Entropy change for spontaneous/ irreversible gas expansion

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Discussion Overview

The discussion revolves around the entropy change associated with irreversible gas expansion, particularly in the context of a gas expanding into a vacuum. Participants explore the implications of the entropy equation dS=dQ/T and the differences between irreversible and reversible processes.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant describes the high-level understanding of why entropy increases during irreversible gas expansion but questions the applicability of the equation dS=dQ/T given that energy and temperature remain constant.
  • Another participant explains that dS=dQ/T pertains to entropy transfer across system boundaries and does not account for entropy generated within the system during irreversible processes, suggesting the need for an alternate reversible process to calculate entropy change.
  • A participant requests clarification on a specific example involving a massless piston chamber, questioning the necessity of an isothermal bath if temperature remains unchanged in both the irreversible and reversible cases.
  • In response, it is noted that in the reversible case, the gas does work, which would change the temperature if not controlled, thus necessitating the isothermal bath to maintain constant temperature.
  • Another participant acknowledges the explanation regarding temperature control by the isothermal bath during the reversible process.

Areas of Agreement / Disagreement

Participants express differing views on the application of the entropy equation in irreversible processes and the necessity of certain conditions in reversible processes. The discussion remains unresolved regarding the specific mechanics of entropy change in these scenarios.

Contextual Notes

Participants highlight the distinction between entropy transfer and entropy generation, as well as the conditions required for reversible processes, indicating potential limitations in understanding the implications of the entropy equation in different contexts.

zanick
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When trying to describe why the entropy goes up for a irreversible process, such as gas expanding into a vacuum, it seems fairly easy at a high level. the valve between the two chambers opens, the free expansion occurs, the pressure drops proportional to the volume change and the temp remains constant because no work was done and the energy of the system didnt change. However, using the equation dS=dQ/T, it doesn't seem to apply , because we know the energy (U) didnt change, and we know the temperature (T) didnt change, but we do know the entropy (S) goes up indicated by that pressure drop. does it have something to do with the (Q) heat per unit volume going down or something like that? or is just as simple as using the wrong equation. if so, it still would be interesting to hear the answer by the experts here.
 
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The equation dS=dQ/T describes the transfer of entropy across the boundary of a system, but it does not account for entropy generated within the system (due to viscous dissipation and finite temperature gradients) during an irreversible process. Such entropy generation is not present during a reversible process. So, to determine the entropy change taking place for an irreversible process, you must devise an alternate reversible process (that may not resemble the irreversible process very much), and calculate dQ/T for that alternate process between the same two end states. For more details on how this all plays out, see my Physics Forums Insights article https://www.physicsforums.com/insights/grandpa-chets-entropy-recipe/. This article presents a simple cookbook recipe on how to determine the entropy change for a reversible process, including worked examples for some typical cases.
 
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Can we review your example 3, and step 3 by the breaking out of the conditions of the free expansion example to a mass less piston chamber that is allows the volume to increase by double to end up at the same final state?
in both cases (original vs step 3), the gas is allowed to expand ( instantaneously in the 1st case) into 2x the original volume. the temp doesn't change because no work is done in the 1st step... but in the 3rd step, the piston is moved slowly as well as the entire chamber placed in a isothermal bath. why is that necessary if the temp will not change in the system in either case?
 

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zanick said:
Can we review your example 3, and step 3 by the breaking out of the conditions of the free expansion example to a mass less piston chamber that is allows the volume to increase by double to end up at the same final state?
in both cases (original vs step 3), the gas is allowed to expand ( instantaneously in the 1st case) into 2x the original volume. the temp doesn't change because no work is done in the 1st step... but in the 3rd step, the piston is moved slowly as well as the entire chamber placed in a isothermal bath. why is that necessary if the temp will not change in the system in either case?
The temperature would change in the reversible case because the gas is doing work in this case. How did you think the gas could be made to expand slowly if the piston were not being controlled? So to prevent the temperature from changing, we need to place the chamber an isothermal bath.
 
makes sense. thanks . the temperature is controlled by the isothermal bath due to the energy used in controlling the expansion. makes sense.
 

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