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A closed container of volume 0.02 m^(2) ...

A closed container of volume 0.02 `m^(2)` contains a mixture of neon and argon gases at `27^(@)C` temperature and `1.0 xx 10^(5) N//m^(2)` pressure. The gram-molecular weights of neon and argon are 20 and 40, respectively. Find the masses of the individual gases in the container. Assuming them to be ideal `(R = 8.314 J//mol-K)`. Total mass of the mixture is 28 g.

Text Solution

Verified by Experts

If the mass of neon is .m., the mass of argon will be (28 - m),
so `n_(Ne)=m/20andn_(Ar)=((28-m))/40`
`thereforen=n_(Ne)+n_(Ar)=m/20+((28-m))/40=(28+m)/40" "...(1)`
`n=(PV)/(RT)=(1xx10^(5)xx0.02)/(8.314xx300)=0.8" "...(2)`
So from equations (1) and (2),
`(28+m)//40 = 0.8,i.e.,m=4gm`
so `m_(Ne)=4gmand m_(Ar)=28-4=24gm`
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