The total kinetic energy of 1 mole of oxygen at $27^{\circ} \mathrm{C}$ is : [Use universal gas constant $(R)=8.31 \mathrm{~J} /$ mole K]

<p>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.</p> <p>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 $\frac{1}{2}RT$.</p> <p>To find the total energy, we multiply the energy per degree of freedom by the number of degrees of freedom for the diatomic gas:</p> $E_{\text{total}} = \text{degrees of freedom} \times \frac{1}{2} R T$ <p>For diatomic oxygen:</p> $E_{\text{total}} = 5 \times \frac{1}{2} R T$ <p>Given that temperature $T$ is $27^{\circ} \mathrm{C}$, we first convert it to kelvins:</p> $T_{\text{K}} = T_{\text{C}} + 273.15 = 27 + 273 = 300 \text{ K}$ <p>Now we plug in the values for $R$ and $T_{\text{K}}$:</p> $$ E_{\text{total}} = 5 \times \frac{1}{2} \times 8.31 \text{ J/mol} \cdot \text{K} \times 300 \text{ K} $$ <p>When we calculate this, we find:</p> <br/>$E_{\text{total}} = \frac{5}{2} \times 8.31 \times 300$ <br/><br/>$E_{\text{total}} = 6232.5 \text{ J/mol}$ <p>So the total kinetic energy of 1 mole of oxygen at $27^{\circ} \mathrm{C}$ is approximately 6232.5 J.</p>