The kinetic enerrgy of 4 mole sof nitrogen gas at `127^(@)C` is `(R=2" cal "mol^(-1)K^(-1))`

Options

A. 4400 cal
B. 3200 cal
C. 4800 cal
D. 1524 cal

Answer

To find the kinetic energy of 4 moles of nitrogen gas at 127°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
- \( R \) = gas constant
- \( T \) = temperature in Kelvin

### Step 1: Convert the temperature from Celsius to Kelvin
To convert the temperature from Celsius to Kelvin, we use the formula:

\[ T(K) = T(°C) + 273 \]

Substituting the given temperature:

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

### Step 2: Substitute the values into the kinetic energy formula
Now we can substitute the values into the kinetic energy formula. We have:
- \( n = 4 \, \text{moles} \)
- \( R = 2 \, \text{cal/mol/K} \)
- \( T = 400 \, K \)

Substituting these values into the formula:

\[ KE = \frac{3}{2} \times 4 \times 2 \times 400 \]

### Step 3: Calculate the kinetic energy
Now we can calculate the kinetic energy step by step:

1. Calculate \( \frac{3}{2} \times 4 = 6 \)
2. Then, calculate \( 6 \times 2 = 12 \)
3. Finally, calculate \( 12 \times 400 = 4800 \)

Thus, the kinetic energy \( KE \) is:

\[ KE = 4800 \, \text{calories} \]

### Final Answer
The kinetic energy of 4 moles of nitrogen gas at 127°C is **4800 calories**.

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