Verified Solution • 2025–26 Session moderate

QPages 34-37: Q9: HC Verma: Concepts of Physics - Volume 2 Kinetic Theory of Gases Pages 34-37: Q9 Solution: The density of an ideal gas is at STP. Calculate the molecular…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students frequently fail to convert the given density from $\text{g cm}^{-3}$ to $\text{kg m}^{-3}$ before substituting into $M = \rho RT / P$, leading to a result that is off by a factor of $1000$.

The density of an ideal gas is 1.25×10−3 g cm−31.25 \times 10^{-3} \text{ g cm}^{-3} at STP. Calculate the molecular weight of the gas. Use R=8.31 J K−1 mol−1R = 8.31 \text{ J K}^{-1} \text{ mol}^{-1}

Step-by-Step Verified Solution

**Answer:** M=28 g/molM = 28 \text{ g/mol} **Step-by-Step Explanation:** ρ=1.25×10−3 g/cm3=1.25 kg/m3\rho = 1.25 \times 10^{-3} \text{ g/cm}^3 = 1.25 \text{ kg/m}^3. At STP, P=1.013×105P = 1.013 \times 10^5 Pa, T=273T = 273 K.
ρ=PMRT⇒M=ρRTP=1.25×8.31×2731.013×105=0.028 kg/mol=28 g/mol\rho = \frac{PM}{RT} \Rightarrow M = \frac{\rho RT}{P} = \frac{1.25 \times 8.31 \times 273}{1.013 \times 10^5} = 0.028 \text{ kg/mol} = 28 \text{ g/mol}

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