Figure show's a vessel portioned by a fixed diathermic separator. Different ideal gases are filled in the two parts. The rms speed of the molecules in the left part equal the mean speed of the molecules in the right part. Calculate the ratio of the mass of a molecule in the left part of the mass of a molecule in the right part.

The separator is diathermal it means the temperature of gas on both parts of vessel should be equal, let this common temperature is T.
Hence rms speed of a molecule of the gas in left part
${\mathrm{v}}_{\mathrm{rms}} = \sqrt{\frac{3\mathrm{kT}}{{\mathrm{m}}_1}}$----------------(i)

${\mathrm{m}}_1$ is the mass of a molecule in left part.
Similarly the mean speed of a molecule of the gas in right part
${\mathrm{v}}_{\mathrm{mean}} = \sqrt{\frac{8\mathrm{kT}}{{\mathrm{\pi m}}_2}}$

${\mathrm{m}}_2$ is the mass of a molecule in right part.
As per question
${\mathrm{r}}_{\mathrm{rms}} = {\mathrm{v}}_{\mathrm{mean}} \Rightarrow \sqrt{\frac{3\mathrm{kT}}{{\mathrm{m}}_1}} = \sqrt{\frac{8\mathrm{kT}}{\mathrm{\pi } {\mathrm{m}}_2}}$

Hence $\frac{{\mathrm{m}}_1}{{\mathrm{m}}_2} = \frac{3\mathrm{\pi }}{8}$