The ratio of rms speeds of the gases in the mixture of nitrogen oxygen will be

The correct option is √8:√7.

To find the ratio of the root mean square (rms) speeds of nitrogen ($N_2$) and oxygen ($O_2$) gases in a mixture, we use the formula for the rms speed of a gas molecule:
$v_{\mathrm{rms}}=\sqrt{\frac{3RT}{M}}$
where:
• $R$ is the universal gas constant,
• $T$ is the absolute temperature of the mixture, and
• $M$ is the molar mass of the gas.

Since both gases are present in the same mixture, they are at the same temperature ($T$). Therefore, the rms speed is inversely proportional to the square root of the molar mass of the gas:
$v_{\mathrm{rms}}\propto \frac{1}{\sqrt{M}}$

Thus, the ratio of the rms speeds of nitrogen ($N_2$) to oxygen ($O_2$) is:
$\frac{v_{\mathrm{rms}\left(N_2\right)}}{v_{\mathrm{rms}\left(O_2\right)}}=\sqrt{\frac{M_{O_2}}{M_{N_2}}}$

Now, let us find the molar masses of both gases:
• The molar mass of nitrogen gas ($N_2$) is $M_{N_2}=2\times 14=28\text{g/mol}$.
• The molar mass of oxygen gas ($O_2$) is $M_{O_2}=2\times 16=32\text{g/mol}$.

Substituting these values into the ratio equation, we get:
$\frac{v_{\mathrm{rms}\left(N_2\right)}}{v_{\mathrm{rms}\left(O_2\right)}}=\sqrt{\frac{32}{28}}$

Simplifying the fraction inside the square root by dividing both the numerator and the denominator by their greatest common divisor, which is 4:
$\frac{v_{\mathrm{rms}\left(N_2\right)}}{v_{\mathrm{rms}\left(O_2\right)}}=\sqrt{\frac{8}{7}}=\frac{\sqrt{8}}{\sqrt{7}}$

Therefore, the ratio of the rms speeds is √8:√7.