Question

In an adiabatic process, 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

The correct option is A $\frac{4}{9}\%$

For an adiabatic process, $P{V}^{\gamma }=K$
Here, $\gamma =\frac{C_P}{C_V}=\frac{3}{2}$ and $K=\text{constant}$
$\Rightarrow P{V}^{3/2}=K$
Taking $\log$ both sides,
$\log P+\frac{3}{2}\log V=\log K$
On differentiating both sides using chain rule,
$\Rightarrow \frac{1}{P}\times dP+\frac{3}{2}\frac{1}{V}\times dV=0$
For finite differences in values of $P$ and $V$, we can write above equation as,
$\Rightarrow \frac{\Delta P}{P}+\frac{3}{2}\frac{\Delta V}{V}=0$
$\Rightarrow \frac{\Delta V}{V}=-\frac{2}{3}\frac{\Delta P}{P}$
$\Rightarrow \frac{\Delta V}{V}\times 100=-\left(\frac{2}{3}\right)(\frac{\Delta P}{P}\times 100)$
$\Rightarrow \frac{\Delta V}{V}\times 100=-\frac{2}{3}\times \frac{2}{3}=-\frac{4}{9}\%$
$\therefore$ Volume decreases by about $\frac{4}{9}\%$