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» <html><body><p></p>Solution :Applying Grahan's law of effusion, <br/> `(r_(He))/(r_(CH_(4)))=sqrt((M_(CH_(4)))/(M_(He)))=sqrt((16)/(4))=sqrt(4)=2`,i.e., He diffuse <a href="https://interviewquestions.tuteehub.com/tag/twotimes-3240288" style="font-weight:bold;" target="_blank" title="Click to know more about TWOTIMES">TWOTIMES</a> faster than `CH_(4)`. <br/>As initially, the mixture <a href="https://interviewquestions.tuteehub.com/tag/contains-11473" style="font-weight:bold;" target="_blank" title="Click to know more about CONTAINS">CONTAINS</a> He and `CH_(4)` in the <a href="https://interviewquestions.tuteehub.com/tag/molar-562965" style="font-weight:bold;" target="_blank" title="Click to know more about MOLAR">MOLAR</a> <a href="https://interviewquestions.tuteehub.com/tag/ratio-13379" style="font-weight:bold;" target="_blank" title="Click to know more about RATIO">RATIO</a> of 4:1, therefore, molar ratio of `He: CH_(4)` effusing out initially =8:1.</body></html>