Why does Callen insist a process must be reversible here?

In summary: But as far as I can tell there is no finite temperature gradient (since the primarhy system is held at the reservoir temperature ##T^r##); that is, all of the processes you have outline there are not quasistatic, right? I'm not sure they then apply as counterexamples?No, they do not apply as counterexamples.
  • #1
EE18
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In a discussion about the (change in the) Helmholtz potential being interpretable as the maximum available amount of work for a system in contact with a thermal reservoir (i.e. the free energy), Callen seems to insist this fact is true only for reversible processes. Why should this be? I understand that any quasistatic process performed in contact/equilibrium with a thermal reservoir (as is the case here) will necessarily be reversible (since heat exchange proceeds at the same temperature), but to my reading Callen's emphasis on reversible seems to suggest there's something extra here I am perhaps missing? After all, as far as I can tell we must have ##dW_{RWS} = -dU - dU^r## by pure energy conservation for any process.
 

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  • #2
My understanding of how this works is as follows: The first law of thermodynamics tells us that $$\Delta U=Q-W$$. For a system that can only exchange heat with an reservoir at temperature T, and for which the initial and final equilibrium states of the system are also at temperature T, the 2nd law of thermodynamics tells us that $$\Delta S=\frac{Q}{T}+\sigma$$where ##\sigma## is the amount of entropy generated (positive) due to irreversibility within the system. If we combine these two equations, we get $$\Delta F=\Delta U-T\Delta S=-W-T\sigma$$or$$W=-\Delta F-T\sigma$$So the maximum work in going from the initial state to the final state is ##W_{max}=-\Delta F##, and occurs when there are no irreversibilities.
 
  • #3
Chestermiller said:
My understanding of how this works is as follows: The first law of thermodynamics tells us that $$\Delta U=Q-W$$. For a system that can only exchange heat with an reservoir at temperature T, and for which the initial and final equilibrium states of the system are also at temperature T, the 2nd law of thermodynamics tells us that $$\Delta S=\frac{Q}{T}+\sigma$$where ##\sigma## is the amount of entropy generated (positive) due to irreversibility within the system. If we combine these two equations, we get $$\Delta F=\Delta U-T\Delta S=-W-T\sigma$$or$$W=-\Delta F-T\sigma$$So the maximum work in going from the initial state to the final state is ##W_{max}=-\Delta F##, and occurs when there are no irreversibilities.
I think I see; if it's OK, I just want to follow-up since Callen has not (yet) introduced explicit notion like ##\sigma## quantifying the extent of the irreversibility. Is it possible Callen is arguing as I have? That because the heat transfer is assumed quasistatic (by writing ##dS + dS^r = 0## in the first place), that then implies the reversibility (since heat transfer is at the same temperature)?
 
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  • #4
EE18 said:
I think I see; if it's OK, I just want to follow-up since Callen has not (yet) introduced explicit notion like ##\sigma## quantifying the extent of the irreversibility. Is it possible Callen is arguing as I have? That because the heat transfer is assumed quasistatic (by writing ##dS + dS^r = 0## in the first place), that then implies the reversibility (since heat transfer is at the same temperature)?
No. Irreversibility within a system is caused by transport processes within a system proceeding at finite rates:
1. Heat conduction occurring at finite temperature gradients
2. Viscous dissipation occurring at finite velocity gradients
3. Molecular diffusion occurring at finite concentration gradients
In addition, irreversibility within a system is caused by chemical reactions occurring at finite reaction rates.

I cannot relate to what is in Callen because I am strongly opposed ti using differentials for describing irreversible processes. In my judgment, it is bad practice to do this, and only leads to confusion.
 
  • #5
Chestermiller said:
No. Irreversibility within a system is caused by transport processes within a system proceeding at finite rates:
1. Heat conduction occurring at finite temperature gradients
2. Viscous dissipation occurring at finite velocity gradients
3. Molecular diffusion occurring at finite concentration gradients
In addition, irreversibility within a system is caused by chemical reactions occurring at finite reaction rates.

