Q. When an ideal gas $(\gamma = 5/3)$ is heated under constant pressure, then what percentage of given heat energy will be utilised in doing external work
A) 40%
B) 30%
C) 60%
D) 20%
Explanation
(a) $\mathrm{\Delta}Q = \mathrm{\Delta}U + \mathrm{\Delta}W \Rightarrow \frac{\mathrm{\Delta}W}{\mathrm{\Delta}Q} = 1 - \frac{\mathrm{\Delta}U}{\mathrm{\Delta}Q} = 1 - \frac{\mu C_{V}dT}{\mu C_{P}dT}$
$\Rightarrow \frac{\mathrm{\Delta}W}{\mathrm{\Delta}Q} = 1 - \frac{C_{V}}{C_{P}} = 1 - \frac{3}{5} = \frac{2}{5} = 0.4$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.