MHTCET 17 Apr 2026 Shift 2 Paper

Concept:For an ideal gas, internal energy depends only on temperature.During an isothermal process, temperature remains constant, so $\Delta U = 0$.Work done under constant external pressure is given by $w = -P_{\text{ext}} \Delta V$.Explanation:Given: $V_1 = 25\ \mathrm{dm}^3$, $V_2 = 10\ \mathrm{dm}^3$, $P_{\text{ext}} = 4\ \mathrm{bar}$.Change in volume:$\Delta V = V_2 - V_1 = 10 - 25 = -15\ \mathrm{dm}^3$.Work done:$w = -P_{\text{ext}} \Delta V = -(4\ \mathrm{bar})(-15\ \mathrm{dm}^3) = 60\ \mathrm{bar\,dm}^3$.Convert using $1\ \mathrm{bar\,dm}^3 = 0.1\ \mathrm{kJ}$:$w = 60 \times 0.1 = 6.0\ \mathrm{kJ}$.For an isothermal process, $\Delta U = 0$.From the first law, $\Delta U = q + w = 0$.Thus, $q = -w = -6.0\ \mathrm{kJ}$.The negative sign means heat is released from the system.Therefore, the heat released is $6.0\ \mathrm{kJ}$.Answer:The quantity of heat released is $6.0\ \mathrm{kJ}$.So, the correct option is D.