A gaseous mixture consists of 16g of helium and 16g of oxygen the ratio of two specific heats of the mixture is

Explanation

Number of moles of helium `(n)=16/4=4,` Number of moles of oxygen `(n^(.))=16/32=1/2`
For For the monoatomic helium gas, degrees of freedom, f=3
So `C_(V)=f/2R=3/2R`
`C_(V)("mixture")=(nC_(V)+n^(.)C_(V))/(n+n^(.))=(4xx3/2R+1/2xx5/2R)/((4+1/2))=29/18R therefore C_(P)-C_(V)=R`
`gamma("mixture")=C_(P)/C_(V)=1+R/(C_(V)("mixture"))=1+R/((29/18R))=1+18/29=47/29`
`gamma("mixture")=1.62`