A gas mixture contains n_1 moles of a monoato | vedclass

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(A) For a gas mixture, the molar heat capacities at constant volume and constant pressure are given by:$C_V = \frac{n_1 C_{V_1} + n_2 C_{V_2}}{n_1 + n

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(A) For a gas mixture, the molar heat capacities at constant volume and constant pressure are given by:$C_V = \frac{n_1 C_{V_1} + n_2 C_{V_2}}{n_1 + n_2}$ and $C_p = \frac{n_1 C_{p_1} + n_2 C_{p_2}}{n_1 + n_2}$For a monoatomic gas, the degrees of freedom $f_1 = 3$, so $C_{V_1} = \frac{3}{2}R$ and $C_{p_1} = \frac{5}{2}R$.For a rigid diatomic gas, the degrees of freedom $f_2 = 5$, so $C_{V_2} = \frac{5}{2}R$ and $C_{p_2} = \frac{7}{2}R$.The adiabatic exponent $\gamma = \frac{C_p}{C_V} = 1.5 = \frac{3}{2}$.Substituting the values:$\frac{C_p}{C_V} = \frac{n_1(\frac{5}{2}R) + n_2(\frac{7}{2}R)}{n_1(\frac{3}{2}R) + n_2(\frac{5}{2}R)} = \frac{5n_1 + 7n_2}{3n_1 + 5n_2} = \frac{3}{2}$Cross-multiplying gives:$2(5n_1 + 7n_2) = 3(3n_1 + 5n_2)$$10n_1 + 14n_2 = 9n_1 + 15n_2$$n_1 = n_2$Therefore, the ratio $\frac{n_1}{n_2} = 1$.

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