Gibb's Paradox, Arrangements and Volume Issues

Traditionally, Maxwell-Boltzmann’s statistical mechanics, i.e. p(ei) = exp(-ei/T) / Sum over i exp(-ei/T), is used to calculate Shannon’s entropy S(Shannon) = - Sum over i p(ei) ln(p(ei)). The problem is that this approach does not lead to additivity of entropy. In particular, if one has a box of size 2V with 2N particles (in equilibrium at temperature T) and partitions it in half (T, N, V) in each side, then S1+S2 not= S unless one adds a term - N ln(N) to Shannon’s entropy. (This term appears in the Sackur-Tetrode equation.) -N ln(N) approx = ln (1/N!), using Stirling’s approximation and so it is traditionally argued that N! is introduced due to the indistinguishability of particles and that this follows from quantum mechanics. In (1), we argued that one may obtain an expression for S directly from the First Law of Thermodynamics, using E=3/2 N kT and PV= NkT. If one wishes to have S1+S2=S, then one may add the term -N ln(N) to an expression for S to achieve this goal. Thus, one does not need to consider Shannon’s entropy at all nor invoke particle indistinguishability. The key idea was that the extra factor - N ln(N) had to be added because of issues with the volume term, Nln(V) of entropy, and not indistinguishability of particles. Here we suggest that the term -N ln(N) may be seen to enter in a natural manner which again does not invoke indistinguishability of particles and again is due to issues with the N ln(V) term of entropy. In particular, we note that one may obtain p(ei) = exp(-ei/T) / Z(T), where 1/Z(T) is a normalization factor using reaction balance. In such a case, entropy does not appear at all. Given this result, one may postulate a math function of p(ei) such that its maximization, subject to Sum over i p(ei) =1 and Sum over i ei p(ei) = Eave, yields p(ei) = exp(-ei/T) / Z(T). The math function is - Sum over i p(ei) ln(p(ei)), (i.e. Shannon’s entropy), but that does not mean that it automatically satisfies the First Law of Thermodynamics as well as S1+S2 = S. In (2), it is argued that Shannon’s entropy yields: S = 3/2 Nk {1 + ln(2*3.14 m kT) } + NkV which is not the Sackur-Tetrode expression. This entropy satisfies the First Law of Thermodynamics for constant N, but does not satisfy S1+S2=S. If one considers - p(volume) ln(1/V) where p(volume) is dV/V, then for a single particle, this expression is ln(V) and for N particles, N ln(V) to be additive. Gibb’s paradox arises because of this volume piece we argue because if one restricts N particles to a compartment of size V, and they have E/2 energy, then there is no issue with the partitioning of energy, but there is an issue with volume as the N particles are restricted to V in a compartment, while for the whole case all 2N particles may be found anywhere in 2V. This is the crux of the Gibb’s paradox, we argue and so one requires a solution to this volume created problem. One may resolve the issue by adding an extra function of N to entropy. If N is constant, this has no effect on the First Law of Entropy. This function is - N ln(N). If one wishes to consider energy changes due to N alone (T,V constant), then one may consider the Sackure-Tetrode equation for S/N and use d(E/N) = 0 = - -d/dN ( ln(N)) = 1/N approx= 0.