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) = (8πν^{2}/c^{3}) 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 λ_{max} (m) = (2.9*10^{-3}
m K)/T.

Derive the Wien's displacement law for the frequency ν_{max} (s^{-1}),
and show that
ν_{max} is NOT equal to c/ λ_{max}. 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/c ^{2}) (ν^{3}/(exp(hν/kT) - 1)).d(ν ^{3}/(exp(hν/kT) - 1))/dν = 0.Let x = hν/kT. d(x ^{3}/(exp(x) – 1))/dx = 0.xe ^{x}/(e^{x} – 1) = 3.Use a calculator to find x ~ 2.82. ν _{max} = (5.88*10^{10 }Hz/K) T(K), ν_{max} is
proportional to the absolute temperature T.c/λ _{max} = (3*10^{8} m/s)*T/(2.898* 10^{-3} m K). c/λ _{max} = ~(1*10^{11} Hz/K) T(K), ν_{max} is NOT
equal to c/λ_{max}. This is because I(λ) is not equal to I(ν). |I(ν)dν| = |I(λ)dλ|, I(λ) = I(ν)|dν/dλ| = (c/λ ^{2})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 V_{A} and
V_{B} 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 = ∫ _{i}^{f} dS = ∫_{i}^{f}
dQ_{r}/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(V _{f}/V_{i}). So for the two
separate gas expansions we have∆S _{A} = n_{A}Rln((V_{A} + V_{B})/V_{A}), ∆S_{B}
= n_{B}Rln((V_{A} + V_{B})/V_{B}).The total entropy change is ∆S = ∆S _{A} + ∆S_{B} = n_{A}Rln((V_{A} + V_{B})/V_{A})
+ n_{B}Rln((V_{A} + V_{B})/V_{B}).PV = nRT implies that the initial volume is directly proportional to the number of moles. Therefore ∆S = n _{A}Rln((n_{A} + n_{B})/n_{A})
+ n_{B}Rln((n_{A} + n_{B})/n_{B}).∆S _{mix} = 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, P _{water} = ρ_{water}gd, neglecting the
atmospheric pressure. Here d is the depth below the surface of the
water. | |

Details of the calculation: Before: P _{water} = ρ_{water}g*200
m = nkT/(A 20 m).Here A is the cross-sectional area of the pipe. nkT/A = ρ _{water}g*4000 m^{2} = constant.After: ρ _{water}g*(200 m - h) = nkT/(A h).nkT/A = ρ _{water}g*(200 m - h)h.4000 m ^{2} = (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(P _{water}(h) - P_{air}(h))/dh must be positive for a
stable equilibrium to exist.d(P _{water} - P_{air})/dh = -ρ_{water}g + nkT/(Ah^{2})
= ρ_{water}g*(4000 m/h^{2} - 1).For h = (100 - 20*√15) m, d(P _{water} - P_{air})/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 = 10^{17} kg/m^{3}. The temperature of the star's
interior is approximately 10^{7} 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 E_{F} = [ħ^{2}/(2m)](3Nπ^{2}/V)^{2/3},
and the average energy per fermion at absolute zero <E> = (3/5) E_{F}.

(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 (E_{F} >> 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:

