One gram mole of an ideal gas A with the rati | vedclass

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(B) For a mixture of gases,the adiabatic index $\gamma_{	ext{mix}}$ is given by the formula:$\gamma_{	ext{mix}} = \frac{n_1 C_{p1} + n_2 C_{p2}}{n_1

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$2$

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(B) For a mixture of gases,the adiabatic index $\gamma_{	ext{mix}}$ is given by the formula:$\gamma_{	ext{mix}} = \frac{n_1 C_{p1} + n_2 C_{p2}}{n_1 C_{v1} + n_2 C_{v2}}$Given $n_1 = 1$ mole,$\gamma_1 = 5/3$ and $n_2 = n$ moles,$\gamma_2 = 7/5$.We know $C_v = \frac{R}{\gamma - 1}$ and $C_p = \frac{\gamma R}{\gamma - 1}$.Substituting these into the formula:$\gamma_{	ext{mix}} = \frac{n_1 \frac{\gamma_1 R}{\gamma_1 - 1} + n_2 \frac{\gamma_2 R}{\gamma_2 - 1}}{n_1 \frac{R}{\gamma_1 - 1} + n_2 \frac{R}{\gamma_2 - 1}} = \frac{n_1 \frac{\gamma_1}{\gamma_1 - 1} + n_2 \frac{\gamma_2}{\gamma_2 - 1}}{n_1 \frac{1}{\gamma_1 - 1} + n_2 \frac{1}{\gamma_2 - 1}}$For gas $A$: $\frac{\gamma_1}{\gamma_1 - 1} = \frac{5/3}{2/3} = 2.5$ and $\frac{1}{\gamma_1 - 1} = \frac{1}{2/3} = 1.5$.For gas $B$: $\frac{\gamma_2}{\gamma_2 - 1} = \frac{7/5}{2/5} = 3.5$ and $\frac{1}{\gamma_2 - 1} = \frac{1}{2/5} = 2.5$.Substituting values: $\frac{19}{13} = \frac{1(2.5) + n(3.5)}{1(1.5) + n(2.5)}$.$19(1.5 + 2.5n) = 13(2.5 + 3.5n)$.$28.5 + 47.5n = 32.5 + 45.5n$.$2n = 4 \Rightarrow n = 2$.

One gram mole of an ideal gas $A$ with the ratio of constant pressure and constant volume specific heats $\gamma_{A} = 5/3$ is mixed with $n$ gram moles of another ideal gas $B$ with $\gamma_{B} = 7/5$. If the $\gamma$ for the mixture is $19/13$,then what will be the value of $n$?

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