Determine the molecular kinetic energy per molecule of nitrogen molecules at `227 ""^(@)C, R = 8.310 J "mol"e^(-1) K^(-1)`,No = `6.03 xx 10^(26)` moleculesKmol`e^(- 1)`. Molecular weight of nitrogen = 28.

Answer

To determine the molecular kinetic energy per molecule of nitrogen at 227 °C, we can follow these steps:

### Step 1: Convert the temperature from Celsius to Kelvin
To convert Celsius to Kelvin, we use the formula:
\[ T(K) = T(°C) + 273 \]
Given \( T = 227 °C \):
\[ T = 227 + 273 = 500 \, K \]

### Step 2: Use the formula for kinetic energy
The kinetic energy (KE) per mole of gas is given by the formula:
\[ KE = \frac{3}{2} RT \]
Where:
- \( R = 8.31 \, \text{J mol}^{-1} \text{K}^{-1} \)
- \( T = 500 \, K \)

### Step 3: Calculate the kinetic energy per mole
Substituting the values into the formula:
\[ KE = \frac{3}{2} \times 8.31 \, \text{J mol}^{-1} \text{K}^{-1} \times 500 \, K \]
\[ KE = \frac{3}{2} \times 8.31 \times 500 \]
\[ KE = \frac{3}{2} \times 4155 = 6232.5 \, \text{J mol}^{-1} \]

### Step 4: Convert kinetic energy per mole to kinetic energy per molecule
To find the kinetic energy per molecule, we divide the kinetic energy per mole by Avogadro's number (\( N_A \)):
\[ KE_{\text{per molecule}} = \frac{KE_{\text{per mole}}}{N_A} \]
Where \( N_A = 6.03 \times 10^{23} \, \text{molecules mol}^{-1} \):
\[ KE_{\text{per molecule}} = \frac{6232.5 \, \text{J mol}^{-1}}{6.03 \times 10^{23} \, \text{molecules mol}^{-1}} \]

### Step 5: Calculate the final result
Now, calculating the above expression:
\[ KE_{\text{per molecule}} = \frac{6232.5}{6.03 \times 10^{23}} \]
\[ KE_{\text{per molecule}} \approx 1.033 \times 10^{-20} \, \text{J} \]

Thus, the molecular kinetic energy per molecule of nitrogen at 227 °C is approximately:
\[ \boxed{1.033 \times 10^{-20} \, \text{J}} \]