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

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Correct answer: A. 1

For a gas mixture, C_V= n_1 C_ V_1 +n_2 C_ V_2 n_1+n_2 and C_p= n_1 C_ p_1 +n_2 C_ p_2 n_1+n_2 For monoatomic gas, n=n_1, C_ V_1 = (3)/(2) R, C_ p_1 = (5)/(2) R For diatomic gas, n=n_2, C_ V_2 = (5)/(2) R and C_ p_2 = (7)/(2) R Now given for gas mixture, gathered (C_p)/(C_V)=1.5 So, 1.5=((n_1)/(n_2)((5)/(2))+(7)/(2))/((n_1)/(n_2)((3)/(2))+(5)/(2)) (9)/(2) ((n_1)/(n_2))+(15)/(2)=((n_1)/(n_2))5+7 (n_1)/(n_2)=1 gathered

A gas mixture contains n_1 moles of a monoatomic gas and n_2 moles...

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