Question

Figure shows a vessel partitioned by a fixed diathermic 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.
Figure

Answer 1

Ratio of the mass of a molecule in the left part to right part = 1.18 : 1

Explanation:

Given that V(rms) = V(avg)

Root of 3KT/ m1 = Root of 8KT/  m2

Where m1 and m2 are mass of gas in left and right chamber respectively.

3m2 = 8m1 / 3.14

m1 / m2 = 1.18

So the ratio is 1.18 : 1.

Answer 2

The ratio of the mass of a molecule in the left part to the mass of a molecule in the right part is $1.175$

Explanation:

Step 1:

Let gas temperature be T in both chambers.

Let the gas's molar mass in chamber left and chamber right be M1 and M2  correspondingly.

Let the gas mass in the chamber at left and right be m1 and m2  correspondingly .

Step 2:

Then ,

 $V_{r m s}=\sqrt{\frac{3 k T}{m_{1}}}=\sqrt{\frac{8 k T}{\pi m_{2}}}$

$3 m_{1}=\frac{8 m_{2}}{3.14}$

$3 \times 3.14=\frac{8 m_{2}}{m_{1}}$

$\frac{m_{2}}{m_{1}}=\frac{3 \times 3.14}{8}$

$\frac{m_{2}}{m_{1}}=\frac{9.42}{8}$

$\frac{m_{2}}{m_{2}}=1.175$