Question
Figure. shows a vessel partitioned by a fixed diathernnc separator. Different ideal gases are filled in the two parts. The rms speed of the molecules in the left part equals the mean speed of the molecules in the right part. Calculate the ratio of the mass of a molecule in the left part to the mass of a molecule in the right part.

Answer

The left side of the container has a gas, let having molecular $wt. M_1$
Right part has $Mol.\ wt = M_2$
Temperature of both left and right chambers are equal as the separating wall is diathermic,
$\sqrt{\frac{3\text{RT}}{\text{M}_1}}=\sqrt{\frac{8\text{RT}}{\pi\text{M}_2}}\Rightarrow\frac{3\text{RT}}{\text{M}_1}=\frac{8\text{RT}}{\pi\text{M}_2}$
$\Rightarrow\frac{\text{M}_1}{\pi\text{M}_2}=\frac{3}{8}$
$\Rightarrow\frac{\text{M}_1}{\text{M}_2}$
$=\frac{3\pi}{8}=1.1778\approx1.18$

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