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.

A closed container of volume 0.02 m^(3) contains a mixture of neon and

1.	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» SOLUTION :Let m be the mass of the neon gas, then mass of argon will be 28 - m. NUMBER of moles of neon, `n_(1)=m/20` Number of moles of argon, `n_(2)=(28-m)/40` Now by IDEAL gas equation, we have PV = nRT Here `n=n_(1)+n_(2)=m/20+(28-m)/40, P=1.0 times 10^(5)N//m^(2)` `T=273+27=300K, V=0.02 m^(3)` `therefore (1.0 times 10^(5)) times 0.02=[m/20+((28-m))/40] times 8.314 times 300` After solving , we get m = 4.07 g , `""` Mass of neon gas = 4.07 g Mass of argon gas = 28 - m = 28 - 4.07 = 23.93 g