$V \rightarrow$ volume of the gas

Van der waals equation $\Rightarrow\left(P+\frac{a n^{2}}{V^{2}}\right)(V-n b)=n R T$

Coefficient a represents the strength of intermolecular attractive forces. We know that molar volume $\left(V_{m}\right)$ is given by $\Rightarrow V_{m}=\frac{V}{n}$. It shows how the molecules experience the attractive force. The force toward the center experienced by a molecule at the edge should be proportional to $\left(\frac{1}{V_{m}}\right)$. Now, number of molecules at the edge of the container should be proportional to $\left(\frac{1}{V_{m}}\right)$. By combining proportionality constants into a single constant $(a)$

we get,

$\left(\frac{a}{V_{m}^{2}}\right)=\left(\frac{a n^{2}}{V^{2}}\right).$

Answer: Proportion in which press ure will decrease is $\left(\frac{n}{v}\right)^{2}$.

Option $(C)$ is the carrect answer.