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`).

Answer

`M_(neon) = 4.074g M_(argon) = 23.926`g

Explanation

Let m = mass of neon gas then n =`((m)/(20) + (28-m)/(40))"from" PV = nRT`
`implies 10^(5) xx 0.2 = ((m)/(20) + (28-m)/(40)) xx 8.314 xx 300`
`m_(Ne) = 4.074 g , m_(Argon) = 23.926` g