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A closed container of volume 0.02 m^(3) contains a mixture of neon and argon gases at 27^(@)C temperature and 1.0 times 10^(5)N/m^(2) pressure. If the gram-molecular weight of neon and argon are 20 and 40 respectively, find the masses of the individual gases in the container. Assuming them to be ideal (R = 8.314 J/mol-K). Total mass of the mixture is 28 g.

Answer» <html><body><p></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :Let m be the mass of the neon gas, then mass of argon will be 28 - m. <br/> <a href="https://interviewquestions.tuteehub.com/tag/number-582134" style="font-weight:bold;" target="_blank" title="Click to know more about NUMBER">NUMBER</a> of moles of neon, `n_(1)=m/20` <br/> Number of moles of argon, `n_(2)=(28-m)/40` <br/> Now by <a href="https://interviewquestions.tuteehub.com/tag/ideal-1035490" style="font-weight:bold;" target="_blank" title="Click to know more about IDEAL">IDEAL</a> gas equation, we have PV = nRT <br/> Here `n=n_(1)+n_(2)=m/20+(28-m)/40, P=1.0 times <a href="https://interviewquestions.tuteehub.com/tag/10-261113" style="font-weight:bold;" target="_blank" title="Click to know more about 10">10</a>^(5)N//m^(2)` <br/> `T=273+27=300K, V=0.02 m^(3)` <br/> `therefore (1.0 times 10^(5)) times 0.02=[m/20+((28-m))/40] times 8.314 times 300` <br/> After solving , we get m = 4.07 g , `""` Mass of neon gas = 4.07 g <br/> Mass of argon gas = 28 - m = 28 - 4.07 = 23.93 g</body></html>


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