The kinetic energy of 1 mole of oxygen molecules in cal `mol^(-1)` at 27°C

Options

A. 300
B. 600
C. 900
D. 800

Answer

To find the kinetic energy of 1 mole of oxygen molecules at 27°C, we can use the formula for the average kinetic energy of a gas, which is given by:

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

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

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

\[ T(K) = T(°C) + 273.15 \]

For 27°C:

\[ T = 27 + 273.15 = 300.15 \approx 300 \, K \]

### Step 2: Identify the values
- Number of moles (\( n \)) = 1 mole
- Universal gas constant (\( R \)) = 2 cal/(mol·K) (approximately)
- Temperature (\( T \)) = 300 K

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

\[ KE = \frac{3}{2} \times n \times R \times T \]

Substituting the values:

\[ KE = \frac{3}{2} \times 1 \times 2 \, \text{cal/(mol·K)} \times 300 \, K \]

### Step 4: Calculate the kinetic energy
Now we perform the calculation:

\[ KE = \frac{3}{2} \times 2 \times 300 \]
\[ KE = 3 \times 300 = 900 \, \text{cal/mol} \]

### Final Answer
The kinetic energy of 1 mole of oxygen molecules at 27°C is **900 cal/mol**.

---