The total kinetic energy of a mole of an ideal gas can be determined by using the equipartition theorem, which states that the energy is equally distributed among degrees of freedom. For a diatomic molecule such as oxygen ($$O_2$$), there are 5 degrees of freedom (3 translational and 2 rotational - assuming the vibrational modes are not excited at room temperature), so each degree of freedom has an average energy of $$\frac{1}{2} kT$$ per molecule, where $$k$$ is the Boltzmann constant and $$T$$ is the temperature in kelvins.

However, since we are dealing with moles, we'll use the universal gas constant $$R$$ instead of the Boltzmann constant $$k$$, because $$R = k \cdot N_A$$ where $$N_A$$ is the Avogadro constant (the number of molecules in a mole). Therefore, the average energy per mole for each degree of freedom is $$\frac{1}{2}RT$$.

To find the total energy, we multiply the energy per degree of freedom by the number of degrees of freedom for the diatomic gas:

$$ E_{\text{total}} = \text{degrees of freedom} \times \frac{1}{2} R T $$

For diatomic oxygen:

$$ E_{\text{total}} = 5 \times \frac{1}{2} R T $$

Given that temperature $$ T $$ is $$ 27^{\circ} \mathrm{C} $$, we first convert it to kelvins:

$$ T_{\text{K}} = T_{\text{C}} + 273.15 = 27 + 273 = 300 \text{ K} $$

Now we plug in the values for $$ R $$ and $$ T_{\text{K}} $$:

$$ E_{\text{total}} = 5 \times \frac{1}{2} \times 8.31 \text{ J/mol} \cdot \text{K} \times 300 \text{ K} $$

When we calculate this, we find:

$$ E_{\text{total}} = \frac{5}{2} \times 8.31 \times 300 $$

$$ E_{\text{total}} = 6232.5 \text{ J/mol} $$

So the total kinetic energy of 1 mole of oxygen at $$27^{\circ} \mathrm{C}$$ is approximately 6232.5 J.