A gaseous mixture enclosed in a vessel of volume V consists of one mole of a gas A with `gamma=(C_p//C_v)=5//3` and another gas B with `gamma=7//5` at a certain temperature T. The relative molar masses of the gasses 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 processes.
(a) Find the number of moles of the gas B in the gaseous mixture.
(b) Compute the speed of sound in the gaseous mixture at `T=300K`.
(c) If T is raised by 1K from 300K, find the `%` change in the speed of sound in the gaseous mixture.
(d) The mixtrue is compressed adiabatically to `1//5` of its initial volume V. Find the change in its adaibatic compressibility in terms of the given quantities.

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`