# 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 $$p_F\sim V^{-5/3}$$).

• 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. Commented Nov 7, 2022 at 19:03
• (This is for an ideal Fermi gas.) Commented Nov 7, 2022 at 19:06
• but where did you get that there is no $O(T)$? Commented Nov 7, 2022 at 19:07
• If you're happy with just the first order terms then I can dig out a derivation if you'd like! Commented Nov 7, 2022 at 19:10
• yes, thank you in advance Commented Nov 7, 2022 at 19:11

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:
• 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. Commented Nov 7, 2022 at 18:57
• When $T$ is much less than the Fermi temperature, the gas would reach degenerate pressure, so $\alpha = -5/3$ and $\beta = 0$. Commented Nov 7, 2022 at 19:09