The kinetic energy for 14 g of nitrogen gas at `127^(@)C` is nearly (mol. Mass of nitrogen = 28 and gas constant = 8.31 J/mol K)

Options

A. 1.0 J
B. 4.15 J
C. 2493 J
D. 3.3 J

Answer

To find the kinetic energy of 14 g of nitrogen gas at 127°C, we can use the formula for kinetic energy of an ideal gas:

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

Where:
- \( KE \) = kinetic energy
- \( n \) = number of moles of the gas
- \( R \) = gas constant (8.31 J/mol·K)
- \( T \) = temperature in Kelvin

### Step 1: Convert the mass of nitrogen to moles
Given:
- Mass of nitrogen = 14 g
- Molar mass of nitrogen = 28 g/mol

To find the number of moles (\( n \)):
\[ n = \frac{\text{mass}}{\text{molar mass}} = \frac{14 \text{ g}}{28 \text{ g/mol}} = 0.5 \text{ moles} \]

### Step 2: Convert the temperature to Kelvin
Given:
- Temperature = 127°C

To convert to Kelvin:
\[ T = 127 + 273 = 400 \text{ K} \]

### Step 3: Substitute values into the kinetic energy formula
Now we can substitute the values of \( n \), \( R \), and \( T \) into the kinetic energy formula:
\[ KE = \frac{3}{2} \times n \times R \times T \]
\[ KE = \frac{3}{2} \times 0.5 \text{ moles} \times 8.31 \text{ J/mol·K} \times 400 \text{ K} \]

### Step 4: Calculate the kinetic energy
Calculating the above expression:
\[ KE = \frac{3}{2} \times 0.5 \times 8.31 \times 400 \]
\[ KE = \frac{3}{2} \times 0.5 \times 3324 \]
\[ KE = \frac{3}{2} \times 1662 = 2493 \text{ J} \]

### Step 5: Convert Joules to kilojoules
To convert Joules to kilojoules:
\[ KE = \frac{2493 \text{ J}}{1000} = 2.493 \text{ kJ} \approx 2.5 \text{ kJ} \]

### Final Answer
The kinetic energy for 14 g of nitrogen gas at 127°C is approximately **2.5 kJ**.

---