JEE Physics 2023 Question with Solution

Given: A gas mixture has $2$ moles of $O_2$ and $4$ moles of $Ne$ at temperature $T$. Vibrational modes are neglected.

Find: The total internal energy of the mixture.

For diatomic gas $O_2$ with vibrational modes neglected:

$$U_{O_2} = \frac{5}{2}nRT = \frac{5}{2} \times 2 \times RT = 5RT$$

For monoatomic gas $Ne$:

$$U_{Ne} = \frac{3}{2}nRT = \frac{3}{2} \times 4 \times RT = 6RT$$

Therefore, total internal energy is:

$$U_{\text{total}} = 5RT + 6RT = 11RT$$

So the computed value is $11RT$. The solution labels the correct option as A, which disagrees with the listed options; among the given options, $11RT$ corresponds to option B.

Given: The mixture contains a diatomic gas $O_2$ and a monoatomic gas $Ne$.

Find: Why their internal energies are added using different coefficients.

Neglecting vibrational modes means $O_2$ has only translational and rotational contributions, so:

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

for the diatomic component.

Neon is monoatomic, so only translational motion contributes:

$$U = \frac{3}{2}nRT$$

Hence,

$$U_{\text{mixture}} = U_{O_2} + U_{Ne} = 5RT + 6RT = 11RT$$

Therefore, the correct option from the listed choices is B.