How Do I Calculate Entropy of Surroundings in Adiabatic Expansion?

AI Thread Summary
Calculating the entropy of the surroundings during an adiabatic expansion can be challenging due to the fact that heat transfer (q) is zero in such processes. For reversible adiabatic processes, the total entropy change is zero, while for irreversible processes, one can analyze the process by substituting it with a series of reversible processes that reach the same end state. The confusion arises when trying to apply the formula ΔS = -w/T, which is typically used for isothermal processes, as it does not directly apply to adiabatic conditions. It is important to understand the assumptions behind these calculations, particularly for ideal gases. Ultimately, the relationships discussed primarily pertain to isothermal processes and may not be suitable for adiabatic scenarios.
pilotpanda
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How do I calculate the entropy of the surroundings for an adiabatic expansion? I know \DeltaSsur = 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 \DeltaSsurr (I know there is a \DeltaS = 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 \DeltaS = -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 \DeltaSsurroundings (-w/T). I think there may be some concepts / formulas that I'm missing here so please help me on that.

pilotpanda
 
Where does \Delta S=-w/T 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 \DeltaS = nRln\frac{V2}{V1} and since work for a reversible process is -nRTln\frac{V2}{V1}...

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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