The pressure exerted by $2.0\mathrm{mol}$ of a van der Waal gas $\left(a=366 \mathrm{kPa} \mathrm{dm} {\mathrm{mol}}^{-2}\right)$and $(b=0.05 {\mathrm{dm}}^3{\mathrm{mol}}^{-1})$ at $T=300\mathrm{K}$ and $V=12.1{\mathrm{dm}}^3$ is $\text{ (use }R=8.32 \mathrm{kPa} {\mathrm{dm}}^3{\mathrm{K}}^{-1}{\mathrm{mol}}^{-1}\text{ ) }$

$p=\frac{nRT}{V-nb}-\frac{n^{2}a}{V^2}$

Now,

 $$\begin{gathered} \frac{nRT}{V-nb}=\frac{\left(2.0\mathrm{mol}\right)\left(8.32\mathrm{kPa}{\mathrm{dm}}^3{\mathrm{K}}^{-1}{\mathrm{mol}}^{-1}\right)\left(300\mathrm{K}\right)}{\left(12.1{\mathrm{dm}}^3\right)-\left(2.0\mathrm{mol}\right)\left(0.05{\mathrm{dm}}^3{\mathrm{mol}}^{-1}\right)} \\ =416.0\mathrm{kPa} \end{gathered}$$

$$\begin{gathered} \frac{n^{2}a}{V^2}=\frac{(2.0\mathrm{mol}{)}^2\left(366{\mathrm{kPadm}}^6{\mathrm{mol}}^{-2}\right)}{{\left(12.1{\mathrm{dm}}^3\right)}^2} \\ =10.0\mathrm{kPa} \end{gathered}$$

Hence,

 $$\begin{gathered} p=(416.0-10.0) \mathrm{kPa} \\ =406.0 \mathrm{kPa} \end{gathered}$$