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When liquid is heat up to a critical temperature $T_{c}$, it starts boiling and converting to gas. In statistical mechanics, we learn that it is a phase transition. We studied all the properties near the critical temperature such as critical exponents.

However, evaporation can occur at all temperature, converting liquid phase to gas phase. So, I wonder if evaporation can be regarded as phase transition even though it seems to have no critical point.

Thanks.

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  • $\begingroup$ Yes it is, you are thinking macroscopically that evaporation occurs at all temperatures, but in the microscopic sense, the molecules evaporating have energies corresponding to the critical temperature. $\endgroup$ – PhysMath May 30 at 0:00
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    $\begingroup$ A phase transition refers to a transition between one state of matter and another state of matter due to a change in temperature, pressure, volume, etc., when the entire system is in thermodynamic equilibrium. When you write "evaporation can occur at all temperature", it sounds like you're talking about some system like an open glass of water that is not really in thermodynamic equilibrium with its surrounding environment because the water is continually evaporating. $\endgroup$ – Samuel Weir May 30 at 1:36
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    $\begingroup$ "Critical temperature" refers to a very specific temperature for all substances that can be made to boil, and that temperature is different for each substance. Note that boiling temperature is distinctly different than critical temperature. $\endgroup$ – David White May 30 at 3:36
  • $\begingroup$ "We studied all the properties near the critical temperature such as critical exponents". There may be some misunderstanding. From the context of your question, we can conclude that $T_c$ is the temperature of the phase transition of the 1st kind (boiling), that is, it is not a point, but the curve $T_c (P)$, which ends at some point $T_c, P_c$ . This is the critical point of the second order phase transition (liquid-vapor), and it is for this point that the so-called critical exponents exist and are studied. $\endgroup$ – Aleksey Druggist May 31 at 8:06
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We should be a bit careful distinguishing the different scenarios.

The phase boundary one usually talks when we talk about the boiling of water is the point of equilibrium between (pure) liquid water and (pure) gaseous water. For this transition, there is a critical point at some $\left(P_c, T_c\right)$, connected to a line of first-order phase transitions $T_{boil}\left(P\right)$.

On the other hand, the situation with evaporation at room temperature is not the conversion between liquid water and gaseous water - indeed, from the phase diagram, we already know that the equilibrium state for pure water is a liquid at this temperature and pressure. If we had an enclosed box full of water and nothing else, with a piston so that it feels atmospheric pressure, it would be a liquid.

If I have a glass of water in my kitchen, however, it is not a system composed purely of water - now the system is composed of both water and air. Let's idealize the air as a single type of molecule, even though it has many. We now have three state variables - pressure, temperature, and water concentration. Since we have a two-component system, Gibbs' phase rule tells us that we can have two phases coexisting over a range of pressures and temperatures.

We can get some idea what these two phases are by considering a simple model of phase coexistence - e.g. something like two types of atoms moving around on a lattice, with some kind of interaction making atoms prefer to be next to atoms of the same type. The conclusion of this kind of analysis is that we can have coexistence between a phase which is mostly water with a small amount of dissolved air, and a phase which is mostly air with a small amount of dissolved water. (See e.g. Jones for this calculation.)

Thus, we conclude that the equilibrium state in my kitchen is a glass of water with a bit of air dissolved in it, and a room full of air with a bit of water dissolved in it. These two phases have the same free energy at this pressure and temperature, so they can coexist, and the relative amount of each is determined by conservation of mass (if I started out with more water in the room before equilibrating, I'll end up with a bigger water-rich phase.)

The usual evaporation scenario occurs because we are not in equilibrium, and the water concentration in the air is below its equilibrium value. So it is not a phase transition, so much as a system out of equilibrium moving toward equilibrium, as the comment by Samuel Weir pointed out.

Incidentally, binary mixture models can also have phase transitions and a critical point. But I think this is not relevant here.

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  • $\begingroup$ It is not uncommon for a first-order phase transition to occur through the formation of nuclei of a new phase (for example, condensation), which sometimes requires a significant deviation of temperature from the equilibrium temperature of the phase transition (overheating or overcooling). Thus, if we consider the kinetic aspects of phase transitions, then it turns out that the phase transition can take place at a non-equilibrium temperature and is suppressed at an equilibrium $\endgroup$ – Aleksey Druggist Jun 2 at 16:04

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