Verified Solution • 2025–26 Session moderate

QPages 62-64: Q16: HC Verma: Concepts of Physics - Volume 2 Laws of Thermodynamics Pages 62-64: Q16 Solution: A gas is initially at a pressure of 100 kPa and its volume is . …

⚠️ Common Exam Traps & Conceptual Pitfalls

When computing work for a process where pressure varies linearly with volume, use the average pressure: $W= rac{P_{ ext{initial}}+P_{ ext{final}}}{2},Delta V$. Remember that $Delta V$ is negative for a compression, so the work is negative. Forgetting the 1/2 factor or using the final pressure instead of the average will give an incorrect sign or magnitude, leading to the wrong heat sign for the cycle.

A gas is initially at a pressure of 100 kPa and its volume is 2.0 m32.0 \text{ m}^3. Its pressure is kept constant and the volume is changed from 2.0 m32.0 \text{ m}^3 to 2.5 m32.5 \text{ m}^3. Its volume is now kept constant and the pressure is increased from 100 kPa to 200 kPa. The gas is brought back to its initial state, the pressure varying linearly with its volume. a) Whether the heat is supplied to or extracted from the gas in the complete cycle? b) How much heat was supplied or extracted?

Step-by-Step Verified Solution

**Answer:** (a) Heat is extracted from the gas. (b) 25 kJ extracted. **Step-by-Step Explanation:** The cycle consists of three processes. Work done is the area enclosed by the cycle on the PP-VV diagram. For the isobaric expansion: W1=PΔV=105×0.5=5×104W_1 = P\Delta V = 10^5 \times 0.5 = 5 \times 10^4 J. For the isochoric pressure increase: W2=0W_2 = 0. For the linear return: W3=−12(105+2×105)×0.5=−7.5×104W_3 = -\frac{1}{2}(10^5 + 2 \times 10^5) \times 0.5 = -7.5 \times 10^4 J. Net W=5×104+0−7.5×104=−2.5×104W = 5 \times 10^4 + 0 - 7.5 \times 10^4 = -2.5 \times 10^4 J. Since ΔU=0\Delta U = 0 for a cycle, Q=W=−25Q = W = -25 kJ. Heat is extracted.

All Questions in this Chapter

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