Q 11-11-153JEE MainJEE Main 2019 (9 Apr, Shift 1)Easy
Following figure shows two processes A and B for a gas. If $\Delta Q_A$ and $\Delta Q_B$ are the amount of heat absorbed by the system in two cases, and $\Delta U_A$ and $\Delta U_B$ are changes in internal energies, respectively, then
Answer: (C) $\Delta Q_A > \Delta Q_B;\ \Delta U_A = \Delta U_B$
Both processes go from the same initial state $i$ to the same final state $f$, so $\Delta U_A = \Delta U_B$.
Work done by the gas is the area under each curve; path A lies above path B, so $W_A > W_B$. From $\Delta Q = \Delta U + W$:
$$\Delta Q_A > \Delta Q_B$$
Solution by Sreeraj P, M.Sc Physics