$\left(p+\frac{an^{2}}{V^{2}}\right)$ $\left(V-nb\right)=nRT$
$\left(p+\frac{a}{V^{2}}\right)\left(V-b\right)=RT$ $\quad$ $\left(\because n=1\right)$
If b is negligible p $=\frac{RT}{V}-\frac{a}{V^{2}}$
The equation is quadratic in $V$ , thus
$V=\frac{+RT\pm\sqrt{R^{2}T^{2}-4ap}}{2p}$
Since $V$ has one value at given $P$ and $T$, thus numericalvalue of discriminant = $0$
$R^{2}T^{2}=4ap$
$p=\frac{R^{2}T^{2}}{4a}$ $=\frac{\left(0.0821\right)^{2}\left(300\right)^{2}}{4\times3.592}$ $\quad$ $\therefore$ $\frac{p}{5.277}=8$