- #1

Sebas4

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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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