Deriving S(T,V) for an ideal gas

1. Apr 17, 2015

DRose87

1. The problem statement, all variables and given/known data
Given: Ideal gas equations:
Find S(T,V) for an ideal gas

2. Relevant equations
Ideal gas equations:
$$\begin{array}{l} {\rm{}}\\ U = \frac{3}{2}N{k_B}{\left( {\frac{N}{V}} \right)^{2/3}}\exp \left[ {\frac{S}{{\left( {3/2} \right)N{k_B}}} - {s_0}} \right]{\rm{ }}\\ T = {\left( {\frac{{\partial U}}{{\partial S}}} \right)_{V,N}} = \frac{U}{{\left( {3/2} \right)N{k_B}}}\\ \\ {\rm{Find: }}\\ {\rm{S = S}}\left( {T,V} \right){\rm{ }}\\ \\\end{array}$$ for an ideal gas

The answer, according to the book (David Goodstein's new book "Thermal Physics: Energy and Entropy")
$$S = \frac{2}{3}N{k_B}\log T{\left( {\frac{V}{N}} \right)^{2/3}} + {s_0} = S\left( {T,V} \right)$$

3. The attempt at a solution
I'm not sure if the answer given in the book is correct and I'm missing something, or if it is an error.
$$\begin{array}{l} \\ U = \frac{3}{2}N{k_B}{\left( {\frac{N}{V}} \right)^{2/3}}\exp \left[ {\frac{S}{{\left( {3/2} \right)N{k_B}}} - {s_0}} \right]{\rm{ }}\\ T = {\left( {\frac{{\partial U}}{{\partial S}}} \right)_{V,N}} = \frac{U}{{\left( {3/2} \right)N{k_B}}} = \frac{{\frac{3}{2}N{k_B}{{\left( {\frac{N}{V}} \right)}^{2/3}}\exp \left[ {\frac{S}{{\left( {3/2} \right)N{k_B}}} - {s_0}} \right]}}{{\frac{3}{2}N{k_B}}}\\ = {\left( {\frac{N}{V}} \right)^{2/3}}\exp \left[ {\frac{S}{{\left( {3/2} \right)N{k_B}}} - {s_0}} \right]\\ \exp \left[ {\frac{S}{{\left( {3/2} \right)N{k_B}}} - {s_0}} \right] = \frac{T}{{{{\left( {\frac{N}{V}} \right)}^{2/3}}}} = T{\left( {\frac{V}{N}} \right)^{2/3}}\\ \frac{S}{{\left( {3/2} \right)N{k_B}}} - {s_0} = \log \left[ {T{{\left( {\frac{V}{N}} \right)}^{2/3}}} \right]\\ \frac{S}{{\left( {3/2} \right)N{k_B}}} = \log \left[ {T{{\left( {\frac{V}{N}} \right)}^{2/3}}} \right] + {s_0}\\ S = \frac{3}{2}N{k_B}\log \left[ {T{{\left( {\frac{V}{N}} \right)}^{2/3}}} \right] + \frac{3}{2}N{k_B}{s_0}\\ \\ \\ \end{array}$$

Last edited: Apr 17, 2015
2. Apr 18, 2015

robphy

3. Apr 19, 2015

BvU

Book is a typo.

I like David Stroud's treatment here (more here, etc.)

4. Apr 20, 2015

DRose87

Thanks BvU and robphy for your opinions and the links you both postsed. I agree that it is a typo. It is kind of funny that this is actually the first problem in the book and there is a typo. I hope that the rest of the book isn't plagued by errors..... fortunately if that turns out to be the case, my class is using a different book, Classical Statistical Thermodynamics by Ashley Carter.

Last edited: Apr 20, 2015
5. Apr 20, 2015

BvU

Goodstein has a mail address; I'm sure he'll appreciate if you point out stuff he can improve for the next edition !