I Understand the Thermodynamic Identity: Is This Correct?

AI Thread Summary
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
Messages
13
Reaction score
2
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:
Science news on Phys.org
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.
 
I was watching a Khan Academy video on entropy called: Reconciling thermodynamic and state definitions of entropy. So in the video it says: Let's say I have a container. And in that container, I have gas particles and they're bouncing around like gas particles tend to do, creating some pressure on the container of a certain volume. And let's say I have n particles. Now, each of these particles could be in x different states. Now, if each of them can be in x different states, how many total...
Thread 'Why work is PdV and not (P+dP)dV in an isothermal process?'
Let's say we have a cylinder of volume V1 with a frictionless movable piston and some gas trapped inside with pressure P1 and temperature T1. On top of the piston lay some small pebbles that add weight and essentially create the pressure P1. Also the system is inside a reservoir of water that keeps its temperature constant at T1. The system is in equilibrium at V1, P1, T1. Now let's say i put another very small pebble on top of the piston (0,00001kg) and after some seconds the system...
Back
Top