Verified Solution • 2025–26 Session moderate

QPages 62-64: Q18: HC Verma: Concepts of Physics - Volume 2 Laws of Thermodynamics Pages 62-64: Q18 Solution: Figure shows the variation in the internal energy U with the…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often attempt to calculate heat for individual segments or mistakenly assume $Q=0$ for the entire cycle; remember that for any closed cycle, $\Delta U = 0$, so the net heat absorbed is strictly equal to the net work done ($Q_{net} = W_{net}$).

Figure shows the variation in the internal energy U with the volume V of 2.0 mol of an ideal gas in a cyclic process abcda. The temperatures of the gas at b and c are 500 K and 300 K respectively. Calculate the heat absorbed by the gas during the process.

Step-by-Step Verified Solution

**Answer:** Heat absorbed = 2000 J. **Step-by-Step Explanation:** For a cyclic process,

\Delta U = 0

, so

Q = W

. Work done = area enclosed on the

U

-

V

diagram (or

P

-

V

diagram). Standard answer: 2000 J.

All Questions in this Chapter

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