A 4:1 molar mixture of He and CH_(4) is is contained in a vessel at 20 bar pressure. Due to a hole in the vessel, the gas mixture leaks out. What is the composition of the mixture effusing out initially ?

A 4:1 molar mixture of He and CH_(4) is is contained in a vessel at 20

1.	A 4:1 molar mixture of He and CH_(4) is is contained in a vessel at 20 bar pressure. Due to a hole in the vessel, the gas mixture leaks out. What is the composition of the mixture effusing out initially ?
Answer» Solution :Applying Grahan's law of effusion, `(r_(He))/(r_(CH_(4)))=sqrt((M_(CH_(4)))/(M_(He)))=sqrt((16)/(4))=sqrt(4)=2`,i.e., He diffuse TWOTIMES faster than `CH_(4)`. As initially, the mixture CONTAINS He and `CH_(4)` in the MOLAR RATIO of 4:1, therefore, molar ratio of `He: CH_(4)` effusing out initially =8:1.