In an adiabatic process, the pressure is increased by$\frac{2}{3}$%. If $\gamma =\frac{3}{2}$, then the volume decreases by nearly:

Given: Increased pressure $=\frac{\Delta P}{P}=\frac{2}{3}$

$\gamma=\frac{c_{p}}{c_{v}}=\frac{3}{2}$

Soln: We know that,

$P V^{\gamma}=c \quad \ldots \quad(1)$

where,

$\gamma=\frac{c_{p}}{c_{v}}$

Put the value of $y$ in $e^{n}(1)$

$\therefore P v^{C_{P} / C_{v}}=c$

Taking $\log$ on both sides,

$\log P+3 / 2 \log v=\log c$

On differentiating,

$\Delta P / P+3 / 2(A V / V)=0$

$\therefore \Delta V / V=-2 / 3(\triangle P / P)$

Negutive sign represents decrement in volume

$\therefore \frac{\Delta v}{v} \times 100=\frac{4}{9} \%$

$\therefore$ Correct Option is (A)