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Question about the fundamental equation of thermodynamics

by Tosh5457
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Tosh5457
#1
Nov3-12, 07:45 PM
P: 238
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?
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Rap
#2
Nov3-12, 09:43 PM
P: 789
Quote Quote by Tosh5457 View Post
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?
This is not an answer to your question, but it might help. U is a thermodynamic potential. To quote from Wikipedia (thermodynamic potential)

"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?"
Chestermiller
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Nov3-12, 11:29 PM
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Quote Quote by Tosh5457 View Post
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?
To do practical problems, you usually need to use both. So neither solely provides all the information you need to know about a system, but the combination does. Why the U equation is called "the fundamental equation," I don't know.

Rap
#4
Nov4-12, 12:14 AM
P: 789
Question about the fundamental equation of thermodynamics

Quote Quote by Chestermiller View Post
To do practical problems, you usually need to use both. So neither solely provides all the information you need to know about a system, but the combination does. Why the U equation is called "the fundamental equation," I don't know.
I don't think that's true. The P(V,T) equation can be derived from the U(S,V) equation, but not vice versa. The "fundamental equation" can take many forms, the usual one is in differential form [tex]dU=T\,dS-P\,dV[/tex] Notice that the three differentials are U, S, and V. U is the thermodynamic potential, S and V are its natural variables. The fundamental equation can be generally written as [tex]d\Phi=x_1\,dy_1+x_2\,dy_2[/tex] where [itex]\Phi[/itex] is any thermodynamic potential, and [itex]y_1[/itex] and [itex]y_2[/itex] are its natural variables. [itex]x_1[/itex] and [itex]y_1[/itex] are called "conjugate variables", as are [itex]x_2[/itex] and [itex]y_2[/itex]. For example, S and T are conjugate variables and so are P and V. The product of two conjugate variables always has the dimension of energy/volume.

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.
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Nov4-12, 06:23 AM
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Quote Quote by Rap View Post
I don't think that's true. The P(V,T) equation can be derived from the U(S,V) equation, but not vice versa. The "fundamental equation" can take many forms, the usual one is in differential form [tex]dU=T\,dS-P\,dV[/tex] Notice that the three differentials are U, S, and V. U is the thermodynamic potential, S and V are its natural variables. The fundamental equation can be generally written as [tex]d\Phi=x_1\,dy_1+x_2\,dy_2[/tex] where [itex]\Phi[/itex] is any thermodynamic potential, and [itex]y_1[/itex] and [itex]y_2[/itex] are its natural variables. [itex]x_1[/itex] and [itex]y_1[/itex] are called "conjugate variables", as are [itex]x_2[/itex] and [itex]y_2[/itex]. For example, S and T are conjugate variables and so are P and V. The product of two conjugate variables always has the dimension of energy/volume.

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.
Oh yeah? Then give one example of how the P = P(V,T) equation can be derived from the U(S,V) equation. In solving the problem of the adiabatic reversible expansion of an ideal gas, both equations are needed.
Rap
#6
Nov4-12, 12:19 PM
P: 789
Quote Quote by Chestermiller View Post
Oh yeah? Then give one example of how the P = P(V,T) equation can be derived from the U(S,V) equation. In solving the problem of the adiabatic reversible expansion of an ideal gas, both equations are needed.
Ok, in the general case, if you have U(S,V) then from the differential fundamental equation [tex]dU=TdS-PdV[/tex] you can see that [tex]\left(\frac{\partial U}{\partial V}\right)_S=-P(S,V)[/tex] and [tex]\left(\frac{\partial U}{\partial S}\right)_V=T(S,V)[/tex] Solve the second one to get S(T,V) and substitute in the first one to get -P(S(T,V),V) or -P(T,V).

As a concrete example, take the Sackur-Tetrode 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(S-S_0)}{3Nk}}[/tex] Differentiating for T as above: [tex]T(S,V)=\frac{2N^{2/3}}{3kV^{2/3}}e^{\frac{2(S-S_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(S-S_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.
Chestermiller
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Nov4-12, 08:36 PM
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Quote Quote by Rap View Post
Ok, in the general case, if you have U(S,V) then from the differential fundamental equation [tex]dU=TdS-PdV[/tex] you can see that [tex]\left(\frac{\partial U}{\partial V}\right)_S=-P(S,V)[/tex] and [tex]\left(\frac{\partial U}{\partial S}\right)_V=T(S,V)[/tex] Solve the second one to get S(T,V) and substitute in the first one to get -P(S(T,V),V) or -P(T,V).

