JEE Physics 2024 Question with Solution

Given: $n = 1$ mole of oxygen and temperature $T = 27^\circ \text{C} = 300 \, \text{K}$.

Find: The total kinetic energy of the gas.

For a diatomic gas like oxygen, the total kinetic energy is

$$E = \frac{5}{2} nRT$$

Using $R = 8.31 \, \text{J mol}^{-1} \text{K}^{-1}$, substitute the values:

$$E = \frac{5}{2} \times 1 \times 8.31 \times 300$$ $$E = 6232.5 \, \text{J}$$

Therefore, the total kinetic energy is $6232.5 \, \text{J}$. The solution marks option A, but the calculated value matches option C in the given options. Hence, the correct option is C.

Given: Oxygen is a diatomic gas at $300 \, \text{K}$.

Find: Total kinetic energy of $1$ mole.

Use the relation

$$E = \frac{f}{2} nRT$$

For a diatomic gas, $f = 5$. So,

$$E = \frac{5}{2} \times 1 \times 8.31 \times 300 = 6232.5 \, \text{J}$$

Therefore, the correct option is C.