QJEEM24Phb8c3: JEE Main 2024 Physics - Class 11 JEEM24Phb8c3 Solution: The total kinetic energy of 1 mole of oxygen at is : [Use…

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:

For diatomic oxygen:

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

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

When we calculate this, we find:

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