Is the thermodynamic identity dU = TdS correct for constant V and N?

[Sebas4]

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.

 

[Chestermiller]

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?

 

[Sebas4]

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?

 

[Chestermiller]

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.

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.