Verified Solution • 2025–26 Session moderate

QPages 34-37: Q34: HC Verma: Concepts of Physics - Volume 2 Kinetic Theory of Gases Pages 34-37: Q34 Solution: A glass tube, sealed at both ends, is 100 cm long. It lies…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often incorrectly assume the air columns remain at their initial 45 cm lengths or fail to recognize that the mercury column's equilibrium requires the pressures on both sides to be equal ($P_1 = P_2$). You must use this equality to establish the relationship $\frac{L_1}{T_1} = \frac{L_2}{T_2}$, rather than solving for each side independently.

A glass tube, sealed at both ends, is 100 cm long. It lies horizontally with the middle 10 cm containing mercury. The two ends of the tube contain air at 27�C and at a pressure 76 cm of mercury. The air column on one side is maintained at 0�C and the other side is maintained at 127�C. Calculate the length of the air column on the cooler side. Neglect the changes in the volume of mercury and of the glass.

Step-by-Step Verified Solution

**Answer:**

L_{cool} = 35.6 \text{ cm}

**Step-by-Step Explanation:** Let

L_1

= length of cooler side (0�C),

L_2

= length of hotter side (127�C).

L_1 + L_2 = 90

cm (since 10 cm is mercury). For each side,

PV = nRT

. Since pressure and moles are same:

\frac{L_1}{T_1} = \frac{L_2}{T_2} \Rightarrow \frac{L_1}{273} = \frac{L_2}{400}

Solving:

L_1 = 35.6

cm.

All Questions in this Chapter

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