What is the average kinetic energy of a gas molecule at `27^(@)C`?

Explanation

To find the average kinetic energy of a gas molecule at \(27^\circ C\), we can follow these steps:

### Step 1: Convert Temperature to Kelvin
The first step is to convert the given temperature from Celsius to Kelvin. The formula for this conversion is:

\[ T(K) = T(°C) + 273 \]

For \(27^\circ C\):

\[ T(K) = 27 + 273 = 300 \, K \]

### Step 2: Use the Formula for Average Kinetic Energy
The average kinetic energy (\(KE\)) of a gas molecule is given by the formula:

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

Where:
- \(k_B\) is the Boltzmann constant, which is \(1.380649 \times 10^{-23} \, J/K\)
- \(T\) is the temperature in Kelvin

### Step 3: Substitute Values into the Formula
Now, we substitute the values of \(k_B\) and \(T\) into the formula:

\[ KE = \frac{3}{2} \times (1.380649 \times 10^{-23} \, J/K) \times (300 \, K) \]

### Step 4: Calculate the Average Kinetic Energy
Calculating the above expression:

\[ KE = \frac{3}{2} \times 1.380649 \times 10^{-23} \times 300 \]

Calculating \(1.380649 \times 300\):

\[ 1.380649 \times 300 = 414.1947 \]

Now, multiplying by \(\frac{3}{2}\):

\[ KE = \frac{3}{2} \times 414.1947 \times 10^{-23} = 621.29205 \times 10^{-23} \, J \]

This can be simplified to:

\[ KE \approx 6.21 \times 10^{-21} \, J \]

### Final Answer
The average kinetic energy of a gas molecule at \(27^\circ C\) is approximately:

\[ KE \approx 6.21 \times 10^{-21} \, J \]

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