Explanation

(a) In an adiabatic process
$pV^{\gamma} = K$ (Poisson’s equation)
Where p is pressure, V the volume and $\gamma$ the ratio of specific heats.
Given, $\gamma = \frac{3}{2}$
$\therefore\ \ \ \ \ \ \ pV^{3/2} = K$
Taking logarithm on both sides, we get
$\log{p + \frac{3}{2}\log{V = \log K}}$
$\therefore\ \ \ \ \ \frac{\mathrm{\Delta}p}{p} + \frac{3}{2}\frac{\mathrm{\Delta}V}{V} = 0$
$\therefore\ \ \ \ \ \ \ \ \ \frac{\mathrm{\Delta}V}{V} = - \frac{2}{3}\frac{\mathrm{\Delta}p}{p}$
$\frac{\mathrm{\Delta}V}{V} \times 100 = - \frac{2}{3}\left( \frac{\mathrm{\Delta}p}{p} \times 100 \right)$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = - \frac{2}{3} \times \frac{2}{3} = - \frac{4}{9}\%$
Minus (-) sign implies that volume decreases by $\frac{4}{9}\%.$