Is the Given Solution for S(T,V) of an Ideal Gas Accurate?

[DRose87]

Homework Statement

Given: Ideal gas equations:
Find S(T,V) for an ideal gas

Homework Equations

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

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

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.
$$\begin{array}{l}<br /> \\<br /> U = \frac{3}{2}N{k_B}{\left( {\frac{N}{V}} \right)^{2/3}}\exp \left[ {\frac{S}{{\left( {3/2} \right)N{k_B}}} - {s_0}} \right]{\rm{ }}\\<br /> T = {\left( {\frac{{\partial U}}{{\partial S}}} \right)_{V,N}} = \frac{U}{{\left( {3/2} \right)N{k_B}}} = \frac{{\frac{3}{2}N{k_B}{{\left( {\frac{N}{V}} \right)}^{2/3}}\exp \left[ {\frac{S}{{\left( {3/2} \right)N{k_B}}} - {s_0}} \right]}}{{\frac{3}{2}N{k_B}}}\\<br /> = {\left( {\frac{N}{V}} \right)^{2/3}}\exp \left[ {\frac{S}{{\left( {3/2} \right)N{k_B}}} - {s_0}} \right]\\<br /> \exp \left[ {\frac{S}{{\left( {3/2} \right)N{k_B}}} - {s_0}} \right] = \frac{T}{{{{\left( {\frac{N}{V}} \right)}^{2/3}}}} = T{\left( {\frac{V}{N}} \right)^{2/3}}\\<br /> \frac{S}{{\left( {3/2} \right)N{k_B}}} - {s_0} = \log \left[ {T{{\left( {\frac{V}{N}} \right)}^{2/3}}} \right]\\<br /> \frac{S}{{\left( {3/2} \right)N{k_B}}} = \log \left[ {T{{\left( {\frac{V}{N}} \right)}^{2/3}}} \right] + {s_0}\\<br /> S = \frac{3}{2}N{k_B}\log \left[ {T{{\left( {\frac{V}{N}} \right)}^{2/3}}} \right] + \frac{3}{2}N{k_B}{s_0}\\<br /> \\<br /> \\<br /> <br /> \end{array}$$

 

[robphy]

Recall $\frac{\partial S}{\partial V}=\frac{P}{T}$...

Compare:

S=(c)*N*k*ln(T*(V/N)^(c)); diff(S,V)
http://www.wolframalpha.com/input/?i=S=(c)*N*k*ln(T*(V/N)^(c));+diff(S,V)
yields c^2*kN/V

S=(1/c)*N*k*ln(T*(V/N)^(c)); diff(S,V)
http://www.wolframalpha.com/input/?i=S=(1/c)*N*k*ln(T*(V/N)^(c));+diff(S,V)
yields kN/V

I think the book's answer is incorrect.

 

[BvU]

Book is a typo.

I like David Stroud's treatment https://www.physics.ohio-state.edu/~stroud/p846/idealgas.pdf (more https://www.physics.ohio-state.edu/~stroud/p846/p846notes3.pdf, etc.)

 

[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.

 

[BvU]

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