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A closed container of volume 0.02m^3cont...

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`).

Text Solution

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Let m gram be the mass of neon. Then, the mass of argon is `(28-m)g`.
Total number of moles of the mixture, `mu = m/20 + (28-m)/40 = (28+m)/40 …. (i)`
Now, `mu` = ` (PV)/(RT) =(1xx10^(5)xx 0.02)/(8.314 xx 300) = 0.8 …. (ii)`
By (i) and (ii) , `(28+m)/40 = 0.8 implies 28+m = 32 implies m = 4` gram or mass of argon =`(28-4)g = 24g`
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