Thermodynamics, on the exam screen.

By the first law of thermodynamics $Q = \Delta U + W$. Differentiating with respect to time, $\frac{dQ}{dt} = \frac{d(\Delta U)}{dt} + \frac{dW}{dt}$, so $\frac{d(\Delta U)}{dt} = 100 - 75 = 25$ W. The distractors correspond to the given work rate (1), heat rate (2), or the incorrect sum (3).

In an adiabatic process $Q = 0$, so the work done by the gas comes entirely out of its internal energy: $W = -\Delta U = nC_V(T_1 - T_2)$. For a monatomic gas $C_V = \frac{3}{2}R$, giving $W = 1 \times \frac{3}{2}(8.3)(10) = 124.5\ \text{J}$. The same follows from $W = \frac{nR(T_1 - T_2)}{\gamma - 1}$ with $\gamma - 1 = \frac{2}{3}$. NCERT Class 11, Thermodynamics.

Over a complete cycle the gas returns to its initial state, so $\Delta U = 0$ and the first law reduces to $Q = W$. The cycle is a rectangle on the $P\!-\!V$ diagram between $P = 100$ and $300\ \text{N m}^{-2}$ and $V = 2$ and $5\ \text{m}^3$, traversed clockwise, so the work done by the gas is the enclosed area $= (300-100)(5-2) = 600\ \text{J}$. Hence 600 J of heat is supplied. NCERT Class 11, Thermodynamics (cyclic process).

Why the other options are wrong:

- (1) miscounts one side of the rectangle, typically taking $\Delta V = 2\ \text{m}^3$
- (2) computes only the net heat of the two isobaric legs and omits the isochoric contributions to the balance
- (4) adds the areas under the two isobaric legs instead of taking their difference, double-counting the return path

The system is thermally insulated, so q = 0; the gas expands freely against vacuum, so w = 0. This is an irreversible adiabatic free expansion (with $\Delta$T = 0 for an ideal gas). Chapter: Thermodynamics.