The average kinetic energy of a gas molecule at `27^(@)C` is `621xx10^(-1)J`. The average kinetic energy of gas molecule at `227^(@)C` will be

Options

A. `52.2xx10^(-21)J`
B. `5.22xx10^(-21)J`
C. `10.35xx10^(-21)J`
D. `11.35xx10^(-21)J`

Explanation

To find the average kinetic energy of a gas molecule at `227°C`, we can use the relationship between kinetic energy and temperature. The average kinetic energy (KE) of a gas molecule is given by the formula:

\[ KE = \frac{3}{2} k T \]

where:
- \( k \) is the Boltzmann constant,
- \( T \) is the absolute temperature in Kelvin.

### Step-by-Step Solution:

1. **Convert the temperatures from Celsius to Kelvin**:
- To convert Celsius to Kelvin, use the formula:
\[ T(K) = T(°C) + 273.15 \]
- For `27°C`:
\[ T_1 = 27 + 273.15 = 300.15 \, K \]
- For `227°C`:
\[ T_2 = 227 + 273.15 = 500.15 \, K \]

2. **Use the ratio of kinetic energies**:
- The kinetic energy at two different temperatures can be related as follows:
\[ \frac{KE_2}{KE_1} = \frac{T_2}{T_1} \]
- Rearranging gives:
\[ KE_2 = KE_1 \times \frac{T_2}{T_1} \]

3. **Substitute the known values**:
- Given that \( KE_1 = 6.21 \times 10^{-21} \, J \):
- Substitute \( T_1 \) and \( T_2 \):
\[ KE_2 = 6.21 \times 10^{-21} \, J \times \frac{500.15}{300.15} \]

4. **Calculate the ratio of temperatures**:
- Calculate \( \frac{500.15}{300.15} \):
\[ \frac{500.15}{300.15} \approx 1.666 \]

5. **Calculate \( KE_2 \)**:
- Now substitute back to find \( KE_2 \):
\[ KE_2 \approx 6.21 \times 10^{-21} \, J \times 1.666 \]
\[ KE_2 \approx 10.35 \times 10^{-21} \, J \]

6. **Final Result**:
- Therefore, the average kinetic energy of the gas molecule at `227°C` is approximately:
\[ KE_2 \approx 10.35 \times 10^{-21} \, J \]