A mixture of `n_1` moles of monoatomic gas and `n_2` moles of diatomic gas has `gamma=1.5`

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

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

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

A mixture of n_1 moles of diatomic gas and n_2 moles of monatomic gas has C_p / C_v = γ = 3/2, then

A mixture of n1 moles of diatomic gas and n2 moles of monatomic gas has Cp/Cv = γ = 3/2, then

A gas mixture contains n_1 moles of a monoatomic gas and n_2 moles of gas of rigid diatomic molecules. Each molecule in monoatomic and diatomic gas has 3 and 5 degrees of freedom respectively. If the adiabatic exponent (C_p)/(C_v) for this gas mixture is 1.5, then the ratio n_1/n_2 will be

Find gamma for a mixture of gases containing n_(1) moles of a monoatomic gas and n_(2) moles of a diatomic gas. Assuming diatomic molecules to be right.