A gaseous mixture enclosed in a vessel o...

A gaseous mixture enclosed in a vessel of volume `V` consists of one gram mole of a gas `A` with `gamm = (C_(p))/(C_(v)) = (5)/(3)` and 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 processes.
(a) Find the number of gram 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 percentage change in the speed of sound in the gaseous mixture.

1mole

3 moles

4 moles

2 moles

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The correct Answer is:

`C_(p)-C_(v)=R c_(V_(A))=R/(5/3-1)=3/2R`
`(C_(p))/(C_(v))=gamma . G_(P_(A))=5/2R`
`C_(v)=R/(gamma-1) C_(v_(B))=R/(7/5-1)=(5R)/2`
`C_(P)=(gamma-1) C_(v_(B))=R/(7/5-1)=(5R)/2`
`C_(P)=(gamma R)/(gamma-1)C_(p_(B))=(7R)/2`
`19/13=((1xxc_(p_(A))+n_(B)Cc_(p_(B)))/(1+n_(B)))/((1xxc_(v_(A))+n_(B)c_(v_(B)))/(1+n_(B)))`
`=L> n_(B)=2` moles

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