Two moles of a monoatomic gas is mixed with six moles of a diatomic gas. The molar specific heat of the mixture at constant volume is :

Find the molar specific heat at constant volume of a mixture of 2 mol monoatomic and 6 mol diatomic gas.

For an ideal gas, the molar specific heat at constant volume is given by $$C_v = \frac{f}{2}R$$, where $$f$$ is the number of degrees of freedom.

Monoatomic gas has $$f = 3$$ (three translational degrees), giving $$C_{v1} = \frac{3}{2}R$$, while diatomic gas has $$f = 5$$ (three translational plus two rotational), giving $$C_{v2} = \frac{5}{2}R$$.

In the mixture, the effective molar specific heat is the mole-fraction-weighted average $$C_{v,\text{mix}} = \frac{n_1 C_{v1} + n_2 C_{v2}}{n_1 + n_2}$$, where $$n_1 = 2$$ mol of monoatomic and $$n_2 = 6$$ mol of diatomic gas.

Substituting these values yields $$C_{v,\text{mix}} = \frac{2 \times \frac{3}{2}R + 6 \times \frac{5}{2}R}{2 + 6} = \frac{3R + 15R}{8} = \frac{18R}{8} = \frac{9}{4}R$$.

The correct answer is Option A: $$\frac{9}{4}R$$.