A gaseous mixture consists of 16,g of helium | vedclass

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(D) The number of moles for helium $(He)$ is $n_1 = \frac{16\,g}{4\,g/mol} = 4\,mol$. Helium is a monoatomic gas,so $f_1 = 3$,$C_{v1} = \frac{3R}{2}$,

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$1.62$

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(D) The number of moles for helium $(He)$ is $n_1 = \frac{16\,g}{4\,g/mol} = 4\,mol$. Helium is a monoatomic gas,so $f_1 = 3$,$C_{v1} = \frac{3R}{2}$,and $C_{p1} = \frac{5R}{2}$.The number of moles for oxygen $(O_2)$ is $n_2 = \frac{16\,g}{32\,g/mol} = 0.5\,mol$. Oxygen is a diatomic gas,so $f_2 = 5$,$C_{v2} = \frac{5R}{2}$,and $C_{p2} = \frac{7R}{2}$.The ratio of specific heats for the mixture is given by $\gamma_{mix} = \frac{n_1 C_{p1} + n_2 C_{p2}}{n_1 C_{v1} + n_2 C_{v2}}$.Substituting the values:$\gamma_{mix} = \frac{4 	imes (\frac{5R}{2}) + 0.5 	imes (\frac{7R}{2})}{4 	imes (\frac{3R}{2}) + 0.5 	imes (\frac{5R}{2})}$$\gamma_{mix} = \frac{10R + 1.75R}{6R + 1.25R} = \frac{11.75R}{7.25R} = \frac{1175}{725} \approx 1.62$.

$A$ gaseous mixture consists of $16\,g$ of helium and $16\,g$ of oxygen. The ratio $(C_p/C_v)$ of the mixture is

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