The kinetic energy of two moles of `N_(2) at 27^(@) C is (R = 8.314 J K^(-1) mol^(-1))`

Options

A. 5491.6 J
B. 6491.6 J
C. 7491.6 J
D. 8882.4 J

Explanation

To find the kinetic energy of two moles of \( N_2 \) at \( 27^\circ C \), we can use the formula for the kinetic energy of an ideal gas:

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

Where:
- \( KE \) = kinetic energy
- \( N \) = number of moles
- \( R \) = universal gas constant
- \( T \) = absolute temperature in Kelvin

### Step 1: Convert the temperature from Celsius to Kelvin
The temperature in Celsius is given as \( 27^\circ C \). To convert this to Kelvin, we use the formula:

\[ T(K) = T(°C) + 273 \]

Substituting the given value:

\[ T = 27 + 273 = 300 \, K \]

### Step 2: Identify the values for \( N \), \( R \), and \( T \)
From the problem statement:
- Number of moles \( N = 2 \)
- Gas constant \( R = 8.314 \, J \, K^{-1} \, mol^{-1} \)
- Temperature \( T = 300 \, K \)

### Step 3: Substitute the values into the kinetic energy formula
Now we can substitute the values into the kinetic energy formula:

\[ KE = \frac{3}{2} NRT = \frac{3}{2} \times 2 \times 8.314 \times 300 \]

### Step 4: Calculate the kinetic energy
Calculating step by step:

1. Calculate \( NRT \):
\[ NRT = 2 \times 8.314 \times 300 = 4988.4 \, J \]

2. Now, calculate \( KE \):
\[ KE = \frac{3}{2} \times 4988.4 = 7482.6 \, J \]

### Final Answer
Thus, the kinetic energy of two moles of \( N_2 \) at \( 27^\circ C \) is approximately:

\[ KE \approx 7482.6 \, J \]