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

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(D) Number of moles of Helium $(n_1)$ = $\frac{16 \, g}{4 \, g/mol} = 4 \, mol$. Helium is monoatomic,so $f_1 = 3$,$C_{v1} = \frac{3}{2}R$,$C_{p1} = \

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$1.62$

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(D) Number of moles of Helium $(n_1)$ = $\frac{16 \, g}{4 \, g/mol} = 4 \, mol$. Helium is monoatomic,so $f_1 = 3$,$C_{v1} = \frac{3}{2}R$,$C_{p1} = \frac{5}{2}R$.Number of moles of Oxygen $(n_2)$ = $\frac{16 \, g}{32 \, g/mol} = 0.5 \, mol$. Oxygen is diatomic,so $f_2 = 5$,$C_{v2} = \frac{5}{2}R$,$C_{p2} = \frac{7}{2}R$.The equivalent $C_v$ of the mixture is $C_{v,mix} = \frac{n_1 C_{v1} + n_2 C_{v2}}{n_1 + n_2} = \frac{4(\frac{3}{2}R) + 0.5(\frac{5}{2}R)}{4 + 0.5} = \frac{6R + 1.25R}{4.5} = \frac{7.25R}{4.5} = \frac{29R}{18}$.The equivalent $C_p$ of the mixture is $C_{p,mix} = \frac{n_1 C_{p1} + n_2 C_{p2}}{n_1 + n_2} = \frac{4(\frac{5}{2}R) + 0.5(\frac{7}{2}R)}{4 + 0.5} = \frac{10R + 1.75R}{4.5} = \frac{11.75R}{4.5} = \frac{47R}{18}$.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 $(C_p / C_v)$ of the mixture is

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