This problem is similar to Problem 7.14 of the Bose gas and can be done the same way.

Parts (i) and (ii) are straightforward. For part (iii), we have to show that

\[
\begin{equation*}
\frac{C_{P}}{C_{\mathrm{V}}}=1+\left(\frac{s}{n}\right)^{2} \frac{C_{\mathrm{V}}}{N k} \frac{f_{(n / s)-1}(z)}{f_{n / s}(z)}=\left(1+\frac{s}{n}\right) \frac{f_{(n / s)+1}(z) f_{(n / s)-1}(z)}{\left\{f_{n / s}(z)\right\}^{2}}, \tag{1}
\end{equation*}
\]

which can be done quite easily; see eqns. (7)-(9) of the solution to Problem 7.14. For part (iv), we observe that, since the quantity \(S / N\) is a function of \(z\) only, an isentropic process implies that \(z=\) const. Accordingly, for such a process,

\[
V T^{n / s}=\text { const. } \quad \text { and } \quad P / T^{(n / s)+1}=\text { const. } ;
\]

see eqns. (2) and (3) of the solution to Problem 7.14. Eliminating \(T\) among these relations, we obtain the desired equation of an adiabat. For part (v), we proceed as follows.

In this limit \(z \rightarrow 0\), eqn. (1) gives

\[
\begin{equation*}
C_{P} / C_{\mathrm{V}} \rightarrow 1+(s / n) . \tag{1a}
\end{equation*}
\]

For \(z \gg 1\), on the other hand, we obtain [see formula (E.17)]

\[
\begin{aligned}
\frac{C_{P}}{C_{\mathrm{V}}} & =\left\{1+\left(\frac{n}{s}+1 \right) \frac{n}{s} \frac{\pi^{2}}{6}(\ln z)^{-2}+\ldots \right\}\left\{1+\left(\frac{n}{s}-1 \right)\left(\frac{n}{s}-2\right) \frac{\pi^{2}}{6}(\ln z)^{-2}+\ldots \right\} \\
& \times\left\{1+\frac{n}{s}\left(\frac{n}{s}-1\right) \frac{\pi^{2}}{6}(\ln z)^{-2}+\ldots\right\} \\
& =1+\frac{\pi^{2}}{3}(\ln z)^{-2}+\ldots \simeq 1+\frac{\pi^{2}}{3}\left(k T / \varepsilon_{F}\right)^{2},
\end{aligned}
\]

regardless of the values of \(s\) and \(n\).