(b) Given, $p_{1} = 100\ mm,\ V_{1} = 200\ mL\ and\ p_{2} = 400\ mm$
From Boyle’ Law
$p_{1}V_{1} = p_{2}V_{2}$
$V_{2} = \frac{p_{1}V_{1}}{p_{2}}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{100 \times 200}{400}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ V_{2} = 50$ mL
Volume of 2 mol gas=$2 \times 50 = 100$ mL