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

`4g ;24g`.

Explanation

Here, `V= 0.02m^(3), T=27^(@)C = (27+273)K`
`=300K`
`m_(N)+m_(A)=28g` ...(i)
`n_(N) = (m_N)/(20)` and `n_(A) = (m_A)/(40)`
As `PV = nRT = (n_N+n_A)RT`
`:.n_(N)+n_(A) = (PV)/(RT) = (10^(5)xx0.02)/(8.314 xx 300) =0.8`
`(m_N)/(20)+(m_A)/(40) = 0.8`
`:. 2m_(N)+m_(A) = 32` ...(ii)
Solving (i) and (ii) , we get
`m_(N) = 4g` and `m_(A) = 24g`.