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 is [2012]

Question

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 $(\frac{v_{rms} (helium)}{v_{rms} (argon)})$ is [2012]

Smart Achievers · Accepted Answer

(4)

$Using V_{rms} = \sqrt{\frac{3 R T}{M}} \Rightarrow V_{rms} \propto \frac{1}{\sqrt{M}}$

$\frac{V_{rms} (helium)}{V_{rms} (argon)} = \sqrt{\frac{M_{argon}}{M_{helium}}} = \sqrt{\frac{40}{4}} = \sqrt{10} \approx 3.16$

Solution

Q.
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 $(\frac{v_{rms} (helium)}{v_{rms} (argon)})$ is [2012]

1 0.32

2 0.45

3 2.24

4 3.16

Ans.
(4)

$Using V_{rms} = \sqrt{\frac{3 R T}{M}} \Rightarrow V_{rms} \propto \frac{1}{\sqrt{M}}$

$\frac{V_{rms} (helium)}{V_{rms} (argon)} = \sqrt{\frac{M_{argon}}{M_{helium}}} = \sqrt{\frac{40}{4}} = \sqrt{10} \approx 3.16$

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