Discussion
The chemical potential \(\mu\) for an indistinguishable ideal monoatomic gas molecule is \(\mu\) = ..... (where, \(\beta = 1/(k_{B}T)\)) :
Answer: B
The chemical potential is \(-k_{B}T\left[\ln\frac{V}{N} + \frac{3}{2}\ln\frac{2\pi m}{\beta h^{2}}\right]\).
For an ideal gas of indistinguishable particles, \(Z = z^{N}/N!\) with \(z = V(2\pi m/\beta h^{2})^{3/2}\). Using Stirling's approximation, \(F = -Nk_{B}T[\ln(z/N) + 1]\), and
\[ \begin{aligned} \mu &= \left(\frac{\partial F}{\partial N}\right)_{T,V} \\ &= -k_{B}T\ln\frac{z}{N} \end{aligned} \]
which expands to the keyed expression. The \(+1\) in \(F\) disappears on differentiating with respect to \(N\).
The option with \(+1\) is \(F/N\), the free energy per particle, not \(\mu\). The option with \(+5/2\) is the Sackur-Tetrode entropy per particle divided by \(k_{B}\). The option with \(\ln V\) alone has lost the \(1/N\) that comes from the \(N!\) of indistinguishable particles, so it is not even intensive.
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