QJEEM24Ph5feb: JEE Main 2024 Physics - Class 11 JEEM24Ph5feb Solution: A gas mixture consists of 8 moles of argon and 6 moles of…

To determine the total internal energy of the gas mixture, we adhere to the equipartition theorem, which dictates that each degree of freedom contributes $\frac{1}{2} RT$ to the internal energy per mole, where $R$ is the gas constant and $T$ is the temperature.

An argon atom, being a noble gas, is monoatomic, with 3 translational degrees of freedom. Since we neglect all vibrational modes, and monoatomic gases have no rotational or vibrational degrees of freedom that contribute to energy at our specified temperature, each mole of argon has $3 \times \frac{1}{2} RT = \frac{3}{2} RT$ of energy.

Oxygen, on the other hand, is a diatomic molecule. This implies that under normal conditions, it has 3 translational and 2 rotational degrees of freedom. So, each mole of oxygen has $5 \times \frac{1}{2} RT = \frac{5}{2} RT$ of energy as we are neglecting vibrational modes which typically become relevant only at higher temperatures.

Hence, the total internal energy ( $U$ ) of the gas mixture is:

$U = (\text{Energy per mole of Ar} \times \text{Number of moles of Ar}) + (\text{Energy per mole of O}_{2} \times \text{Number of moles of O}_{2})$

$U = \left(\frac{3}{2} RT \times 8\right) + \left(\frac{5}{2} RT \times 6\right)$

$U = \left(12 RT + 15 RT\right)$

$U = 27 RT$

Thus, the total internal energy of the system is $27 RT$ .