If the average kinetic energy of gas molecule at `27^(@)C` is `6.21xx10^(-21)J`, then the average kinetic energy at `227^(@)C` will be

Options

A. `9.35xx10^(-21)J`
B. `10.35xx10^(-21)J`
C. `11.35xx10^(-21)J`
D. `12.35xx10^(-21)J`

Answer

To find the average kinetic energy of gas molecules at `227°C`, we can use the relationship between kinetic energy and temperature. The average kinetic energy (KE) of gas molecules is directly proportional to the absolute temperature (T) in Kelvin.

### Step-by-Step Solution:

1. **Identify the given values**:
- Average kinetic energy at `27°C` (T1) = `6.21 × 10^(-21) J`
- Temperature T1 = `27°C`
- Temperature T2 = `227°C`

2. **Convert temperatures from Celsius to Kelvin**:
- To convert Celsius to Kelvin, we use the formula:
\[ T(K) = T(°C) + 273 \]
- For T1:
\[ T1 = 27 + 273 = 300 \, K \]
- For T2:
\[ T2 = 227 + 273 = 500 \, K \]

3. **Use the relationship between kinetic energy and temperature**:
- The average kinetic energy is proportional to the temperature:
\[ \frac{KE_1}{KE_2} = \frac{T_1}{T_2} \]
- Rearranging gives:
\[ KE_2 = KE_1 \times \frac{T_2}{T_1} \]

4. **Substitute the known values**:
- Substitute KE1, T1, and T2 into the equation:
\[ KE_2 = 6.21 \times 10^{-21} \times \frac{500}{300} \]

5. **Calculate the ratio**:
- Simplify the fraction:
\[ \frac{500}{300} = \frac{5}{3} \]

6. **Calculate KE2**:
- Now substitute back:
\[ KE_2 = 6.21 \times 10^{-21} \times \frac{5}{3} \]
- Calculate:
\[ KE_2 = 6.21 \times 10^{-21} \times 1.6667 \approx 10.35 \times 10^{-21} \, J \]

### Final Answer:
The average kinetic energy at `227°C` is approximately:
\[ KE_2 \approx 10.35 \times 10^{-21} \, J \]