A gaseous mixture consists of 16 g of helium | vedclass

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(B) Number of moles of Helium $(n_1)$ = $\frac{16 \ g}{4 \ g/mol} = 4 \ mol$. Helium is a monoatomic gas,so its degrees of freedom $f_1 = 3$.Number of

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$1.62$

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(B) Number of moles of Helium $(n_1)$ = $\frac{16 \ g}{4 \ g/mol} = 4 \ mol$. Helium is a monoatomic gas,so its degrees of freedom $f_1 = 3$.Number of moles of Oxygen $(n_2)$ = $\frac{16 \ g}{32 \ g/mol} = 0.5 \ mol$. Oxygen is a diatomic gas,so its degrees of freedom $f_2 = 5$.The molar heat capacity at constant volume for the mixture is $C_{v,mix} = \frac{n_1 C_{v1} + n_2 C_{v2}}{n_1 + n_2} = \frac{n_1 (\frac{f_1}{2}R) + n_2 (\frac{f_2}{2}R)}{n_1 + n_2}$.The molar heat capacity at constant pressure for the mixture is $C_{p,mix} = \frac{n_1 C_{p1} + n_2 C_{p2}}{n_1 + n_2} = \frac{n_1 (\frac{f_1+2}{2}R) + n_2 (\frac{f_2+2}{2}R)}{n_1 + n_2}$.Calculating $C_{v,mix} = \frac{4(\frac{3}{2}R) + 0.5(\frac{5}{2}R)}{4 + 0.5} = \frac{6R + 1.25R}{4.5} = \frac{7.25R}{4.5} = \frac{29R}{18}$.Calculating $C_{p,mix} = \frac{4(\frac{5}{2}R) + 0.5(\frac{7}{2}R)}{4 + 0.5} = \frac{10R + 1.75R}{4.5} = \frac{11.75R}{4.5} = \frac{47R}{18}$.The ratio $\gamma = \frac{C_{p,mix}}{C_{v,mix}} = \frac{47R/18}{29R/18} = \frac{47}{29} \approx 1.62$.

$A$ gaseous mixture consists of $16 \ g$ of helium and $16 \ g$ of oxygen. The ratio $\frac{C_p}{C_v}$ of the mixture is

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