Question

For an isothermal expansion of a perfect gas, the value of $\frac{\Delta P}{P}$ is equal of

• A. $- \gamma^{1\backslash2} \frac{\Delta V}{V}$
• B. $- \frac{\Delta V}{V}$
• C. $- \gamma\frac{\Delta V}{V}$
• D. $- \gamma^{2}\frac{\Delta V}{V}$

Answer 1

Answer: (B)

Differentiate $PV =$ constant w.r.t $V$
$\Rightarrow P \Delta V+V \Delta P=0$
$ \Rightarrow \frac{\Delta P}{P}=-\frac{\Delta V}{V}$