is brought into the vertical position, its closed end resting at the bottom of the lake.
What is the height of the air column inside the vertical pipe?
Neglect the atmospheric pressure and assume the water temperature is the same throughout the lake.

Solution:

 Concepts:The ideal gas law, pressure in a liquid as a function of depth Reasoning:In equilibrium the air pressure in the pipe equals the water pressure, Pwater = ρwatergd, neglecting the atmospheric pressure.  Here d is the depth below the surface of the water. Details of the calculation:Before: Pwater = ρwaterg*200 m = nkT/(A 20 m). Here A is the cross-sectional area of the pipe. nkT/A = ρwaterg*4000 m2 = constant. After: ρwaterg*(200 m - h) = nkT/(A h). nkT/A = ρwaterg*(200 m - h)h. 4000 m2 = (200 m - h)h. h = (100 ± √6000) m. h = (100 - 20*√15) m = 22.5 m. Only the solution h = 100 - 20*√15 represents a stable equilibrium. d(Pwater(h) - Pair(h))/dh must be positive for a stable equilibrium to exist. d(Pwater - Pair)/dh = -ρwaterg + nkT/(Ah2) = ρwaterg*(4000 m/h2 - 1). For h = (100 - 20*√15) m, d(Pwater - Pair)/dh is positive but for h = (100 + 20*√15) m it is negative.

Problem 4:

Consider a neutron star, a macroscopic body composed of neutrons, at a density ρ = 1017 kg/m3. The temperature of the star's interior is approximately 107 K.  For this problem, consider the star to be a non-interacting Fermi gas of neutrons.
For an ideal non-relativistic 3D Fermi gas comprising N non-interacting fermions, the Fermi energy (the energy difference between the highest and lowest occupied single-particle states at T = 0) is given by EF = [ħ2/(2m)](3Nπ2/V)2/3, and the average energy per fermion at absolute zero <E> = (3/5) EF.
(a)  Determine the Fermi energy of the neutrons in the neutron star.  Are the neutrons relativistic or nonrelativistic?
(b)  Determine whether or not the neutrons are reasonably well considered to be a zero temperature (EF >> kT) Fermi gas.
(c)  Estimate the pressure in the neutron star P = -∂U/∂V.  (quasi-static, adiabatic, fixed # of particles)  In equilibrium this pressure balances the pressure due to gravity.
(d)  Use (c) to estimate the mass of this neutron star.

Solution:

 Concepts Fermi energy, thermodynamic pressure, gravitational pressure Reasoning: For a T = 0 Fermi gas P = -∂U/∂V = -N∂/∂V = (3/5)N ∂EF/∂V. Details of the calculation: (a)  EF = [ħ2/(2mn)](3Nπ2/V)2/3. N/V = ρ/mn = (1017 kg/m3)/(1.67*10-27 kg). EF ≈ 5*10-12 J,  kT = 1.38*10-23*107 J ≈ 10-16 J, mnc2 = 1.5*10-10 J. mnc2 >> EF >>kT.  The neutrons are nonrelativistic. (b)  EF >> kT, the neutrons are reasonably well considered to be a zero temperature Fermi gas. (c)  dU = -PdV for a gas isolated from is environment. P = -∂U/∂V.  U = N = (3/5)N EF. P = (2/3)(3/5) N EF/V = (2/5) N EF/V = [ħ2/(5mn)](3π2)2/3(N/V)5/3 = [ħ2/(5mn8/3)](3π2)2/3ρ5/3. (d)  The gravitational energy of the neutron star is Eg = -∫0RGMinside4πr2ρdr/r = -∫0RG(4πr3ρ/3)4πr2ρdr/r = -(4π)2Gρ2R5/15 = -3GM2/(5R). Eg = -(3/5)G(Nmn)2(4π/3)⅓V-⅓. Pg = -∂Eg/∂V = -(1/5)G(Nmn)2(4π/3)⅓V-4/3 = -(1/5)Gρ2(4π/3)R2. Pg + P = 0 in equilibrium. [ħ2/(5mn8/3)](3π2)2/3ρ5/3 - (1/5)Gρ2(4π/3)R2 = 0. [ħ2/mn8/3]35/3π⅓ = 4Gρ⅓R2. R = 1.4*104 m = 14 km. M = (1017 kg/m3)(4πR3/3) ≈ 1030 kg, on the order of the mass of the sun.

Problem 5:

The operation of a gasoline engine is (roughly) similar to the Otto cycle.  A S-V diagram is shown.

A → B:  Gas compressed adiabatically.
B → C:  Gas heated isochorically (constant volume; corresponds to combustion of gasoline).
C → D:  Gas expanded adiabatically (power stroke).
D → A:  Gas cooled isochorically.
Compute the efficiency of the Otto cycle for an ideal gas (with temperature-independent heat capacities) as a function of the compression ratio VA/VB, and the heat capacity of the gas, CV .

Solution:

 Concepts: Heat engines Reasoning: Efficiency = net work done by the engine divided by the heat absorbed.  We use the ideal gas law and the definition of the heat capacity CV to find these quantities. Details of the calculation: Energy conservation:  dU = dQ – PdV,  dS = dQ/T,  dU = TdS – PdV. For an ideal gas we have  PV = NkBT,  CVdT = dU. Work done by the engine during one cycle:  Wtot = WA-->B + WC-->D = ∫A-->BPdV + ∫C-->DPdV. Heat absorbed by the engine during one cycle:  Qin = QB-->C = ∫B-->CTdS. Efficiency: eff = Wtot/Qin. Finding Wtot: WA-->B = ∫A-->BPdV = -NkB∫VBVA (T/V)dV. S = S2 = constant along the path.  We need to express T in terms of S and V. dS = dU/T + PdV/T = CVdT/T + NkBdV/V. This integrates to S = CV lnT + NkB lnV + constant. Solving for T we find:  T = α*exp(S/CV)*V^(-NkB/CV), where α is a constant. WA-->B = -αNkBexp(S2/CV)∫VBVA V^(-1 - NkB/CV))dV = -α exp(S2/CV) CV (VB^(-NkB/CV) - VA^(-NkB/CV)). Similarly, WC-->D = α exp(S1/CV) CV (VB^(-NkB/CV) - VA^(-NkB/CV)). Wtot = α CV ((VB^(-NkB/CV) - VA^(-NkB/CV))(exp(S1/CV) - exp(S2/CV)). Finding Qin: Qin = ∫S2S1 TdS = α*VB^(-NkB/CV)∫S2S1 exp(S/CV)dS = α*CV*VB^(-NkB/CV)(exp(S1/CV) - exp(S2/CV)). eff = 1 - (VA/VB)^(NkB/CV). For a monatomic ideal gas dU = (3/2)NkBdT = CVdT, NkB/CV = 2/3, e = 1 - (VA/VB)-2/3.