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(D) The molar specific heat at constant volume for a mixture is given by the formula: $C_{V_{mix}} = \frac{n_1 C_{V_1} + n_2 C_{V_2}}{n_1 + n_2}$.For

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$3$

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(D) The molar specific heat at constant volume for a mixture is given by the formula: $C_{V_{mix}} = \frac{n_1 C_{V_1} + n_2 C_{V_2}}{n_1 + n_2}$.For a monoatomic gas,$C_{V_1} = \frac{3}{2} R$ and $n_1 = 1$.For a diatomic gas without vibrational modes,$C_{V_2} = \frac{5}{2} R$ and $n_2 = 3$.Substituting these values into the formula:$C_{V_{mix}} = \frac{1 \cdot (\frac{3}{2} R) + 3 \cdot (\frac{5}{2} R)}{1 + 3} = \frac{\frac{3}{2} R + \frac{15}{2} R}{4} = \frac{\frac{18}{2} R}{4} = \frac{9 R}{4}$.Given that $C_{V_{mix}} = \frac{\alpha^2}{4} R$,we equate the two expressions:$\frac{9 R}{4} = \frac{\alpha^2}{4} R$.This implies $\alpha^2 = 9$,so $\alpha = 3$ (taking the positive value).

One mole of a monoatomic gas is mixed with three moles of a diatomic gas. The molecular specific heat of the mixture at constant volume is $\frac{\alpha^{2}}{4} R \ J/mol \ K$; then the value of $\alpha$ will be $.......$ (Assume that the given diatomic gas has no vibrational mode.)

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