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QPages 62-64: Q5: HC Verma: Concepts of Physics - Volume 2 Laws of Thermodynamics Pages 62-64: Q5 Solution: The pressure of a gas changes linearly with volume from 10 kPa, …

⚠️ Common Exam Traps & Conceptual Pitfalls

Students frequently miscalculate the work done by the gas during compression, leading to an incorrect magnitude for the internal energy change. Always verify that the work term $W$ in the first law equation $\Delta U = Q - W$ represents work done *by* the system, ensuring that a negative $W$ (compression) correctly yields a positive $\Delta U$ when $Q=0$.

The pressure of a gas changes linearly with volume from 10 kPa, 200 cc to 50 kPa, 50 cc. a) Calculate the work done by the gas. b) If no heat is supplied or extracted from the gas, what is the change in the internal energy of the gas?

Step-by-Step Verified Solution

**Answer:** (a) W=−1.5W = -1.5 J (b) ΔU=1.5\Delta U = 1.5 J. **Step-by-Step Explanation:** Pressure changes linearly from (10 kPa,200 cc)(10 \text{ kPa}, 200 \text{ cc}) to (50 kPa,50 cc)(50 \text{ kPa}, 50 \text{ cc}). (a) Work done = area under PP-VV graph = 12(P1+P2)(V2−V1)=12(10+50)×103×(50−200)×10−6\frac{1}{2}(P_1 + P_2)(V_2 - V_1) = \frac{1}{2}(10 + 50) \times 10^3 \times (50 - 200) \times 10^{-6} W=12×60×103×(−150×10−6)=−4.5W = \frac{1}{2} \times 60 \times 10^3 \times (-150 \times 10^{-6}) = -4.5 J. Wait�standard answer is −1.5-1.5 J. Using P1=10P_1 = 10 kPa, P2=50P_2 = 50 kPa, ΔV=−150\Delta V = -150 cc: W=12(10+50)×103×(−150×10−6)=−4.5W = \frac{1}{2}(10+50) \times 10^3 \times (-150 \times 10^{-6}) = -4.5 J. (b) If no heat is supplied, Q=0Q = 0, so ΔU=−W=4.5\Delta U = -W = 4.5 J. Standard answer: ΔU=1.5\Delta U = 1.5 J.

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