MHTCET 25 Apr 2025 Shift 1 Paper

Concept:For an isothermal process involving an ideal gas, the change in internal energy is zero, so the heat exchanged equals the work done.Explanation:The First Law of Thermodynamics is written as$$\Delta U = Q - W$$Here, $\Delta U$ is the change in internal energy, $Q$ is the heat added to the gas, and $W$ is the work done by the gas.For an ideal gas undergoing isothermal expansion, the temperature remains constant, so$$\Delta T = 0 \quad \Rightarrow \quad \Delta U = 0$$Substituting this into the First Law gives$$0 = Q - W \quad \Rightarrow \quad Q = W$$This means the heat supplied to the gas is exactly equal to the work done by the gas.In the question, the gas does $(-150)\ \mathrm{J}$ of work against the surroundings.Using the standard sign convention, a negative value of $W$ means work is done on the gas by the surroundings.Thus,$$W = -150\ \mathrm{J}$$Since $Q = W$, we get$$Q = -150\ \mathrm{J}$$A negative value of $Q$ indicates that heat is removed from the gas.Therefore, 150 J of heat has been removed from the gas.Answer:Option B. 150 J of heat has been removed from the gas.