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I still cannot fully comprehend the essence of a critical point on phase diagrams.

It is usually said in textbooks that the difference between liquid and gaseous state of a substance is quantitative rather than qualitative. While it is easy to understand for a liquid-solid transition (symmetry breaking is a qualitative change), it is unclear to me what meaning does it have for a liquid and its gas: there is always a quantitative difference between a gas at 300 $K$ and at 400 $K$.

  1. Is it correct to say just "this substance is in gaseous state"? Shouldn't we also specify the path on the phase diagram by which the substance got in its current state? Did it cross the boiling curve or went over the critical point and never boiled?

  2. Why does a critical point even exist? Blindly, I would assume that either there is no boiling curve at all - since the difference is quantitative, the density of a substance smoothly decreases with temperatures and increases with pressure; or that the boiling curve goes on to "infinity" (to as high pressures and temperatures as would remain molecules intact). Why does it stop?

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    $\begingroup$ Possible duplicates: physics.stackexchange.com/q/246505/2451 , physics.stackexchange.com/q/19815/2451 and links therein. $\endgroup$
    – Qmechanic ♦
    Commented Jul 19, 2016 at 16:04
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    $\begingroup$ The answer given in physics.stackexchange.com/q/246505 does not actually answer my question. It rather describes what happens near critical point, but doesn't give an explanation to why it exists etc $\endgroup$
    – xaxa
    Commented Jul 19, 2016 at 18:18
  • $\begingroup$ That the answer to the other question is insufficient does not mean this is not a duplicate of the question, which it is, since both questions ask why the critical point exists. If you are not satisfied with answers to questions we already have, you are not supposed to duplicate the question, but start a bounty instead. $\endgroup$
    – ACuriousMind ♦
    Commented Jul 20, 2016 at 13:43

5 Answers 5

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I will try to answer these questions from different views.

Macroscopic view

The "quantitative" rather than qualitative difference in a liquid-gas phase transition is due to the fact that the molecules arrangement does not change so much (there is no qualitative difference) but the value of the compressibility changes a lot (quantitative difference). This can be easily seen in the Van der Waals isotherms below the critical temperature,

enter image description here

The phase transition occurs at the dashed line $AD$. For volumes smaller than $V_D$, the high slope of the curve means that one needs a huge amount of pressure in order to decrease a small amount of volume. This characterizes a liquid phase which has a very low compressibility. For the slope is much lesser and the compressibility is high, which characterizes a gas. In between $V_D$ and $V_A$ there is a mixed phase characterizes by a divergent compressibility, i.e. the volume changes at constant pressure.

Above the critical temperature there is no longer such a radical change in the compressibility. The Van der Waals isotherm is the following

enter image description here

As you mentioned the density continually increase with the pressure. You can also see from the Van der Waals equation, when written as $$p=\frac{NkT}{V-Nb}-a\frac{N^2}{V^2},$$ that at very high temperatures it behaves like $$p\rightarrow \frac{NkT}{V-Nb},$$ which is not qualitatively different from an ideal gas isotherm. There is no liquid phase.

Microscopic view

Let us consider a substance below its critical temperature. After a phase transition from gas to liquid, a meniscus (interface) appears between the liquid portion and a vapor (gas) portion which is present due to the kinetic distribution of velocities. The vapor has much smaller density so a molecule in the bulk of the liquid have more bonds than a molecule in the surface (interface). Each bond has a negative binding energy (bonded states) so the molecules in the surface have an excess of energy.

enter image description here

This gives rise to a (positive) surface energy density which is nothing but the surface tension of the interface. When we increase the temperature, the vapor density increases and at some point it equals the liquid density. At this point the number of bonds for the molecules in the bulk and in the surface equals so that there is no surface tension. This means there is no meniscus and no phase transition. There must be a critical point therefore.

