A vessel contains a mixture of one mole of oxygen and two moles of nitrogen at 300K. The ratio of the average rorational kinetic energy per `O_2` molecules to that per `N_2` molecules is

Explanation

To find the ratio of the average rotational kinetic energy per \( O_2 \) molecules to that per \( N_2 \) molecules, we can follow these steps:

### Step 1: Understand the Degrees of Freedom
Both oxygen (\( O_2 \)) and nitrogen (\( N_2 \)) are diatomic molecules. For diatomic molecules, the degrees of freedom associated with rotational motion are 2.

### Step 2: Use the Equipartition Theorem
According to the Maxwell's law of Equipartition of energy, the average kinetic energy per molecule can be expressed as:
\[ \text{Average Kinetic Energy} = \frac{f}{2} k T \]
where \( f \) is the number of degrees of freedom, \( k \) is the Boltzmann constant, and \( T \) is the temperature in Kelvin.

### Step 3: Calculate Average Rotational Kinetic Energy for \( O_2 \)
For \( O_2 \):
- Degrees of freedom \( f = 2 \) (for rotational motion)
- Therefore, the average rotational kinetic energy per \( O_2 \) molecule is:
\[ KE_{O_2} = \frac{2}{2} k T = k T \]

### Step 4: Calculate Average Rotational Kinetic Energy for \( N_2 \)
For \( N_2 \):
- Similarly, the average rotational kinetic energy per \( N_2 \) molecule is:
\[ KE_{N_2} = \frac{2}{2} k T = k T \]

### Step 5: Find the Ratio of Average Rotational Kinetic Energies
Now, we can find the ratio of the average rotational kinetic energy per \( O_2 \) molecule to that per \( N_2 \) molecule:
\[ \text{Ratio} = \frac{KE_{O_2}}{KE_{N_2}} = \frac{k T}{k T} = 1 \]

### Conclusion
The ratio of the average rotational kinetic energy per \( O_2 \) molecules to that per \( N_2 \) molecules is:
\[ \text{Ratio} = 1 : 1 \]

---