The average kinetic energy of an ideal gas per molecule in SI units at `25^(@)C` will be

Options

A. `6.17 xx 10^(-21) kJ`
B. `6.17 xx 10^(-21)J`
C. `6.17 xx 10^(-20)J`
D. `6.17 xx 10^(-20) kJ`

Explanation

To find the average kinetic energy of an ideal gas per molecule at \(25^\circ C\), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Formula**:
The average kinetic energy (\(KE_{avg}\)) of an ideal gas per molecule is given by the formula:
\[ KE_{avg} = \frac{3}{2} k T \]
where \(k\) is the Boltzmann constant and \(T\) is the temperature in Kelvin.

2. **Convert Temperature to Kelvin**:
The temperature is given as \(25^\circ C\). To convert this to Kelvin, use the formula:
\[ T(K) = T(°C) + 273 \]
Therefore:
\[ T = 25 + 273 = 298 \, K \]

3. **Substitute Values into the Formula**:
The value of the Boltzmann constant \(k\) is:
\[ k = 1.38 \times 10^{-23} \, \text{J/K} \]
Now substitute \(k\) and \(T\) into the kinetic energy formula:
\[ KE_{avg} = \frac{3}{2} \times (1.38 \times 10^{-23} \, \text{J/K}) \times (298 \, K) \]

4. **Calculate the Average Kinetic Energy**:
Now, perform the multiplication:
\[ KE_{avg} = \frac{3}{2} \times 1.38 \times 10^{-23} \times 298 \]
First, calculate \(1.38 \times 298\):
\[ 1.38 \times 298 \approx 411.24 \]
Now multiply by \(\frac{3}{2}\):
\[ KE_{avg} = \frac{3}{2} \times 411.24 \times 10^{-23} \approx 618.86 \times 10^{-23} \, \text{J} \]
Simplifying this gives:
\[ KE_{avg} \approx 6.19 \times 10^{-21} \, \text{J} \]

5. **Final Result**:
The average kinetic energy of an ideal gas per molecule at \(25^\circ C\) is approximately:
\[ KE_{avg} \approx 6.17 \times 10^{-21} \, \text{J} \]

### Conclusion:
The average kinetic energy of an ideal gas per molecule at \(25^\circ C\) is \(6.17 \times 10^{-21} \, \text{J}\).