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 $\frac{4}{9}\%$.