At which temperature the r.m.s. velocity of a hydrogen molecule equal to that of an oxygen molecule at $$47°C$$?

The r.m.s. velocity of a gas molecule is given by:

$$v_{rms} = \sqrt{\frac{3RT}{M}}$$

where $$R$$ is the gas constant, $$T$$ is the absolute temperature, and $$M$$ is the molar mass.

We need the r.m.s. velocity of hydrogen ($$M_{H_2} = 2$$ g/mol) to equal that of oxygen ($$M_{O_2} = 32$$ g/mol) at $$47°C = 320$$ K. Setting the two velocities equal:

$$\sqrt{\frac{3RT_{H_2}}{M_{H_2}}} = \sqrt{\frac{3RT_{O_2}}{M_{O_2}}}$$

$$\frac{T_{H_2}}{M_{H_2}} = \frac{T_{O_2}}{M_{O_2}}$$

$$T_{H_2} = T_{O_2} \times \frac{M_{H_2}}{M_{O_2}} = 320 \times \frac{2}{32}$$

$$T_{H_2} = 320 \times \frac{1}{16} = 20 \text{ K}$$

The correct answer is $$20 \text{ K}$$.