The average kinetic energy of gas molecule at `27^(@)C` is `6.21xx10^(-21)` J. Its average kinetic energy at `127^(@)C` will be

Options

A. `12.2xx10^(-21)J`
B. `8.28xx10^(-21)J`
C. `10.35xx10^(-21)J`
D. `11.35xx10^(-21)J`

Explanation

To find the average kinetic energy of a gas molecule at \(127^\circ C\), we can use the relationship between the average kinetic energy and the temperature of the gas. The average kinetic energy \(K\) of gas molecules is directly proportional to the absolute temperature \(T\) in Kelvin.

### Step 1: Convert the temperatures from Celsius to Kelvin
First, we need to convert the given temperatures from Celsius to Kelvin.

- For \(27^\circ C\):
\[ T_1 = 27 + 273 = 300 \, K \]

- For \(127^\circ C\):
\[ T_2 = 127 + 273 = 400 \, K \]

### Step 2: Use the proportionality of kinetic energy and temperature
The average kinetic energy \(K\) is proportional to the absolute temperature \(T\):
\[ K \propto T \]
This can be expressed as:
\[ \frac{K_1}{K_2} = \frac{T_1}{T_2} \]
Where:
- \(K_1\) is the average kinetic energy at \(27^\circ C\) (which is \(6.21 \times 10^{-21} \, J\))
- \(K_2\) is the average kinetic energy at \(127^\circ C\)

### Step 3: Substitute the known values into the equation
Substituting the known values into the equation:
\[ \frac{6.21 \times 10^{-21}}{K_2} = \frac{300}{400} \]

### Step 4: Solve for \(K_2\)
Now, we can solve for \(K_2\):
\[ K_2 = 6.21 \times 10^{-21} \times \frac{400}{300} \]
\[ K_2 = 6.21 \times 10^{-21} \times \frac{4}{3} \]
\[ K_2 = 6.21 \times 10^{-21} \times 1.3333 \]
\[ K_2 = 8.28 \times 10^{-21} \, J \]

### Final Answer
The average kinetic energy of the gas molecule at \(127^\circ C\) is approximately:
\[ K_2 \approx 8.28 \times 10^{-21} \, J \]

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