A gaseous mixture enclosed in a vessel consists of one gram mole of a gas A with `gamma=(5/3)` and some amount of gas B with `gamma=7/5` at a temperature T.
The gases A and B do not react with each other and are assumed to be ideal. Find the number of gram moles of the gas B if `gamma` for the gaseous mixture is `(19/13)`.

Explanation

Here, `n_(1) = 1 , gamma_(1)= 5//3, n_(2)=? Gamma_(2) = 7/5`
`gamma_(mix) = 19//13`
For gas A, `C_(upsilon) = (R)/(gamma -1) = (R)/((5//3-1)) = 3/2 R`
for gas B, `C_(upsilon)^(') = (R)/(gamma-1) = (R)/((7//5 -1)) = 5/2 R`
for mixture, `C_(upsilon)^('')=(R)/(gamma-1) = (R)/((19//13-1)) = 13/6 R`
As `C_(upsilon)^('') = (n_(1)xxC_(upsilon)+n_(2)C_(upsilon)^('))/(n_(1)+n_(2))`
`:. 13/6 R = (1 xx 3/2 R +n_(2) xx 5/2 R)/(1+n_(2))`
or, ` 13/6 = (3+5n_(2))/(2(1+n_(2)))`
`18 + 30 n_(2) = 26 +26 n_(2)`
`4n_(2)= 26 - 18 = 8, n_(2) =2`.