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The ratio between the RMS velocity of $\mathrm{N}_2$ at $200 \mathrm{~K}$ and that of $\mathrm{CO}$ at $800 \mathrm{~K}$ is (molecular mass of $\mathrm{N}_2=28 \mathrm{~g} \mathrm{~mol}^{-1}$, molecular mass of $\mathrm{CO}=28 \mathrm{~g} \mathrm{~mol}^{-1}$ )
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The correct answer is:
0.50
RMS velocity $\left(v_{\mathrm{rms}}\right)=\sqrt{3 R T / M}$
$$
\begin{aligned}
& \text { where, } R=\text { molar gas constant } \\
& T=\text { temperature in Kelvin } \\
& \begin{aligned}
& M_m=\text { molar mass in kilograms } / \mathrm{mol} \\
& {\left[\because M_{\left(\mathrm{N}_2\right)}=M_{(\mathrm{CO})}\right] } \\
& v_{\mathrm{rms}}\left(\mathrm{N}_2\right)=\sqrt{3 \times 8.31 \times 200} \\
& v_{\mathrm{rms}}(\mathrm{CO})=\sqrt{3 \times 8.31 \times 800} \\
& \therefore \frac{v_{\mathrm{rms}}\left(\mathrm{N}_2\right)}{v_{\mathrm{rms}}(\mathrm{CO})}=\frac{\sqrt{200}}{\sqrt{800}} \\
&=\frac{14.14}{28.28}=0.50
\end{aligned}
\end{aligned}
$$
$$
\begin{aligned}
& \text { where, } R=\text { molar gas constant } \\
& T=\text { temperature in Kelvin } \\
& \begin{aligned}
& M_m=\text { molar mass in kilograms } / \mathrm{mol} \\
& {\left[\because M_{\left(\mathrm{N}_2\right)}=M_{(\mathrm{CO})}\right] } \\
& v_{\mathrm{rms}}\left(\mathrm{N}_2\right)=\sqrt{3 \times 8.31 \times 200} \\
& v_{\mathrm{rms}}(\mathrm{CO})=\sqrt{3 \times 8.31 \times 800} \\
& \therefore \frac{v_{\mathrm{rms}}\left(\mathrm{N}_2\right)}{v_{\mathrm{rms}}(\mathrm{CO})}=\frac{\sqrt{200}}{\sqrt{800}} \\
&=\frac{14.14}{28.28}=0.50
\end{aligned}
\end{aligned}
$$
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