Thermodynamics Part 5

Concept:The root mean square speed of a gas depends on its absolute temperature and molar mass.Formula:$$v_{\mathrm{rms}} = \sqrt{\frac{3RT}{M}}$$Solution:Given that $v_{\mathrm{rms}}$ of oxygen ($\mathrm{O}_2$) at $47^{\circ}\mathrm{C}$ equals that of hydrogen ($\mathrm{H}_2$).Convert oxygen temperature to Kelvin: $T_{\mathrm{O}_2} = 47 + 273 = 320\ \mathrm{K}$.Molar masses (ratio): $M_{\mathrm{O}_2} = 32$ and $M_{\mathrm{H}_2} = 2$ (in g/mol).Since $v_{\mathrm{rms}} \propto \sqrt{T/M}$, equality gives $\frac{T_{\mathrm{O}_2}}{M_{\mathrm{O}_2}} = \frac{T_{\mathrm{H}_2}}{M_{\mathrm{H}_2}}$.Substitute: $\frac{320}{32} = \frac{T_{\mathrm{H}_2}}{2}$ → $10 = \frac{T_{\mathrm{H}_2}}{2}$ → $T_{\mathrm{H}_2} = 20\ \mathrm{K}$.Convert to Celsius: $T_{\mathrm{H}_2} = 20 - 273 = -253^{\circ}\mathrm{C}$.Answer:The required temperature is $-253^{\circ}\mathrm{C}$, which corresponds to option (B).