1.

Find the adiabatic exponent `gamma` for a mixture consisting of `v_1` moles of a monatomic gas and `v_2` moles of gas of rigid diatomic molecules.

Answer» Obviously
`(1)/(R ) C_V = (3)/(2) gamma_1 + (5)/(2) gamma_2`
(Since a monoatomic gas has `C_V = (3)/(2) R` and a diotomic gas has `C_V = (5)/(2) R`. [The diatomic molecule is rigid so no vibration])
`(1)/(R) C_p = (3)/(2) gamma_1 + (5)/(2) gamma_2 + gamma_1 + gamma_2`
Hence `gamma = (C_p)/(C_V) = (5 gamma_1 + 7 gamma_2)/(3 gamma_1 + 5 gamma_2)`.


Discussion

No Comment Found

Related InterviewSolutions