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

JEE Advanced · 2012 · 08 Apr 2012 · Paper 1 · Heat & Thermodynamics

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

(\frac{v _{r m s} ( helium )}{v _{r m s} ( argon )})

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

Answer: (D)

Solution

The rms speed is given by

v_{r m s} = \frac{3 R T}{M}

. The ratio is

\frac{v _{r m s} ( He )}{v _{r m s} ( Ar )} = \frac{M _{A r}}{M _{H e}} = \frac{40}{4} = 10 \approx 3.16

JEE Advanced · 2012 · 08 Apr 2012 · Paper 1 · Single correct option

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