If one mole of an ideal monoatomic gas left(g | vedclass

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(A) For a monoatomic gas,the molar heat capacity at constant volume is $C_{v1} = \frac{3}{2}R$.For a diatomic gas,the molar heat capacity at constant

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(A) For a monoatomic gas,the molar heat capacity at constant volume is $C_{v1} = \frac{3}{2}R$.For a diatomic gas,the molar heat capacity at constant volume is $C_{v2} = \frac{5}{2}R$.For a mixture of $n_1$ moles of gas $1$ and $n_2$ moles of gas $2$,the equivalent $C_v$ is given by $C_{v,mix} = \frac{n_1 C_{v1} + n_2 C_{v2}}{n_1 + n_2}$.Substituting the values: $C_{v,mix} = \frac{1 \cdot \frac{3}{2}R + 1 \cdot \frac{5}{2}R}{1 + 1} = \frac{4R}{2} = 2R$.Similarly,the equivalent $C_p$ is $C_{p,mix} = \frac{n_1 C_{p1} + n_2 C_{p2}}{n_1 + n_2}$. Since $C_p = C_v + R$,we have $C_{p1} = \frac{5}{2}R$ and $C_{p2} = \frac{7}{2}R$.$C_{p,mix} = \frac{1 \cdot \frac{5}{2}R + 1 \cdot \frac{7}{2}R}{2} = \frac{6R}{2} = 3R$.The adiabatic index for the mixture is $\gamma_{mix} = \frac{C_{p,mix}}{C_{v,mix}} = \frac{3R}{2R} = 1.5$.

If one mole of an ideal monoatomic gas $\left(\gamma = \frac{5}{3}\right)$ is mixed with one mole of a diatomic gas $\left(\gamma = \frac{7}{5}\right)$,the value of $\gamma$ for the mixture is:

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