Suppose a container is evacuated to leave just one molecule of a gas in it. Let `v_(mp)` and `v_(av)` represent the most probable speed and the average speed of the gas, then

Options

A. `v_(mp) gt v_(av)`
B. `v_(mp) lt v_(av)`
C. `v_(mp) = v_(av)`
D. none of these

Explanation

To solve the problem, we need to analyze the definitions of the most probable speed (\(v_{mp}\)) and the average speed (\(v_{av}\)) of a gas molecule in a container that has been evacuated to leave just one molecule.

### Step-by-Step Solution:

1. **Understanding the Scenario**:
- We have a container that has been evacuated to leave just one molecule of gas in it.

2. **Defining Average Speed**:
- The average speed (\(v_{av}\)) of a gas is defined as the total speed of all molecules divided by the number of molecules. Mathematically, it is given by:
\[ v_{av} = \frac{v_1 + v_2 + ... + v_n}{n} \]
- In our case, since there is only one molecule, let’s denote its speed as \(v\). Therefore:
\[ v_{av} = \frac{v}{1} = v \]

3. **Defining Most Probable Speed**:
- The most probable speed (\(v_{mp}\)) is defined as the speed at which the maximum number of molecules are found. It is typically calculated from the Maxwell-Boltzmann distribution.
- However, since we only have one molecule, this molecule's speed is also \(v\). Thus:
\[ v_{mp} = v \]

4. **Comparing the Speeds**:
- Now we compare the two speeds:
\[ v_{mp} = v \quad \text{and} \quad v_{av} = v \]
- Since both speeds are equal, we conclude that:
\[ v_{mp} = v_{av} \]

5. **Final Conclusion**:
- Therefore, in this scenario, both the most probable speed and the average speed are equal.

### Final Answer:
Both speeds are equal: \(v_{mp} = v_{av}\).

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