What is the $C_{P}$ (specific heat under constant pressure) for a Fermi gas? For some reason I can't find $C_{P}$ for Fermi (electron) gas anywhere; there is only $C_{V}$ quoted everywhere.  However, I definitely can (as I think) change temperature and compensate the small increase in pressure

by increasing the volume (because $p_F\sim V^{-5/3}$).
 A: This question is a little tricky to answer because the equation of state of the Fermi gas will change depending on the temperature and interactions involved in the gas. Therefore, I will answer this question by generalizing to any non-ideal equation of state:
$$p = \kappa V^{\alpha} T^{\beta}$$
where $\alpha, \beta, \kappa \in \mathbb R$. Let $S (p, V)$ be the entropy of the gas. From the multivariate chain rule, we can see
$$S(p, V) = \left(\frac{\partial S}{\partial p}\right)_V \text{d}p + \left(\frac{\partial S}{\partial V}\right)_p \text{d}V$$
Furthermore, if we define the index $\gamma$ as $\frac{C_p}{C_V}$, then we can write
$$\gamma = \frac{C_p}{C_V} = \frac{T\left(\frac{\partial S}{\partial T}\right)_p}{T\left(\frac{\partial S}{\partial T}\right)_V}$$
Notice that
\begin{align*}
\left (\frac{\partial S}{\partial T}\right)_V&=\left(\frac{\partial S}{\partial p}\right)_V \left(\frac{\partial p}{\partial T}\right)_V \\
\left(\frac{\partial S}{\partial T}\right)_p&=\left(\frac{\partial S}{\partial V}\right)_p \left(\frac{\partial V}{\partial T}\right)_p
\end{align*}
where
\begin{align*}
\left(\frac{\partial p}{\partial T}\right)_V &= \beta (\kappa V^{\alpha} T^{\beta - 1}) = \frac{\beta p}{T} \\
\left(\frac{\partial p}{\partial T}\right)_T &= -\frac{\beta V}{\alpha T}
\end{align*}
Therefore, going back to $\gamma$ allows us to write
$$\gamma = \frac{C_p}{C_V} = -\frac{V}{\alpha p} \frac{\left(\frac{\partial S}{\partial V} \right)_p}{\left(\frac{\partial S}{\partial p}\right)_V}.$$
Furthermore, you can use the Maxwell relation
$$\left(\frac{\partial S}{\partial P}\right)_V = -\left(\frac{\partial V}{\partial T}\right)_p$$
and the equation
$$\left(\frac{\partial S}{\partial p}\right)_T= \left(\frac{\partial S}{\partial p}\right)_V + \left(\frac{\partial S}{\partial V}\right)_p \left(\frac{\partial V}{\partial p}\right)_T$$
to solve for the numerator and denominator of $\gamma$. You will soon find that
$$\gamma = \frac{\alpha +\beta}{\alpha (1-\beta)} = \frac{C_p}{C_V}$$
Therefore if you know $C_V$ and the equation of state of the gas, you can find $C_p$.
So as @xzd209 pointed out, in the degenerate limit, $\beta = 0$, so $C_V \approx C_p$.
References:

*

*Wikipedia

*Blundell Concepts in Thermal Physics

*Online Physics Olympiad
