At room temperature a diatomic gas is found to have an r.m.s. speed of 1930 ms ⁻¹. The gas is: (a) H2 (b) Cl2 (c) O2 (d) F2

Answer: (a). vrms = √(3RT/M) ⇒ M = (3RT)/(v²) Taking T = 300 K: M = (3 × 8.314 × 300)/((1930)²) = 74833.72 × 10⁶ M ≈ 2 × 10⁻³ kg/mol = 2 g/mol So the answer is (a) H₂.

At room temperature a diatomic gas is found to have an r.m.s. speed… | JEE Main 2014 (12 Apr, Shift 1)

Q86JEE Main 2014 · 12 APR · SHIFT 1Easy

Molecular Speeds & Maxwell Distribution

At room temperature a diatomic gas is found to have an r.m.s. speed of 1930 ms ${}^{-1}$ . The gas is:

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Answer: (a)

$v_{rms} = \sqrt{\frac{3RT}{M}} \Rightarrow M = \frac{3RT}{v^2}$
Taking $T = 300\text{ K}$ : $M = \frac{3 \times 8.314 \times 300}{(1930)^2} = \frac{7483}{3.72 \times 10^{6}}$
$M \approx 2 \times 10^{-3}\text{ kg/mol} = 2\text{ g/mol}$
So the answer is (a) $\text{H}_2$ .

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