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

Let mass of neon is `x g`, mass of argon `= (28 - x) g`
Number of moles of neon, `n_(1) = x//20`
Number of moles argon `n_(2) = (28 - x)//40`
`V = 0.5 m^(3), T = 27 + 273 = 300 K`
`P = p_(1) + p_(2) = (n_(1) + n_(2)) (R T)/(V)`
`10^(5) = ((x)/(20) + (28 - x)/(40)) ((8.3 xx 300)/(0.02))`
`= ((28 + x))/(40) xx (8.3 xx 300)/(0.02) = (28 + x) xx (3112.5)`
`28 + x = 32`
`x = 4 g, 28 - x = 24 g`
Mass of neon `= 4 g`
Mass of argon `= 24 g`