JEE Main 2014 · 12 Apr · Shift 1 · Physics Q13

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PhysicsSingle correct+4 / -1Question 13 of 90

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

\ce{H2}

\ce{Cl2}

\ce{O2}

\ce{F2}

Answer and worked solution

Official answer: A

$v_{rms}=\sqrt{\dfrac{3RT}M}$ with $T\approx300$ K:
$M=\frac{3\times8.314\times300}{1930^2}\approx2.0\times10^{-3}\text{ kg/mol},$
which is $\ce{H2}$ .

Topics:Molecular speeds

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JEE Main 2014 · 12 Apr · Shift 1, Physics question 13