Is dn = dS = 0 when two phases are equilibrium?

[Philethan]

Hello, every one! I'm studying the Section 9.2 - Phase Equilibrium of "Classical and Statistical Thermodynamics" written by Carter.

In equation (9.21) and (9.22)

He said:

[itex]dS_{A}=\frac{1}{T_{A}}(dU_{A}+P_{A}dV_{A}-\mu _{A}dn_{A})[/itex]-----(9.21)

[itex]dS_{B}=\frac{1}{T_{B}}(dU_{B}+P_{B}dV_{B}-\mu _{B}dn_{B})[/itex]-----(9.22)

I think dS and differentials of all other state variables should be zero! That is, dS_A=dS_B=dn_A=dn_B=dV_A=dV_B=...=0

Is it correct? If it's correct, then I cannot get the following conclusions which needs to assume those increments dU_A,dV_A,dn_A,... could be arbitrary.

T_A = T_B (thermal equilibrium)
P_A = P_B (mechanical equilibrium)
μ_A = μ_B (diffusive equilibrium)

Thanks in advance!

 

[DrDu]

No, why should these differential be constant?

 

[Philethan]

No, why should these differential be constant?

It's due to the definition of equilibrium. In other words, if those macroscopic state variables are not constant, then it's not at equilibrium.
For example, if the color of the solution is changing, then it's not at equilibrium.

I don't know why I was wrong..

 

[DrDu]

But e.g. you can have a mixture of ice and water and even if some of the ice is melting (i.e. both the amount of ice and water is changing), the two are in equilibrium.

 

[Philethan]

Really? Even if some of the ice is melting, the two are still in equilibrium?
So... What's the definition of equilibrium? It seems that the exact definition of equilibrium is "the system is at equilibrium if and only if its entropy is a maximum".

Is that correct?

Thank you very much :)

 

[BvU]

I think DD is straying somewhat. A phase equilibrium is not enough for a thermodynamic equilibrium: matter and energy can still flow macroscopically.
So I'm siding with PhE post #5.
However, physics isn't a democratic thingy, so if someone knows better, I'm eager to be put right !

 

[DrDu]

I think DD is straying somewhat. A phase equilibrium is not enough for a thermodynamic equilibrium: matter and energy can still flow macroscopically.

I'd rather say that phase equilibrium is a special case of thermodynamic equilibrium.

 

[DrDu]

Really? Even if some of the ice is melting, the two are still in equilibrium?

At equilibrium, the system is insensitive to changes in particle number of the two phases.

So... What's the definition of equilibrium? It seems that the exact definition of equilibrium is "the system is at equilibrium if and only if its entropy is a maximum".

Is that correct?

Thank you very much :)

This depends on the system. Is it thermally isolated or constant temperature, constant volume or constant pressure. E.g. at constant temperature and pressure, free enthalpy has to be maximal, not entropy. Which case is your book considering?

 

[BvU]

We aren't making progress here if I rather say that thermodynamic equilibrium is a special case of phase equilibrium, namely phase equilibrium plus constant P, U, G, A, H, S etc...

 

[DrDu]

I don't know what you have in mind, but in equilibrium thermodynamics (I suppose we are talking about this) we are considering only thermodynamic equilibrium states. E.g. considering the state diagram of water as a function of p and T, there is one line where both phases are in equilibrium. In this sense, phase equilibria are a subset of thermodynamic equilibria.

 

[BvU]

in equilibrium thermodynamics (I suppose we are talking about this) we are considering only thermodynamic equilibrium states

Somewhat circular, to me that seems.

The 'subset' issue is a matter of language, it seems to me. I think we agree: phase equilibrium isn't thermodynamic equilibrium, but thermodynamic equilibrium is phase equilibrium.

(The "fruit isn't orange but orange is fruit" comes to mind. For me that makes oranges a subset of fruit. Not the other way around).

Yet another approach: there are a lot more phase equilibria than thermodynamic equilibria. A TE is a PE with some extra restrictions. So TE is a subset of PE.

considering the state diagram of water as a function of p and T, there is one line where both phases are in equilibrium

If a system is moving back and forth over this line where both phases are in equilibrium, I can not consider that system to be in thermodynamic equilibrium: T changes. [edit] woops, V changes.

 

[DrDu]

It seems that the exact definition of equilibrium is "the system is at equilibrium if and only if its entropy is a maximum".

This is true if the system is isolated and there is also no work done on the overall system. Then it is clear that ##dU_B=-dU_A##, ##dV_B=-dV_A## and ##dn_B=-dn_A## as total internal energy, volume and particle number of the two phases taken together is constant. As total S is maximal, we can write ##0=\partial S/\partial U_A=1/T_A-1/T_B##, so that ##T_A=T_B##. Analogous equations hold for p and ##
mu##.

 

[Chestermiller]

This is true if the system is isolated and there is also no work done on the overall system. Then it is clear that ##dU_B=-dU_A##, ##dV_B=-dV_A## and ##dn_B=-dn_A## as total internal energy, volume and particle number of the two phases taken together is constant. As total S is maximal, we can write ##0=\partial S/\partial U_A=1/T_A-1/T_B##, so that ##T_A=T_B##. Analogous equations hold for p and ##
mu##.

Yes. I think this is more like what the original question was getting at. The idea is to look at the combined system in a situation where the combination is isolated.