Question

For an adiabatic expansion of a perfect gas, $\frac{\Delta P}{P}$ is equal to:

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

Answer 1

Answer: (C)

$-\gamma \frac{\Delta V}{V}$

${PV}^{\gamma }=constant$

$P.{\gamma V}^{Y-1}dV+V^Y.dP=0$

$-\gamma .V^{Y-1}dV=V^{\gamma }\frac{dP}{P}$

or $\frac{dP}{P}=\gamma -\frac{V^{Y-1}}{V^y}.dV=-\gamma \frac{dV}{V}$

$\therefore \frac{\Delta P}{P}=-\gamma \frac{\Delta V}{V}$