Q. The ratio of specific heats of a gas is $\gamma.$ The change in internal energy of one mole of the gas, when the volume changes from V to 2V at constant pressure p is
A) $\frac{\gamma - 1}{pV}$
B) $pV$
C) $\frac{pV}{\gamma - 1}$
D) $\frac{pV}{\gamma}$
Explanation
(c) Change in internal energy
$dU = dQ - dW$
At constant pressure
$dU = C_{p}dT - p\ dV$
$= C_{p}dT - R\ dT$
$= \left( C_{p} - R \right)\ dT$
$= C_{v}\ dT$
$= \frac{R}{\gamma - 1}dT$
$= \frac{R}{\gamma - 1} \times \frac{pV}{R}$
$= \frac{pV}{\gamma - 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.