One mole of a monoatomic gas is mixed with three moles of a diatomic gas. The molecular specific heat of mixture at constant volume is $$\frac{\alpha^2}{4}R$$ J mol$$^{-1}$$ K$$^{-1}$$; then the value of $$\alpha$$ will be _____. (Assume that the given diatomic gas has no vibrational mode.)

For a monoatomic gas, the molar specific heat at constant volume is $$C_{v1} = \dfrac{3}{2}R$$ (3 degrees of freedom).

For a diatomic gas with no vibrational mode, $$C_{v2} = \dfrac{5}{2}R$$ (5 degrees of freedom: 3 translational + 2 rotational).

For a mixture of $$n_1 = 1$$ mole of monoatomic gas and $$n_2 = 3$$ moles of diatomic gas, the molar specific heat at constant volume is:

$$C_{v,\text{mix}} = \dfrac{n_1 C_{v1} + n_2 C_{v2}}{n_1 + n_2} = \dfrac{1 \cdot \dfrac{3}{2}R + 3 \cdot \dfrac{5}{2}R}{1 + 3}$$

$$= \dfrac{\dfrac{3R}{2} + \dfrac{15R}{2}}{4} = \dfrac{\dfrac{18R}{2}}{4} = \dfrac{9R}{4}$$

We are given that $$C_{v,\text{mix}} = \dfrac{\alpha^2}{4}R$$. Comparing: $$\dfrac{\alpha^2}{4} = \dfrac{9}{4}$$, so $$\alpha^2 = 9$$, giving $$\alpha = 3$$.

The answer is $$3$$.