A vessel is partitioned in two equal halves by a fixed diathermic separator. Two different ideal gases are filled in left (L) and right (R) halves. The rms speed of the molecules in L part is equal to the mean speed of molecules in the R part. Then the ratio of the mass of a molecule in L part to that of a molecule in R part is

The correct answer is 3π/8.

Let us solve this problem step-by-step by analyzing the conditions given in the question.

Step 1: Thermal Equilibrium
The vessel is partitioned into two equal halves by a fixed diathermic separator. A diathermic wall allows heat to flow freely between the two compartments until they reach thermal equilibrium.
Therefore, the temperature of both the left (L) and right (R) halves must be the same:
$T_L=T_R=T$

Step 2: Expressions for molecular speeds
Let $m_L$ be the mass of a single molecule of the gas in the left part, and $m_R$ be the mass of a single molecule of the gas in the right part.
The root-mean-square (rms) speed of the molecules in the left compartment (L) is given by:
$v_{\mathrm{rms},L}=\sqrt{\frac{3kT}{m_L}}$
where $k$ is the Boltzmann constant.
The mean (average) speed of the molecules in the right compartment (R) is given by:
$v_{\mathrm{mean},R}=\sqrt{\frac{8kT}{\pi m_R}}$

Step 3: Equating the two speeds
According to the problem, the rms speed in the L part is equal to the mean speed in the R part:
$v_{\mathrm{rms},L}=v_{\mathrm{mean},R}$
Substituting the expressions, we get:
$\sqrt{\frac{3kT}{m_L}}=\sqrt{\frac{8kT}{\pi m_R}}$

Step 4: Finding the ratio of molecular masses
Squaring both sides of the equation:
$\frac{3kT}{m_L}=\frac{8kT}{\pi m_R}$
We can cancel the common terms $k$ and $T$ from both sides:
$\frac{3}{m_L}=\frac{8}{\pi m_R}$
Rearranging the equation to find the ratio of the mass of a molecule in the L part to that of a molecule in the R part ($\frac{m_L}{m_R}$):
$\frac{m_L}{m_R}=\frac{3\pi }{8}$