[Pw_forum] a questoin about pressure and bulk modulus of system

[Stefano Baroni]

Folks: a distinguished colleague of mine just made me notice that if I were a student I would probably fail the first-year thermodynamics exam. Some of my statements are right, but my claims of Meyasam being wrong are partially wrong (and he was therefore partially right!). Let me point out my own mistakes:

>> I have another question :Is these terms are right?
>> E=-PV+uN+TS

Contrary to my claims, this equation is correct, as it is a consequence of Euler's theorem on homogeneous functions [E(λS,λV,λN)=λE(S,V,N) implies E= S ∂E/∂S + V ∂E/∂V +N ∂E/∂N] - Notice that you need Euler's theorem *AND* the definition of T,P,μ as energy derivatives to establish the equation.

>> => dE=-PdV-VdP+udN+Ndu+Tds+SdT
> 
> NO - this is nonsense. Nor the energy, nor its differential are what you write. As said in my previous post, the energy is an extensive function of the three extensive quantities S,V,N: E=E(S,V,N) ≠ -PV+uN+TS. In calculating its differential, you should vary only the variables it depends upon (i.e., S,V,N).

MY comment is wrong. As said, E=-PV+uN+TS is a consequence of Euler's theorem, and it is legitimate to take the formal differential the way Meyasam did.

> 
>> and we know from first law of thermodynamics that dE=-PdV+udN+Tds
> 
> This is correct, in fact it is different from the expression you gave before. Form this relation you can conclude that P=-∂E/∂V; T=∂E/∂S, μ=∂E/∂N. The first relation is what you need
> 
>> and we conclude that -VdP+Ndu+SdT=0
> 
> NO - see above

RIGHT for a more subtle reason than I was able to guess at first sight
the above equation even has a distinguished name: its is know as the Gibbs-Duhem equation

> 
>> and in zero temrature and fixed number of atoms
>> we have dE=-PdV and P=-dE/dV
> 
> right, for the wrong reason (i.e. your conclusion is correct, the argument you use to derive it is not)

my last negative statement is less wrong than others, in the sense that you do not need the Gibbs-Duhem equation to establish that P=-∂E/∂V. This relation is instead a definition of the pressure, as the derivative of the energy with respect to one of its natural variables, and it is indeed needed to establish the Gibbs-Duhem equation.

Sorry for the mess I may have created. I can only conclude that I could have directed the recommendation to study more thermodynamics to myself in the first place ... In any event, I now know more than I did before, and so I hope will some of you. Thank you for making me (re-) learn something that I had long forgotten ...

Stefano B

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Stefano Baroni - SISSA  &  DEMOCRITOS National Simulation Center - Trieste
http://stefano.baroni.me [+39] 040 3787 406 (tel) -528 (fax) / stefanobaroni (skype)

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