Calculate the average kinetic energy of neon molecule at `27^@ C`. Given `k_(B) = 1.38 xx 10^(-23) JK^(-1)`.

Explanation

To calculate the average kinetic energy of a neon molecule at 27°C, we can follow these steps:

### Step 1: Convert the temperature from Celsius to Kelvin
The formula to convert Celsius to Kelvin is:
\[ T(K) = T(°C) + 273 \]
Given that the temperature is 27°C:
\[ T = 27 + 273 = 300 \, K \]

### Step 2: Use the formula for average kinetic energy
The average kinetic energy \(E\) of a molecule is given by the formula:
\[ E = \frac{3}{2} k_B T \]
where \(k_B\) is the Boltzmann constant.

### Step 3: Substitute the values into the formula
We know:
- \(k_B = 1.38 \times 10^{-23} \, J/K\)
- \(T = 300 \, K\)

Now substituting these values into the formula:
\[ E = \frac{3}{2} \times (1.38 \times 10^{-23} \, J/K) \times (300 \, K) \]

### Step 4: Calculate the average kinetic energy
Now we can perform the calculation:
\[ E = \frac{3}{2} \times 1.38 \times 10^{-23} \times 300 \]
Calculating step-by-step:
1. First, calculate \(1.38 \times 300 = 414\).
2. Now multiply by \(\frac{3}{2}\):
\[ E = \frac{3}{2} \times 414 \times 10^{-23} = 621 \times 10^{-23} \, J \]
3. This simplifies to:
\[ E = 6.21 \times 10^{-21} \, J \]

### Final Answer
Thus, the average kinetic energy of a neon molecule at 27°C is:
\[ E = 6.21 \times 10^{-21} \, J \]

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