Mixing of gases 1: Gibbs’ Paradox If N molecules of an ideal monatomic gas make a free expansion from a V to 2V, ΔU = 0, so that ΔS = Sf – Si = Nk ln(2V/V) = Nk ln 2. If the container of volume 2V is now redivided into equal parts, so that (N/2) molecules are in each half, then ΔS’ = Sf’ – Si’ = 2(N/2)k ln(V/2V) = – Nk ln 2. This is Gibbs’ paradox, which is removed by assuming that the molecules are indistiguishable; i.e. that ω(U,V,N) = B(N) U3N/2 VN/N!. The term Nk lnV in the expression for S is replaced by Nk ln(V/N!), where (by Stirling’s theorem), Nk ln(V/N!) ≈ Nk [ln(V/N) + 1].

Entropy of an Ideal Monatomic Gas 5 If S is assumed to be an extensive quantity, Gibbs’ paradox may be avoided. Letting S = Ns, with s(U,V) = k [lnC + lnv +(3/2) lnu], where C = B(N)/N, v = V/N, and u = U/N. Thus, S(U,V,N) = Nk [lnC + (3/2)ln(U/N) + ln(V/N)]. Temperature is defined as 1/T  (S/U)V,N = (3/2)Nk/U, since U = (3/2)NkT, so that S(U,V,N) = Nk [D + (3/2)lnT + ln(V/N)]. Note that P/T = (S/V)T,N = Nk/V, so that PV = NkT. Mixing gases Gas A in the left compartment, a different gas B in the right compartment. ΔStot = ΔS1 + ΔS2. b. Gas A in the left compartment, the same gas B in the right compartment. ΔStot = Sfinal – S1initial – S2initial.