Two vessels A and B are identical. A has 1 g hydrogen at `0^(@)C` and B has 1 g oxygen at `0^(@)C`. Vessel A contains x molecules and B contains y molecules. The average kinetic energy per molecules in A is 'n' times the average kinetic energy per molecule in B. The value of 'n' is

Options

A. 16
B. 8
C. 32
D. 1

Answer

To solve the problem, we need to find the ratio of the average kinetic energy per molecule of hydrogen (in vessel A) to the average kinetic energy per molecule of oxygen (in vessel B).

### Step-by-Step Solution:

1. **Understanding Average Kinetic Energy**:
The average kinetic energy (KE) per molecule of an ideal gas can be expressed using the formula:
\[ KE = \frac{f}{2} k T \]
where \( f \) is the degrees of freedom, \( k \) is the Boltzmann constant, and \( T \) is the absolute temperature in Kelvin.

2. **Identify the Gases**:
- For hydrogen (H₂), which is a diatomic gas, the degrees of freedom \( f \) is 5.
- For oxygen (O₂), which is also a diatomic gas, the degrees of freedom \( f \) is similarly 5.

3. **Temperature Conversion**:
Both gases are at \( 0^\circ C \), which is equivalent to \( 273 \, K \).

4. **Calculate Average Kinetic Energy for Each Gas**:
- For hydrogen (A):
\[ KE_A = \frac{5}{2} k T \]
- For oxygen (B):
\[ KE_B = \frac{5}{2} k T \]

5. **Finding the Ratio of Kinetic Energies**:
Since both gases have the same degrees of freedom and are at the same temperature, their average kinetic energies will be equal:
\[ KE_A = KE_B \]
Therefore, the ratio \( n \) of the average kinetic energy per molecule in A to that in B is:
\[ n = \frac{KE_A}{KE_B} = \frac{\frac{5}{2} k T}{\frac{5}{2} k T} = 1 \]

6. **Conclusion**:
The value of \( n \) is \( 1 \).

### Final Answer:
The value of \( n \) is \( 1 \).

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