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A 4.1 molar mixutre of He and CH(4) is c...

A 4.1 molar mixutre of He and `CH_(4)` is contained in a vessel at 20 bar pressure. Due to a hole in the vessel, the gas mixture leeks out. What is the composition of the mixutre effusing out initially ?

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Molar ratio of He and `CH_(4)` is 4:1
`therefore` Partial pressure ratio of He and `CH_(4)` is 16:4 (`because` total pressure =20 bar)
`(n_(He))/(n_(CH_(4)))=sqrt((M_(CH_(4)))/(M_(He)))xx(P_(He))/(P_(CH_(4)))` (`because` time of diffusion for both is same)
`=sqrt((16)/(4) xx(16)/(4))=8:1`
The composition of mixutre initially gone out for He and `CH_(4)` is 8:1
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