这是一份liverpool利物浦大学MATH221的成功案例

Think of $E$ as a function of $T, V ;$ that is, $E=E(S(T, V), V)$, where $S(T, V)$ means $S$ as a function of $T, V$. Then
$$
\left(\frac{\partial E}{\partial T}\right){V}=\frac{\partial E}{\partial S}\left(\frac{\partial S}{\partial T}\right){V}=T\left(\frac{\partial S}{\partial T}\right){V}=C{V} .
$$
Likewise, think of $H=H(S(T, P), P)$. Then
$$
\left(\frac{\partial H}{\partial T}\right){P}=\frac{\partial H}{\partial S}\left(\frac{\partial S}{\partial T}\right){P}=T\left(\frac{\partial S}{\partial T}\right){P}=C{P}
$$
where we used (20) from $\S \mathrm{D}$.

MATH221 COURSE NOTES ：

(i) $C_{V}$ is a function of $T$ only:
$$
C_{V}=C_{V}(T) .
$$
(ii) $E$ as a function of $(T, V)$ is:
$$
E=E(T, V)=\int_{T_{0}}^{T} C_{V}(\theta) d \theta-\frac{a}{V}+E_{0} .
$$
(iii) $S$ as a function of $(T, V)$ is
$$
S=S(T, V)=R \log (V-b)+\int_{T_{0}}^{T} \frac{C_{V}(\theta)}{\theta} d \theta+S_{0} .
$$