ConceptsFermi energy, thermodynamic pressure, gravitational pressure | |

Reasoning: For a T = 0 Fermi gas P = -∂U/∂V = -N∂<E>/∂V = (3/5)N ∂E _{F}/∂V. | |

Details of the calculation: (a) E _{F} = [ħ^{2}/(2m_{n})](3Nπ^{2}/V)^{2/3}.N/V = ρ/m _{n} = (10^{17} kg/m^{3})/(1.67*10^{-27}
kg).E _{F} ≈ 5*10^{-12} J, kT = 1.38*10^{-23}*10^{7}
J ≈ 10^{-16} J, m_{n}c^{2} = 1.5*10^{-10} J.m _{n}c^{2} >> E_{F} >>kT. The neutrons are
nonrelativistic.(b) E _{F} >> 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<E> = (3/5)N E _{F}.P = (2/3)(3/5) N E _{F}/V = (2/5) N E_{F}/V = [ħ^{2}/(5m_{n})](3π^{2})^{2/3}(N/V)^{5/3}
= [ħ ^{2}/(5m_{n}^{8/3})](3π^{2})^{2/3}ρ^{5/3}.(d) The gravitational energy of the neutron star is E _{g} = -∫_{0}^{R}GM_{inside}4πr^{2}ρdr/r
= -∫_{0}^{R}G(4πr^{3}ρ/3)4πr^{2}ρdr/r =
-(4π)^{2}Gρ^{2}R^{5}/15 = -3GM^{2}/(5R).E _{g} = -(3/5)G(Nm_{n})^{2}(4π/3)^{⅓}V^{-⅓}.P _{g} = -∂E_{g}/∂V = -(1/5)G(Nm_{n})^{2}(4π/3)^{⅓}V^{-4/3}
= -(1/5)Gρ^{2}(4π/3)R^{2}.P _{g} + P = 0 in equilibrium.[ħ ^{2}/(5m_{n}^{8/3})](3π^{2})^{2/3}ρ^{5/3}
- (1/5)Gρ^{2}(4π/3)R^{2} = 0.[ħ ^{2}/m_{n}^{8/3}]3^{5/3}π^{⅓} =
4Gρ^{⅓}R^{2}.R = 1.4*10 ^{4} m = 14 km.M = (10 ^{17} kg/m^{3})(4πR^{3}/3) ≈ 10^{30}
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
V_{A}/V_{B}, and the heat capacity of the gas, C_{V} .

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 C _{V} 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 = Nk _{B}T, C_{V}dT = dU.Work done by the engine during one cycle: W _{tot} = W_{A-->B}
+ W_{C-->D} = ∫_{A-->B}PdV + ∫_{C-->D}PdV.Heat absorbed by the engine during one cycle: Q _{in }= Q_{B-->C
}= ∫_{B-->C}TdS.Efficiency: eff = W _{tot}/Q_{in}._{
}Finding W_{tot}:W _{A-->B} = ∫_{A-->B}PdV = -Nk_{B}∫_{VB}^{VA}
(T/V)dV.S = S _{2} = constant along the path. We need to express T in
terms of S and V.dS = dU/T + PdV/T = C _{V}dT/T + Nk_{B}dV/V.This integrates to S = C _{V} lnT + Nk_{B }lnV + constant.Solving for T we find: T = α*exp(S/C _{V})*V^(-Nk_{B}/C_{V}),
where α is a constant.W _{A-->B} = -αNk_{B}exp(S_{2}/C_{V})∫_{VB}^{VA}
V^(-1 - Nk_{B}/C_{V}))dV = -α exp(S _{2}/C_{V})
C_{V }(V_{B}^(-Nk_{B}/C_{V}) - V_{A}^(-Nk_{B}/C_{V})).Similarly, W _{C-->D} = α exp(S_{1}/C_{V}) C_{V
}(V_{B}^(-Nk_{B}/C_{V}) - V_{A}^(-Nk_{B}/C_{V})).W _{tot} = α C_{V} ((V_{B}^(-Nk_{B}/C_{V})
- V_{A}^(-Nk_{B}/C_{V}))(exp(S_{1}/C_{V})
- exp(S_{2}/C_{V})).Finding Q _{in}:Q _{in }= ∫_{S2}^{S1 }TdS = α*V_{B}^(-Nk_{B}/C_{V})∫_{S2}^{S1
}exp(S/C_{V})dS = α*C _{V}*V_{B}^(-Nk_{B}/C_{V})(exp(S_{1}/C_{V})
- exp(S_{2}/C_{V})).eff = 1 - (V _{A}/V_{B})^(Nk_{B}/C_{V}).For a monatomic ideal gas dU = (3/2)Nk _{B}dT = C_{V}dT, Nk_{B}/C_{V}
= 2/3, e = 1 - (V_{A}/V_{B})^{-2/3}. |