Calculate the average kinetic energy of neon molecule at `27^(@)C`. Given `k_(B)=1.38xx10^(-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
To convert the temperature from Celsius to Kelvin, we use the formula:
\[ T(K) = T(°C) + 273 \]

Given:
\[ T(°C) = 27 \]
So,
\[ T(K) = 27 + 273 = 300 \, K \]

### Step 2: Identify the degrees of freedom for a neon molecule
Neon (Ne) is a monoatomic gas, which means it has three degrees of freedom (all translational). Therefore:
\[ F = 3 \]

### Step 3: Use the formula for average kinetic energy
The average kinetic energy (KE) of a molecule of gas can be calculated using the formula:
\[ KE = \frac{F}{2} k_B T \]

Where:
- \( k_B \) is the Boltzmann constant, given as \( 1.38 \times 10^{-23} \, J/K \)
- \( T \) is the temperature in Kelvin

### Step 4: Substitute the values into the formula
Now substituting the values into the formula:
\[ KE = \frac{3}{2} \times (1.38 \times 10^{-23} \, J/K) \times (300 \, K) \]

### Step 5: Calculate the average kinetic energy
Calculating the above expression:
\[ KE = \frac{3}{2} \times 1.38 \times 10^{-23} \times 300 \]
\[ KE = \frac{3 \times 1.38 \times 300}{2} \times 10^{-23} \]
\[ KE = \frac{1242}{2} \times 10^{-23} \]
\[ KE = 621 \times 10^{-23} \, J \]
\[ KE = 6.21 \times 10^{-21} \, J \]

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

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