Does the Max. Work Theorem Contradict the Reversible Work Source Statement?

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SUMMARY

The discussion centers on the Max. Work Theorem and its relationship with the Reversible Work Source (RWS) statement. It concludes that the apparent contradiction arises from the misunderstanding of work and heat transfer in different thermodynamic processes. Specifically, while the process along the adiabat (dQ=0) appears to deliver less work, the additional work in the isochoric and isobaric paths is actually compensated by heat transfer, maintaining the equivalence of work delivered to the RWS. The insights are drawn from thermodynamic principles outlined in Callen's textbook.

PREREQUISITES
  • Understanding of thermodynamic processes, specifically adiabatic, isochoric, and isobaric processes.
  • Familiarity with the concepts of work and heat transfer in thermodynamics.
  • Knowledge of P-V diagrams and their interpretation in thermodynamic cycles.
  • Basic grasp of reversible and irreversible processes in thermodynamics.
NEXT STEPS
  • Study the principles of thermodynamics as outlined in "Thermodynamics" by Callen.
  • Explore the mathematical formulation of work done in adiabatic and isochoric processes.
  • Learn about the implications of the First and Second Laws of Thermodynamics on reversible processes.
  • Investigate the role of heat transfer in different thermodynamic cycles and its impact on work output.
USEFUL FOR

Students and professionals in thermodynamics, particularly those studying energy systems, mechanical engineering, and physical chemistry, will benefit from this discussion.

derrickb
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Homework Statement


A system can be taken from state A to state B where SA = SB either by (a) directly along the adiabat S = constant, or (b) first along an isochore A to C and then along the isobar C to B. The difference in the work done by the system is simply the area enclosed between the two paths in a P–V diagram. Does this contradict the statement that the work delivered to a reversible work source (RWS) is the same for every reversible process? Explain.

The Attempt at a Solution


I don't think it contradicts the statement because although process b seems to involve a greater amount of work, it would also have a dQ which cannot be seen on the P-V diagram. If the path is taken along the adiabat, dQ=0. I think(and I could be wrong here) that the same amount of work is being delivered to the renewable work source, and the "extra" work of traveling along the isochor and isobar is actually in the form of heat.

I'm a bit shaky on this topic, and Callen's book can be very hard to follow at some points. I would really appreciate it if someone could help me out with understanding this. Thank you
 
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You appear to be saying, basically, that one of the processes is not reversible.
Is that correct?
 

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