IIT-JEE 2012 · Paper 1 · Physics Q8

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A mixture of 2 moles of helium gas (atomic mass = 4 amu) and 1 mole of argon gas (atomic mass = 40 amu) is kept at 300 K in a container. The ratio of the rms speeds $\left(\dfrac{v_{rms}(\text{helium})}{v_{rms}(\text{argon})}\right)$ is

0.32

0.45

2.24

3.16

Answer and worked solution

Official answer: D

rms speed. $v_\text{rms}=\sqrt{3RT/M}$ , the same $T$ for both gases (the amounts do not matter):
$\frac{v_\text{He}}{v_\text{Ar}}=\sqrt{\frac{40}4}=\sqrt{10}\approx3.16.$

(D)

Topics:Molecular speeds

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IIT-JEE 2012 · Paper 1, Physics question 8