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

`4.072, 23.926`

Explanation

Given, temperature of the mixture, `T = 27^@ C = 300 K`
`"Let"^m` be the mass of the neon gas in the mixture. Then, mass of argon would be (28 -m) Number of gram moles of neon, `n_1 = m/20`
Number of gram moles of argon, ` n_2 = ((28-m))/40`
From Dalton’s law of partial pressures.
Total pressure of the mixture (P)= Pressure due to neon `(P_1)` + Pressure due to argon `(P_2)`
or `P = P_1 + P_2 = (n_1 RT)/V + (n_2 RT)/V = (n_1 + n_2) (RT)/V`
Substituting the values `1.0 xx10^5 =(m/20 + (28-m)/40)((8.314 )(300))/(0.02)`
Solving this equation, we get
` m = 4.074 g` and `28 - m = 23.926 g`
Therefore, in the mixture ,4.074g neon is present and the rest
i.e., 23.926 g argon is present.