A mixture of 2 moles of helium gas ( (at...

A mixture of 2 moles of helium gas (` (atomic mass)=4a.m.u` ) and 1 mole of argon gas ( `(atomic mass)=40a.m.u` ) is kept at 300K in a container. The ratio of the rms speeds `((v_(rms)(helium))/((v_(rms)(argon))` is

`0.32`

`0.45`

`2.24`

`3.16`

Text Solution

Verified by Experts

The correct Answer is:

`(V_(RmS_(He)))/(V_(RmS_(Ar)))=sqrt((3RT)/(m_(He)))/sqrt((3RT)/(m_(Ar)))`
`= sqrt((m_(Ar))/(m_(He))) = sqrt((40)/(4)) = sqrt(10)~~3.16`

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