A cotainer contains CO and `H_(2)` in equal molar concentrations at the same temperature. What is the ratio of the average molar kinetic energy of the two gases ?

Options

A. Depends upon volume
B. `1 : 1`
C. `KE_(CO) gt KE_(H_(2))`
D. `KE_(H_(2)) gt KE_(CO)`

Explanation

To find the ratio of the average molar kinetic energy of CO and H₂ gases in a container where they are present in equal molar concentrations at the same temperature, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Concept of Average Molar Kinetic Energy**:
The average molar kinetic energy (KE) of a gas is given by the formula:
\[ KE = \frac{3}{2} RT \]
where \( R \) is the universal gas constant and \( T \) is the absolute temperature.

2. **Identify the Given Information**:
- We have two gases: Carbon Monoxide (CO) and Hydrogen (H₂).
- Both gases are at the same temperature \( T \).
- They are present in equal molar concentrations.

3. **Calculate the Average Molar Kinetic Energy for Each Gas**:
- For CO:
\[ KE_{CO} = \frac{3}{2} RT \]
- For H₂:
\[ KE_{H_2} = \frac{3}{2} RT \]

4. **Find the Ratio of the Average Molar Kinetic Energies**:
Since both gases have the same expression for average molar kinetic energy:
\[ \frac{KE_{CO}}{KE_{H_2}} = \frac{\frac{3}{2} RT}{\frac{3}{2} RT} = 1 \]

5. **Conclusion**:
The ratio of the average molar kinetic energy of CO to H₂ is:
\[ \text{Ratio} = 1:1 \]

### Final Answer:
The ratio of the average molar kinetic energy of CO to H₂ is \( 1:1 \).

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