Question

At what temperature will the root mean square velocity of the molecule quarter that if it is at a temperature of 32 degree Celsius?

Answer 1

$$v_\text{rms1}=\sqrt\frac{3RT_1}{\Mu},\\\space\\ v_\text{rms2}=\frac14v_\text{rms1},\\\space\\ \sqrt\frac{3RT_2}{\Mu}=\frac14\sqrt\frac{3RT_1}{\Mu},\\\space\\ T_2=\frac{T_1}{16}=\frac{273+32}{16}=19\text{ K}.$$

Or -254°C.