Calculate `gamma` of a gaseous mixture consisting of 3 moles of nitrogen and 2 moles of carbon dioxide.

Explanation

The `gamma` of a mixture is given by
`(n_(1) + n_(2))/(gamma -1) = (n_(1))/(gamma_(1) - 1) + (n_(2))/(gamma_(2) - 1)`
Here `n_(1) = 3, n_(2) = 3, gamma_(1) = 7//5` (as nitrogen is diatomic and `gamma_(2) = 4//3` (as carbon dioxide is triatomic)
Therefore, `(5)/(gamma -1) = (3)/((7)/(5) - 1) + (2)/((4)/(3) - 1) implies gamma = 1.37`