Fundamental thermodynamic relation confusion.

In summary, the conversation discusses the relationships between energy, heat, and work in different processes, focusing on reversible and irreversible processes. The concept of state variables is introduced and used to explain why the equation dE = TdS - PdV holds for all processes. The confusion around this equation is clarified, and it is shown that for an isothermal transformation, ΔF = -PdV = ΔWrev ≤ ΔW. Overall, the conversation highlights the importance of understanding the differences between reversible and irreversible processes in thermodynamics.
  • #1
AntiElephant
25
0
dE = [STRIKE]d[/STRIKE]Q + [STRIKE]d[/STRIKE]W = [STRIKE]d[/STRIKE]Qrev + [STRIKE]d[/STRIKE]Wrev = [STRIKE]d[/STRIKE]Qirev + [STRIKE]d[/STRIKE]Wirev.

We have for an reversible process, [STRIKE]d[/STRIKE]Qrev = TdS and [STRIKE]d[/STRIKE]Wrev = -PdV. So;

dE = TdS - PdV

So this relation is for all changes (irreversible or reversible) since dS and dV are state functions. What doesn't make sense to me is the next part when Helmholtz free energy is defined;

F = E - TS

Then dF = -PdV - SdT.

Another relation for all changes. I'm told and shown that for a system at constant temperature, then ΔW ≥ ΔF, with equality for reversible processes. However how is this equality true ONLY for reversible processes? If it's at constant temperature, then dT = 0. So dF = -PdV = dW no matter what process is?
 
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  • #2
But -PdV = dW only for reversible processes (as you said yourself in the second line of your post).
 
  • #3
mr. vodka said:
But -PdV = dW only for reversible processes (as you said yourself in the second line of your post).

That's true, but then surely dE = TdS - PdV does not hold for all processes as I'm told it does?
 
  • #4
I'm not sure if I understand your problem, but it seems as if you're under the impression that dE = Q + W = TdS - PdV implies that Q = TdS and W - PdV? Is that correct? If so, that is where your mistake is.
 
  • #5
mr. vodka said:
I'm not sure if I understand your problem, but it seems as if you're under the impression that dE = Q + W = TdS - PdV implies that Q = TdS and W - PdV? Is that correct? If so, that is where your mistake is.

I guess. It personally doesn't make sense to me how dE = TdS - PdV holds for anything other than a reversible process. E is a state function, and if we go through a reversible process then dE = [STRIKE]d[/STRIKE]Qrev + [STRIKE]d[/STRIKE]Wrev = TdS - PdV. However if we go through an irreversible process, then how is dE = TdS - PdV still? - I'm told that the relationship still holds because the equation dE = TdS - PdV only depends on state variables.
 
  • #6
The idea is as follows: take any two states of a system. No matter what process you use to go from the first state to the second state, dE, dS and dV are always the same, by definition of state variables (i.e. they only depend on the state, not on the process). So TdS - PdV is a quantity that only depends on the initial and the final state, not on the process that connects these two states. And since we know that dE = TdS - PdV is true for at least one process (i.e. the reversible one), and knowing that this equality cannot depend on the process, we know that it most be true for all the processes.

Does that clarify matters?
 
  • #7
mr. vodka said:
The idea is as follows: take any two states of a system. No matter what process you use to go from the first state to the second state, dE, dS and dV are always the same, by definition of state variables (i.e. they only depend on the state, not on the process). So TdS - PdV is a quantity that only depends on the initial and the final state, not on the process that connects these two states. And since we know that dE = TdS - PdV is true for at least one process (i.e. the reversible one), and knowing that this equality cannot depend on the process, we know that it most be true for all the processes.

Does that clarify matters?

Thanks for the reply. EDIT: Actually you have clarified it. As you said. -PdV = W only for a reversible process, for an irreversible process -PdV ≤ W. So of course for an isothermal transformation ΔF = -PdV = ΔWrev ≤ ΔW.

Deep down what confused me is that my notes say this;

"For an infinitesimal reversible process;

dF = -PdV - SdT"


Although this is true, it's actually for all processes, not just reversible ones. My notes saying "for an infinitesimal reversible process" made me think it was suggesting that this relationship only holds for reversible processes and not irreversible ones.
 
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  • #8
I think what you're saying is correct yes.

As a caveat, do note that I'm still somewhat confused myself about some of the details related to this discussion (cf. a thread I started in this same subforum inspired by yours). That being said, I'm fairly sure the answers I gave to your questions were correct, and anyway you have your own brain that can judge the arguments accordingly :)
 

1. What is the fundamental thermodynamic relation?

The fundamental thermodynamic relation is a mathematical equation that describes the relationship between the internal energy, entropy, and pressure of a system. It is also known as the first law of thermodynamics and is expressed as dU = TdS - PdV, where dU is the change in internal energy, T is the temperature, dS is the change in entropy, P is the pressure, and dV is the change in volume.

2. How is the fundamental thermodynamic relation derived?

The fundamental thermodynamic relation is derived from the first law of thermodynamics, which states that energy cannot be created or destroyed, only transferred or converted. By considering the different forms of energy in a system, such as internal energy, heat, and work, and their relationships, the fundamental thermodynamic relation is derived.

3. What is the significance of the fundamental thermodynamic relation?

The fundamental thermodynamic relation is significant because it provides a fundamental understanding of the behavior of thermodynamic systems and allows for the prediction of their properties and behavior. It is the basis for many other thermodynamic equations and principles, making it a crucial concept in the field of thermodynamics.

4. How is the fundamental thermodynamic relation applied in real-world scenarios?

The fundamental thermodynamic relation is applied in a variety of real-world scenarios, including the design and operation of power plants, refrigeration systems, and chemical reactions. It is also used in the development of new materials and processes, such as in the field of renewable energy.

5. What are some common sources of confusion when dealing with the fundamental thermodynamic relation?

Some common sources of confusion when dealing with the fundamental thermodynamic relation include understanding the difference between heat and temperature and the role of work in the equation. Additionally, the use of different thermodynamic variables and units can also lead to confusion when applying the fundamental thermodynamic relation.

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