Free (unrestrained) expansion of an ideal gas:\nWhy does it not cool during Joule expansion into a vacuum?

Introduction / Context:
In a Joule (free) expansion, an ideal gas expands into a vacuum without doing work and without heat exchange. Many learners expect cooling, but an ideal gas does not cool in this process. Understanding why sharpens the distinction between ideal and real gases in throttling/expansion devices.

Given Data / Assumptions:

• Ideal-gas behavior: intermolecular potential energy is negligible.
• Process is adiabatic (Q = 0) and does no external work (W = 0).
• Internal energy U of an ideal gas depends only on temperature.

Concept / Approach:
From the first law of thermodynamics, ΔU = Q − W. For free expansion of an ideal gas, Q = 0 and W = 0, so ΔU = 0. Since U = U(T) for an ideal gas, ΔU = 0 implies ΔT = 0, hence no cooling. The microscopic reason is the absence of intermolecular attractions that would convert potential energy to kinetic energy changes during expansion.

Step-by-Step Solution:
Apply first law: ΔU = 0 → ΔT = 0 (ideal gas).Link to molecular model: no attractive forces → U independent of volume.Therefore, temperature remains constant; no cooling occurs.

Verification / Alternative check:
Real gases may warm or cool depending on the Joule effect, but for the idealized case, caloric equation of state shows U = f(T) only, confirming constant T when ΔU = 0.

Why Other Options Are Wrong:
Collisions do not dissipate energy in an ideal gas (elastic), but that alone is not the defining reason. Inversion temperature refers to throttling (Joule–Thomson), not free expansion. No external work is done in free expansion, so option (d) is inaccurate. Heat leakage (e) is outside the stated idealized setup.

Common Pitfalls:
Confusing Joule free expansion with Joule–Thomson throttling through a valve.

Final Answer:
molecules do not exert attractive forces (U independent of volume)