One mole of a gas having gamma = frac{7}{5} i | vedclass

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(C) Given: $n_1 = 1 	ext{ mole}$,$n_2 = 1 	ext{ mole}$,$\gamma_1 = \frac{7}{5}$,$\gamma_2 = \frac{4}{3}$.For a mixture of gases,the equivalent adiab

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$\frac{15}{11}$

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(C) Given: $n_1 = 1 	ext{ mole}$,$n_2 = 1 	ext{ mole}$,$\gamma_1 = \frac{7}{5}$,$\gamma_2 = \frac{4}{3}$.For a mixture of gases,the equivalent adiabatic index $\gamma_{	ext{mix}}$ is given by the formula:$\frac{n_1 + n_2}{\gamma_{	ext{mix}} - 1} = \frac{n_1}{\gamma_1 - 1} + \frac{n_2}{\gamma_2 - 1}$Substituting the values:$\frac{1 + 1}{\gamma_{	ext{mix}} - 1} = \frac{1}{\frac{7}{5} - 1} + \frac{1}{\frac{4}{3} - 1}$$\frac{2}{\gamma_{	ext{mix}} - 1} = \frac{1}{\frac{2}{5}} + \frac{1}{\frac{1}{3}}$$\frac{2}{\gamma_{	ext{mix}} - 1} = \frac{5}{2} + 3 = \frac{5 + 6}{2} = \frac{11}{2}$$\gamma_{	ext{mix}} - 1 = \frac{4}{11}$$\gamma_{	ext{mix}} = 1 + \frac{4}{11} = \frac{15}{11}$.

One mole of a gas having $\gamma = \frac{7}{5}$ is mixed with one mole of a gas having $\gamma = \frac{4}{3}$. The value of $\gamma$ for the mixture is ($\gamma$ is the ratio of the specific heats of the gas).

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