Determine the molecular kinetic energy per mole 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 mole of nitrogen molecules at 227 °C, we can use the formula for the kinetic energy of an ideal gas:

\[ KE = \frac{3}{2} RT \]

Where:
- \( KE \) is the kinetic energy per mole,
- \( R \) is the universal gas constant,
- \( T \) is the absolute temperature in Kelvin.

### Step 1: Convert the temperature from Celsius to Kelvin
The temperature in Kelvin can be calculated using the formula:
\[ T(K) = T(°C) + 273.15 \]
Given \( T = 227 °C \):
\[ T = 227 + 273.15 = 500.15 K \approx 500 K \]

### Step 2: Substitute the values into the kinetic energy formula
Now we can substitute the values into the kinetic energy formula:
- \( R = 8.310 \, J \, mol^{-1} \, K^{-1} \)
- \( T = 500 \, K \)

\[ KE = \frac{3}{2} \times 8.310 \, J \, mol^{-1} \, K^{-1} \times 500 \, K \]

### Step 3: Calculate the kinetic energy
Now we perform the multiplication:
\[ KE = \frac{3}{2} \times 8.310 \times 500 \]
Calculating \( 8.310 \times 500 \):
\[ 8.310 \times 500 = 4155 \]
Now, calculating \( \frac{3}{2} \times 4155 \):
\[ KE = \frac{3 \times 4155}{2} = \frac{12465}{2} = 6232.5 \, J \, mol^{-1} \]

### Final Answer
The molecular kinetic energy per mole of nitrogen molecules at 227 °C is approximately:
\[ KE \approx 6232.5 \, J \, mol^{-1} \]

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