If one mole of a mono-atomic gas `(gamma=5//3)` is mixed with one mole of a diatomic gas `(gamma=7//5)`, the value of `gamma` for the mixture is :

Options

A. `1.40`
B. `1.50`
C. `1.53`
D. `3.07`

Answer

B

Explanation

`gamma_(1)=5//3` means gas is monatomic or `C_(v_(1))=(3)/(2)R`
`gamma_(2)=7//5` means gas is diatomic or `C_(v_(2))=5//2R`
`C_(v)` (of the mixture)
`=(n_(1)C_(v_(1))+n_(2)C_(v_(2)))/(n_(1)+n_(2))=((1)((3)/(2)R)+(1)((5)/(2))R)/(1+1)=2R`
`C_(p)` (of the mixture) `=C_(v)+R=3R`
`gamma_("mixture")=(C_(p))/(C_(v))=(3R)/(2R)=1.5`