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)`.

Options

A. 2
B. 12
C. 16
D. 8

Answer

A

Explanation

Let the mixture contain `eta` moles of gas B
As `C_(P) - C_(v) = R` and `lambda = C_(P)/C_(v)`
`therefore C_(V) =R/(lambda-1)`
For gas A, `C_(V) = R/(5/3 -1) = 3/2 R`
For gas B, `C_(V) = R/(7/2-1) = 5/2 R`
For the mixture, `C_(V) = R/(19/12-1) = 13/6 R`
By conservation of energy, molar specific heat of the mixture is
`C_(V) = (n_(A)(C_(V))_(A) + n_(B)(C_(V))_(B))/(n_(A) + n_(B))`
`therefore 13/R = (1 xx 3/2 R + n xx 5/2 R)/(1 + n) = (3+5n)R/(2(1+n))`
`therefore n =2`