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. 3
B. 4
C. 2
D. 5

Answer

C

Explanation

Let there be n gram moles of the gas B in the mixture for one gram mole of the gas A.
`becauseC_(P)-C_(V)=R`
`thereforeC_(P)/C_(V)-C_(V)/C_(V)=R/C_(V)`
`thereforegamma-1=R/C_(V)" "thereforeC_(V)=R/(gamma-1)`
For the gas A, `C_(V)=R/(5/3-1)=3/2R`
For the gas B, `C_(V)=R/(7/5-1)=5/2R`
and for the mixture, `C_(V)=R/(19/13-1)=13/6R`
`therefore` The molar specific heat of the mixture is given by
`C_(V)=(n_(A)(C_(V))_(A)+n_(B)(C_(V))_(B))/(n_(A)+n_(B))`
`therefore13/6R=(1xx3/2R+n(5/2R))/(1+n)=((3+5n)R)/(2(n+1))`
`therefore26(n+1)=6(3+5n)or13(n+1)=3(3+5n)`
`therefore13n+13=9+15n" "therefore2n=4orn=2`