Three closed vessels `A, B` and `C` are at the same temperature T and contain gases which obey the Maxwellian distribution of velocities. Vessel A contains only `O_(2), B` only `N_(2)` and `C` a mixture of equal quantities of `O_(2)` and `N_(2)`. If the average speed of the `O_(2)` molecules in vessel `A` is `V_(1)`, that of the `N_(2)` molecules in vessel `B` is `V_(2)`, the average speed of the `O_(2)` molecules in vessel `C` is (where `M` is the mass of an oxygen molecules)

Options

A. `(V_(1)+V_(2))/2`
B. `V_(1)`
C. `(V_(1)V_(2))^(1//2)`
D. `(V_(1))/(2)`

Explanation

To solve the problem, we need to analyze the average speeds of the gas molecules in the three vessels based on the information given.

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have three vessels: A, B, and C.
- Vessel A contains only \( O_2 \) (oxygen).
- Vessel B contains only \( N_2 \) (nitrogen).
- Vessel C contains a mixture of equal quantities of \( O_2 \) and \( N_2 \).

2. **Average Speed Formula**:
- The average speed of gas molecules can be expressed using the formula derived from the Maxwell-Boltzmann distribution:
\[ V_{\text{avg}} = \sqrt{\frac{8kT}{\pi m}} \]
- Here, \( k \) is the Boltzmann constant, \( T \) is the temperature, and \( m \) is the mass of a gas molecule.

3. **Identifying Variables**:
- Let \( V_1 \) be the average speed of \( O_2 \) in vessel A.
- Let \( V_2 \) be the average speed of \( N_2 \) in vessel B.
- We need to find the average speed of \( O_2 \) in vessel C.

4. **Temperature Consideration**:
- All vessels are at the same temperature \( T \). Therefore, the temperature factor cancels out when comparing average speeds.

5. **Mass Consideration**:
- The mass of \( O_2 \) in vessel A is equal to the mass of \( O_2 \) in vessel C because it is given that vessel C contains the same amount of oxygen as vessel A.
- Thus, the mass \( m \) for \( O_2 \) in both vessels A and C is the same.

6. **Conclusion**:
- Since the average speed \( V_{\text{avg}} \) is directly proportional to \( \frac{1}{\sqrt{m}} \) and the mass of \( O_2 \) in vessels A and C is the same, we can conclude:
\[ V_{\text{avg}}(O_2 \text{ in C}) = V_1 \]
- Therefore, the average speed of \( O_2 \) molecules in vessel C is equal to the average speed of \( O_2 \) molecules in vessel A, which is \( V_1 \).

### Final Answer:
The average speed of the \( O_2 \) molecules in vessel C is \( V_1 \).