A gaseous mixture enclosed in a vessel of volume `V` consists of one 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 moles of the gas `B` in the gaseous mixture.

Explanation

(i) `(n_A+n_B)(RT)/(gamma_m-1)=n_A(RT)/(gamma_A-1)+n_B(Rt)/(gamma_B-1)[therefore gamma_m=19//13]`
`implies n_B=2` (ii) `V=sqrt((gamma_mRT)/(M_m))`
`implies M_m=(n_AM_A+n_B+M_b)/(n_A+n_B)=22.67 and V_m 19/13, T=300k`
`V=401 m//s` (iii) Percentage change of velocity of sound is given by
`(triangleV)/(V)xx100=1/2 (triangleT)//Txx100=0.167%` (iv) Compressibility (i)`=(1)/("Bulk modulus")`,let initial is p then
`k=(1)/(gamma_mP)`
And `k'=(1)/(gamma P')`

`therefore Since Pv'=P'(v//5)^(gamma)implies P'=P(5)^(gamma)`
`implies trianglek=k-k'=(1)/(gamma_mP)[1-(1/5)^(gamma)]`
`implies `trianglek=(V)/(gamma_m(n_A+n_B)RT)[1-(1/5)^(gamma)]=(v)/(19/13xx3xxRT)[1-(1/5)^(19//13)]`
`triangle k=8.27xx10^(-5)V`