1. 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].

2. 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.