Explanation

(b) From $p_{2}V_{2}^{\gamma} = p_{1}V_{1}^{\gamma}\ \Rightarrow \ \ p_{2} = p_{1}\left( \frac{V_{1}}{V_{2}} \right)^{\gamma}$
$p_{2} = \left( \frac{100}{124} \right)^{5/3}p_{1}$
$p_{2} = 0.6985\ p_{1}$
$\therefore$ % decrease in pressure
$= \frac{p_{1} - p_{2}}{p_{1}} \times 100\%$
$= \frac{p_{1} - 0.6985p_{1}}{p_{1}} \times 100$
$= \frac{0.3015p_{1}}{p_{1}} \times 100\%$
$= 30.15\% = 30\%$