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Take this phase diagram :

enter image description here

At the point "A" I will have a liquid, at point B a gas.

But what I want to know is : which phase do I have at point C.

I know that this is precisely where the phase transition occurs, but imagine I put the pressure and temperature at $(P_c,T_c)$. Will my system have a liquid phase with the "first molecule of gas close to appear", or will it be a gas will the "first molecule of liquid close to appear" ? (I puted quotes because I don't know what is the exact word in english for this). How can I know ?

Does it in fact depend of where I was previously ?

Like if I start from A, go to C then on C it will still be a liquid. But if I started at B and changed temperature and pressure to go to C I would have a gas on C ?

[Edit because of the comment] : To be more accurate in my question, I know from here https://www.uam.es/personal_pdi/ciencias/evelasco/master/tema_III.pdf that two phases coexisting are not stable thermodynamically.

Here is the curve they have :

enter image description here

And what they say :

Inside the spinodal line the fluid is unstable, since $\kappa_T^{-1}<0$ , which implies $\frac{\partial P}{\partial V} > 0$, which is not possible for a stable material. If we disregard these unstable states (between ‘a’ and ‘b’), predicted by the theory, we are left with the gas and liquid branches, from small volumes up to ‘a’, and from large volumes up to ‘b’ (i.e. the regions where $\kappa_T^{-1}>0$).

So in the context of thermodynamic we are looking for stability. Thus between between $V_1$ and $a$, $b$ and $V_2$, we have metastability and between a and b we have unstability.

So in conclusion we don't have thermodynamic equilibrium (because we are looking for stable states) when we have two phases (liq+gas).

Thus how can we have a mixture at thermodynamic equilibrium with this explanation, I don't get...

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    $\begingroup$ At the point C you will actually have BOTH in general. You'll have two coexisting states, and the quantities are such that the chemical potential or molar gibbs potential is the same. $\endgroup$ – FGSUZ Jan 5 '18 at 0:25
  • $\begingroup$ At the point C, as FGSUZ said, you have both liquid and vapor phase. They co-exist. Now, you might further ask how much is the liquid phase or how much is the vapor phase. The term for this is quality. It doesn't depend on where it is previously. But quality is change when you inject or retract heat from the mixture. $\endgroup$ – user115350 Jan 5 '18 at 5:20
  • $\begingroup$ @FGSUZ I am not sure to understand because I thought that when I have two phases in equilibrium, the system is not thermodynamically stable and it will go either to full liquid or to full gas : uam.es/personal_pdi/ciencias/evelasco/master/tema_III.pdf beginning of page 64. They say that on figure 3.10 between a and b the system is unstable. Thus I can't have thermo equilibrium between gas and liquid ? $\endgroup$ – StarBucK Jan 5 '18 at 9:14
  • $\begingroup$ @user115350 To understand physically. Imagine I push on a piston to compress my experiment. So if I start at A, I can compress it until I reach the transition curve (vertical projection of A on the curve). Then at this moment my piston will be "stucked" and all the energy I spend on pushing on it will be transformed into heat to do the phase transition. Then when I have full gas, my piston is not stucked anymore and the energy I spend pushing on it will be indeed converted into mechanical motion (the piston will move). Is that it ? $\endgroup$ – StarBucK Jan 5 '18 at 9:16
  • $\begingroup$ @StarBucK Sorry I don't see any reference to what you say. Those lines separate two states. The lines are actually coexistence lines. $\endgroup$ – FGSUZ Jan 5 '18 at 10:33
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I'm completing my comment posting an answer.

As I said, you have both states along the coexistence line.

enter image description here

Outside the line, the main minimum in Gibbs potential pulls the system to the corresponding state (liquid upside, vapor below, and so on), however, tehre's still another secondary minimun in general, when you are near the line. The secondary minimum is the metastable state, and you can have liquid in the gas region if you're near the line, but this is much less stable than the main state.

The quantities are those who equate the chemical potential $\mu$. That's like the "equation of the lineas", if you're on the lines, the system is forced to satisfy that equality, so the components will react and change phase until both potentials are the same.

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  • $\begingroup$ Consider we are only interested in the thermodynamic equilibrium, thus we forget about the metastable because they are not the global minimum of Gibbs. On the $(P,T)$ diagram, on the coexistence curve it is not possible to know what would be the liquid and gas proportion. Thus we look at a $(P,V)$ diagram where we can see those (we see the phase transition on it). In this diagram, again if we focus on thermodynamic equilibrium thus only the stable ones, we realise that only pure liquid or pure gas are possible. This is why I misunderstand. $\endgroup$ – StarBucK Jan 5 '18 at 10:55
  • $\begingroup$ But indeed if we consider the metastable states then I understand that we can have liquid and gas in different proportion on the coexistence curve, but I read that to have those metastable states we have to do "tricks" because they are not the real stable states. Thus I avoided them for a first comprehension. Thank you a lot for your time. $\endgroup$ – StarBucK Jan 5 '18 at 10:57
  • $\begingroup$ You'll have some parts goign to liquid and some parts going to gas. The molar fraction $\chi$ of each one is what you've got to find with math. $\endgroup$ – FGSUZ Jan 5 '18 at 11:01
  • $\begingroup$ Hey, how comes they are not real stable states? They are, they exist, but only locally stable. A perturbation might ruin that. $\endgroup$ – FGSUZ Jan 5 '18 at 11:02
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    $\begingroup$ It's all about the time of relaxation times. You can do it with a fast cooling, so that the system has no time to rearrange, or also really really carefully. The thing is that perturbations move it outside the minimum, and once it's out it will fall in the deepest pit. $\endgroup$ – FGSUZ Jan 5 '18 at 20:04

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