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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

A

`0.32`

B

`0.45`

C

`2.24`

D

`3.16`

Text Solution

Verified by Experts

The correct Answer is:
D
`V_("rm sAs")/V_("rm sAs") = sqrt((3RT)/(h_(He)))/sqrt((3RT)/(h_(Ar)))=sqrt((m_(Ar))/(m_(He))) = sqrt(40/4) = sqrt10 ~~3.16`
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