For any distribution of speeds `v_(rms) ge v_(av)` Is this statement true or false ?

Explanation

To determine whether the statement "For any distribution of speeds, \( v_{rms} \geq v_{av} \)" is true or false, we can analyze the definitions of the root mean square (RMS) speed and the average speed.

### Step-by-Step Solution:

1. **Definitions**:
- The average speed (\( v_{av} \)) for a distribution of speeds can be defined as:
\[ v_{av} = \frac{n_1 v_1 + n_2 v_2}{n_1 + n_2} \]
where \( n_1 \) and \( n_2 \) are the number of molecules with speeds \( v_1 \) and \( v_2 \), respectively.

- The root mean square speed (\( v_{rms} \)) is defined as:
\[ v_{rms} = \sqrt{\frac{n_1 v_1^2 + n_2 v_2^2}{n_1 + n_2}} \]

2. **Analyzing the Relationship**:
- To compare \( v_{rms} \) and \( v_{av} \), we can consider the case where the speeds are different. If we have two speeds \( v_1 \) and \( v_2 \), we can see how the squares of these speeds influence the RMS value.

- If \( v_1 \) and \( v_2 \) are of opposite signs, the average speed can be zero, while the RMS speed will still be positive since it involves squaring the speeds.

3. **Special Cases**:
- If all molecules have the same speed (i.e., \( v_1 = v_2 \)), then:
\[ v_{av} = v_{rms} = v_1 \]
- In this case, \( v_{rms} = v_{av} \).

4. **General Case**:
- For any distribution of speeds, the RMS speed will always be greater than or equal to the average speed. This can be shown mathematically using the Cauchy-Schwarz inequality, which states that:
\[ \left( \sum a_i^2 \right) \left( \sum b_i^2 \right) \geq \left( \sum a_i b_i \right)^2 \]
- Applying this to our speeds confirms that:
\[ v_{rms}^2 \geq v_{av}^2 \]
- Taking the square root gives:
\[ v_{rms} \geq v_{av} \]

5. **Conclusion**:
- Therefore, the statement "For any distribution of speeds, \( v_{rms} \geq v_{av} \)" is **true**.