Throttle (Joule–Thomson) expansion: During Joule–Thomson expansion of real gases through a valve or porous plug, which thermodynamic property remains constant?

Introduction / Context:Joule–Thomson (J–T) expansion is a throttling process widely used in refrigeration, natural gas processing, and cryogenics. Characterizing the conserved quantity helps predict temperature change (cooling or heating) across the restriction and informs selection of pre-cooling and pressure levels.

Given Data / Assumptions:

• Steady-flow throttling with large pressure drop across a valve/porous plug.
• No shaft work; negligible changes in kinetic and potential energies.
• Adiabatic surroundings (no heat exchange).

Concept / Approach:The steady-flow energy equation reduces to h1 ≈ h2 for a throttling device under the listed assumptions; hence the process is isenthalpic. Temperature does not in general remain constant; the change is determined by the J–T coefficient μ_JT = (∂T/∂P)_H. Entropy increases due to irreversibility; the process is not isentropic.

Step-by-Step Solution:Apply steady-flow energy balance: h1 + ke1 + pe1 + q − w = h2 + ke2 + pe2.Set q ≈ 0, w = 0, and ke, pe changes negligible → h1 ≈ h2.Therefore, enthalpy remains constant during J–T expansion.Temperature change sign depends on μ_JT and inlet state relative to inversion curve.

Verification / Alternative check:Real-gas property charts (h–P diagrams) display vertical throttling lines, visually confirming constant enthalpy paths.

Why Other Options Are Wrong:Entropy remains constant only for reversible adiabatic (isentropic) expansions, not for throttling. Temperature remaining constant would imply μ_JT = 0, which is not generally true except at specific conditions.

Common Pitfalls:

• Assuming all expansions are isentropic; throttling is highly irreversible.
• Ignoring inversion temperature; some gases warm on J–T expansion above inversion.

Final Answer:enthalpy remains constant