A mixture of 2 moles of helium gas (atomic mass = 4 u), and 1 mole of argon gas (atomic mass = 40 u) is kept at 300 K in a container. The ratio of their rms speeds [ V_rms\,(helium) \over V_rms\,(\arg on) ], is close to :

Heat and Thermodynamics PYQ — JEE Main 2019 · 9 Jan · Shift 1 · Q72 with Solution

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Correct answer: A

For any ideal gas,

V_{\text{rms}} = \sqrt{\frac{3RT}{M}}

where $M$ is the molar mass.

So, at the same temperature,

V_{\text{rms}} \propto \frac{1}{\sqrt{M}}

\frac{V_{\text{rms}}(\text{He})}{V_{\text{rms}}(\text{Ar})} = \sqrt{\frac{M_{\text{Ar}}}{M_{\text{He}}}}

Helium: $M_{\text{He}} = 4\,\text{u}$

Argon: $M_{\text{Ar}} = 40\,\text{u}$

Thus,

\frac{V_{\text{rms}}(\text{He})}{V_{\text{rms}}(\text{Ar})} = \sqrt{\frac{40}{4}} = \sqrt{10}

\sqrt{10} \approx 3.16

Therefore, the correct option is:

\boxed{\text{A: } 3.16}

Stored correct answer: A

Our derived answer: A

So, they agree.