Two moles of an ideal gas with frac{C_{P}}{C_ | vedclass

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(B) For gas $1$: $n_1 = 2$,$\gamma_1 = \frac{5}{3}$. Since $\gamma = 1 + \frac{2}{f}$,$f_1 = 3$. Thus,$C_{V_1} = \frac{3}{2}R$ and $C_{P_1} = \frac{5}

Vedclass · Accepted Answer

$1.42$

Vedclass · Answer

(B) For gas $1$: $n_1 = 2$,$\gamma_1 = \frac{5}{3}$. Since $\gamma = 1 + \frac{2}{f}$,$f_1 = 3$. Thus,$C_{V_1} = \frac{3}{2}R$ and $C_{P_1} = \frac{5}{2}R$.For gas $2$: $n_2 = 3$,$\gamma_2 = \frac{4}{3}$. Since $\gamma = 1 + \frac{2}{f}$,$f_2 = 6$. Thus,$C_{V_2} = \frac{6}{2}R = 3R$ and $C_{P_2} = 4R$.The molar heat capacity at constant volume for the mixture is $C_{V_{mix}} = \frac{n_1 C_{V_1} + n_2 C_{V_2}}{n_1 + n_2} = \frac{2(\frac{3}{2}R) + 3(3R)}{2+3} = \frac{3R + 9R}{5} = \frac{12R}{5} = 2.4R$.The molar heat capacity at constant pressure for the mixture is $C_{P_{mix}} = \frac{n_1 C_{P_1} + n_2 C_{P_2}}{n_1 + n_2} = \frac{2(\frac{5}{2}R) + 3(4R)}{2+3} = \frac{5R + 12R}{5} = \frac{17R}{5} = 3.4R$.The ratio $\gamma_{mix} = \frac{C_{P_{mix}}}{C_{V_{mix}}} = \frac{17R/5}{12R/5} = \frac{17}{12} \approx 1.4167 \approx 1.42$.

Two moles of an ideal gas with $\frac{C_{P}}{C_{V}}=\frac{5}{3}$ are mixed with $3$ moles of another ideal gas with $\frac{C_{P}}{C_{V}}=\frac{4}{3}$. The value of $\frac{C_{P}}{C_{V}}$ for the mixture is

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