Verified Solution • 2025–26 Session moderate

QPages 34-37: Q49: HC Verma: Concepts of Physics - Volume 2 Kinetic Theory of Gases Pages 34-37: Q49 Solution: Pure water vapour is trapped in a vessel of volume . The…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often confuse relative humidity with a volume fraction or fail to recognize that it directly defines the initial partial pressure as a fraction of the saturated vapor pressure ($P_{initial} = RH \times P_{sat}$). To avoid this, explicitly convert the percentage to a pressure ratio before applying Boyle's Law, ensuring the final pressure is set to $P_{sat}$ rather than an arbitrary value.

Pure water vapour is trapped in a vessel of volume

10 \text{ cm}^3

. The relative humidity is 40%. The vapour is compressed slowly and isothermally. Find the volume of the vapour at which it will start condensing.

Step-by-Step Verified Solution

**Answer:** Volume at which condensation starts = 4 cm_ **Step-by-Step Explanation:** Initial volume

V_1 = 10

cm_, RH = 40%. Vapour pressure

P_1 = 0.40 P_{sat}

. At constant temperature,

P_1 V_1 = P_2 V_2

. Condensation starts when

P_2 = P_{sat}

:

0.40 P_{sat} \times 10 = P_{sat} \times V_2

V_2 = 4 \text{ cm}^3

All Questions in this Chapter

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