If one mole of a monoatomic 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. `15/23`
B. `23/15`
C. `7/3`
D. `3/7`

Answer

B

Explanation

For a monoatomic gas,
`gamma_(m)=1+2/f=1+2/3=5/3`, where f = 3 and for a diatomic gas,
`gamma_(d)=1+2/f=1+2/5=7/5`, where f = 5
`thereforegamma_("mixture")` is the weighted average of `gamma`
`gamma_("mixture")=(n_(1)gamma_(m)+n_(2)gamma_(d))/(n_(1)+n_(2))`
where `n_(1) and n_(2)` are the number of moles of monoatomic and diatomic gases. In this case `n_(1)=n_(2)=1`
`thereforegamma_(m)=(1xx5/3+1+7/5)/(1+1)=(5/3+7/5)/2=46/30=23/15`