`E_(0)` and `E_(n)` respectively represent the average kinetic energy of a molecule of oxygen and hydrogen. If the two gases are at the same temperature, which of the following statements is true?

Options

A. `E_O gt E_H`
B. `E_O = E_H`
C. `E_O lt E_H`
D. Data insufficient

Answer

To solve the problem, we need to analyze the average kinetic energy of the molecules of oxygen (O2) and hydrogen (H2) at the same temperature.

### Step-by-Step Solution:

1. **Understanding Average Kinetic Energy**:
The average kinetic energy (E) of a molecule in a gas is given by the formula:
\[ E = \frac{F}{2} k T \]
where \( F \) is the degrees of freedom, \( k \) is the Boltzmann constant, and \( T \) is the absolute temperature.

2. **Identifying Degrees of Freedom**:
For diatomic gases like O2 and H2, the degrees of freedom include translational and rotational motions.
- Translational degrees of freedom: 3 (movement in x, y, z directions)
- Rotational degrees of freedom: 2 (rotation about two axes)
- Total degrees of freedom for diatomic gases = 3 (translational) + 2 (rotational) = 5.

3. **Applying the Formula to Both Gases**:
Since both gases are diatomic and at the same temperature, we can express their average kinetic energies as:
\[ E_O = \frac{5}{2} k T \quad \text{(for O2)} \]
\[ E_H = \frac{5}{2} k T \quad \text{(for H2)} \]

4. **Comparing the Average Kinetic Energies**:
Since both expressions for \( E_O \) and \( E_H \) are equal:
\[ E_O = E_H \]
This indicates that the average kinetic energy of a molecule of oxygen is equal to that of a molecule of hydrogen when both are at the same temperature.

5. **Conclusion**:
Therefore, the correct statement is that the average kinetic energy of a molecule of oxygen (E_O) is equal to the average kinetic energy of a molecule of hydrogen (E_H) when both gases are at the same temperature.

### Final Answer:
The average kinetic energy of a molecule of oxygen (E_O) is equal to the average kinetic energy of a molecule of hydrogen (E_H) when both are at the same temperature.

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