Verified Solution • 2025–26 Session moderate

QPages 34-37: Q8: HC Verma: Concepts of Physics - Volume 2 Kinetic Theory of Gases Pages 34-37: Q8 Solution: 2 g of hydrogen is sealed in a vessel of volume and is…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students frequently fail to convert the molar mass of hydrogen from grams per mole to kilograms per mole when using the mass-based ideal gas law, leading to a pressure error of a factor of 1000. Always ensure mass is in kilograms and molar mass is in kg/mol to maintain consistency with SI units for Joules.

2 g of hydrogen is sealed in a vessel of volume 0.02 m30.02 \text{ m}^3 and is maintained at 300 K. Calculate the pressure in the vessel. Use R=8.3 J K−1 mol−1R = 8.3 \text{ J K}^{-1} \text{ mol}^{-1}

Step-by-Step Verified Solution

**Answer:** P=1.23×105 PaP = 1.23 \times 10^5 \text{ Pa} **Step-by-Step Explanation:** m=2m = 2 g, M=2M = 2 g/mol, n=1n = 1 mol. V=0.02 m3V = 0.02 \text{ m}^3, T=300T = 300 K.
P=nRTV=1×8.3×3000.02=1.245×105 Pa≈1.23×105 PaP = \frac{nRT}{V} = \frac{1 \times 8.3 \times 300}{0.02} = 1.245 \times 10^5 \text{ Pa} \approx 1.23 \times 10^5 \text{ Pa}

All Questions in this Chapter

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