Q. For an isothermal expansion of a perfect gas, the value of $\frac{\mathrm{\Delta}P}{P}$ is equal
A) $- \gamma^{1/2}\frac{\mathrm{\Delta}V}{V}$
B) $- \frac{\mathrm{\Delta}V}{V}$
C) $- \gamma\frac{\mathrm{\Delta}V}{V}$
D) $- \gamma^{2}\frac{\mathrm{\Delta}V}{V}$
Explanation
(b) Differentiate $PV =$ constant $w.r.t.V$
$\Rightarrow P\mathrm{\Delta}V + V\mathrm{\Delta}P = 0 \Rightarrow \frac{\mathrm{\Delta}P}{P} = - \frac{\mathrm{\Delta}V}{V}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.