Explanation

(a)
Work obtained is isothermal
$W = - 2.303\ nRT\ \log\frac{p_{1}}{p_{2}}$
Given, $p_{1} = 500\ kPa,\ p_{2} = 200\ kPa$
$(W) = - 2.303\ nRT = \log\frac{500}{200}$
$= - 2.303\ nRT \times 0.3980\ nRT$
$\therefore$ Minimum work will be obtained when number of moles (n) is minimum.
Moles of substances, mass=1 kg=1000 g
Moles of $Cl_{2} = \frac{1000}{71}$
Moles of $O_{2} = \ \frac{1000}{32}$
Moles of $N_{2} = \ \frac{1000}{28}$
Moles of $CH_{4} = \ \frac{1000}{16}$
$\because$ Moles of $Cl_{2}$ are least.
Minimum work is obtained when 1 kg of chlorine gas expands.