Q67 / 75 · VITEEE 2021 · Physics
The root mean square velocity of hydrogen molecules at $300 \mathrm{~K}$ is $1930 \mathrm{metre} / \mathrm{sec}$. Then the r.m.s velocity of oxygen molecules at $1200 \mathrm{~K}$ will be
- A482.5 metre/sec
- Correct answerB$965 \mathrm{metre} / \mathrm{sec}$
- C1930 metre/sec
- D3860 metre/sec
Solution
Root-mean square-velocity is given by
$$
\begin{aligned}
& \mathrm{v}_{\mathrm{rms}}=\sqrt{\frac{3 R T}{M}} \text { i.e., } \mathrm{v}_{\mathrm{rms}} \propto \sqrt{\left(\frac{\mathrm{T}}{\mathrm{M}}\right)} \\
& \frac{\left(\mathrm{v}_{\mathrm{rms}}\right) \mathrm{O}_2}{\left(\mathrm{v}_{\mathrm{rms}}\right) \mathrm{H}_2}=\sqrt{\left[\frac{\mathrm{T}_{\mathrm{O}_2}}{\mathrm{~T}_{\mathrm{H}_2}} \times \frac{\mathrm{M}_{\mathrm{H}_2}}{\mathrm{M}_{\mathrm{O}_2}}\right]} \\
& =\sqrt{\left[\left(\frac{1200}{300}\right) \times\left(\frac{2}{32}\right)\right]}=\frac{1}{2} \\
& \therefore \quad\left(\mathrm{v}_{\mathrm{rms}}\right) \mathrm{O}_2=\left(\mathrm{v}_{\mathrm{rms}}\right) \mathrm{H}_2 \times \frac{1}{2}=\frac{1930}{2}=
\end{aligned}
$$
965m/s
Hence, the correct answer is option: 'B'.