Question

The total kinetic energy of 1 mole of oxygen at 27°C is :
[Use universal gas constant (R)= 8.31 J/mole K]

• A. 6232.5 J
• B. 6845.5 J
• C. 5942.0 J
• D. 5670.5 J

Answer 1

The Correct Option is A

The total kinetic energy of 1 mole of oxygen at 27°C is calculated using the kinetic theory of gases. For a diatomic gas, the kinetic energy per mole is expressed as:

\[\text{Total Kinetic Energy} = \frac{5}{2} nRT\]

Where:

• \(n\) = 1 mole
• \(R\) = 8.31 J/mole K
• \(T\) = Absolute temperature in Kelvin

The temperature is converted from Celsius to Kelvin: T = 27°C + 273.15 = 300.15 K. Approximating to 300 K for simplicity.

Substituting these values into the formula:

\[\text{Total Kinetic Energy} = \frac{5}{2} \times 1 \times 8.31 \times 300\]

The calculation yields:

\[\text{Total Kinetic Energy} = 6232.5 \text{ J}\]

Therefore, the total kinetic energy of 1 mole of oxygen at 27°C is 6232.5 J.

This value is derived directly from the kinetic theory of gases, making other options incorrect.