Given a van der Waal's gas with state equation, $\left(p+\frac{n^2 a}{V^2}\right)(V-n b)=n R T$

For adiabatic process, $d W=0$

So, from first law of thermodynamics, we have

$d U=d Q+d W$

$\Rightarrow \quad d U-d W=0$

$\Rightarrow d U+p d V=0$

$\Rightarrow\left(\frac{\partial U}{\partial T}\right)_V d T+\left(\frac{\partial U}{\partial V}\right)_T$

$d V+\left(p+\frac{n^2 a}{V^2}\right) d V=0$

$\Rightarrow \quad C_V d T+\left(\frac{n k T}{V-n b}\right) d V=0$

$\Rightarrow \quad C_V \frac{d T}{T}+\left(\frac{n k}{V-n b}\right) d V=0$

Integrating above equation, we get

$\Rightarrow \quad C_V \int \frac{d T}{T}+n k \int \frac{d V}{V-n b}=\text { constant }$

$\Rightarrow \quad \log T^{C_V}+\log (V-n b)^{n k}=C$

$\Rightarrow \quad(V-n b)^R \cdot T^{C_V}=e^C$

$\Rightarrow \quad(V-n b)^{R / C_V \cdot T}=e^{C / C_V}=\text { constant }$

$\therefore \text { Equation for quasistatic adiabatic }$

$\text { process is }(V-n b)^{R / C_V \cdot T}=\text { constant. }$