Amazon.com let me read the pages in question by searching for "the equilibrium era"
(first I looked for pg 109, saw that that was in the chapte equilibrium era, then I searched for equilibrium era).
Weinberg is considering a non-relativistic gas of baryons and radiation. He's asking the question : "When we have only photons, temperature falls as 1/a(t). But if we had only baryons, he argues that temperature would fall as 1/a(t)^2. So he wants to know which dominates when we have thermal equilibrium, and he proceeds to derive it from thermodynamics.
The sorts of non-relativistic particles in this gas are protons, helium nuclei, and electrons. N is a number of order unity, which gives the number of particles / number of baryons.
So n_B is the number density of baryons, N n_b is the number density of particles
(3/2) (N n_b) k T is just the total energy, which by the equipartition theorem for the non-relativistic gas in equilibrium is (1/2) kT per particle.
Weinberg uses k_B where I write k - it's just Boltman's constant. And E isn't quite right for the symbol he uses , but it's close enough I hope to not be confusing if I use it rather than his symbol.
Weinberg wants to calculate the
change in entropy, delta-S, which he writes as k \sigma, where sigma is the entropy per unit baryon.
Thermo isn't my forte, so I'll consult another textbook, MTW, to see if it tells us the same thing. (Which it does)
MTW argues that in curves space-time we wite
d(energy in a volume containing a constant number of baryons ) =
- p d(volume) + T * d(entropy)
This should be equivalent to Weinberg's 2.21, i.e.
d(k sigma) = [d (E/n_b) + p d(1/n_b)] / T
which is equivalent to MTW's formulation ( a few pages more in the textbook)
d(rho/n) + P d(1/n) - T ds = 0, MTW uses the more familar rho for the energy density rather than the E.
Therefore, we apparently do need to include the pressure-volume terms in computing the entropy change, the P d(1/n) terms. In flat space-time we interpret them as "work done by the expanding gas", in curved space-time the interpretation is a bit hazier, but we still need them. Maybe someone better than I am with thermo can give a more convincing explanation of why they're needed, other than "two textbooks say so".