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

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SUMMARY

In free expansion, the entropy change can be calculated using the equation △S=ncᵥln(T_f/T_i)+nRln(V_f/V_i), despite the process being irreversible. This is valid because entropy is a state function, which means it is independent of the path taken to reach the final state. To determine the entropy change for irreversible processes, a reversible process is designed that connects the same initial and final equilibrium states. The validity of this approach relies on the initial and final states being thermodynamic equilibrium states.

PREREQUISITES
  • Understanding of thermodynamic concepts, particularly entropy and state functions.
  • Familiarity with the laws of thermodynamics, especially the second law.
  • Knowledge of reversible and irreversible processes in thermodynamics.
  • Basic grasp of equations related to entropy, such as dS = δQrev/T.
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  • Study the implications of the second law of thermodynamics on entropy changes.
  • Learn about the derivation and applications of the equation △S=ncᵥln(T_f/T_i)+nRln(V_f/V_i).
  • Explore examples of reversible processes and how they relate to irreversible processes in thermodynamics.
  • Investigate the concept of thermodynamic equilibrium and its significance in entropy calculations.
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Students and professionals in thermodynamics, physicists, and engineers focusing on energy systems and entropy analysis will benefit from this discussion.

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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