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`).

Options

A. `4g,24g`
B. `8g,20g`
C. `12,16g`
D. `6g,22g`

Answer

A

Explanation

Let `m_(2)m_(2)` be masses of Neon and Argon respectively `m_(1)+m_(2)=28`
`PV=nRT PV =(m_(1)/(m_(1))+(m_(2))/(m_(2))) RT (or)`
`PV=((m_(1))/(20)+m_(2)/(40)) RT`
`m_(1)/(20)+m_(2)/40= (1xx10^(5)0.02)/8.314xx300implies2m_(1)=32`
`m_(1)+m_(2)=28`