A mixture of 2 moles of helium gas (atomic mass $=4 \mathrm{u}$), and 1 mole of argon gas (atomic mass $=40 \mathrm{u}$) is kept at 300 K in a container. The ratio of their rms speeds $\left[\frac{\mathrm{V}_{\text {rms }} \text { (helium) }}{\mathrm{V}_{\text {rms }} \text { (argon) }}\right]$, is close to:

JEE Main · 2019 · 09 Jan 2019 · Shift 1 · Heat & Thermodynamics

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

[\frac{V _{rms} (helium)}{V _{rms} (argon)}]

, is close to:

(A)3.16✓ Correct
(B)0.32
(C)0.45
(D)2.24

Answer: (A)

Solution

Root-mean-square speed depends only on temperature and molar mass,

v_{rms} = \frac{3 R T}{M}

, so at the common

300

K the mole numbers drop out and only the masses matter. The ratio is

\frac{v _{rms} ( He )}{v _{rms} ( Ar )} = \frac{M _{Ar}}{M _{He}} = \frac{40}{4} = 10 \approx 3.16

JEE Main · 2019 · 09 Jan 2019 · Shift 1 · Single correct option

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