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

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(A) For a mixture of gases,the equivalent adiabatic exponent $\gamma_{	ext{mix}}$ is given by the formula: $\frac{n_1 + n_2}{\gamma_{	ext{mix}} - 1}

Vedclass · Accepted Answer

$3/2$

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(A) For a mixture of gases,the equivalent adiabatic exponent $\gamma_{	ext{mix}}$ is given by the formula: $\frac{n_1 + n_2}{\gamma_{	ext{mix}} - 1} = \frac{n_1}{\gamma_1 - 1} + \frac{n_2}{\gamma_2 - 1}$.Given: $n_1 = 1$ mole,$\gamma_1 = 5/3$ (monoatomic); $n_2 = 1$ mole,$\gamma_2 = 7/5$ (diatomic).Substituting the values:$\frac{1 + 1}{\gamma_{	ext{mix}} - 1} = \frac{1}{5/3 - 1} + \frac{1}{7/5 - 1}$$\frac{2}{\gamma_{	ext{mix}} - 1} = \frac{1}{2/3} + \frac{1}{2/5} = \frac{3}{2} + \frac{5}{2} = \frac{8}{2} = 4$.$\frac{2}{\gamma_{	ext{mix}} - 1} = 4 \implies \gamma_{	ext{mix}} - 1 = 2/4 = 1/2$.$\gamma_{	ext{mix}} = 1 + 0.5 = 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 the value of $\gamma$ for the mixture? Here,$\gamma$ denotes the ratio of specific heat at constant pressure to that at constant volume.

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