1. Use the kinetic theory result

For an ideal gas, the total translational kinetic energy of $n$ moles is

$K = \frac{3}{2}nRT$

Here,

• $n = 1$ mole
• $R = 8.31\,\text{J mol}^{-1}\text{K}^{-1}$
• $T = 27^\circ \text{C} = 300\,\text{K}$

2. Substitute the values

$K = \frac{3}{2}(1)(8.31)(300)$

3. Calculate step-by-step

First,

$8.31 \times 300 = 2493$

Then,

$K = \frac{3}{2} \times 2493 = 1.5 \times 2493 = 3739.5\,\text{J}$

4. But this value is not among the options

So the question is likely asking for the total kinetic energy of all degrees of freedom of oxygen molecules.

Since oxygen is a diatomic gas, at ordinary temperature its molar kinetic energy is

$K = \frac{5}{2}nRT$

(because it has 3 translational and 2 rotational degrees of freedom).

5. Now calculate using $\frac{5}{2}RT$

$K = \frac{5}{2}(1)(8.31)(300)$

$K = 2.5 \times 2493 = 6232.5\,\text{J}$

6. Match with the options

$6232.5\,\text{J}$

This corresponds to Option A.

Final Answer

Option A: $6232.5\,\text{J}$