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.

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

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.

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

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

Find the molar specific heat of mixture at constnat volume. It one mole of a monoatomic gas is mixed with three moles of a diatomic gas.