(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

From Hill's "Introduction to Statistical Thermodynamics", question 3-4 reads:

(note "STR" denotes the case of most probable distribution and should be read as C Show that the substitution of of the most probable distribution in

[itex]S=k*ln(\Omega(C^{str}))[/itex] leads to

[itex]A=-NkT*ln(\sum e^{\beta \epsilon_j})[/itex]^{*})

2. Relevant equations

The most probable distribution for a system of independent indistinguishable molecules:

[itex]\eta_i = \bar{C_j} /N =C^{str}/N = e^{-\beta \epsilon_j} / \sum_i e^{-\beta \epsilon_i}[/itex]

Formula for accessible quantum states for an ensemble of independent indistinguishable molecules:

[itex]\Omega(C) = N! /( \prod_j C_j) ![/itex]

Relationship between S and A

S = [itex]- (\partial A / \partial E) = kT ( \partial ln(Q) / \partial T )+ k ln(Q) [/itex]

3. The attempt at a solution

First, I computed C in terms of the given equations. This is pretty straight forward:

[itex] C^{str}_j = Ne^{-\beta \epsilon_j} /{ \sum_i e^{-\beta \epsilon} }[/itex]

Then, I computed the number of accessible quantum states with the above expression for C

[itex] \Omega (C^{str}) = { N! }/{\prod C^{str} } [/itex]

Skipping a lot of algebra I arrived as this expression for ln(Ω(C)):

[itex] ln(C^{str}) = ln(N!)- \sum(ln(e^{-\beta \epsilon}) )- ln(\sum(e^{-\beta \epsilon}) [/itex]

Which can be reduced to:

[itex]ln(C^{str}) = ln(N!) + \sum(\beta \epsilon) - ln(Q) = ln(N!) + E \beta -ln(Q)[/itex]

I am not sure exactly where I went wrong, as the formula I derived is very similar to the relationship between A and S. I think an approximation may have to be applied somewhere or perhaps the natural logs should not have been broken down into their operations. I am not sure, hence why I am posting here!

Thanks for the help.

PS, if needed I can post a small nomenclature of the variables.

EDIT:

Just realized I did not take into account the N factor in the equation for C, I believe this makes N! become (N-1)! in the following equations. Also, I did not show the additional formula if one were to expand [itex] \partial ln(Q) / \partial T[/itex]

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