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

Options

A. 1.25
B. `1.40`
C. `1.50`
D. `1.60`

Answer

C

Explanation

For the first gas, `gamma=5/3`. Hence it is a monoatomic gas having 3 degrees of freedom and `(C_(V))_(1)=3/2R`
For the second gas, `gamma=7/5`. It is a diatomic gas, having 5 degreed of freedom and `(C_(V))_(2)=5/2R`
`therefore` For the mixture,
`(C_(V))_("mixture")=(n_(1)C_(V_(1))+n_(2)C_(V_(2)))/(n_(1)+n_(2))`
`=((1xx3/2R)+(1xx5/2R))/(1+1)=2R`
`thereforeC_(P)=C_(V)+R=2+R=3R`
`thereforegamma=C_(P)/C_(V)=(3R)/(2R)=3/2=1.5`