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

To determine the total kinetic energy of 1 mole of oxygen at 27°C, we can use the formula for the kinetic energy of gases as derived from the kinetic theory of gases. The kinetic energy of one mole of a diatomic gas like oxygen can be given by:

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

where:

• \(n\) is the number of moles, which is 1 mole in this case.
• \(R\) is the universal gas constant, given as 8.31 J/mole K.
• \(T\) is the absolute temperature in Kelvin.

First, convert the given temperature from Celsius to Kelvin:

T = 27 + 273.15 = 300.15 \text{ K}

For simplicity, we round off to 300 K. Substitute these values into the kinetic energy formula:

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

Calculating this gives:

\[\text{Total Kinetic Energy} = \frac{5}{2} \times 8.31 \times 300 = 6232.5 \text{ J}\]

Thus, the total kinetic energy of 1 mole of oxygen at 27°C is 6232.5 J. This confirms that the correct answer is

6232.5 J

.

 

The other options are incorrect as they do not match the calculated value based on the kinetic theory of gases.

Answer 2

The kinetic energy of a gas is given by:

\(E = \frac{f}{2} nRT,\)

where \(f\) is the degrees of freedom.

For a diatomic gas like oxygen:
- \(f = 5\),
- \(n = 1\),
- \(R = 8.31 \, \text{J/molK}\),
- \(T = 27^\circ \text{C} = 300 \, \text{K}\).

Substituting the values:

\(E = \frac{5}{2} \times 1 \times 8.31 \times 300 = 6232.5 \, \text{J}.\)

The correct option is (A) : 6232.5 J