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

(a) As for ideal gas `C_(P) - C_(V) = R`and `gamma = (C_(P)//C_(V))`,
So, `gamma-1 =(R )/(C_(V))`
or `C_(V) = (R )/((gamma-1))`
`(C_(V))_(1) = (R )/((5//3)-1) = (3)/(2)R`,
`(C_(V))_(2) = (R )/((7//5)-1) = (5)/(2)R`
and `(C_(V))_(mix) = (R )/((19//13)-1) = (13)/(6)R`
Now from conservation of energy, i.e., `DeltaU = DeltaU_(1) +DeltaU_(2)`,
`(n_(1)+n_(2)) (C_(V))_(mix) DeltaT = [n_(1)(C_(V))_(1) +n_(2) (C_(V))_(2)] DeltaT`
i.e., `(C_(V))_(mix) = (n_(1)(C_(V))_(1)+n_(2)(C_(V))_(2))/(n_(1)+n_(2))`
We have `(13)/(6)R = (1xx(3)/(2)R+nxx(5)/(2)R)/(1+n)=((3+5n))/(2(1+n))`
or, `13 + 13n = 9 +15n`,
(b) Molecular weight of the mixture will be given by
`M = (n_(A)M_(A)+n_(B)M_(B))/(n_(A)+n_(B)) = ((1)(4)+2(32))/(1+2)`
`M = 22.67`
Speed of sound in a gas is given by
`v = sqrt((gamma RT)/(M))`
Therefore, in the mixture of the gas
`v = sqrt(((19//13)(8.31)(300))/(22.67 xx 10^(-3)))m//s`
`v ~~ 401 m//s`
(c) `v prop sqrt(T)`
or `v = KT^(1//2).....(2)`
`rArr (dv)/(dT) = (1)/(2) KT^(-1//2)`
`rArr dv = K ((dT)/(2sqrt(T)))`
`rArr (dv)/(v) = (k)/(v) ((dT)/(2sqrt(T)))`
`rArr (dv)/(v) = (1)/(sqrt(T)) ((dT)/(2sqrt(T))) = (1)/(2) ((dT)/(T))`
`rArr (dv)/(v) xx 100 =(1)/(2) ((dT)/(T)) xx 100 = (1)/(2) ((1)/(300)) xx 100 = 0.167 = (1)/(6)`
Therefore, percentage change in speed is `0.167%`.