Let γ _ 1 be the ratio of molar specific heat at constant pressure...

As we know, ${C}_{v} = f\dfrac{R}{2}$

and ${C}_{p} = {C}_{v} + R = f\dfrac{R}{2} + R$

Therefore, $\gamma = \dfrac{{C}_{p}}{{C}_{v}} = 1 + \dfrac{2}{f}$

For monatomic gas, $f = 3$ and hence ${\gamma}_{1} = \dfrac{5}{3}$

For diatomic gas, $f = 5$ and hence ${\gamma}_{2} = \dfrac{7}{5}$

Required ratio, $\dfrac{{\gamma}_{1}}{{\gamma}_{2}} = \dfrac{25}{21}$