Verified Solution • 2025–26 Session moderate

QPages 77-80: Q26: HC Verma: Concepts of Physics - Volume 2 Specific Heat Capacities of Gases Pages 77-80: Q26 Solution: Figure shows a cylindrical tube with adiabatic walls and fitted …

⚠️ Common Exam Traps & Conceptual Pitfalls

Students frequently invert the volume ratio when applying the relation $TV^{\gamma-1} = \text{constant}$, leading to $T_1/T_2 = \sqrt{V_1/V_2}$ instead of the correct $\sqrt{V_2/V_1}$. Always verify that the gas occupying the smaller volume (compressed) corresponds to the higher temperature to ensure the ratio is physically consistent.

Figure shows a cylindrical tube with adiabatic walls and fitted with an adiabatic separator. The separator can be slid into the tube by an external mechanism. An ideal gas

\gamma = 1.5

) is injected in the two sides at equal pressures and temperatures. The separator remains in equilibrium at the middle. It is now slid to a position where it divides the tube in the ratio 1 : 3. Find the ratio of the temperatures in the two parts of the vessel.

Step-by-Step Verified Solution

**Answer:**

T_1/T_2 = 3^{0.5} = \sqrt{3}

. **Step-BY-Step Explanation:** Given:

\gamma = 1.5

. Separator slid to divide tube in ratio 1:3. For adiabatic process:

TV^{\gamma-1} = \text{constant}

.

T_1V_1^{0.5} = T_2V_2^{0.5}

\frac{T_1}{T_2} = \sqrt{\frac{V_2}{V_1}} = \sqrt{3}

All Questions in this Chapter

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