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(B) For a mixture of gases,the adiabatic index $\gamma_{	ext{mix}}$ is given by the formula:$\gamma_{	ext{mix}} = \frac{\frac{\mu_1 \gamma_1}{\gamma

Vedclass · Accepted Answer

$1.34$

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(B) For a mixture of gases,the adiabatic index $\gamma_{	ext{mix}}$ is given by the formula:$\gamma_{	ext{mix}} = \frac{\frac{\mu_1 \gamma_1}{\gamma_1 - 1} + \frac{\mu_2 \gamma_2}{\gamma_2 - 1}}{\frac{\mu_1}{\gamma_1 - 1} + \frac{\mu_2}{\gamma_2 - 1}}$Given: $\mu_1 = 3$ (for $CO_2$),$\gamma_1 = 1.3$,$\mu_2 = 2$ (for $O_2$),$\gamma_2 = 1.4$.Substituting the values:$\gamma_{	ext{mix}} = \frac{\frac{3 	imes 1.3}{1.3 - 1} + \frac{2 	imes 1.4}{1.4 - 1}}{\frac{3}{1.3 - 1} + \frac{2}{1.4 - 1}}$$\gamma_{	ext{mix}} = \frac{\frac{3.9}{0.3} + \frac{2.8}{0.4}}{\frac{3}{0.3} + \frac{2}{0.4}} = \frac{13 + 7}{10 + 5} = \frac{20}{15} = 1.333... \approx 1.34$.

Considering the gases to be ideal,the value of $\gamma = \frac{C_P}{C_V}$ for a gaseous mixture consisting of $3$ moles of carbon dioxide and $2$ moles of oxygen will be $(\gamma_{O_2} = 1.4, \gamma_{CO_2} = 1.3)$.

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