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

here, `V= 0.02 m^(3)`,
`T=27^(@) C = (27 +273)K = 300 K`
`P=10^(5) N//m^(2), M_(1)= 20, M_(2) = 40`
If `m_(1), m_(2)` are masses of neon and argon respectively , then
`m_(1)+m_(2) = 29 gram` ...`(i)`
If `mu_(1).mu_(2)` are number of moles of noen and argon then `mu_(1)=(m_1)/(20) , mu_(2) = (m_2)/(40)`
According to perfect gas equation
`Pv = (mu_(1) +mu_(2))RT`
`mu_(1)+mu_(2) = (PV)/(RT) = (10^(5)xx0.02)/(8.31 xx 300)`
`(m_1)/(20) +(m_2)/(40) = 0.80`
`2 m_(1)+m_(2) = 32` ...(ii)
Subtact (i) from (ii)
`m_(1) = 4 g` from (i) `m_(2) = 28 -4 =24g`.