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A 4: 1 molar mixture of He and CH4 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><br/></p>Solution :Total <a href="https://interviewquestions.tuteehub.com/tag/pressure-1164240" style="font-weight:bold;" target="_blank" title="Click to know more about PRESSURE">PRESSURE</a> of the mixture = 20 bar Since He and `CH_4` are <a href="https://interviewquestions.tuteehub.com/tag/present-1163722" style="font-weight:bold;" target="_blank" title="Click to know more about PRESENT">PRESENT</a> in the molar <a href="https://interviewquestions.tuteehub.com/tag/ratio-13379" style="font-weight:bold;" target="_blank" title="Click to know more about RATIO">RATIO</a> <a href="https://interviewquestions.tuteehub.com/tag/4-311707" style="font-weight:bold;" target="_blank" title="Click to know more about 4">4</a>:1, we have <br/>partial pressure of Help `He(P_(He)) = 4/5xx 20 = 16` bar <br/>and partial pressure of `CH4(P_(Ch_4)) = 1/5 xx 20 = 4` bar <br/>If t is the time taken in effusion, rate of diffusion of <br/> `He(r_(He)) = (""^nHe)/t` <br/> and rate of diffusion of `CH_4(r_(CH_4)) =(""^nCH_4)/t` <br/>where `n""^Heand n""^CH_4` are respectively the number of moles of He and `CH_4` effused initially. <br/> According to Graham.s law of diffusion, for two gases at ifferent pressures, we have <br/> `r_(He)/r_(CH_4)=sqrt(M_(CH_4)/M_(He))xxP_1/P_2` <br/> `:. ""(""^nHe//t)/(""^nCH_4//t)=sqrt(16/4)xx16/4` <br/> or`(""^nHe)/(""^nCH_4)=8/1` <br/>Hence, the initially effused mixture contains the moles of helium and methane in the ratio 8: 1</body></html>


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