Verified Solution • 2025–26 Session moderate

QPages 34-37: Q13: HC Verma: Concepts of Physics - Volume 2 Kinetic Theory of Gases Pages 34-37: Q13 Solution: A sample of 0.177 g of an ideal gas occupies at STP. Calculate…

⚠️ Common Exam Traps & Conceptual Pitfalls

When using $v_{ ext{rms}}=sqrt{dfrac{3RT}{M}}$, the molar mass $M$ must be in kilograms per mole. If you leave $M$ in grams per mole, the result will be too large by a factor of $sqrt{1000}approx31.6$. Always convert $M$ to kg mol⁻¹ before computing the rms speed.

A sample of 0.177 g of an ideal gas occupies 1000 cm31000 \text{ cm}^3 at STP. Calculate the rms speed of the gas molecules.

Step-by-Step Verified Solution

**Answer:** vrms=1302 m/sv_{rms} = 1302 \text{ m/s} **Step-by-Step Explanation:** m=0.177m = 0.177 g, V=1000 cm3=10−3 m3V = 1000 \text{ cm}^3 = 10^{-3} \text{ m}^3, T=273T = 273 K, P=1.013×105P = 1.013 \times 10^5 Pa.
n=PVRT=1.013×105×10−38.314×273=0.0446 moln = \frac{PV}{RT} = \frac{1.013 \times 10^5 \times 10^{-3}}{8.314 \times 273} = 0.0446 \text{ mol}
M=m/n=0.177/0.0446=3.97M = m/n = 0.177/0.0446 = 3.97 g/mol ≈4\approx 4 g/mol (helium).
vrms=3RTM=3×8.314×2734×10−3=1302 m/sv_{rms} = \sqrt{\frac{3RT}{M}} = \sqrt{\frac{3 \times 8.314 \times 273}{4 \times 10^{-3}}} = 1302 \text{ m/s}

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