Five moles of helium are mixed with two moles | vedclass

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

Vedclass · Accepted Answer

$1.56$

Vedclass · Answer

(C) For a mixture of gases,the equivalent molar heat capacity at constant volume is given by $C_{V,mix} = \frac{n_1 C_{V1} + n_2 C_{V2}}{n_1 + n_2}$.Helium is a monoatomic gas,so $C_{V1} = \frac{3}{2}R$. Hydrogen is a diatomic gas,so $C_{V2} = \frac{5}{2}R$.Given $n_1 = 5 \ mol$ and $n_2 = 2 \ mol$,we have $C_{V,mix} = \frac{5(\frac{3}{2}R) + 2(\frac{5}{2}R)}{5 + 2} = \frac{7.5R + 5R}{7} = \frac{12.5R}{7}$.The equivalent molar heat capacity at constant pressure is $C_{P,mix} = \frac{n_1 C_{P1} + n_2 C_{P2}}{n_1 + n_2}$.For helium,$C_{P1} = \frac{5}{2}R$. For hydrogen,$C_{P2} = \frac{7}{2}R$.$C_{P,mix} = \frac{5(\frac{5}{2}R) + 2(\frac{7}{2}R)}{5 + 2} = \frac{12.5R + 7R}{7} = \frac{19.5R}{7}$.The adiabatic index $\gamma_{mix} = \frac{C_{P,mix}}{C_{V,mix}} = \frac{19.5R/7}{12.5R/7} = \frac{19.5}{12.5} = 1.56$.

Five moles of helium are mixed with two moles of hydrogen to form a mixture. Take the molar mass of helium $M_1 = 4 \ g/mol$ and that of hydrogen $M_2 = 2 \ g/mol$. The equivalent value of $\gamma$ for the mixture is:

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