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)`.

Answer

B

Explanation

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