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    $\begingroup$ Second part is the most detailed and clear answer for me. I've got one more question. If we start from a gas below a critical point and decrease its temperature at a constant pressure. It is known that, being careful, one can prepare a supersaturated vapour and that it's imperfections like droplets, dust, etc that become a core on which liquid starts to condensate. From your description the reason for that seems to be that the probability of multiple collisions needed to form a few bonded molecules is very low. Is my understanding correct? $\endgroup$
    – xaxa
    Commented Jul 20, 2016 at 10:45
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    $\begingroup$ Same reasoning, but in reverse, applies to overheated liquids. It is very hard for a molecule to detach from all bonds if it is closely surrounded by a lot of other molecules. $\endgroup$
    – xaxa
    Commented Jul 20, 2016 at 10:47
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For a pure substance that can exist in the solid, liquid, and vapor states (i.e., wood is not in this category), let's assume that a closed container is half full of liquid and half full of vapor. As the temperature rises, the liquid expands and the liquid density falls. Also, as the temperature rises, the pressure in the container rises due to the vapor pressure of this material, so the vapor density rises. At some point, the vapor density becomes equal to the liquid density, and only one phase can exist. This occurs at the critical temperature and the critical pressure.

The most common example of a material above its critical temperature is air. No matter how much you compress air, it will not condense at room temperature. This may just be a personal preference, but I would call air at room temperature a gas (as opposed to a vapor).

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    $\begingroup$ Whoa, I never realized that air itself was a supercritical fluid! That's neat. $\endgroup$
    – knzhou
    Commented Jul 19, 2016 at 16:55
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    $\begingroup$ @knzhou Well, room temperature is well above the critical temperature for most of what's in air, but standard pressure is well below the critical pressure, so one can debate how supercritical air is. $\endgroup$
    – user10851
    Commented Jul 19, 2016 at 16:59
  • $\begingroup$ OK, but why at lower temperatures there are two separate phases? Why does a transition (boiling/condensation) curve exist? I'd consider that a homogeneous "liquid-air" inside a container would be much simpler, if simple is a right word here. It seems clear to me what is the difference between solid and liquid phases, and how I can tell a solid from a liquid. But gas and liquid seem to be about the same thing, yet they do not mix below critical point and occupy separate volumes with a boundary in between. $\endgroup$
    – xaxa
    Commented Jul 19, 2016 at 18:32
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    $\begingroup$ There is a boundary :) Even if I compress the gas to a density just little below that of the liquid, there still will be a boundary. If I walk to a beach, there is sand everywhere. Someplace it's thick, someplace it's thin, but the boundary does not form. So in case of a gas/liquid there's something special going on below the critical point, and not above the critical point $\endgroup$
    – xaxa
    Commented Jul 19, 2016 at 18:45
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    $\begingroup$ @xaxa Basically, the water is always evaporating from the pool, and the water vapor is always condensing from the air. But under unchanging conditions, an equilibrium forms where both are perfectly balanced - so both the pool content and the air vapour content stays the same on average. But it's still a statistical process, just like all thermodynamics - if you look at the individual atoms, everything is jiggling all the time. Some molecules have more momentum, some have less - and sometimes, through random collisions, a molecule has enough momentum to get loose. $\endgroup$
    – Luaan
    Commented Jul 19, 2016 at 19:48
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Good question. I don't have my Widom around, but I'll try to answer from memory.

  1. I think the consensus is to say a substance is at its gas state if it could be a liquid at the same temperature. This, as opposed to same pressure, same volume, etc. If the temperature is supercritical, there is no transition between liquid and gas, and the generic term "fluid" should be used.

  2. You can imagine that at really high temperatures the fluid should be in a highly disordered state. If you start with a dilute gas and pressurize it, you should be able to make it as dense as desired. Hence, there should be no liquid-gas transition at high temperatures. Therefore, if one exists at low temperatures, it must end. Notice a liquid-gas transition could be preempted by a solid-gas one and not exist at all. Also, there is always a solid-gas transition, because that one is due to the molecule's hard cores (ultimately quantum in origin), and largely unaffected by temperature.

