Q. An ideal gas is heated at constant pressure and absorbs amount of heat $Q$. If the adiabatic exponent is $\gamma$, then the fraction of heat absorbed in raising the internal energy and performing the work, in
A) $1 - \frac{1}{\gamma}$
B) $1 + \frac{1}{\gamma}$
C) $1 - \frac{2}{\gamma}$
D) $1 + \frac{2}{\gamma}$
Explanation
(a) Heat absorbed by the system at constant pressure
$Q = nc_{p}\mathrm{\Delta}T$
Change in internal energy $\mathrm{\Delta}U = nc_{v}\mathrm{\Delta}T$
$W = Q - \mathrm{\Delta}U$
$\therefore\ \ \frac{W}{Q} = \frac{Q - \mathrm{\Delta}U}{Q} = 1 - \frac{\mathrm{\Delta}U}{Q}$
$= 1 - \frac{nc_{v}\mathrm{\Delta}T}{nc_{p}\mathrm{\Delta}T} = 1 - \frac{c_{v}}{c_{p}}$
$= \left( 1 - \frac{1}{\gamma} \right)$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.