I Understand the Thermodynamic Identity: Is This Correct?

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The Thermodynamic Identity, expressed as dU = TdS - PdV + μ dN, simplifies to dU = TdS under constant volume and particle number conditions. When heat is added, dU equals dQ, leading to increases in entropy and temperature. The integral ΔU = ∫TdS describes the energy increase, assuming temperature varies with entropy changes. The equation remains valid in reversible processes and for closely neighboring thermodynamic equilibrium states, regardless of the irreversibility of the path taken. Understanding the thermodynamic identity is crucial for analyzing energy changes in various thermodynamic processes.
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.
 
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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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