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(B) For a monoatomic gas,the molar heat capacity at constant volume is $C_{v1} = \frac{3}{2}R$ and at constant pressure is $C_{p1} = \frac{5}{2}R$.For

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$1.5$

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(B) For a monoatomic gas,the molar heat capacity at constant volume is $C_{v1} = \frac{3}{2}R$ and at constant pressure is $C_{p1} = \frac{5}{2}R$.For a diatomic gas,the molar heat capacity at constant volume is $C_{v2} = \frac{5}{2}R$ and at constant pressure is $C_{p2} = \frac{7}{2}R$.When $n_1 = 1 	ext{ mole}$ and $n_2 = 1 	ext{ mole}$ are mixed,the equivalent molar heat capacity at constant volume for the mixture is:$C_{v,mix} = \frac{n_1 C_{v1} + n_2 C_{v2}}{n_1 + n_2} = \frac{1(\frac{3}{2}R) + 1(\frac{5}{2}R)}{1 + 1} = \frac{4R}{2} = 2R$.The equivalent molar heat capacity at constant pressure for the mixture is:$C_{p,mix} = \frac{n_1 C_{p1} + n_2 C_{p2}}{n_1 + n_2} = \frac{1(\frac{5}{2}R) + 1(\frac{7}{2}R)}{1 + 1} = \frac{6R}{2} = 3R$.The adiabatic index $\gamma_{mix}$ is given by:$\gamma_{mix} = \frac{C_{p,mix}}{C_{v,mix}} = \frac{3R}{2R} = 1.5$.

If $1 \text{ mole}$ of a monoatomic gas $\left[ \gamma = \frac{5}{3} \right]$ is mixed with $1 \text{ mole}$ of a diatomic gas $\left[ \gamma = \frac{7}{5} \right]$,then the $\gamma$ of the mixture is ..........?

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