A closed container of volume `0.02m^3`contains a mixture of neon and argon gases, at a temperature of `27^@C` and pressure of `1xx10^5Nm^-2`. The total mass of the mixture is 28g. If the molar masses of neon and argon are `20 and 40gmol^-1` respectively, find the masses of the individual gasses in the container assuming them to be ideal (Universal gas constant `R=8.314J//mol-K`).

Explanation

Here, `V= 0.02 m^(3)`
`T= 27^(@)C=(27+273)K =300K`
`P=10^(5) N//m^(2) = 28 g, M_(1) = 20 g , M_(2) = 40 g`.
Let m be mass of neon gas, therefore, mass of arogan gas = `(28-m) g`
Number of moles of neon , `n_(1) =(m)/(20)` ltbr. Number of moles of argon , `n_(2) =(28-m)/(40)`
From standard gas eqn, for the mixture,
`PC = (n_(1)+n_(2))RT`
`10^(5) xx 0.02 = ((m)/(20)+(28-m)/(40)) xx 8.314 (300)`
` 2 xx 10^(3) = ((2m+28-m)/(40)) xx 8.314 xx 3 xx 10^(2)`
On solving, we get `m=4.07 g`
This is the mass of neon.
Mass of argon = `28-m = 28-4.07 = 23.93 g`.