Thermodynamics state equations

[Cathr]

Homework Statement

I am always confused about how to correctly write the functions U, H, F, G when they're not depending on the usual variables p, V, T, S - same question for Q and W.

For example, we have to calculate the temperature variation of a small surface of water when we isentropically change the surface. The variables are A=a-bT (superficial tension constant), T (temperature) and E (surface). It is given that the elemeptary work dW=AdE.

Homework Equations

dU=TdS+AdE (why don't we add -pdV too? sometimes we do)
dF=-SdT+AdE
dH=TdS-EdA
dG=-SdT-EdA

The Attempt at a Solution

Now we know that, to solve this, we must find the right state function between the 4.
I am really confused about the EdA and AdE part. How to know how to write these equations correctly, without doubting if I'm right?

This is just an example, but this is the kind of problem I have for more cases. If there is a generalised way of reasoning to understand them all?

For the classic 4 variables there is the mnemotic scheme of Born so there's no problem.

 

[Chestermiller]

Homework Statement

I am always confused about how to correctly write the functions U, H, F, G when they're not depending on the usual variables p, V, T, S - same question for Q and W.

For example, we have to calculate the temperature variation of a small surface of water when we isentropically change the surface. The variables are A=a-bT (superficial tension constant), T (temperature) and E (surface). It is given that the elemeptary work dW=AdE.

Homework Equations

dU=TdS+AdE (why don't we add -pdV too? sometimes we do)
dF=-SdT+AdE
dH=TdS-EdA
dG=-SdT-EdA

The Attempt at a Solution

Now we know that, to solve this, we must find the right state function between the 4.
I am really confused about the EdA and AdE part. How to know how to write these equations correctly, without doubting if I'm right?

This is just an example, but this is the kind of problem I have for more cases. If there is a generalised way of reasoning to understand them all?

For the classic 4 variables there is the mnemotic scheme of Born so there's no problem.

This is exactly the same kind of problem you have already been doing on gases with volume changes, except here you are dealing with a liquid surface, with area changes. So, in this development A takes the place of P and E takes the place of V. The only difference is that A is a tensile force per unit length, and P is a compressive force per unit area; so there is a sign difference. Instead of -PdV being the work done by the surroundings on the system, you have AdE. Once you have this and dU=TdS+AdE, what do you get for dF=dU-d(TS), dH=dU-d(AE), and dG=dF-d(AE)?

Now, if S(T,E) is the entropy as a function of temperature T and area E, using the mathematical rules for partial differentiation, what is dS in terms of dT and dE?

To be continued.

 

[Cathr]

This is exactly the same kind of problem you have already been doing on gases with volume changes, except here you are dealing with a liquid surface, with area changes. So, in this development A takes the place of P and E takes the place of V. The only difference is that A is a tensile force per unit length, and P is a compressive force per unit area; so there is a sign difference. Instead of -PdV being the work done by the surroundings on the system, you have AdE. Once you have this and dU=TdS+AdE, what do you get for dF=dU-d(TS), dH=dU-d(AE), and dG=dF-d(AE)?

Now, if S(T,E) is the entropy as a function of temperature T and area E, using the mathematical rules for partial differentiation, what is dS in terms of dT and dE?

To be continued.

I see, but why is it dU=TdS+AdE and not -AdE, like in -pdV? For dF, dH, dG I get the right answers, thank you!

I think dS(T, E)=(dS/dT)dT+(dS/dE)dE.

 

[Chestermiller]

I see, but why is it dU=TdS+AdE and not -AdE, like in -pdV? For dF, dH, dG I get the right answers, thank you!

Because P is compressive and A is tensile. So they have opposite signs in the equations.

I think dS(T, E)=(dS/dT)dT+(dS/dE)dE.

Yes. This is correct. More precisely, $$dS=\left(\frac{\partial S}{\partial T}\right)_EdT+\left(\frac{\partial S}{\partial E}\right)_TdE$$From this equation, if S is constant, what is $(\partial T/\partial E)_S$?

 

[Cathr]

Because P is compressive and A is tensile. So they have opposite signs in the equations.

Yes. This is correct. More precisely, $$dS=\left(\frac{\partial S}{\partial T}\right)_EdT+\left(\frac{\partial S}{\partial E}\right)_TdE$$From this equation, if S is constant, what is $(\partial T/\partial E)_S$?

Then dS is zero, so I guess it would be the same to write partial derivatives as dT/dE, but I guess you were looking for another answer. I don't really know.

 

[Chestermiller]

Then dS is zero, so I guess it would be the same to write partial derivatives as dT/dE, but I guess you were looking for another answer. I don't really know.

Set dS equal to zero in the equation and solve for dT/dE.