Verified Solution • 2025–26 Session moderate

QPages 62-64: Q6: HC Verma: Concepts of Physics - Volume 2 Laws of Thermodynamics Pages 62-64: Q6 Solution: An ideal gas is taken from an initial state i to a final state…

⚠️ Common Exam Traps & Conceptual Pitfalls

When the ratio $P/T$ is kept constant, the ideal‑gas law gives $V= rac{nR}{P/T}=nR imes rac{T}{P}= ext{constant}$. Students often mistake this for an isothermal or isobaric process and assume $V$ changes, but in fact $Delta V=0$, so $W=int P,dV=0$. Always derive $V$ from the given condition before evaluating work.

An ideal gas is taken from an initial state i to a final state f in such a way that the ratio of the pressure to the absolute temperature remains constant. What will be the work done by the gas?

Step-by-Step Verified Solution

**Answer:** W=0W = 0. **Step-by-Step Explanation:** If PT\frac{P}{T} remains constant, then from PV=nRTPV = nRT, V=nRTP=nR×TP=nR×constantV = \frac{nRT}{P} = nR \times \frac{T}{P} = nR \times \text{constant}. So VV is constant, meaning ΔV=0\Delta V = 0. Hence W=PΔV=0W = P\Delta V = 0.

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