A gas mixture consists of 16 g of helium and | vedclass

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(D) For helium $(He)$,molar mass $M_1 = 4 \ g/mol$. Number of moles $\mu_1 = 16/4 = 4 \ mol$.Helium is a monoatomic gas,so degrees of freedom $f_1 = 3

Vedclass · Accepted Answer

$1.62$

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(D) For helium $(He)$,molar mass $M_1 = 4 \ g/mol$. Number of moles $\mu_1 = 16/4 = 4 \ mol$.Helium is a monoatomic gas,so degrees of freedom $f_1 = 3$. $C_{V1} = \frac{3}{2}R$ and $C_{P1} = \frac{5}{2}R$.For oxygen $(O_2)$,molar mass $M_2 = 32 \ g/mol$. Number of moles $\mu_2 = 16/32 = 0.5 \ mol$.Oxygen is a diatomic gas,so degrees of freedom $f_2 = 5$. $C_{V2} = \frac{5}{2}R$ and $C_{P2} = \frac{7}{2}R$.The equivalent molar heat capacity at constant volume for the mixture is $C_{V,mix} = \frac{\mu_1 C_{V1} + \mu_2 C_{V2}}{\mu_1 + \mu_2} = \frac{4(\frac{3}{2}R) + 0.5(\frac{5}{2}R)}{4 + 0.5} = \frac{6R + 1.25R}{4.5} = \frac{7.25R}{4.5} = \frac{29}{18}R$.The equivalent molar heat capacity at constant pressure for the mixture is $C_{P,mix} = \frac{\mu_1 C_{P1} + \mu_2 C_{P2}}{\mu_1 + \mu_2} = \frac{4(\frac{5}{2}R) + 0.5(\frac{7}{2}R)}{4 + 0.5} = \frac{10R + 1.75R}{4.5} = \frac{11.75R}{4.5} = \frac{47}{18}R$.The ratio $\gamma_{mix} = \frac{C_{P,mix}}{C_{V,mix}} = \frac{47/18}{29/18} = \frac{47}{29} \approx 1.62$.

$A$ gas mixture consists of $16 \ g$ of helium and $16 \ g$ of oxygen. For this mixture,the ratio $C_P/C_V$ is equal to:

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