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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

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If the mass of neon is .m., the mass of argon wiU be (28- m) ,
so `n_(Ne) = (m)/(20) and n_(Ar) = ((28 - m))/(40)`
`therefore n = n_(Ne) + n_(Ar) = (m)/(20) + ((28 - m))/(40) = (28 + m)/(40)` ..... (1)
n = `(PV)/(RT) = (1 xx 10^(5) xx 0.02 )/(8.314 xx 300) = 0.8 `... (2)
So from equation (1) and (2),
`(28 + m )`40 = 0.8 , i.e., m = 4 gm
so `m_(Ne) = `4 gm and `m_(Ar)` = 28 - 4 = 24 gm
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