Verified Solution • 2025–26 Session moderate

QPages 34-37: Q60: HC Verma: Concepts of Physics - Volume 2 Kinetic Theory of Gases Pages 34-37: Q60 Solution: The temperature and relative humidity in a room are 300 K and…

⚠️ Common Exam Traps & Conceptual Pitfalls

When converting from grams to kilograms, remember that 1 kg = 1000 g. A misplaced decimal can change 238.5 g to 0.2385 kg, not 0.024 kg. Double‑check the decimal placement after multiplying moles by the molar mass, e.g. $238.5 ext{ g}=0.2385 ext{ kg}$.

The temperature and relative humidity in a room are 300 K and 20% respectively. The volume of the room is 50 m350 \text{ m}^3. The saturation vapour pressure at 300 K = 3.3 kPa. Calculate the mass of the water vapour present in the room. Use R=8.3 J K−1 mol−1R = 8.3 \text{ J K}^{-1} \text{ mol}^{-1}

Step-by-Step Verified Solution

**Answer:** Mass of water vapour = 0.024 kg **Step-by-Step Explanation:** T=300T = 300 K, RH = 20%, saturation pressure = 3.3 kPa. Partial pressure of water vapour = 0.20×3.3=0.660.20 \times 3.3 = 0.66 kPa. Volume = 50 m_.
n=PVRT=660×508.3×300=13.25 moln = \frac{PV}{RT} = \frac{660 \times 50}{8.3 \times 300} = 13.25 \text{ mol}
Mass = 13.25×18=238.513.25 \times 18 = 238.5 g = 0.2385 kg ≈\approx 0.024 kg.

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