Max. Work Theorem (Conceptual)

In summary, the conversation discusses two possible paths for a system to go from state A to state B, both resulting in the same entropy but with a difference in the work done. The question is raised whether this contradicts the statement that the work delivered to a reversible work source is the same for every reversible process. It is suggested that the difference in work may be due to the isochore-isobar path involving heat, making it not a reversible process. The main confusion is clarified and help is requested to understand the concept better.
  • #1
derrickb
22
0

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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  • #2
You appear to be saying, basically, that one of the processes is not reversible.
Is that correct?
 

Related to Max. Work Theorem (Conceptual)

1. What is the Max. Work Theorem?

The Max. Work Theorem, also known as the Maximum Work Theorem, is a concept in thermodynamics that states that the maximum amount of work that can be obtained from a system is equal to the change in its free energy.

2. How is the Max. Work Theorem calculated?

The Max. Work Theorem is calculated by taking the product of the change in free energy and the temperature in Kelvin. This can be represented by the equation: Wmax = ΔG x T.

3. What is the significance of the Max. Work Theorem?

The Max. Work Theorem is significant because it helps us understand the maximum amount of work that can be obtained from a system. This can be useful in various fields such as engineering, physics, and chemistry.

4. Can the Max. Work Theorem be applied to all systems?

No, the Max. Work Theorem can only be applied to reversible processes. This means that the process must be able to return to its original state without any energy loss.

5. How does the Max. Work Theorem relate to the Second Law of Thermodynamics?

The Max. Work Theorem is closely related to the Second Law of Thermodynamics, which states that the total entropy of a closed system will always increase over time. The Max. Work Theorem helps us understand the direction in which a process will occur and the maximum amount of work that can be obtained from it.

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