JEE Main · 2023 · 31 Jan 2023 · Shift 1 · Heat & Thermodynamics

With no heat exchanged, $Q=0$, the first law $Q=\Delta U+W$ reduces to $\Delta U=-W$, where $W$ is the work done by the gas. The gas moves from A at $(200\text{cc},10\text{kPa})$ to B at $(50\text{cc},50\text{kPa})$, so its volume shrinks and the work done by it is negative. The magnitude of that work is the area of the trapezium under the straight line, $\frac{1}{2}(P_A+P_B)|\Delta V|=\frac{1}{2}(10+50)\times 10^3\times 150\times 10^{-6}=4.5$ J, so $W=-4.5$ J. The change in internal energy is therefore $\Delta U=-W=+4.5$ J.