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.
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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