Physics 4311: Thermal Physics – HW Solution 8


due date: Wednesday, Apr 1, 2026

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Problem 1: Entropy of the ideal gas (12 points)

a)

π‘‘π‘ˆ = T𝑑𝑆 βˆ’ p𝑑𝑉 .𝑑𝑆 = 1 Tπ‘‘π‘ˆ + p T𝑑𝑉

use 𝑝𝑉 = NkBT and U = (3βˆ•2)NkBT

𝑑𝑆 = 3NkB 2T 𝑑𝑇 + NkB V 𝑑𝑉

S = 3 2NkB 𝑙𝑛 ( T T0 ) + NkB ln ( V V 0 )

b)

for T β†’ 0, entropy S β†’βˆ’βˆž, this contradicts condition S β‰₯ 0 which follows fromS = kB 𝑙𝑛Ω.
β‡’ ideal gas laws cannot hold down to T = 0

Problem 2: Maxima of entropy (16 points)

a)

define

f = βˆ’kB βˆ‘ ipi lnpi βˆ’ Ξ» (βˆ‘ ipi βˆ’ 1)

βˆ‚π‘“ βˆ‚pj = βˆ’kB(𝑙𝑛pj + 1) βˆ’ Ξ» = 0

pj = e=βˆ’1βˆ’Ξ»βˆ•kB independentΒ of 𝑗

βˆ‘ jpj = 1 β‡’pj = 1 N

b)

f = βˆ’kB βˆ‘ ipi lnpi βˆ’ Ξ» (βˆ‘ ipi βˆ’ 1) βˆ’ ΞΌ (βˆ‘ ipiEi βˆ’βŸ¨E⟩)

βˆ‚π‘“ βˆ‚pj = βˆ’kB(𝑙𝑛pj + 1) βˆ’ Ξ» βˆ’ ΞΌEj = 0

pj = e=βˆ’1βˆ’Ξ»βˆ•kBβˆ’ΞΌEjβˆ•kB structureΒ ofΒ BoltzmannΒ factor

now find Ξ» and ΞΌ from βˆ‘ ⁑ ipi = 1 and βˆ‘ ⁑ ipiEi = ⟨E⟩

Problem 3: Additivity of the entropy (12 points)

S = βˆ’kB βˆ‘ i=1n βˆ‘ j=1mp(i,j)lnp(i,j)

if subsystems are statistically independent, p(i,j) = pA(i)pB(j)

S = βˆ’kB βˆ‘ i=1n βˆ‘ j=1mp A(i)pB(j)ln[pA(i)pB(j)]

S = βˆ’kB βˆ‘ i=1n βˆ‘ j=1mp A(i)pB(j)[lnpA(i) + lnpB(j)]

S = βˆ’kB βˆ‘ i=1np A(i)lnpA(i)βˆ‘ j=1mp B(j) βˆ’ kB βˆ‘ j=1mp B(j)lnpB(j)βˆ‘ i=1np A(i)

S = SA + SB