A mixture of one mole of monoatomic gas and one mole of a diatomic gas (rigid) are kept at room temperature $$(27°C)$$. The ratio of specific heat of gases at constant volume respectively is:

We need to find the ratio of specific heats at constant volume for a monoatomic gas and a diatomic (rigid) gas.

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

For a monoatomic gas, $$f = 3$$ (3 translational), so $$C_{v_1} = \frac{3}{2}R$$. For a diatomic gas (rigid, i.e., no vibrational modes), $$f = 5$$ (3 translational + 2 rotational), so $$C_{v_2} = \frac{5}{2}R$$.

Therefore, $$\frac{C_{v_1}}{C_{v_2}} = \frac{\frac{3}{2}R}{\frac{5}{2}R} = \frac{3}{5}$$.

The correct answer is Option (2): $$\frac{3}{5}$$.