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A gaseous mixture enclosed in a vessel c...

A gaseous mixture enclosed in a vessel consists of one gram mole of a gas A with `gamma=(5/3)` and some amount of gas B with `gamma=7/5` at a temperature T.
The gases A and B do not react with each other and are assumed to be ideal. Find the number of gram moles of the gas B if `gamma` for the gaseous mixture is `(19/13)`.

Text Solution

Verified by Experts

As for ideal gas `C_(P) = C_(V) = R` and `gamma = (C_(P)//C_(v))`,
`gamma - 1 = (R )/(C_(V))` or `C_(V) = (R )/((gamma - 1))`
`(C_(V))_(1) = (R )/((5//3) - 1) = (3)/(2) R`
`(C_(V))_(2) = (R )/((7//5) - 1) = (5 R)/(2) `
and `(C_(V))_(mix) = (R )/((19//13) - 1) = (13)/(6) R`
Now form the conservation of energy, i.e., `Delta U = DeltaU_(1) + Delta_(2)`, we get
`(mu_(1) + mu_(2)) (C_(V))_(mix) Delta T = [mu_(1) (C_(V))_(1) + mu_(2) (C_(V))_(2)] Delta T`
`(C_(V))_(mix) = (mu_(1) (C_(V))_(1) + mu_(2) (C_(V))_(2))/(mu_(1) + mu_(2))`
`(13)/(6) mu_(2) = (1 xx (3)/(2) R + mu_(2) xx (5)/(2) R)/(1 + mu_(2)) = ((3 + 5 mu_(2)) R)/(2(1 + mu_(2))`
or `13 + 13 mu_(2) = 9 + 15 mu_(2)`, i.e., `mu_(2) = 2 g "mole"`
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