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Question: Answered & Verified by Expert
Root mean square $(\mathrm{rms})$ speed of $\mathrm{O}_2$ is $500 \mathrm{~m} / \mathrm{s}$ at a constant temperature. Calculate the rms speed and the average kinetic energy of $\mathrm{H}_2$ at the same temperature. (Consider, $\left.R=8.33 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\right)$
ChemistryStates of MatterTS EAMCETTS EAMCET 2020 (10 Sep Shift 1)
Options:
  • A $500 \mathrm{~m} / \mathrm{s}$ and $4.0 \mathrm{~kJ} / \mathrm{mol}$
  • B $2000 \mathrm{~m} / \mathrm{s}$ and $4.0 \mathrm{~kJ} / \mathrm{mol}$
  • C $500 \mathrm{~m} / \mathrm{s}$ and $4.7 \mathrm{~kJ} / \mathrm{mol}$
  • D $2000 \mathrm{~m} / \mathrm{s}$ and $4.7 \mathrm{~kJ} / \mathrm{mol}$
Solution:
2996 Upvotes Verified Answer
The correct answer is: $2000 \mathrm{~m} / \mathrm{s}$ and $4.0 \mathrm{~kJ} / \mathrm{mol}$
Given, root mean square $(\mathrm{rms})$ speed of $\mathrm{O}_2$ is
$500 \mathrm{~m} / \mathrm{s}$ at a constant temperature.
Root mean square speed in given by the following expression
$u_{\mathrm{rms}}=\sqrt{\frac{3 R T}{M w}}$
For $\mathrm{H}_2$ gas, $u_{\mathrm{Tms}\left(\mathrm{H}_2\right)}=\sqrt{\frac{3 R T}{M w_{\mathrm{H}_2}}}$
$\left(M w\right.$ of $\left.\mathrm{H}_2=2 \mathrm{~g} / \mathrm{mol}\right)$
For $\mathrm{O}_2$ gas, $u_{\mathrm{rms}\left(\mathrm{O}_2\right)}=\sqrt{\frac{3 R T}{M w_{\mathrm{O}_2}}}$
$\left(M w\right.$ of $\left.\mathrm{O}_2=32 \mathrm{~g} / \mathrm{mol}\right)$
$\frac{u_{\mathrm{rms}\left(\mathrm{H}_2\right)}}{u_{\mathrm{rms}\left(\mathrm{O}_2\right)}}=\sqrt{\frac{3 R T}{M w_{\mathrm{H}_2}}} \times \sqrt{\frac{M w_{\mathrm{O}_2}}{3 R T}}$
(Temperature constant, $R=8.33 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$ )
$\begin{aligned}
u_{\mathrm{rms}\left(\mathrm{H}_2\right)} & =\sqrt{\frac{M w_{\mathrm{O}_2}}{M w_{\mathrm{H}_2}}} \times u_{\mathrm{rms}\left(\mathrm{O}_2\right)}=\sqrt{\frac{32}{2}} \times 500 \\
& =2000 \mathrm{~m} / \mathrm{s}
\end{aligned}$
Average kinetic energy of $\mathrm{H}_2$ can be calculate as
$=\frac{1}{2} m u^2$
$\begin{aligned}
{\left[u=2000 \mathrm{~m} / \mathrm{s}, m=\mathrm{H}_2=2 \mathrm{~g} / \mathrm{mol}=0.002 \mathrm{~kg} / \mathrm{mol}\right] } \\
=\frac{1}{2} \times 0.002 \times(2000)^2 \\
=\frac{1}{2} \times \frac{2}{1000} \times 2000 \times 2000 \\
=4000 \mathrm{~J} \mathrm{~mol}^{-1} \text { or } 4 \mathrm{~kJ} \mathrm{~mol}^{-1}
\end{aligned}$

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