The equation of state for Van der Waals gas is,
$$
P=\frac{R T}{(v-b)}-\frac{a}{v^{2}}
$$
Therefore, $\left(\frac{\partial P}{\partial T}\right){v}=\frac{R}{(v-b)}$ and $\left(\frac{\partial P}{\partial v}\right){T}=\frac{2 a(v-b)^{2}-R T v^{3}}{(v-b)^{2} v^{3}}$
$c_{P}-c_{v}=-T \frac{\left[\left(\frac{\partial P}{\partial T}\right){v}\right]^{2}}{\left(\frac{\partial P}{\partial v}\right){T}}=\frac{R}{1-2 a(v-b)^{2} / R T v^{3}}$
Also,
$$
\kappa_{T}=-\frac{1}{v}\left(\frac{\partial v}{\partial P}\right){T}=-\frac{1}{v}\left(\frac{\partial P}{\partial v}\right){T}^{-1}=\frac{(v-b)^{2} v^{2}}{R T v^{3}-2 a(v-b)^{2}}
$$
But $\kappa_{s}=\frac{\kappa_{T}}{\gamma}$ and $\gamma=1.66$ for a monatomic gas
Hence,
Similarly,
$$
\begin{aligned}
&\kappa_{s}=0.6 \kappa_{T}=0.6 \frac{(v-b)^{2} v^{2}}{R T v^{3}-2 a(v-b)^{2}} \
&\beta=\frac{1}{v}\left(\frac{\partial v}{\partial T}\right)_{v}=\frac{R v^{2}(v-b)}{R T v^{3}-2 a(v-b)^{2}}
\end{aligned}
$$
PHS2061 COURSE NOTES ：
Since
$$
d F=d U-T d s-S d T
$$
The difference in the value of $F$ of two nearby equilibrium states of an open system may be given by
$$
d F=-P d V-S d T+\mu d \mathbb{N}
$$
$7.61$
Also
$$
d F=\left(\frac{\partial F}{\partial V}\right){T, \mathbb{N}} d V+\left(\frac{\partial F}{\partial T}\right){V, \mathbb{N}} d T+\left(\frac{\partial F}{\partial \mathbb{N}}\right){V, T} d \mathbb{N} $$ Comparison of $$ \mu=\left(\frac{\partial F}{\partial \mathbb{N}}\right){V, T}
$$