Free expansion (Joule experiment) – is it a constant enthalpy process? Assess the statement: “The free expansion process is a constant enthalpy (isenthalpic) process.”

Introduction / Context:Free expansion (Joule experiment) involves a gas expanding into a vacuum from an insulated container with no external work. It is often contrasted with throttling, which is an isenthalpic process.

Given Data / Assumptions:

• Insulated system (adiabatic).
• No shaft work and no boundary work (expansion into vacuum).
• Real or ideal gas as working substance.

Concept / Approach:First law for a closed, adiabatic system: ΔU = Q − W = 0 − 0 = 0, so internal energy remains constant. Enthalpy H = U + pV generally changes in free expansion because pV changes and there is no constraint keeping H constant. By contrast, throttling (flow through a restriction at steady state) satisfies h_in = h_out (constant specific enthalpy).

Step-by-Step Solution:Identify process: free expansion into vacuum → Q = 0, W = 0.Apply first law (closed system): ΔU = 0.Conclude: U is constant, but H = U + pV need not be constant since p and V change.Therefore, the statement “constant enthalpy” is incorrect for free expansion (it is constant internal energy).

Verification / Alternative check:For an ideal gas, U depends only on temperature, so ΔU = 0 implies ΔT = 0; hence ΔH = 0 as well. However, this special case does not make free expansion generally isenthalpic for real gases; the standard isenthalpic process is throttling.

Why Other Options Are Wrong:

• “Correct”: conflates free expansion with throttling.
• “Correct only for throttling” (option d) actually describes a different process.
• Heat transfer is not present in free expansion; adding heat changes the process class.

Common Pitfalls:Generalizing ideal-gas behavior to real-gas free expansions; real gases may change temperature even when ΔU = 0 due to intermolecular effects.

Final Answer:Incorrect