A mixture of `n_(1)` moles of monoatomic gas and `n_(2)` moles of diatomic gas has `(C_(p))/(C_(V))=gamma=1.5`

Options

A. `n_(1)=2n_(2)`
B. `n_(1)=n_(2)`
C. `n_(2)=2n_(1)`
D. `n_(1)=3/2n_(2)`

Answer

B

Explanation

For a diatomic gas, `C_(P_(2))=7/2RandC_(V_(2))=5/2R`
For a monoatomic gas,
`C_(P_(1))=5/2RandC_(V_(1))=3/2R`
`gamma_("mix.")=(C_(P)/C_(V))_("mix.")`
`=(n_(1)C_(P_(1))+n_(2)C_(P_(2)))/(n_(1)+n_(2))xx(n_(1)+n_(2))/(n_(1)C_(V_(1))+n_(2)C_(V_(2)))`
`therefore3/2=(n_(1)xx5/2R+n_(2)(7/2R))/(n_(1)xx3/2R+n_(2)(5/2R))`
`therefore3[3/2n_(1)+5/2n_(2)]=2[5/2n_(1)+7/2n_(2)]`
`therefore9n_(1)+15n_(2)=10n_(1)+14n_(2)`
`thereforen_(2)=n_(1)`