Question

In an adrabatic process wherein pressure is increased by $\frac{2}{3} \%$. If $\frac{C_{p}}{C_{V}}=\frac{3}{2}$, then the volume decreases by about

• A. $\frac{4}{9} \%$
• B. $\frac{2}{3}\%$
• C. $4\%$
• D. $\frac{9}{4}\%$

Answer 1

Answer: (A)

For an adiabatic process,
$p V^{\gamma}=$ constant $($ say $C)$
Here, $\gamma=\frac{C_{p}}{C_{V}}=\frac{3}{2}$
$\therefore p V^{3 / 2}=C$
$\Rightarrow \log p+\frac{3}{2} \log V=\log C$
$\Rightarrow \frac{\Delta \pi}{p}+\frac{3}{2} \frac{\Delta V}{V}=0$
$\therefore \frac{\Delta V}{V}=\frac{-2}{3} \frac{\Delta \pi}{p}$
$\frac{\Delta V}{V} \times 100=-\left(\frac{2}{3}\right)\left(\frac{\Delta \pi}{p} \times 100\right)$
$=-\frac{2}{3} \times \frac{2}{3} \%=-\frac{4}{9} \%$
Thus, volume decreases by $\frac{4}{9} \%$.