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    $\begingroup$ What is a Widom? $\endgroup$
    – Phiteros
    Commented Jul 19, 2016 at 16:03
  • $\begingroup$ I find point (2) unsatisfying. Sure, at high temperatures, it should be harder to impose liquid order by pressurizing. But I don't see why it should be impossible. The required pressure could just be enormous. $\endgroup$
    – knzhou
    Commented Jul 19, 2016 at 16:05
  • $\begingroup$ Even less intuitively, there exists a temperature (the critical temperature) where liquid-gas phase transitions suddenly go from possible to impossible. But what's special about that temperature? $\endgroup$
    – knzhou
    Commented Jul 19, 2016 at 16:06
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    $\begingroup$ @Phiteros stat. mech. guy from ~1950s I think, he wrote a few books and papers which are were quite hard to track down during my BSc... $\endgroup$
    – Mark K Cowan
    Commented Jul 20, 2016 at 10:00
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    $\begingroup$ Yes, sorry, by Widom I meant mostly "Molecular theory of capillarity", a very good book. store.doverpublications.com/0486425444.html . There's also amazon.es/Statistical-Mechanics-Concise-Introduction-Chemists/… , a great book imho. $\endgroup$
    – Daniel Duque
    Commented Jul 20, 2016 at 19:21
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Attempting to answer the "why" question intuitively:

In a liquid, the molecules experience significant intermolecular force - so much so, that the average energy of the molecules is insufficient to escape the attractive force of the surrounding materials. The result is that it energetically favorable for them to remain close together, even if that means they do not fill all available space (space above the liquid).

As temperature rises, the vapor pressure of the liquid increases as a greater number of molecules reach "escape velocity". In the process they remove energy from the liquid (greater than average velocity needed to escape). However, if you continue to increase the temperature, you will eventually reach a point where the increase in entropy offsets the loss of energy due to vaporization - in other words, there is no longer a "penalty" for a molecule to go from one state to the other, and the distinction between the two phases disappears.

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  • $\begingroup$ Good point, but from the second paragraph of your answer I'd figure that there should be a temperature difference between a colder liquid (deprived of fast molecules) and a hotter gas. However, since the thermodynamical equilibrium should be reached anyway, I'd think that eventually kinetic energies would equalize and a single homogeneous phase would occur. As it does indeed above the critical point. $\endgroup$
    – xaxa
    Commented Jul 20, 2016 at 6:53
  • $\begingroup$ @xaxa Not a temperature difference. Don't forget that temperature isn't the same thing as energy. The matter is quite complicated, but consider that there may be additional "jiggly" modes depending on the state. For example, in liquid water, the individual molecules are slightly bound together, so they have more energy at the same temperature compared to the gaseous state. This is not surprising - it's the difference between the energy of liquid water at 0 °C and gaseous water at 0 °C (and analogously, solid water). $\endgroup$
    – Luaan
    Commented Jul 20, 2016 at 8:26
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    $\begingroup$ @xaxa the concept of the "wet bulb temperature" is precisely that - when the vapor pressure is less than 100% there will indeed be a difference between the temperature of a liquid and its vapor. This is how relative humidity can be determined with a thermometer. $\endgroup$
    – Floris
    Commented Jul 20, 2016 at 14:15
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Why does a critical point even exist?

I think this question is equal to this one:

"Why the width of the two phase region is bigger at lower temperatures and pressures?" enter image description here Specific volume of liquids mostly depends on the temperature of them in comparing with their pressure. This means, for a well-defined increment of the pressure, we can neglect its effect on the specific volume of the liquid with respect to increment of the temperature of the liquid. So, by increasing the temperature, specific volume of the saturated liquid will increase.

On the other hand, specific volume of gases mostly depends on the pressure of them in comparing with their temperature. For a well-defined increment of the temperature, the influence of the pressure increment will be more much than temperature. So, by increasing the temperature, specific volume of the saturated vapor will decrease.

Thus, by increasing the temperature, the width of the two phase region will decrease and because of continuity of the properties of the substances, finally this width will be eliminated. I.e. a critical point certainly will exist.

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