Verified Solution • 2025–26 Session moderate

QPages 34-37: Q38: HC Verma: Concepts of Physics - Volume 2 Kinetic Theory of Gases Pages 34-37: Q38 Solution: One mole of an ideal gas undergoes a process where and are…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students frequently drop the parentheses and evaluate $(1+V/V_0)^2$ as $1+(V/V_0)^2$, which gives $P=P_0/2$ at $V=V_0$ instead of the correct $P=P_0/4$. Always square the whole sum before substituting the volume.

One mole of an ideal gas undergoes a process P=P01+V/V0)2P = \frac{P_0}{1 + V/V_0)^2} where P0P_0 and V0V_0 are constants. Find the temperature of the gas when V=V0V = V_0.

Step-by-Step Verified Solution

**Answer:** T=P0V04RT = \frac{P_0 V_0}{4R} **Step-by-Step Explanation:** Given: P=P0(1+V/V0)2P = \frac{P_0}{(1 + V/V_0)^2}. At V=V0V = V_0:
P=P0(1+1)2=P04P = \frac{P_0}{(1 + 1)^2} = \frac{P_0}{4}
Using ideal gas law for n=1n = 1 mole:
T=PVnR=(P0/4)⋅V01⋅R=P0V04RT = \frac{PV}{nR} = \frac{(P_0/4) \cdot V_0}{1 \cdot R} = \frac{P_0 V_0}{4R}

All Questions in this Chapter

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