Free energy in the free expansion of an ideal gas

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Discussion Overview

The discussion revolves around the free expansion of an ideal gas, specifically examining the implications of the change in Gibbs free energy (ΔG) during this process. Participants explore the relationship between ΔG, work done by the system, and the interpretation of ΔG as a measure of useful work in the context of a spontaneous process at constant temperature.

Discussion Character

  • Conceptual clarification, Debate/contested, Technical explanation

Main Points Raised

  • One participant notes that during the free expansion of an ideal gas, ΔU = 0, Q = 0, and W = 0, leading to a negative ΔG when calculating ΔG = nRT ln(P2/P1) for P2 < P1.
  • Another participant states that ΔG indicates the difference in free energy between two states but does not specify the path or work done.
  • A third participant reiterates that ΔG is commonly interpreted as the opposite of the maximum possible useful work for a flow process, questioning whether ΔG remains a physical quantity when no work is done.
  • A later reply acknowledges that a negative change in free energy does not require the system to perform work.

Areas of Agreement / Disagreement

Participants express differing views on the interpretation of ΔG in the context of free expansion, with some suggesting it remains a useful theoretical measure while others question its physical significance when no work is done. The discussion does not reach a consensus on these interpretations.

Contextual Notes

Participants highlight the lack of specification regarding the path taken during the process and the implications of zero work done on the interpretation of ΔG.

MexChemE
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Hello PF! Consider the free expansion of an ideal gas. The process occurs at constant temperature, therefore, ΔU = 0, Q = 0, and W = 0. Suppose we are given the initial and final pressures of the gas, and we calculate ΔG = nRT ln(P2/P1). As P2 < P1, ΔG < 0. This is intuitive, as a free expansion is clearly spontaneous. My question is, ΔG is commonly (physically) interpreted as the opposite of the "useful work" done by the system, but in this case the system does zero work. Does this mean ΔG is not an actual physical quantity, but a helpful theoretical measure of the useful work that can be extracted from the system?

Thanks in advance for any input!
 
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All it really tells you is the difference in free energy between the two states. It does not include any specification of path, or work done.
 
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MexChemE said:
My question is, ΔG is commonly (physically) interpreted as the opposite of the "useful work" done by the system, but in this case the system does zero work. Does this mean ΔG is not an actual physical quantity, but a helpful theoretical measure of the useful work that can be extracted from the system?
ΔG is commonly interpreted as the opposite of the maximum possible useful work (for a flow process) between the initial and final states.

Chet
 
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Got it, so the system doesn't have to do work for it to have a negative change in free energy.
 

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