Q. For an adiabatic expansion of a perfect gas, the value of $\frac{\mathrm{\Delta}P}{P}$ is equal to
A) $- \sqrt{\gamma}\frac{\mathrm{\Delta}V}{V}$
B) $- \frac{\mathrm{\Delta}V}{V}$
C) $- \gamma\frac{\mathrm{\Delta}V}{V}$
D) $- \gamma^{2}\frac{\mathrm{\Delta}V}{V}$
Explanation
(c) $PV^{\gamma} =$ constant : Differentiating both sides
$P_{\gamma}V^{\gamma - 1}dV + V^{\gamma}dP = 0 \Rightarrow \frac{dP}{P} = - \gamma\frac{dV}{V}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.