(c)

Process is quasistatic adiabatic.

$\text { So, } d Q =0$

$\Rightarrow d W =-d U$

$\Rightarrow p \Delta V =-n C_V d T \quad \dots(i)$

$U =C T-\frac{n^2 a }{V} \quad \dots(ii)$

and $\quad p=\frac{n R T}{V-n b}-\frac{n^2 a}{V^2} \quad \dots(iii)$

From Eq. $(ii)$, $d U=C d T+\frac{n^2 a }{V^2} d V$

So, from Eq. $(i)$, we have

$d W=-d U$

$\Rightarrow \quad p d V=-\left(C d T+\frac{n^2}{V^2} d V\right)$

Substituting for $p$ from Eq. $(iii)$ in above equation, we have

$\Rightarrow\left(\frac{n R T}{V-n b}-\frac{n^2 a}{V^2}\right) d V$

$=-C d T-\frac{n^2 \alpha }{V^2} \cdot d V$

$\Rightarrow\left(\frac{n R T}{V-n b}\right) d V=-C d T$

$\Rightarrow \quad \frac{d V}{V-n b}=\frac{-C}{n R} \cdot \frac{d T}{T}$

$\Rightarrow \quad \int \frac{d V}{V-n b}=-\frac{C}{n R} \int \frac{d T}{T}$

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

$\Rightarrow \log \left(T^{C / n R}  \cdot(V-n b)\right)=$ constant

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