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 teh container assuming them to be ideal (Universal gas constant `R=8.314J//mol-K`).

Answer

A, B, D

Explanation

The total pressure exerted by the mixture `P=10^5 Nm^-2`
`Temperature T=300K, Volume=0.02m^3`
Let there be x gram of Ne. Then mass of Ar will be `28-x.`
`Number of moles of Neon =x/20,`
`Number of moles of Argon=(28-x)/40`
Partial pressure due to Neon,
`p_1=((x//20)RT)/V`
Partial pressure due to Argon
`p_2=([(28-x)//40]RT)/V`
But according to Dalton's law of partial pressure
`P=p_1+p_2`
`10^5=(xRT)/(20V)+((28-x)RT)/(40V)`
`rArr(10^5xx40xx0.02)/(8.314xx300)=x+28 rArr x=4g`
`rArrMass of Neon=4g`
`:. Mass of Argon=24g`