Explanation

(b)
For $'n'$ moles, the van der Waals’ equation is
$\left( p + \frac{an^{2}}{V^{2}} \right)\ (V - nb) = nRT$
where, $n = 2\ moles$
$R = 0.0821\ L\ atm\ K^{- 1}\ {mol}^{- 1}$
$T = 27 + 273 = 300\ K$
$V = 5\ L$
$a = 4.17$
$b = 0.03711$
$so\ \ \ \ \ \ \ \ \ \ p = \frac{nRT}{V - nb\ } - \frac{an^{2}}{V^{2}}$
$= \frac{2 \times 0.0821 \times 300}{(5 - 2 \times 0.3711)} - \frac{4.17 \times (12)^{2}}{(5)^{2}}\$
$= \frac{49.26}{4.926} - \frac{16.68}{25}$
$= 10 - 0.66$
$= 9.33\ atm$