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.

Options

A. `-0.025V/Tp_(a)^(-1)`
B. `-0.050 V/T p_(a)^(-1)`
C. `-0.075V/Tp_(a)^(-1)`
D. `-0.100V/Tp_(a)^(-1)`

Answer

A

Explanation

`:.` Bulk Modulus `B=gamma P`
`:.` Compressibiity `K=1/B=1/(gammaP)`
`:. K_(1)=1/(gammaP_(2))-1/(gammaP_(1))=1/(gamma)(1/(P_(2))-1/(P_(1)))`
Since the process is adiabatic
`P_(2)V_(2)^(gamma)=P_(1)V_(2)^(gamma)`
`:.P_(2)=P_(1)((V_(1))/(V_(2)))^(gamma)=P_(1)((V_(1))/(V_(1)//5))^(gamma)=P_(1)5^(gamma)`
`:.DeltaK=1/(gamma)(1/(P_(1)5^(gamma))-1/(P_(1)))=1/(gammaP_(1))(1/(5^(gamma))-1)`
`P_(1)=((n_(A)+n_(g))RT)/V=((1+2)xx8.31xxT)/V=(24.93T)/V`
`impliesDeltaK=1/(19/13xx24.93xxT/V)(1/(5^(19/13))-1)=-[0.025V/T]Pa^(-1)`