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DRose87
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Homework Statement
Given: Ideal gas equations:
Find S(T,V) for an ideal gas
Homework Equations
Ideal gas equations:
[tex]\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}[/tex] for an ideal gas
The answer, according to the book (David Goodstein's new book "Thermal Physics: Energy and Entropy")
[tex]S = \frac{2}{3}N{k_B}\log T{\left( {\frac{V}{N}} \right)^{2/3}} + {s_0} = S\left( {T,V} \right)[/tex]
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.
[tex]\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}[/tex]
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