Answer
$19.8\ \mathrm{K}$
Work Step by Step
The rms (root mean square) speed of the molecules in a gas at the Kelvin temperature $T$ is
$v_{\mathrm{r}\mathrm{m}\mathrm{s}}=\sqrt{\frac{3kT}{m}}=\sqrt{\frac{3RT}{M}}\qquad( 17- 13 )$
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From the periodic table at the back of the book (Appendix E)
$\mathrm{m}_{\mathrm{H}_{2}}=2(1.01\mathrm{g}/\mathrm{mol}),\quad \mathrm{m}_{\mathrm{O}_{2}}=2( 16.00\mathrm{g}/\mathrm{mol})$
Equating the $v_{\mathrm{r}\mathrm{m}\mathrm{s}}$ for $\mathrm{O}_{2}$ and $\mathrm{H}_{2}:$
$\sqrt{\frac{3kT_{\mathrm{O}_{2}}}{m_{\mathrm{O}_{2}}}}=\sqrt{\frac{3kT_{\mathrm{H}_{2}}}{m_{\mathrm{H}_{2}}}}\qquad$ ... squaring,
$\displaystyle \frac{T_{\mathrm{O}_{2}}}{m_{\mathrm{O}_{2}}}=\frac{T_{\mathrm{H}_{2}}}{m_{\mathrm{H}_{2}}},\qquad$ ... and solving for $T_{\mathrm{H}_{2}}$
$T_{\mathrm{H}_{2}}=\displaystyle \frac{T_{\mathrm{O}_{2}\cdot m_{\mathrm{H}_{2}}}}{m_{\mathrm{O}_{2}}}$
$= \displaystyle \frac{(313\mathrm{K})2(1.01\ \mathrm{g}/\mathrm{mol})}{2(16.00\ \mathrm{g}/\mathrm{mol})}=19.8\ \mathrm{K}$
