One mole of an ideal monoatomic gas (gamma = | vedclass

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(A) For a mixture of gases,the adiabatic index $\gamma_{	ext{mix}}$ is given by the formula:$\gamma_{	ext{mix}} = \frac{n_1 C_{p,1} + n_2 C_{p,2}}{n

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$3/2$

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(A) For a mixture of gases,the adiabatic index $\gamma_{	ext{mix}}$ is given by the formula:$\gamma_{	ext{mix}} = \frac{n_1 C_{p,1} + n_2 C_{p,2}}{n_1 C_{v,1} + n_2 C_{v,2}}$Alternatively,using the relation $C_v = \frac{R}{\gamma - 1}$,we have:$\gamma_{	ext{mix}} = \frac{n_1 + n_2}{\frac{n_1}{\gamma_1 - 1} + \frac{n_2}{\gamma_2 - 1}}$Given $n_1 = 1, \gamma_1 = 5/3$ and $n_2 = 1, \gamma_2 = 7/5$:$\gamma_{	ext{mix}} = \frac{1 + 1}{\frac{1}{5/3 - 1} + \frac{1}{7/5 - 1}} = \frac{2}{\frac{1}{2/3} + \frac{1}{2/5}} = \frac{2}{3/2 + 5/2} = \frac{2}{8/2} = \frac{2}{4} = 1/2$ (Wait,re-calculating).Correct calculation:$C_{v,1} = \frac{R}{5/3 - 1} = \frac{3R}{2}$,$C_{v,2} = \frac{R}{7/5 - 1} = \frac{5R}{2}$.$C_{v,	ext{mix}} = \frac{n_1 C_{v,1} + n_2 C_{v,2}}{n_1 + n_2} = \frac{1(1.5R) + 1(2.5R)}{2} = 2R$.$C_{p,	ext{mix}} = C_{v,	ext{mix}} + R = 2R + R = 3R$.$\gamma_{	ext{mix}} = \frac{C_{p,	ext{mix}}}{C_{v,	ext{mix}}} = \frac{3R}{2R} = 1.5 = 3/2$.

One mole of an ideal monoatomic gas $(\gamma = 5/3)$ is mixed with one mole of a diatomic gas $(\gamma = 7/5).$ What is $\gamma$ for the mixture? $\gamma$ denotes the ratio of specific heat at constant pressure to that at constant volume.

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