Question

N molecules each of mass m of gas A and 2N molecules each of mass 2m of gas B are contained in a vessel which is maintained at a temperature T. The mean square velocity of the molecules of gas B is denoted by V₂² and the mean square of the x-component velocity of the molecules of gas B is denoted by V₁², then V₁\V₂ is: 

• A. 2
• B. 1
• C. 2/3
• D. √2/3

Answer 1

The Correct Option is D

To solve the problem, we need to find the ratio $ \frac{V_1}{V_2} $, where $V_1$ is the root of the mean square of the x-component velocity of gas A molecules, and $V_2$ is the root mean square velocity of gas B molecules.

1. Understanding the Relation for RMS Velocity:
The root mean square (RMS) speed of gas molecules is given by:
$ V_{\text{rms}} = \sqrt{ \frac{3kT}{m} } $
Where:
- $k$ is Boltzmann constant
- $T$ is temperature
- $m$ is the mass of one molecule

2. Details from the Question:
- Gas A: N molecules, each of mass = $m$
- Gas B: 2N molecules, each of mass = $2m$
- Temperature = $T$ (same for both gases)
- Mean square velocity of gas B = $V_2^2$
- Mean square of x-component of velocity of gas A = $V_1^2$

3. Mean Square of One Component of Velocity:
Since the velocity components are isotropic (equal in all directions), we know:
$ V_{x}^2 = \frac{1}{3} V_{\text{rms}}^2 $
So,
$ V_1^2 = \frac{1}{3} \cdot \frac{3kT}{m} = \frac{kT}{m} $
$ V_2^2 = \frac{3kT}{2m} $

4. Take Ratio of Speeds:
We want $ \frac{V_1}{V_2} = \sqrt{ \frac{V_1^2}{V_2^2} } $
Substitute the values:
$ \frac{V_1^2}{V_2^2} = \frac{ \frac{kT}{m} }{ \frac{3kT}{2m} } = \frac{1}{1} \cdot \frac{1}{\frac{3}{2}} = \frac{2}{3} $
So,
$ \frac{V_1}{V_2} = \sqrt{ \frac{2}{3} } $

Final Answer:
$ \frac{V_1}{V_2} = \sqrt{ \frac{2}{3} } $