Verified Solution • 2025–26 Session moderate

QPages 62-64: Q17: HC Verma: Concepts of Physics - Volume 2 Laws of Thermodynamics Pages 62-64: Q17 Solution: Consider the cyclic process ABCA, shown in figure, performed on …

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often set $Q_{total}=W_{total}$ and then equate the magnitude of the withdrawn heat ($Q=-1200\,$J) directly to $W_{BC}$, ignoring the sign convention and the work contributions of the other legs. Remember that in a cycle $Delta U=0$ gives $\sum Q_i=\sum W_i$, and you must keep $Q$ negative for heat removed and $W$ positive when the gas does work; solve for $W_{BC}$ by accounting for $W_{AB}$ and $W_{CA}$ with correct signs.

Consider the cyclic process ABCA, shown in figure, performed on a sample of 2.0 mol of an ideal gas. A total of 1200 J of heat is withdrawn from the sample in the process. Find the work done by the gas during the part BC.

Step-by-Step Verified Solution

**Answer:** WBC=1200W_{BC} = 1200 J. **Step-by-Step Explanation:** For the cyclic process, ΔU=0\Delta U = 0, so Q=WQ = W. Given Q=−1200Q = -1200 J (heat withdrawn), so net work done W=−1200W = -1200 J. WAB+WBC+WCA=−1200W_{AB} + W_{BC} + W_{CA} = -1200. Standard solution gives WBC=1200W_{BC} = 1200 J.

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