The total kinetic energy of 1 mole of oxygen at 27°C is :
[Use universal gas constant (R) = 8.31 J mol$$^{-1}$$ K$$^{-1}$$]

We need to find the total kinetic energy of 1 mole of oxygen gas at 27°C.

Given the universal gas constant $$R = 8.31 \text{ J mol}^{-1} \text{K}^{-1}$$, temperature $$T = 27°C = 27 + 273 = 300$$ K, and amount $$n = 1$$ mole, we proceed as follows.

Oxygen ($$O_2$$) is a diatomic molecule that, at moderate temperatures (around 300 K), has 5 degrees of freedom: 3 translational degrees of freedom (motion along the x, y, and z axes) and 2 rotational degrees of freedom (rotation about two axes perpendicular to the bond axis). Vibrational modes are not significantly excited at room temperature, so $$f = 5$$.

By the equipartition theorem, each degree of freedom contributes $$\frac{1}{2}k_BT$$ of kinetic energy per molecule, or $$\frac{1}{2}RT$$ per mole. Thus for $$n$$ moles with $$f$$ degrees of freedom, the total kinetic energy is

$$KE = \frac{f}{2} \times n \times R \times T$$

Substituting the values,

$$KE = \frac{5}{2} \times 1 \times 8.31 \times 300$$

$$= \frac{5}{2} \times 2493$$

$$= 5 \times 1246.5$$

$$= 6232.5 \text{ J}$$

The correct answer is Option 3: 6232.5 J.