Explanation

(a)
$V_{1} = 100$ mL
$V_{2} = 250$ mL
Pressure $p = 2\ atm\ or\ 2 \times 1.01 \times 10^{5}\ Nm^{- 2}$
Work done by the gas
$W = p\mathrm{\Delta}V\ or\ p(V_{2} - V_{1})$
Put the value in given formula
$W = 2 \times 1.01 \times 10^{5}(0.250 \times 10^{- 3} - 0.100 \times 10^{- 3})$
$= 2 \times 1.01 \times 10^{5} \times 0.15 \times 10^{- 3}$
= 30.30 J