$\left(P+\frac{a}{V^{2}}\right)(V-b)=R T$

Where, $a$ and $b$ are constant Solving above equation, $P=\frac{R T}{V-b}-\frac{a}{V^{2}}$

Taking derivative of $P$ $w.r.t$ volume

$\frac{\partial P}{\partial V}=0$

$\frac{\partial^{2} P}{\partial V^{2}}=0$

So, $P$ becomes$:$

$\frac{\partial P}{\partial V}=-\frac{R T}{(V-b)^{2}}+\frac{2 a}{V^{3}}=0$

$\frac{2 a}{V^{3}}=\frac{R T}{(V-b)^{2}}$$.....(1)$

$\frac{a}{V^{4}}=\frac{R T}{2 V(V-b)^{2}}$$.....(2)$

Taking double derivative again,

$\frac{\partial^{2} P}{\partial V^{2}}=\frac{2 R T}{(V-b)^{3}}-\frac{6 a}{V^{4}}=0$

or

$\frac{R T}{(V-b)^{3}}=\frac{3 a}{V^{4}}$

Put equation $( 2 )$ in above equation

$\frac{R T}{(V-b)^{3}}=\frac{3 R T}{2 V(V-b)^{2}}$

On rearranging,

$3 V-3 b=2 V$

$V_{c}=3 b$

$\mathrm{V}_{\mathrm{C}}$ is critical volume

Use this value in equation $(1)$

$\frac{R T}{4 b^{2}}=\frac{2 a}{27 b^{3}}$

$T_{c}=\frac{8 a}{27 R b}$

$\mathrm{T}_{\mathrm{C}}$ is critical temperature.