Q. For a monoatomic gas, work done at constant pressure is $W$. The heat supplied at constant volume for the same rise in temperature of the gas is
A) $W/2$
B) $3W/2$
C) $5W/2$
D) $W$
Explanation
(b) For the process at constant pressure
$dQ = C_{p}dT + dw$
$dT = \frac{dQ - dW}{C_{p}}$
For the process at constant volume,
$dQ = C_{v}dT\ \ \ \ \ (\therefore\ dW = 0)$
$= C_{v}\left( \frac{dQ - dW}{C_{p}} \right) = \frac{dQ - dW}{C_{p}/C_{v}} = \frac{dQ - dW}{\gamma}$
or $(\gamma - 1)dQ = dW$
$\left( \frac{5}{3} - 1 \right)dQ = W,\ dQ = \frac{3}{2}W$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.