Let $$\gamma_1$$ be the ratio of molar specific heat at constant pressure and molar specific heat at constant volume of a monoatomic gas and $$\gamma_2$$ be the similar ratio of diatomic gas. Considering the diatomic gas molecule as a rigid rotator, the ratio $$\frac{\gamma_1}{\gamma_2}$$ is:

Find the ratio $$\frac{\gamma_1}{\gamma_2}$$ where $$\gamma_1$$ corresponds to a monoatomic gas and $$\gamma_2$$ refers to a diatomic gas treated as a rigid rotator.

Since a monoatomic gas has three degrees of freedom (translational only), its molar heat capacity at constant volume is $$C_V = \frac{3}{2}R$$. Therefore, $$C_P = C_V + R = \frac{5}{2}R$$, giving

$$\gamma_1 = \frac{C_P}{C_V} = \frac{5/2}{3/2} = \frac{5}{3}\,.$$

Now for a diatomic gas modeled as a rigid rotator, there are five degrees of freedom (three translational and two rotational, with vibrational modes frozen out). Thus, $$C_V = \frac{5}{2}R$$ and $$C_P = C_V + R = \frac{7}{2}R$$, which leads to

$$\gamma_2 = \frac{C_P}{C_V} = \frac{7/2}{5/2} = \frac{7}{5}\,.$$

Substituting these values into the desired ratio yields

$$\frac{\gamma_1}{\gamma_2} = \frac{5/3}{7/5} = \frac{5}{3} \times \frac{5}{7} = \frac{25}{21}\,.$$

Therefore, the answer is Option C: $$\frac{25}{21}\,.$$