Verified Solution • 2025–26 Session moderate

QPages 34-37: Q47: HC Verma: Concepts of Physics - Volume 2 Kinetic Theory of Gases Pages 34-37: Q47 Solution: The condition of air in a closed room is described as follows.…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often treat relative humidity as the fraction of the total pressure contributed by water vapour, using $p_{H_2O}=RH\times P_{total}$. In reality $RH= p_{H_2O}/p_{sat}$, so the correct vapour pressure is $p_{H_2O}=RH\times p_{sat}$; only this amount should be subtracted from the initial pressure.

The condition of air in a closed room is described as follows. Temperature = 25�C, relative humidity = 60%, pressure = 104 kPa. If all the water vapour is removed from the room without changing the temperature, what will be the new pressure? The saturation vapour pressure at 25�C = 3.2 kPa.

Step-by-Step Verified Solution

**Answer:** New pressure = 102.08 kPa **Step-by-Step Explanation:** Total pressure = 104 kPa. RH = 60%. Saturation vapour pressure at 25�C = 3.2 kPa. Partial pressure of water vapour = 0.60×3.2=1.920.60 \times 3.2 = 1.92 kPa. Partial pressure of dry air = 104−1.92=102.08104 - 1.92 = 102.08 kPa. After removing all water vapour, only dry air remains: Pnew=102.08P_{new} = 102.08 kPa.

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