Verified Solution • 2025–26 Session moderate

QPages 34-37: Q56: HC Verma: Concepts of Physics - Volume 2 Kinetic Theory of Gases Pages 34-37: Q56 Solution: 50 cc of oxygen is collected in an inverted gas jar over water. …

50 cc of oxygen is collected in an inverted gas jar over water. The atmospheric pressure is 99.4 kPa and the room temperature is 27�C. The water level in the jar is same as the level outside. The saturation vapour pressure at 27�C is 3.4 kPa. Calculate the number of moles of oxygen collected in the jar.

Step-by-Step Verified Solution

**Answer:** Moles of oxygen = 1.92×10−31.92 \times 10^{-3} mol **Step-by-Step Explanation:** Atmospheric pressure = 99.4 kPa. Saturation vapour pressure at 27�C = 3.4 kPa. Partial pressure of oxygen = 99.4−3.4=96.099.4 - 3.4 = 96.0 kPa. Volume = 50 cc = 50×10−650 \times 10^{-6} m_, T=300T = 300 K.
n=PVRT=96×103×50×10−68.314×300=1.92×10−3 moln = \frac{PV}{RT} = \frac{96 \times 10^3 \times 50 \times 10^{-6}}{8.314 \times 300} = 1.92 \times 10^{-3} \text{ mol}

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