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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

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by increasing the volume (because $p_F\sim V^{-5/3}$).

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    $\begingroup$ As $T\to 0$ the ratio of the specific heats tends towards 1, i.e. $C_P=C_V+\mathcal{O}(T^2)$ (I think that the next term will actually be $\mathcal{O}(T^3)$). If you really want the higher order terms in $T$ too then you will have to use the Sommerfeld expansion and some complicated asymptotic methods. $\endgroup$
    – xzd209
    Commented Nov 7, 2022 at 19:03
  • $\begingroup$ (This is for an ideal Fermi gas.) $\endgroup$
    – xzd209
    Commented Nov 7, 2022 at 19:06
  • $\begingroup$ but where did you get that there is no $O(T)$? $\endgroup$
    – Aslan Monahov
    Commented Nov 7, 2022 at 19:07
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    $\begingroup$ If you're happy with just the first order terms then I can dig out a derivation if you'd like! $\endgroup$
    – xzd209
    Commented Nov 7, 2022 at 19:10
  • $\begingroup$ yes, thank you in advance $\endgroup$
    – Aslan Monahov
    Commented Nov 7, 2022 at 19:11

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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:

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  • $\begingroup$ But what are the coefficients $\alpha$ and $\beta$ for $T \rightarrow 0$? I assume that in the first order $\beta=0$ but cannot derive it from integral equation for pressure. $\endgroup$
    – Aslan Monahov
    Commented Nov 7, 2022 at 18:57
  • $\begingroup$ When $T$ is much less than the Fermi temperature, the gas would reach degenerate pressure, so $\alpha = -5/3$ and $\beta = 0$. $\endgroup$
    – Ashmit Dutta
    Commented Nov 7, 2022 at 19:09

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