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.

Answer

`2`mole

Explanation

`gamma_("mtx") = (n_(A)C_(PA) + n_(B)C_(PB))/(n_(A)C_(VA) + n_(B)C_(VD))`
`(19)/(13) = ((1)((5)/(2)R) + n_(B)((7)/(2)R))/((1)((3)/(2)R) + n_(B)((5)/(2)R)) implies n_(B) = 2 "mole"`