How Do I Calculate Entropy of Surroundings in Adiabatic Expansion?

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Discussion Overview

The discussion revolves around calculating the entropy of the surroundings during an adiabatic expansion, exploring the implications of adiabatic processes on entropy changes and the applicability of various formulas.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • Some participants note that for a reversible adiabatic process, the total entropy change is zero, while others suggest analyzing irreversible processes by substituting them with a combination of reversible processes.
  • One participant expresses confusion regarding the application of entropy formulas in adiabatic processes, particularly questioning the use of \DeltaS = -w/T when q=0.
  • Another participant challenges the origin of the formula \Delta S=-w/T, stating that reversible work does not carry entropy.
  • There is a mention that isothermal processes have specific relationships for entropy change that may not apply to adiabatic processes, particularly for ideal gases.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the applicability of certain formulas for calculating entropy in adiabatic processes, indicating that multiple competing views remain regarding the correct approach.

Contextual Notes

The discussion highlights limitations in understanding the assumptions behind various entropy calculations, particularly in distinguishing between reversible and irreversible processes and the specific conditions of isothermal versus adiabatic processes.

pilotpanda
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How do I calculate the entropy of the surroundings for an adiabatic expansion? I know [tex]\Delta[/tex]Ssur = qrev/T for reversible process and qirrev/T for irreversible process. But q=0 for adiabatic process, so are there any other formulas that I could use to calculate [tex]\Delta[/tex]Ssurr (I know there is a [tex]\Delta[/tex]S = w/T formula but I'm not sure if it applies here)?

Thanks,

pilotpanda
 
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Hi pilotpanda, welcome to PF!

A reversible adiabatic process is isentropic (total entropy change is zero), under the reasoning you describe. If you're analyzing an irreversible adiabatic process, try replacing the process with a combination of reversible processes that achieves the same end state (but these reversible processes will likely not be adiabatic). Does this answer your question?
 
Hi Mapes

Thanks! I guess I'm just really confused about the adiabatic expansion. If the process is isothermal, I know how to approach the problem since qsys=qsur=-w, I can use [tex]\Delta[/tex]S = -w/T to calculate entropy of the surroundings and compare that with the system (either given or calculated) to determine whether I have a reversible or irreversible process. But for an adiabatic process, since q=0, I'm not sure if I can still use the abovementioned formulas to calculate [tex]\Delta[/tex]Ssurroundings (-w/T). I think there may be some concepts / formulas that I'm missing here so please help me on that.

pilotpanda
 
Where does [itex]\Delta S=-w/T[/itex] come from? Reversible work doesn't carry entropy.

Also, isothermal only implies constant energy for an ideal gas. Definitely make sure you're familiar with the assumptions that go into these calculations.
 
Well, for an isothermal process [tex]\Delta[/tex]S = nRln[tex]\frac{V2}{V1}[/tex] and since work for a reversible process is -nRTln[tex]\frac{V2}{V1}[/tex]...

After some struggle, I think I know the answer now. Thanks for your help!
 
OK, cool, but note that those relationships only apply for an isothermal process on an ideal gas. They arise because as you increase or decrease volume, it's necessary to heat or cool the ideal gas to keep the temperature constant.
 

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