Question
A gaseous mixture contains $16g$ of helium and $16g$ of oxygen, then calculate the ratio of $\frac{\text{C}_{\text{p}}}{\text{C}_{\text{V}}}$ of the mixture.

Answer

Moles of helium $(\mu_{\text{He}})=\frac{16}{4}=4$ Moles of oxygen $(\mu_{\text{O}_2})=\frac{16}{32}=\frac{1}{2}$As helium is monoatomic, so degrees of freedom of helium, f = 3, so $\text{C}_{\text{V}_{\text{He}}}=\frac{\text{f}}{2}\text{R}=\frac{3}{2}\text{R}$
As oxygen is diatomic, so degrees of freedom of oxygen f = 5, so$\text{C}_{\text{V}_{\text{O}_2}}=\frac{\text{f}}{2}\text{R}=\frac{5}{2}\text{R}$
$\therefore\text{C}_{\text{V mixture}}=\frac{\mu_{\text{He}}\text{C}_{\text{V}_{\text{He}}}+\mu_{\text{O}_2}\text{C}_{\text{V}_{\text{O}_2}}}{\mu_{\text{He}}+\mu\text{O}_2}$
$=\frac{4\times\frac{3}{2}\text{R}+\frac{1}{2}\times\frac{5}{2}\text{R}}{4+\frac{1}{2}}=\frac{29}{18}\text{R}$
$\gamma=\frac{\text{C}_{\text{P}}}{\text{C}_{\text{V}}}$ [of mixture]
$\gamma_{\text{mixture}}=1+\frac{\text{R}}{\text{C}_{\text{V}_{\text{mixture}}}}$
$=1+\frac{\text{R}}{\frac{29}{18}\text{R}}=1.62$ [as $C_P - C_V = R$]

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