Entropy Change in Free Expansion: Why △S=ncᵥln(T_f/T_i)+nRln(V_f/V_i)?

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    Entropy Thermodynaics
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Discussion Overview

The discussion revolves around the calculation of entropy change during free expansion, specifically questioning the validity of using the equation △S=ncᵥln(T_f/T_i)+nRln(V_f/V_i) in this context. The scope includes theoretical considerations of entropy as a state function and its implications for irreversible processes.

Discussion Character

  • Technical explanation
  • Debate/contested

Main Points Raised

  • Some participants assert that in free expansion, the equation dS=dQ/T cannot be used, suggesting instead that dS>dQ/T applies.
  • It is proposed that entropy is a state function, meaning the entropy change is independent of the path taken to reach the final state.
  • Participants mention that to calculate the entropy change for an irreversible process, a reversible process is designed that connects the same initial and final states, allowing the use of dS = δQrev/T.
  • One participant emphasizes that the reversible process does not need to resemble the actual irreversible process, as long as the initial and final states are matched.
  • There is a question raised about whether the initial and final states must be equilibrium states for the methods discussed to be valid.

Areas of Agreement / Disagreement

Participants generally agree on the concept that entropy is a state function and that the change in entropy can be calculated using reversible processes. However, there is contention regarding the necessity of equilibrium states for the validity of these calculations, indicating unresolved disagreement on this point.

Contextual Notes

The discussion highlights the dependence of entropy calculations on the definitions of states and the conditions under which the equations are applied. There are unresolved questions about the assumptions regarding equilibrium states in the context of irreversible processes.

spideyjj
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In a free expansion, I know that we cannot use the equation dS=dQ/T...(1). Instead we use dS>dQ/T...(2).

The question is that why we can use △S=ncᵥln(T_f/T_i)+nRln(V_f/V_i) , which is derived from the equation(1), to calculate the entropy change? Shouldn’t it be a inequality too?
 
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Entropy is a state function and doesn’t depend on the path by which the system arrived at its present state. The entropy change on going from an initial state to a final state is thus independent of how the final state is arrived at. To calculate the entropy change dS for an irreversible process, one generally designs a reversible process - linking the same two endpoints - by means of which dS can be calculated using dS = δQrev/T.
 
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Lord Jestocost said:
Entropy is a state function and doesn’t depend on the path by which the system arrived at its present state. The entropy change on going from an initial state to a final state is thus independent of how the final state is arrived at. To calculate the entropy change dS for an irreversible process, one generally designs a reversible process - linking the same two endpoints - by means of which dS can be calculated using dS = δQrev/T.
I would add that the designed reversible process does not need to bear any resemblance whatsoever to the actual irreversible process, except insofar as matching the initial and final thermodynamic equilibrium states.
 
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Don't the beginning and ending states have to be equilibrium states for the scheme described in Posts #2 and #3 to be valid ways of finding the change in entropy?
 
Mister T said:
Don't the beginning and ending states have to be equilibrium states for the scheme described in Posts #2 and #3 to be valid ways of finding the change in entropy?
Isn't that what I said?
 

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