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

Consider a mixture of 2 mole helium and 4 mole of oxygen. Compute the

1.	Consider a mixture of 2 mole helium and 4 mole of oxygen. Compute the speed of sound in this gas mixture at 300 K.
Answer» Solution :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`