The kinetic energy for 14 grams of nitrogen gas at `127^@C` is nearly (mol. mass of nitrogen = 28 and gas constant = `8.31 JK^(-1) mol^(-1)`)

Options

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

Answer

To solve the problem of calculating the kinetic energy of 14 grams of nitrogen gas at 127°C, we will follow these steps:

### Step 1: Convert Temperature to Kelvin
To use the gas laws, we need to convert the temperature from Celsius to Kelvin. The formula for this conversion is:

\[ T(K) = T(°C) + 273 \]

Substituting the given temperature:

\[ T(K) = 127 + 273 = 400 \, K \]

### Step 2: Calculate the Number of Moles of Nitrogen Gas
We can calculate the number of moles using the formula:

\[ n = \frac{mass}{molar \, mass} \]

Given:
- Mass of nitrogen gas = 14 g
- Molar mass of nitrogen (N₂) = 28 g/mol

Substituting the values:

\[ n = \frac{14 \, g}{28 \, g/mol} = \frac{1}{2} \, mol = 0.5 \, mol \]

### Step 3: Use the Kinetic Energy Formula
The kinetic energy (KE) of an ideal gas is given by the formula:

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

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

Substituting the values we have:

\[ KE = \frac{3}{2} \times 0.5 \, mol \times 8.31 \, J/(K·mol) \times 400 \, K \]

### Step 4: Calculate the Kinetic Energy
Now, we perform the calculations step by step:

1. Calculate \( \frac{3}{2} \times 0.5 \):

\[ \frac{3}{2} \times 0.5 = \frac{3}{4} = 0.75 \]

2. Multiply by \( R \):

\[ 0.75 \times 8.31 = 6.2325 \, J/K \]

3. Finally, multiply by \( T \):

\[ KE = 6.2325 \, J/K \times 400 \, K = 2493 \, J \]

### Conclusion
The kinetic energy for 14 grams of nitrogen gas at 127°C is approximately **2493 Joules**.

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