Question

For a reversible adiabatic ideal gas expansion, the value of $\frac{dP}{P}$ is equal to:

• A. $\gamma \frac{dV}{V}$
• B. $-\gamma \frac{dV}{V}$
• C. $\left(\frac{\gamma }{\gamma -1}\right)\frac{dV}{V}$
• D. $\frac{dV}{V}$

Answer 1

The correct option is B $-\gamma \frac{dV}{V}$

We know that for a reversible adiabatic process,
$P{V}^{\gamma }$ = constant
So, on differentiating with respect to $V$ we get,
$P.\gamma .V^{\gamma -1}.dV+V^{\gamma }.dP=0$
On rearranging, $\therefore \frac{dP}{P}=-\gamma \frac{dV}{V}$
hence, $\frac{dP}{P}=-\gamma \frac{dV}{V}$