Animated Solution for Physics - Thermodynamics: From the following statements concerning ideal gas at any given temperature $T$, select the correct one (s).

This problem is a fantastic conceptual check on the Kinetic Theory of Gases. It tests your fundamental understanding of how macroscopic properties like volume expansion and pressure relate to microscopic properties like kinetic energy and mean free path. Let's break down each statement systematically.

The first statement asks us about the coefficient of volume expansion at constant pressure, denoted by $\gamma$. By definition, this coefficient measures the fractional change in volume per degree change in temperature:

For one mole of an ideal gas, the equation of state is $pV=RT$. If we rearrange this to solve for volume, we get $V=\frac{RT}{p}$. Differentiating this with respect to temperature $T$ while holding pressure $p$ constant yields:

Substituting this back into our definition for $\gamma$, we find:

Since $pV=RT$, we can replace the denominator to get:

Notice the elegance of this result! The coefficient of volume expansion for an ideal gas depends only on its absolute temperature $T$. It does not depend on the molar mass, the atomicity, or the specific identity of the gas. Therefore, at any given temperature, $\gamma$ is identical for all ideal gases. Statement (a) is correct.

Statement (b) claims that the average translational kinetic energy per molecule of oxygen gas is $3kT$. To verify this, we turn to the Equipartition Theorem, which states that every active degree of freedom contributes $\frac{1}{2}kT$ to the average energy of a molecule.

Regardless of whether a gas is monatomic (like Helium), diatomic (like Oxygen), or polyatomic, it can only move in three independent spatial directions ($x$, $y$, and $z$). Thus, every molecule has exactly 3 translational degrees of freedom.

The average translational kinetic energy is therefore:

Since $\frac{3}{2}kT eq 3kT$, statement (b) is incorrect.

Statement (c) discusses the mean free path, $\lambda$, which is the average distance a molecule travels between successive collisions. The formula for the mean free path is:

where $d$ is the collision diameter of the molecule and $p$ is the pressure of the gas.

Looking at the formula, we can clearly see that at a constant temperature $T$, the mean free path is inversely proportional to the pressure:

Physically, this makes perfect sense. If you decrease the pressure (while keeping temperature constant), the gas expands. The molecules are spread further apart, meaning a molecule can travel a longer distance before bumping into another one. Thus, the mean free path increases as pressure decreases. Statement (c) is correct.

Finally, statement (d) suggests that in a gaseous mixture, the average translational kinetic energy differs for each component.

As we derived earlier, the average translational kinetic energy for any gas molecule is strictly $\frac{3}{2}kT$. In a gaseous mixture that has reached thermal equilibrium, all components share the exact same temperature $T$. Because the translational kinetic energy depends exclusively on $T$, every single component in the mixture will possess the exact same average translational kinetic energy.

Heavier molecules will simply move with a slower root-mean-square velocity to compensate for their larger mass, ensuring the kinetic energy remains balanced. Therefore, statement (d) is incorrect.

Final Conclusion: The correct statements are (a) and (c).