Q12 / 160 · WBJEE 2015 · Chemistry
The RMS velocity of CO gas molecules at $27^{\circ} \mathrm{C}$ is approximately $1000 \mathrm{m} / \mathrm{s}$. For molecules at $600 \mathrm{K}$, the RMS velocity is approximately
- A$2000 \mathrm{m} / \mathrm{s}$
- Correct answerB$1414 \mathrm{m} / \mathrm{s}$
- C$1000 \mathrm{m} / \mathrm{s}$
- D$1500 \mathrm{m} / \mathrm{s}$
Solution
As trom the formula, $v_{\text {ms }}=\sqrt{\frac{3 R T}{M}}$
Given that $\left(v_{\text {rms }}\right)_{CO}=1000 \mathrm{m} / \mathrm{s}$
$(Temp )_{CO}=27^{\circ} \mathrm{C}=300 \mathrm{K}$
$(Temp.) _{\mathrm{N}_{\mathrm{2}}}=600 \mathrm{K}$
Now putting the values, we get
$\frac{\left(V_{\operatorname{mot}}\right)_{CO}}{\left(v_{m}\right)_{N_{2}}}=\sqrt{\frac{3 R}{3 R} \times \frac{T_{CO}}{M_{CO}} \times \frac{M_{N_{2}}}{T_{N_{2}}}}$
$\frac{1000}{\left(v_{\text {rad }}\right)_{N_{2}}}=\sqrt{\frac{300}{28} \times \frac{28}{600}}=\frac{1}{\sqrt{2}}$
or. $\left(v_{\text {rms }}\right)_{N_{2}}=1000 \times 1414$ $=1414 \mathrm{m} / \mathrm{s}$
Hence, the correct answer is option: 'B'.