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 `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 xx 10^(5) N//m^(2)`
`T = 273 + 27 = 300 K, V = 0.02 m^(3)`
`:. (1.0 xx 10^(5)) xx 0.02 = [(m)/(20) + ((28 - m))/(40)] xx 8.314 xx 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`