Understand the Thermodynamic Identity: Is This Correct?

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Discussion Overview

The discussion revolves around the Thermodynamic Identity, specifically its application and validity under different conditions, such as constant volume and particle number, as well as the implications of adding heat to a system. Participants explore the nuances of the identity in both reversible and irreversible processes.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Conceptual clarification

Main Points Raised

  • One participant presents the Thermodynamic Identity and simplifies it under the assumption of constant volume and particle number, questioning its correctness.
  • Another participant raises the issue of non-uniform temperature when heat is added irreversibly, questioning how the identity applies in such cases.
  • Some participants assert that the identity may not hold if the temperature is not uniform, prompting a discussion about the conditions under which the identity is valid.
  • There is a suggestion that the identity is valid in the reversible limit and for closely neighboring thermodynamic equilibrium states, regardless of the irreversibility of the path taken.

Areas of Agreement / Disagreement

Participants express differing views on the applicability of the Thermodynamic Identity under various conditions, particularly regarding uniformity of temperature and the nature of heat addition. No consensus is reached on the broader implications of the identity.

Contextual Notes

Participants highlight limitations related to assumptions about uniform temperature and the nature of heat addition, as well as the distinction between reversible and irreversible processes. These factors remain unresolved in the discussion.

Sebas4
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I have a question about the Thermodynamic Identity.
The Thermodynamic Identity is given by
dU = TdS - PdV + \mu dN.
We assume that the volume V and that the number of particles N is constant.
Thus the Thermodynamic Identity becomes
dU = TdS.
Assume that we add heat to the system (we see that dU = dQ because dQ = TdS and the work done is 0, because dV=0).
We see that the entropy and the temperature of the system increase.
The increase of energy in the system is given by
\Delta U = \int TdS,
with T the temperature of the system (which is not constant) and dS the change in entropy of the system.
Is this correct?

I am not trying to calculate anything. I just want to know if this is correct or not.

Thanks in advance.

- Sebas4.
 
Last edited:
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If the heat is added irreversibly (i.e., very rapidly), the temperature of the system is not uniform spatially (and neither is the internal energy per unit mass and the entropy per unit mass). How would you use this equation under those circumstances?
 
Not, because the temperature of the system is not uniform (or equal). Is it possible to add heat in such a way that the process in quasistatic? In real life it is not possible but theoretically? Is the equation then valid?

Maybe I have to change my question. My question is what's is the use of the thermodynamic identity?
 
Sebas4 said:
Not, because the temperature of the system is not uniform (or equal). Is it possible to add heat in such a way that the process in quasistatic? In real life it is not possible but theoretically? Is the equation then valid?
Yes, in the reversible limit, the equation is valid.
Sebas4 said:
Maybe I have to change my question. My question is what's is the use of the thermodynamic identity?
The equation is valid in the reversible limit, and is also valid for two closely neighboring (differentially separated) thermodynamic equilibrium states, irrespective of how tortuous and irreversible the path between these states had been, provided only that they are differentially separated and each at thermodynamic equilibrium. Of course, a reversible path is a continuous sequence of thermodynamic equilibrium states.
 

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