If two moles of diatomic gas and one mole of | vedclass

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(C) For a mixture of gases,the equivalent molar heat capacity at constant volume is given by $C_{V,eq} = \frac{n_1 C_{V1} + n_2 C_{V2}}{n_1 + n_2}$.Fo

Vedclass · Accepted Answer

$19/13$

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(C) For a mixture of gases,the equivalent molar heat capacity at constant volume is given by $C_{V,eq} = \frac{n_1 C_{V1} + n_2 C_{V2}}{n_1 + n_2}$.For a diatomic gas,$C_{V1} = \frac{5}{2}R$ and $C_{P1} = \frac{7}{2}R$. For a monoatomic gas,$C_{V2} = \frac{3}{2}R$ and $C_{P2} = \frac{5}{2}R$.Given $n_1 = 2$ moles (diatomic) and $n_2 = 1$ mole (monoatomic).The equivalent molar heat capacity at constant volume is $C_{V,eq} = \frac{2(\frac{5}{2}R) + 1(\frac{3}{2}R)}{2+1} = \frac{5R + 1.5R}{3} = \frac{6.5R}{3} = \frac{13R}{6}$.The equivalent molar heat capacity at constant pressure is $C_{P,eq} = \frac{n_1 C_{P1} + n_2 C_{P2}}{n_1 + n_2} = \frac{2(\frac{7}{2}R) + 1(\frac{5}{2}R)}{2+1} = \frac{7R + 2.5R}{3} = \frac{9.5R}{3} = \frac{19R}{6}$.The ratio of specific heats $\gamma_{eq} = \frac{C_{P,eq}}{C_{V,eq}} = \frac{19R/6}{13R/6} = \frac{19}{13}$.

If two moles of diatomic gas and one mole of monoatomic gas are mixed,then the ratio of specific heats is:

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