A gaseous mixture enclosed in a vessel of volume `V` consists of one gram mole of gas `A` with `gamma = (C_(P))/(C_(V)) = (5)/(3)`an another gas `B` with `gamma = (7)/(5)` at a certain temperature `T`. The gram molecular weights of the gases `A` and `B` are `4` and `32` respectively. The gases `A` and `B` do not react with each other and are assumed to be ideal. The gaseous mixture follows the equation `PV^(19//13)` = constant , in adiabatic process. Find the number of gram moles of the gas `B` in the gaseous mixture.

Options

A. 2
B. 3
C. 4
D. 5

Answer

A

Explanation

Let the number of moles of gas B=n The number of moles of gas A=1
Since `U=(nRT)/(gamma-1)` For mixture, `U=U_(m),gamma=(19)/(13)`
`therefore U_(m)=U_(A)+U_(B)`
`therefore (n_(A)+n_(B)RT)/(gamma_(m)-1)=(n_(A)RT)/(gamma_(A)-1)+(n_(B)RT)/(gamma_(B)-1)`
or `(1+n)/((19)/(13)-1)=(1)/((5)/(3)-1)+(n)/((7)/(5)-1)`
or `(13(1+n))/(6)=(3)/(2)+(5n)/(2)or 13+13n=9+15n`
or 4=2n or n=2