Free (Joule) Expansion of an Ideal Gas – Energy Transfers and State Change During free expansion of an ideal gas into a vacuum, there is no boundary work and no heat transfer; internal energy remains constant and temperature is unchanged.

Difficulty: Easy

Correct Answer: all of these

Explanation:


Introduction / Context:
Free expansion (Joule expansion) is a classic thought/bench experiment where a gas expands into a vacuum. It is invaluable for clarifying the roles of work, heat, and internal energy in non-quasi-equilibrium processes.


Given Data / Assumptions:

  • Ideal gas expands into an evacuated volume (no external resistance).
  • Adiabatic, rigid container boundaries aside from the partition removal.
  • No shaft work or other interactions.


Concept / Approach:

With no external pressure opposing expansion, boundary work is zero: W = ∫ p_ext dV = 0 (p_ext = 0). If the vessel is insulated, Q = 0. For an ideal gas, internal energy depends only on temperature; since ΔU = Q − W = 0, we have ΔU = 0 and therefore ΔT = 0. Pressure and specific volume change, but temperature does not for an ideal gas.


Step-by-Step Solution:

Remove partition between gas and vacuum.No resisting pressure → W1-2 = 0.Insulated system → Q1-2 = 0.First Law: ΔU = Q − W = 0 → dU = 0; for ideal gas, T2 = T1.


Verification / Alternative check:

Joule's vacuum bulb experiment confirms no temperature change for ideal gases. Real gases may cool or warm slightly (Joule–Thomson effects), but at low pressures the ideal result is approached.


Why Other Options Are Wrong:

Any claim of expansion work on surroundings contradicts p_ext = 0; asserting heat transfer violates the insulated assumption; stating ΔU ≠ 0 contradicts First Law for this setup.


Common Pitfalls:

Confusing free expansion with throttling (which imposes flow and maintains h constant, not u); assuming temperature must drop whenever gas expands (not true here for ideal gas).


Final Answer:

all of these

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