Verified Solution • 2025–26 Session moderate

QPages 34-37: Q29: HC Verma: Concepts of Physics - Volume 2 Kinetic Theory of Gases Pages 34-37: Q29 Solution: A vessel contains 1.60 g of oxygen and 2.80 g of nitrogen. The…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students frequently fail to convert the given masses from grams to kilograms or forget that molar mass in $\text{kg/mol}$ is required when using SI units for $R$, leading to a pressure error by a factor of 1000. Always ensure mass is in kg or molar mass is in kg/mol to match the Joule definition in $R$.

A vessel contains 1.60 g of oxygen and 2.80 g of nitrogen. The temperature is maintained at 300 K and the volume of the vessel is 0.166 m30.166 \text{ m}^3. Find the pressure of the mixture. 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=2250 PaP = 2250 \text{ Pa} **Step-by-Step Explanation:** nO2=1.60/32=0.05n_{O_2} = 1.60/32 = 0.05 mol, nN2=2.80/28=0.1n_{N_2} = 2.80/28 = 0.1 mol.
P=(nO2+nN2)RTV=0.15×8.3×3000.166=2250 PaP = \frac{(n_{O_2} + n_{N_2})RT}{V} = \frac{0.15 \times 8.3 \times 300}{0.166} = 2250 \text{ Pa}

All Questions in this Chapter

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