Q 11-12-049JEE MainJEE Main 2024 (27 Jan, Shift 2)Easy
The total kinetic energy of 1 mole of oxygen at $27\,^\circ\text{C}$ is: [Use universal gas constant $R = 8.31\ \text{J mol}^{-1}\,\text{K}^{-1}$]
Answer: (C) $6232.5\ \text{J}$
Oxygen is diatomic with 5 degrees of freedom (3 translational + 2 rotational):
$$E = \frac52 nRT = \frac52\times1\times8.31\times300 = 6232.5\ \text{J}$$
Solution by Sreeraj P, M.Sc Physics