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Old Feb 7th 2018, 08:50 AM   #1
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Thermodynamic Potential

Hello,

I came across the Lagrange Transformation explaining the Thermodynamic Potential. Textbooks show, how the math is executed.

Can someone explain the practical use of transforming the energies pdV and TdS into Vdp and TdS.


Regards

Lothar
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Old Feb 7th 2018, 09:46 AM   #2
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Do you mean Legendre transformations?

The value is that they change the variables of state from variables that we do not normally encounter to variables that we do.
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Old Feb 7th 2018, 10:21 AM   #3
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Hello,
Legendre of course.

Thanks for the answer. It seems to have to do with what I can determine easier by an experiment?? Still fuzzy in m mind.

Can you give me perhaps a simple example?

Lothar
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Old Feb 8th 2018, 04:00 AM   #4
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You can take expressions for any of the four basic 'thermodynamic potentials' and derive expressions for the others from there. For instance you can start with internal energy (U) and derive Helmoltz Free Energy (F), Gibbs Free energy (G) and Enthalpy (H).

However this can also be done by other means which are normally taught first.

So the method tends to be used for more modern (I hesitate to say more advanced) purposes involving quantum theory, statistical mechanics etc.

Here is a typical example

In a crystal where the ions have only two energy states called 0 and e the Helmholtz free energy was shown by Reidi in 1988 to be

F = NRT {1 + exp(-U/RT)}

Taking U as constant find the entropy relation and the molar heat capacity as a function of temperature.



Would working this example be of interest?

The molar heat capacity leads to a solid state effect known as the Schottky Anomaly

https://en.wikipedia.org/wiki/Schottky_anomaly

Last edited by studiot; Feb 8th 2018 at 01:17 PM.
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