Two gases `O_2` and `H_2` are at the same temperature . If `E_o` is the average kinetic energy of a molecule of oxygen sample , and `E_H` is the average kinetic energy of a molecule of hydrogen sample , then

Options

A. `E_o=1/16E_H`
B. `E_o=16E_H`
C. `E_o gt E_H`
D. `E_o=E_H`

Answer

To solve the problem, we need to analyze the average kinetic energy of gas molecules. The average kinetic energy of a molecule in a gas is given by the formula:

\[ E = \frac{3}{2} k_B T \]

where:
- \( E \) is the average kinetic energy,
- \( k_B \) is the Boltzmann constant,
- \( T \) is the absolute temperature of the gas.

Given that we have two gases, oxygen (\( O_2 \)) and hydrogen (\( H_2 \)), both at the same temperature, we can denote their average kinetic energies as follows:

- For oxygen, the average kinetic energy is \( E_O \).
- For hydrogen, the average kinetic energy is \( E_H \).

Since both gases are at the same temperature, we can express their average kinetic energies using the formula mentioned above:

1. For oxygen:
\[ E_O = \frac{3}{2} k_B T \]

2. For hydrogen:
\[ E_H = \frac{3}{2} k_B T \]

Since both \( E_O \) and \( E_H \) depend on the same temperature \( T \) and the same Boltzmann constant \( k_B \), we can equate the two expressions:

\[ E_O = E_H \]

Thus, the relationship between the average kinetic energies of the two gases is:

\[ E_O = E_H \]

**Final Answer:**
The average kinetic energy of a molecule of oxygen is equal to the average kinetic energy of a molecule of hydrogen at the same temperature, i.e., \( E_O = E_H \).

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