Explanation

(c) Here, $V_{1} = 1\ L = 10^{- 3}\ m^{3},\ V_{2} = 3L = 3 \times 10^{- 3}m^{3}$
$p_{1} = 1$ atm $= 1.013 \times 10^{5}Nm^{- 2},\ \gamma = 1.40,\ W = ?$
As changes are adiabatic,
$p_{1}V_{1}^{\gamma} = p_{2}V_{2}^{\gamma}$
$\frac{p_{1}}{p_{2}} = \left( \frac{V_{2}}{V_{1}} \right)^{\gamma} = (3)^{1.4} = 4.6555$
$\therefore\ \ p_{2} = \frac{p_{1}}{4.6555} = \frac{1.013 \times 10^{5}}{4.6555}$
$= 0.217 \times 10^{5}\ Nm^{- 2}$
Work done$= \frac{p_{1}V_{1} - p_{2}V_{2}}{\gamma - 1}$
$= \frac{1.013 \times 10^{5} \times 10^{- 3} - 0.217 \times 10^{5} \times 3 \times 10^{- 3}}{1.4 - 1} = 90.5\ J$