Question

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 [ Vᵣₘₛ (helium) / Vᵣₘₛ (argon) ]
is close to:
(A) 3.16 (B) 0.32
(C) 0.45 (D) 2.24
[JEE Main 2019]

Answer 1

(A) 3.16 according to root square mean velocity

Answer 2

The ratio of their rms speeds is close to (A) 3.16

Explanation:

Given :

The no. of moles of helium gas = 2

The no. of mole of argon gas = 1

Atomic mass of helium gas = 4 u

Atomic mass of argon gas = 40 u

The temperature of container = 300 K

From the formula of rms speed, ∴ $V_{rms} = \sqrt{\frac{3RT}{M} }$

⇒ $V_{rms}$ ∝ $\frac{1}{\sqrt{M} }$

We know that, $V_{rms}$ is inversely proportional to the square root of mass.

So from that we can write,

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

   $\frac{V_{rms(He)} }{V_{rms(Ar)} } = 3.16$

Thus, the ratio of their rms speeds is close to 3.16