- #1
Sebas4
- 13
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I have a question about the Thermodynamic Identity.
The Thermodynamic Identity is given by
[tex] dU = TdS - PdV + \mu dN [/tex].
We assume that the volume [itex]V[/itex] and that the number of particles [itex]N[/itex] is constant.
Thus the Thermodynamic Identity becomes
[tex] dU = TdS [/tex].
Assume that we add heat to the system (we see that [itex]dU = dQ[/itex] because [itex]dQ = TdS[/itex] and the work done is 0, because [itex]dV=0[/itex]).
We see that the entropy and the temperature of the system increase.
The increase of energy in the system is given by
[tex] \Delta U = \int TdS [/tex],
with [itex]T[/itex] the temperature of the system (which is not constant) and [itex]dS[/itex] 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.
The Thermodynamic Identity is given by
[tex] dU = TdS - PdV + \mu dN [/tex].
We assume that the volume [itex]V[/itex] and that the number of particles [itex]N[/itex] is constant.
Thus the Thermodynamic Identity becomes
[tex] dU = TdS [/tex].
Assume that we add heat to the system (we see that [itex]dU = dQ[/itex] because [itex]dQ = TdS[/itex] and the work done is 0, because [itex]dV=0[/itex]).
We see that the entropy and the temperature of the system increase.
The increase of energy in the system is given by
[tex] \Delta U = \int TdS [/tex],
with [itex]T[/itex] the temperature of the system (which is not constant) and [itex]dS[/itex] 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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