A gaseous mixture consists of 16 g of helium | vedclass

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(A) The number of moles of Helium $(He)$ is $n_1 = \frac{16}{4} = 4 \ mol$. Helium is a monoatomic gas,so its degrees of freedom $f_1 = 3$ and $C_{v1}

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$1.46$

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(A) The number of moles of Helium $(He)$ is $n_1 = \frac{16}{4} = 4 \ mol$. Helium is a monoatomic gas,so its degrees of freedom $f_1 = 3$ and $C_{v1} = \frac{3}{2}R$.The number of moles of Hydrogen $(H_2)$ is $n_2 = \frac{16}{2} = 8 \ mol$. Hydrogen is a diatomic gas,so its degrees of freedom $f_2 = 5$ and $C_{v2} = \frac{5}{2}R$.The equivalent $C_v$ of the mixture is given by $C_{v,mix} = \frac{n_1 C_{v1} + n_2 C_{v2}}{n_1 + n_2} = \frac{4(\frac{3}{2}R) + 8(\frac{5}{2}R)}{4 + 8} = \frac{6R + 20R}{12} = \frac{26R}{12} = \frac{13}{6}R$.The equivalent $C_p$ of the mixture is given by $C_{p,mix} = C_{v,mix} + R = \frac{13}{6}R + R = \frac{19}{6}R$.The ratio $\gamma = \frac{C_{p,mix}}{C_{v,mix}} = \frac{19/6 R}{13/6 R} = \frac{19}{13} \approx 1.46$.

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

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