As a concrete example, take the Sackur-Tetrode 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(S-S_0)}{3Nk}}[/tex] Differentiating for T as above: [tex]T(S,V)=\frac{2N^{2/3}}{3kV^{2/3}}e^{\frac{2(S-S_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(S-S_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.
I'm still a little confused. You chose an equation for S that applies specifically to the case of a monatomic ideal gas. It seems to me this is the same as replacing P = P (V, T) by this alternate equation S (V,U). So you are still using two equations. For a gas beyond the ideal gas range, you will still have a functionality P = P (V, T) which can presumably be replaced by an equation such as S (V,U), but again, there will be two equations required. The functionality P = P (V,T) provides the specificity for the particular material involved.
Rap
#8
Nov4-12, 09:03 PM
P: 789
Quote Quote by Chestermiller View Post
I'm still a little confused. You chose an equation for S that applies specifically to the case of a monatomic ideal gas. It seems to me this is the same as replacing P = P (V, T) by this alternate equation S (V,U). So you are still using two equations. For a gas beyond the ideal gas range, you will still have a functionality P = P (V, T) which can presumably be replaced by an equation such as S (V,U), but again, there will be two equations required. The functionality P = P (V,T) provides the specificity for the particular material involved.
I'm not sure what you mean. The only equation used was the S(U,V) equation, the P(V,T) equation was never assumed in the derivation, it was derived from the S(U,V) equation. The S(U,V) equation (or, equivalently, the U(S,V) equation) does not replace the P(V,T) equation. It implies the P(V,T) equation, and more. In fact, it implies every other equation you can write about that ideal gas, which is why it is called a fundamental equation. The functionality U=U(S,V) provides the same specificity for the particular material involved as does the P(V,T) equation - it must, because it implies the P(V,T) equation. I think it makes the math easier to use both equations from the beginning of a particular analysis, but this does not mean they are independently true.
Tosh5457
#9
Nov5-12, 11:51 AM
P: 238
Quote Quote by Rap View Post
Ok, in the general case, if you have U(S,V) then from the differential fundamental equation [tex]dU=TdS-PdV[/tex] you can see that [tex]\left(\frac{\partial U}{\partial V}\right)_S=-P(S,V)[/tex] and [tex]\left(\frac{\partial U}{\partial S}\right)_V=T(S,V)[/tex] Solve the second one to get S(T,V) and substitute in the first one to get -P(S(T,V),V) or -P(T,V).

As a concrete example, take the Sackur-Tetrode 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(S-S_0)}{3Nk}}[/tex] Differentiating for T as above: [tex]T(S,V)=\frac{2N^{2/3}}{3kV^{2/3}}e^{\frac{2(S-S_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(S-S_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.
Nice, I tried to do that but I got stuck at how to put S in function of T and V
Rap
#10
Nov5-12, 11:55 AM
P: 789
Quote Quote by Tosh5457 View Post
Nice, I tried to do that but I got stuck at how to put S in function of T and V
You mean getting S(T,V) from T(S,V)? Or plugging in S(T,V) into the P(S,V) equation?
Tosh5457
#11
Nov5-12, 05:52 PM
P: 238
Quote Quote by Rap View Post
You mean getting S(T,V) from T(S,V)? Or plugging in S(T,V) into the P(S,V) equation?
Getting S(T,V) from T(S,V). I was looking for a way to express P(V,T) analytically (in terms of derivatives of U(S,V) and such) so I didn't think of that But I guess T(S,V) would have to be injective and in general it isn't...
Rap
#12
Nov5-12, 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 Sackur-Tetrode 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.
Tosh5457
#13
Nov7-12, 12:19 PM
P: 238
Quote Quote by Rap View Post
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 Sackur-Tetrode 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.
For a one real variable function to be injective a sufficient condition is that its derivative function is always positive. I don't know if that applies to multivariate functions though (using partial derivatives). If it does apply, and dT/dS > 0 and dT/dV < 0 the function T(S,V) is injective and you can get P(V,T) analytically. But even if that is a sufficient condition to prove T(S,V) is injective, I'm pretty sure that to prove dT/dS > 0 and dT/dV < 0 in general, statistical mechanics is needed...
Hetware
#14
Nov7-12, 09:10 PM
P: 125
Quote Quote by Tosh5457 View Post
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?
It would really be a good idea to define your variables. Not everybody uses the same symbols to represent the same quantities.
Rap
#15
Nov7-12, 11:31 PM
P: 789
Quote Quote by Hetware View Post
It would really be a good idea to define your variables. Not everybody uses the same symbols to represent the same quantities.
U=internal energy, T=temperature, S=entropy, P=pressure, V=volume, N=number of particles, k=Boltzmann's constant

Quote Quote by Tosh5457 View Post
For a one real variable function to be injective a sufficient condition is that its derivative function is always positive. I don't know if that applies to multivariate functions though (using partial derivatives). If it does apply, and dT/dS > 0 and dT/dV < 0 the function T(S,V) is injective and you can get P(V,T) analytically. But even if that is a sufficient condition to prove T(S,V) is injective, I'm pretty sure that to prove dT/dS > 0 and dT/dV < 0 in general, statistical mechanics is needed...
I don't think stat mech is needed to prove anything in thermodynamics. Stat mech explains thermodynamics, and extends analysis of systems beyond thermodynamics. But I guess I am not clear why we need to prove that T(S,V) is injective. The goal is to express P(V,T) in terms of U(S,V) analytically, in the case of an ideal gas.


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