Average K.E. of `CO_(2)` at `27^(@)C` is E. the average kinetic energy of `N_(2)` at the same temperature will be

Options

A. E
B. 22E
C. E/22
D. `E//sqrt2`

Answer

To solve the problem regarding the average kinetic energy of \( N_2 \) at \( 27^\circ C \) given that the average kinetic energy of \( CO_2 \) at the same temperature is \( E \), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Concept of Average Kinetic Energy**:
The average kinetic energy (K.E.) of one mole of an ideal gas is given by the formula:
\[ \text{K.E.} = \frac{3}{2}RT \]
where \( R \) is the universal gas constant and \( T \) is the absolute temperature in Kelvin.

2. **Convert Temperature to Kelvin**:
The given temperature is \( 27^\circ C \). To convert this to Kelvin:
\[ T(K) = 27 + 273 = 300 \, K \]

3. **Kinetic Energy of \( CO_2 \)**:
According to the problem, the average kinetic energy of \( CO_2 \) at \( 27^\circ C \) is \( E \). Using the formula:
\[ E = \frac{3}{2}R(300) \]

4. **Kinetic Energy of \( N_2 \)**:
Since the average kinetic energy of an ideal gas depends only on the temperature and not on the type of gas, the average kinetic energy of \( N_2 \) at the same temperature (300 K) will also be:
\[ \text{K.E. of } N_2 = \frac{3}{2}R(300) = E \]

5. **Conclusion**:
Therefore, the average kinetic energy of \( N_2 \) at \( 27^\circ C \) is also \( E \).

### Final Answer:
The average kinetic energy of \( N_2 \) at \( 27^\circ C \) will be \( E \).

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