Verified Solution • 2025–26 Session moderate

QPages 34-37: Q32: HC Verma: Concepts of Physics - Volume 2 Kinetic Theory of Gases Pages 34-37: Q32 Solution: A container of volume 50 cc contains air mean molecular weight…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students frequently confuse the state of the gas in part (c) by assuming the mass remains constant from part (a) or that the pressure stays at 100 kPa. You must recognize that the container was open during the heating phase, so the mass trapped in part (c) is the reduced mass from part (b), not the initial mass from part (a).

A container of volume 50 cc contains air mean molecular weight = 28.8 g) and is open to atmosphere where the pressure is 100 kPa. The container is kept in a bath containing melting ice 0�C). a) Find the mass of the air in the container when thermal equilibrium is reached. b) The container is now placed in another bath containing boiling water 100�C). Find the mass of air in the container. c) The container is now closed and placed in the melting-ice bath. Find the pressure of the air when thermal equilibrium is reached. Use R=8.3 J K−1 mol−1R = 8.3 \text{ J K}^{-1} \text{ mol}^{-1}c) The container is now closed and placed in the melting-ice bath. Find the pressure of the air when thermal equilibrium is reached. Use R=8.3 J K−1 mol−1R = 8.3 \text{ J K}^{-1} \text{ mol}^{-1}

Step-by-Step Verified Solution

**Answer:** (a) m=0.0635 gm = 0.0635 \text{ g} (b) m=0.0465 gm = 0.0465 \text{ g} (c) P=73 kPaP = 73 \text{ kPa} **Step-by-Step Explanation:** (a) At 0�C: n=PVRT=105×50×10−68.3×273=2.21×10−3n = \frac{PV}{RT} = \frac{10^5 \times 50 \times 10^{-6}}{8.3 \times 273} = 2.21 \times 10^{-3} mol. Mass =2.21×10−3×28.8=0.0635= 2.21 \times 10^{-3} \times 28.8 = 0.0635 g. (b) At 100�C: n=105×50×10−68.3×373=1.61×10−3n = \frac{10^5 \times 50 \times 10^{-6}}{8.3 \times 373} = 1.61 \times 10^{-3} mol. Mass =0.0465= 0.0465 g. (c) Closed at 0�C: P=nRTV=1.61×10−3×8.3×27350×10−6=73P = \frac{nRT}{V} = \frac{1.61 \times 10^{-3} \times 8.3 \times 273}{50 \times 10^{-6}} = 73 kPa.

All Questions in this Chapter

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