Verified Solution • 2025–26 Session moderate

QPages 34-37: Q11: HC Verma: Concepts of Physics - Volume 2 Kinetic Theory of Gases Pages 34-37: Q11 Solution: Figure shows a cylindrical tube with adiabatic walls and fitted …

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often incorrectly apply the adiabatic relation $PV^\gamma = \text{constant}$ to the gas segments because the outer tube walls are adiabatic. However, the internal separator is diathermic, ensuring both gas segments remain at the same temperature, so the isothermal relation $P \propto 1/V$ must be used instead.

Figure shows a cylindrical tube with adiabatic walls and fitted with a diathermic separator. The separator can be slid in the tube by an external mechanism. An ideal gas is injected into the two sides at equal pressures and equal temperatures. The separator remains in equilibrium at the middle. It is now slid to a position where it divides the tube in the ratio of 1:3. Find the ratio of the pressures in the two parts of the vessel. Use

R = 8.314 \text{ J K}^{-1} \text{ mol}^{-1}

Step-by-Step Verified Solution

**Answer:**

P_1 : P_2 = 3 : 1

**Step-by-Step Explanation:** For an ideal gas,

P \propto 1/V

at constant temperature. If volume is divided in ratio 1:3, then

V_1 : V_2 = 1 : 3

.

P_1 : P_2 = V_2 : V_1 = 3 : 1

All Questions in this Chapter

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