I cannot relate to what is in Callen because I am strongly opposed ti using differentials for describing irreversible processes. In my judgment, it is bad practice to do this, and only leads to confusion.
But as far as I can tell there is no finite temperature gradient (since the primarhy system is held at the reservoir temperature ##T^r##); that is, all of the processes you have outline there are not quasistatic, right? I'm not sure they then apply as counterexamples?

I will abstain from using differentials to your point, but I think the question still stands, no? How come quasistatic does not imply reversible here (when heat transfer is at the same temperature).
 
  • #6
In the irreversible case, even though the part of the system in contact with the reservoir is at the reservoir temperature (which is the same as the initial and final temperatures of the system), the temperature within the system can vary with spatial position during the process, and there can be finite temperature gradients within the system. If the process takes place rapidly, this will happen with compression heating or expansion cooling. The interior will be at a higher or lower average temperature than the boundary. For the process to be reversible, it must be carried out very slowly in order for the interior temperatures to equilibrate with the boundary.
 
  • #7
EE18 said:
I understand that any quasistatic process performed in contact/equilibrium with a thermal reservoir (as is the case here) will necessarily be reversible (since heat exchange proceeds at the same temperature), but to my reading Callen's emphasis on reversible seems to suggest there's something extra here I am perhaps missing? After all, as far as I can tell we must have ##dW_{RWS}=−dU−dU^r## by pure energy conservation for any process.
I think all he means is that you're not losing work to dissipative processes, like friction sapping some of the available work. In other words, even if the heat exchange is done reversibly and the system evolves quasi-statically, the entire process would still be irreversible if some work is lost to friction.
 
  • #8
vela said:
I think all he means is that you're not losing work to dissipative processes, like friction sapping some of the available work. In other words, even if the heat exchange is done reversibly and the system evolves quasi-statically, the entire process would still be irreversible if some work is lost to friction.
Frictional work will typically affect the amount of heat transferred.
 
  • #9
I strongly recommend you see Moran, et al, Fundamentals of Engineering Thermodynamics, I believe available online. See Chapter 6 in particular. They show that there are only 2 ways that the entropy of a closed system can change:

1. By heat flow from the surrounding to the system Q at the temperature of the interface between the system and surroundings through which the heat flows. This mechanism occurs in both reversible and irreversible processes.

2. By entropy generation within the system by the mechanisms I mentioned in an earlier post. This occurs only in irreversible processes.
 

FAQ: Why does Callen insist a process must be reversible here?

Why does Callen insist a process must be reversible in thermodynamics?

Callen insists on reversibility because reversible processes are idealized processes that maximize efficiency and provide a clear path to understanding the fundamental limits of thermodynamic systems. They allow for the precise calculation of work and heat transfer, which are essential for formulating the laws of thermodynamics.

What is the significance of reversible processes in Callen's approach?

In Callen's approach, reversible processes are significant because they represent the most efficient way to convert energy. They help in defining the state functions and in deriving relations such as the Carnot efficiency. Reversible processes serve as a benchmark against which real processes can be compared.

How does reversibility relate to entropy in Callen's framework?

In Callen's framework, reversibility is directly related to entropy because, during a reversible process, the entropy change of the system and its surroundings is zero. This concept is crucial for understanding the second law of thermodynamics and for calculating changes in entropy in various thermodynamic processes.

Can you explain why irreversibility leads to inefficiency according to Callen?

According to Callen, irreversibility leads to inefficiency because it involves dissipative effects such as friction, unrestrained expansion, and heat transfer across finite temperature differences. These effects cause the system to lose energy in forms that cannot be fully recovered as useful work, thereby reducing the overall efficiency of the process.

What role do reversible processes play in the formulation of thermodynamic cycles in Callen's work?

Reversible processes play a crucial role in the formulation of thermodynamic cycles in Callen's work because they provide an idealized model for understanding the maximum possible efficiency of these cycles. For example, the Carnot cycle, which is composed entirely of reversible processes, sets the upper limit on the efficiency of any heat engine operating between two temperature reservoirs.

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