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

Given `V =0.02m^(3),T = 300 K,P =1 xx 10^(5) Nm^(-2)`
`R =8.314 J`
Let a nad b g be mass of Ne and Ar respectively Thus `a + b =28`
Also total mole of Ne and `Ar = (a)/(20) + (b)/(40)`
Thus from `PV =nRT`
`:. 1 xx 10^(5) xx 0.02 = [(a)/(20) +(b)/(40)] xx 8.314 xx 300`
`2a + b = 32.0`
By eqns (i) and (ii)
` a = 4g, b =24g` .