Question

The relation between root mean square speed \((v_{rms})\) and most probable speed (\(v_p\)) for the molar mass M of oxygen gas molecule at the temperature of 300 K will be

• A. \(v_{rms} = \sqrt {\frac 23}v_p\)
• B. \(v_{rms} = \sqrt {\frac 32}v_p\)
• C. \(v_{rms} = v_p\)
• D. \(v_{rms} = \sqrt {\frac 13}v_p\)

Answer 1

The Correct Option is B

The problem at hand requires us to express the relationship between the root mean square speed \((v_{rms})\) and the most probable speed \((v_p)\) of gas molecules, specifically oxygen in this case. These speeds are characteristic of the Maxwell-Boltzmann distribution describing the velocity of gas molecules at a given temperature. Let's analyze this relationship step by step.

1. **Understanding Root Mean Square Speed (\(v_{rms}\))**:

The root mean square speed is a measure of the average speed of particles in a gas, given by the formula:

v_{rms} = \sqrt{\frac{3kT}{m}}

where \(k\) is the Boltzmann constant, \(T\) is the temperature in Kelvin, and \(m\) is the mass of a single molecule of the gas.

2. **Understanding Most Probable Speed (\(v_p\))**:

The most probable speed is the speed at which the largest number of molecules is moving, given by the formula:

v_p = \sqrt{\frac{2kT}{m}}

3. **Finding the Relationship**:

Given the equations for \(v_{rms}\) and \(v_p\), we can express the ratio between these two speeds:

\frac{v_{rms}}{v_p} = \frac{\sqrt{\frac{3kT}{m}}}{\sqrt{\frac{2kT}{m}}}

Simplifying under the square root:

v_{rms} = \sqrt{\frac{3}{2}} \cdot v_p

4. **Conclusion**:

The correct relationship between the root mean square speed and the most probable speed for a gas molecule at a given temperature (in this case, Oxygen at 300 K, although temperature is not a factor influencing the ratio directly) is:

v_{rms} = \sqrt{\frac{3}{2}} \cdot v_p

This matches the given correct option: v_{rms} = \sqrt{\frac{3}{2}}v_p.