A mixture of ideal gases consists of 7 , kg o | vedclass

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(D) Number of moles of $N_2$: $n_{N_2} = \frac{7 \, kg}{28 \, kg/kmol} = 0.25 \, kmol$.Degrees of freedom for $N_2$: $f_{N_2} = \frac{2}{\gamma_{N_2}-

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Both $(A)$ and $(B)$.

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(D) Number of moles of $N_2$: $n_{N_2} = \frac{7 \, kg}{28 \, kg/kmol} = 0.25 \, kmol$.Degrees of freedom for $N_2$: $f_{N_2} = \frac{2}{\gamma_{N_2}-1} = \frac{2}{1.4-1} = 5$.Number of moles of $CO_2$: $n_{CO_2} = \frac{11 \, kg}{44 \, kg/kmol} = 0.25 \, kmol$.Degrees of freedom for $CO_2$: $f_{CO_2} = \frac{2}{\gamma_{CO_2}-1} = \frac{2}{1.3-1} = \frac{20}{3}$.Equivalent molecular weight $M_{mix} = \frac{m_1 + m_2}{n_1 + n_2} = \frac{7 + 11}{0.25 + 0.25} = \frac{18}{0.5} = 36 \, kg/kmol$.Equivalent degrees of freedom $f_{mix} = \frac{n_1 f_1 + n_2 f_2}{n_1 + n_2} = \frac{0.25(5) + 0.25(20/3)}{0.25 + 0.25} = \frac{5 + 20/3}{2} = \frac{35/3}{2} = \frac{35}{6}$.Equivalent adiabatic index $\gamma_{mix} = 1 + \frac{2}{f_{mix}} = 1 + \frac{2}{35/6} = 1 + \frac{12}{35} = \frac{47}{35}$.Thus,both $(A)$ and $(B)$ are correct.

$A$ mixture of ideal gases consists of $7 \, kg$ of nitrogen and $11 \, kg$ of $CO_2$. Then (Take $\gamma$ for nitrogen and $CO_2$ as $1.4$ and $1.3$ respectively):

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