Q.
Figure shows two processes $a$ and $b$ for a given sample of a gas, if $\mathrm{\Delta}Q_{1},\ \mathrm{\Delta}Q_{2}$ are the amounts of heat absorbed by the system in the two cases, and $\mathrm{\Delta}U_{1},\ \mathrm{\Delta}U_{2}$ are changes in internal energies respectively, then
A) $\mathrm{\Delta}Q_{1} = \mathrm{\Delta}Q_{2},;\ \mathrm{\Delta}U_{1} = \mathrm{\Delta}U_{2}$
B) $\mathrm{\Delta}Q_{1} > \mathrm{\Delta}Q_{2};\ \mathrm{\Delta}U_{1} > \mathrm{\Delta}U_{2}$
C) $\mathrm{\Delta}Q_{1} < \mathrm{\Delta}Q_{2};\ \mathrm{\Delta}U_{1} < \mathrm{\Delta}U_{2}$
D) $\mathrm{\Delta}Q_{1} > \mathrm{\Delta}Q_{2};\ \mathrm{\Delta}U_{1} = \mathrm{\Delta}U_{2}$
Explanation
(d) As initial and final states in the two processes are same. Therefore, $\mathrm{\Delta}U_{1} = \mathrm{\Delta}U_{2}$. As area under curve $a >$ area under curve $b$, therefore, $\mathrm{\Delta}W_{1} > \mathrm{\Delta}W_{2}$
As $\mathrm{\Delta}Q = \mathrm{\Delta}U + \mathrm{\Delta}W$
$\therefore\ \mathrm{\Delta}Q_{1} > \mathrm{\Delta}Q_{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.