Consider a mixture of 2 mole helium and 4 mole of oxygen. Compute the speed of sound in this gas mixture at 300 K.

Explanation

Number of molecules of helium=2
Number of molecules of oxygen =4
When helium and oxygen are mixed, hence molecular weight of the mixture of gases is given by
`M_("mix") [(n_(1)M_(1)+n_(2)M_(2))/(n_(1)+n_(2))]`
`=[(2xx4+4xx32)/(2+4)] kg //"mol"`
`=(8+128)/(6)=(136)/(6)` kg/mol
`=22.6 xx10^(-3)` kg/ mol
In addition, helium is monoatomic,
`C_(v_(2))=(3R)/(2)`
OXygen is diatomic
`C_(v_(1))=(5R)/(2)`
`:.` For example mixture `(C_(v))_("mix")=(n_(1)C_(v_(1))+n_(2)C_(v_(2)))/(n_(1)+n+(2))`
`(C_(v))_("mix")=(2xx(3)/(2)R+4xx(5)/(2)R)/(2+4)=(13R)/(6)`
From Meyor's relation
`(C_(p))_("mix")=(C_(v))_("mix")+R`
`(C_(P))_("mix")=(13)/(6)+R=(13R+6R)/(6)=(19R)/(6)`
Ratio of specific heat capacitors of a mixture of gases is
`y_("mix")(C_(p))/(C_(v))=((19R)/(6))/((13R)/(6))=(19)/(13)`
According to Laplace, the speed of sound in a gas is
`v= sqrt((gamma RT)/(M))`
`v= sqrt((19)/(3)xx(8.31xx300)/(22.6xx10^(-3)))`
`=sqrt((28420.2)/(1768)xx10^(4))`
`=4.009xx10^(-2) m//s`
`=400.9 m/s`
`:.` The speed of sound `=400.9 m//s`