The ratio C_p/C_v for a gas mixture consistin | vedclass

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(C) For helium $(He)$,molar mass $M_1 = 4 \ g/mol$. Number of moles $n_1 = 8 \ g / 4 \ g/mol = 2 \ mol$. Since $He$ is monoatomic,$C_{v1} = \frac{3}{2

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$27/17$

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(C) For helium $(He)$,molar mass $M_1 = 4 \ g/mol$. Number of moles $n_1 = 8 \ g / 4 \ g/mol = 2 \ mol$. Since $He$ is monoatomic,$C_{v1} = \frac{3}{2}R$.For oxygen $(O_2)$,molar mass $M_2 = 32 \ g/mol$. Number of moles $n_2 = 16 \ g / 32 \ g/mol = 0.5 \ mol$. Since $O_2$ is diatomic,$C_{v2} = \frac{5}{2}R$.The equivalent molar heat capacity at constant volume for the mixture is given by $C_{v,mix} = \frac{n_1 C_{v1} + n_2 C_{v2}}{n_1 + n_2}$.$C_{v,mix} = \frac{2 	imes (\frac{3}{2}R) + 0.5 	imes (\frac{5}{2}R)}{2 + 0.5} = \frac{3R + 1.25R}{2.5} = \frac{4.25R}{2.5} = 1.7R = \frac{17}{10}R$.Using the relation $C_{v,mix} = \frac{R}{\gamma_{mix} - 1}$,we have $\frac{17}{10}R = \frac{R}{\gamma_{mix} - 1}$.$\gamma_{mix} - 1 = \frac{10}{17} \Rightarrow \gamma_{mix} = 1 + \frac{10}{17} = \frac{27}{17}$.

The ratio $C_p/C_v$ for a gas mixture consisting of $8 \ g$ of helium and $16 \ g$ of oxygen is

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