The r.m.s. speed of oxygen molecules at $47^{\circ} \mathrm{C}$ is equal to that of the hydrogen mol

Question

The r.m.s. speed of oxygen molecules at $47^{\circ} \mathrm{C}$ is equal to that of the hydrogen molecules kept at $\_\_\_\_$ ${ }^{\circ}$ C. (Mass of oxygen molecule/mass of hydrogen molecule $=32 / 2$ )

Professor PC Thomas Classes · Answer

The root mean square speed is given by $v_{rms} = \sqrt{\frac{3RT}{M}}$. Given $v_{O_2} = v_{H_2}$, we have $\sqrt{\frac{T_{O_2}}{M_{O_2}}} = \sqrt{\frac{T_{H_2}}{M_{H_2}}}$. Converting temperature to Kelvin: $T_{O_2} = 47 + 273 = 320	ext{ K}$. Substituting the values: $\frac{320}{32} = \frac{T_{H_2}}{2} \Rightarrow 10 = \frac{T_{H_2}}{2} \Rightarrow T_{H_2} = 20	ext{ K}$. Converting back to Celsius: $T = 20 - 273 = -253^\circ	ext{C}$.