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I'd like to derive the $S(U,V,N)$ function. In the lecture, we were using the $S=k_b \log(\Omega)$, some combinatorics, approximation, $6N$ dimensional spheres, etc. But I'd like to avoid that if it's possible. So I think I should do it from the first and second law of thermodynamics: $$\mathrm{d}U=T\mathrm{d}S-p\mathrm{d}V+\mu\mathrm{d}N$$ $$\mathrm{d}S=\frac{1}{T}\mathrm{d}U+\frac{p}{T}\mathrm{d}V-\frac{\mu}{T}\mathrm{d}N$$ And with $pV=Nk_bT$ and $U=\frac{f}{2}Nk_b T$, I could get that $$\mathrm{d}S=\frac{f}{2}k_b \frac{N}{U}\mathrm{d}U+k_b\frac{N}{V}\mathrm{d}V-\frac{\mu}{T}\mathrm{d}N$$ But I can't get rid of the $\frac{\mu}{T}$. How could I do it?

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  • $\begingroup$ Just compute the partition function. For an ideal gas it’s a trivial Gaussian integration. $\endgroup$
    – JamalS
    Commented Jun 12, 2018 at 9:53
  • $\begingroup$ @JamalS I never heard of that. I think I will learn about that in the statistical physics class. $\endgroup$
    – Botond
    Commented Jun 12, 2018 at 10:05
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    $\begingroup$ Try to use some Maxwell relations maybe? $\endgroup$ Commented Jun 12, 2018 at 11:24
  • $\begingroup$ That third equation is likely derived from, or maybe just a differnt form of the Gibbs-Durham equation - it can further be modified for heat per unit particle.. When you do this properly, you can also find the state of the entropy. $\endgroup$ Commented Apr 13, 2019 at 2:12

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From the definition of chemical potential:

\begin{equation}\mu = \left(\frac{\partial U}{\partial N}\right) = cRT \end{equation} where c = 3/2 for a monatomic ideal gas, c = 5/2 for a diatomic ideal gas.

To verify the consistency of the derivation:

From the “fundamental equation” of an ideal gas:

\begin{equation} PV = N RT \end{equation}

\begin{equation} U=c N RT\end{equation}

if you rewrite the equations of state as

\begin{equation} \frac{1}{T}=\frac{cN R}{U}=(∂S/∂U)_{V,N}\end{equation}

\begin{equation} PT=\frac{NR}{V}=(∂S/∂V)_{U,N} \end{equation}

By integrating both, the first equation gives:

\begin{equation} S(U, V, N) =c N R \ln(U) +f(V, N)\end{equation} and the second

\begin{equation} S(U, V, N) =N R \ln(V) +g(U, N) \end{equation}

Which are consistent only if we can write S in the form:

\begin{equation}S(U, V, N) =cN R \ln(U) +N R \ln(V) +f(N) \end{equation}

where $f(N)$ is a function of N alone namely:

\begin{equation}f(n) = c RN \end{equation}

which accounts also for the fact that $S$ must be extensive in $U, V$, and $N$.

the arguments of the logarithms must be dimensionless (otherwise, the logarithm doesn’t make sense)

\begin{equation}S(U, V, N) =cN R \ln \frac{U}{Nk_1} +N R \ln \frac{V}{Nk_k2} +f(N) \end{equation}

Here $k_1$ and $k_2$, are constants with the dimensions of energy per mole, volume per mole

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