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Question about the fundamental equation of thermodynamics 
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#1
Nov312, 07:45 PM

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In a PVT system, knowing the function U(S,V) (or S(U,V) ) allows us to know all the information about the system, so it's called the fundamental equation of thermodynamics (and it makes sense to me because it's derived from the 1st and 2nd law). But if we know the function P(V,T) for example we also know all the information (all the state variables) about the system, so why isn't that the fundamental equation? I guess what I really want to know is what is "knowing all the information about the system"...
The difference I see is that knowing U(S,V) allows us to know what will be the new state variables if there is work or heat transferred to the system, while only knowing P(V,T) as far as I know doesn't allows us to see that. So I guess the difference between the 2 is that knowing U(S,V) allows us to know how the system will react to thermodynamic processes, while P(V,T) doesn't? 


#2
Nov312, 09:43 PM

P: 789

"The variables that are held constant in this process are termed the natural variables of that potential. The natural variables are important not only for the above mentioned reason, but also because if a thermodynamic potential can be determined as a function of its natural variables, all of the thermodynamic properties of the system can be found by taking partial derivatives of that potential with respect to its natural variables and this is true for no other combination of variables. On the converse, if a thermodynamic potential is not given as a function of its natural variables, it will not, in general, yield all of the thermodynamic properties of the system." S and V are the natural variables of U. P is not a thermodynamic potential. So the expanded question is "why is the above true?" 


#3
Nov312, 11:29 PM

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#4
Nov412, 12:14 AM

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Question about the fundamental equation of thermodynamics
You can integrate the general differential form to get [itex]\Phi(y_1,y_2)[/itex] which is also called a fundamental equation. You can derive any fundamental equation from any other fundamental equation. For the potential U, you get U(S,V). If you express [itex]\Phi[/itex] in terms of any other variables than its natural variables, you cannot integrate it. Again, this doesn't answer the OP's question. The answer has something to do with Pfaffian forms and integrability, and I don't quite get it yet. 


#5
Nov412, 06:23 AM

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#6
Nov412, 12:19 PM

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As a concrete example, take the SackurTetrode equation for a monatomic ideal gas: [tex]S=kN\ln\left(\frac{V}{N}\left(\frac{U}{N}\right)^{3/2}\right)+S_0[/tex] Solving for U(S,V) gives [tex]U(S,V)=\frac{N^{5/3}}{V^{2/3}}e^{\frac{2(SS_0)}{3Nk}}[/tex] Differentiating for T as above: [tex]T(S,V)=\frac{2N^{2/3}}{3kV^{2/3}}e^{\frac{2(SS_0)}{3Nk}}[/tex] which you can solve for S(T,V) as [tex]S(T,V)=S_0\frac{3}{2}Nk\ln\left(\frac{2(N/V)^{2/3}}{3kT}\right)[/tex] Differentiating for P as above gives [tex]P(S,V)=\frac{2N^{5/3}}{3V^{5/3}}e^{\frac{2(SS_0)}{3Nk}}[/tex] Substituting S(T,V), you get [tex]P(S(T,V),V)=P(T,V)=\frac{NkT}{V}[/tex] If you start out with P(T,V), you cannot derive the U(S,V) equation because there is no form of the fundamental equation where [itex]dP=x_1 dT + x_2 dV[/itex] which will allow you to derive it. That's because P is not a thermodynamic potential. Only thermodynamic potentials, of which U is one, can be expressed this (fundamental) way, and when it is, it is in terms of the natural variables of the potential  S and V in the case of U. (EDIT) Actually, I think that is not correct. P is not a thermodynamic potential, because its not an energy, but you don't need to restrict yourself to thermodynamic potentials. The bottom line is that you are looking for an expression of the fundamental equation as [itex]dP=x_1 dT+x_2 dV[/itex] and you will not find it. The fundamental equation using the thermodynamic potential known as enthalpy (H) is: [tex]dH=T dS+V dP[/tex] which you could rewrite as: [tex]dP=\frac{1}{V}dH\frac{T}{V}dS[/tex] which would be a correct form of the fundamental equation. That would mean that the natural variables of P are H and S. So the reason you cannot derive the U(S,V) equation from the P(T,V) equation is because T and V are not the natural variables of P, and so P(T,V) does not contain all the information about the system. 


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Nov412, 08:36 PM

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#8
Nov412, 09:03 PM

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#9
Nov512, 11:51 AM

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#10
Nov512, 11:55 AM

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#11
Nov512, 05:52 PM

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#12
Nov512, 11:12 PM

P: 789

There are restrictions on S, N, k, and V. They are all real and N, k, and V are greater than zero. Under those conditions, I think T(S,V) is injective.
Physically, you cannot have negative entropy, which means SackurTetrode falls apart for temperatures below a certain point, but I think that we can ignore that for the moment, not get sidetracked. I wonder if its even possible to express P(V,T) analytically in terms of U(S,V). This is where my understanding gets less than clear. I tentatively think of an analogy like y>0, x=2y which implies x>y. But you cannot derive x=2y from x>y. x>y contains less information than x=2y, just like P(V,T) contains less information than U(S,V). I mean, if I had an analytical expression for P(V,T) in terms of U(S,V) I would want to try to invert it, and I expect there would be trouble. Thermodynamic mathematics always gives me trouble, and I would like to really understand it at some point. It has to do with Pfaffian equations and integrability  try googling "Pfaffian thermodynamics" without the quotes and look at the google book entry (second one down, by Sychev, I think). If you come up with any insights, let me know. Also, one of these days I have to understand Caratheodory. From what I understand, he nailed it, but he wasn't good at simple expositions. 


#13
Nov712, 12:19 PM

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#14
Nov712, 09:10 PM

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#15
Nov712, 11:31 PM

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