If two moles of diatomic gas and one mole of | vedclass

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(C) For monoatomic gas: $n_1 = 1$,$\gamma_1 = \frac{5}{3}$.For diatomic gas: $n_2 = 2$,$\gamma_2 = \frac{7}{5}$.The formula for the adiabatic index of

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$\frac{19}{13}$

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(C) For monoatomic gas: $n_1 = 1$,$\gamma_1 = \frac{5}{3}$.For diatomic gas: $n_2 = 2$,$\gamma_2 = \frac{7}{5}$.The formula for the adiabatic index of a mixture is given by $\gamma_{mix} = \frac{n_1 C_{p1} + n_2 C_{p2}}{n_1 C_{v1} + n_2 C_{v2}}$.Using $C_v = \frac{R}{\gamma - 1}$ and $C_p = \frac{\gamma R}{\gamma - 1}$,we have:$\gamma_{mix} = \frac{\frac{n_1 \gamma_1}{\gamma_1 - 1} + \frac{n_2 \gamma_2}{\gamma_2 - 1}}{\frac{n_1}{\gamma_1 - 1} + \frac{n_2}{\gamma_2 - 1}}$.Substituting the values:$\gamma_{mix} = \frac{\frac{1 	imes (5/3)}{(5/3) - 1} + \frac{2 	imes (7/5)}{(7/5) - 1}}{\frac{1}{(5/3) - 1} + \frac{2}{(7/5) - 1}} = \frac{\frac{5/3}{2/3} + \frac{14/5}{2/5}}{\frac{1}{2/3} + \frac{2}{2/5}} = \frac{2.5 + 7}{1.5 + 5} = \frac{9.5}{6.5} = \frac{19}{13}$.

If two moles of diatomic gas and one mole of monoatomic gas are mixed,then the ratio of specific heats $\gamma$ for the mixture is:

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