What I've understood is no matter how much pressure we apply to a gas present at its critical temperature, it will never liquefy, and that there will be no 'distinction' between the 2 phases, liquid & gas. What I don't understand is, what does it mean that there's no distinction between the two?

What I've also read about is that the heat required to evaporate that liquid at the critical temperature or above is zero. Does that mean that the specific enthalpy gap between the liquid and gaseous states is also zero? If that's true, then isn't that 'liquid' phase already turned into gas? Or is it still in a liquefied state with identical properties to that of the gaseous? Thanks

  • $\begingroup$ I always had/have this kind of practical visualisation problem at CP. If I assume to stay right at CP is tricky tor think of just one phase by definition. But the same is for two undistinguisheable ones. Practically we shall see a sort of fog so fine droplets should be there --> 2 phases easily transitioning as for all others properties are the same, specific Enthalpy in primis. And surface tension should be disrupted too, already near CT, so that liquid droplets can evaporate/condense easily but I am not sure is relevant. Hopefully someone can answer other than pointing to known refs. $\endgroup$ – Alchimista Jan 11 '19 at 10:05
  • $\begingroup$ All reduces down to "can be two phases be truly indistinguishable? ". In my opinion no as it cuts the legs of the chair we are sitting on. $\endgroup$ – Alchimista Jan 11 '19 at 10:21
  • $\begingroup$ Above the critical point there is not 'gas' vs 'liquid' distinction, which is why it is often called a 'fluid'. Yes, the two have merged into one, and will remain so until you move into a different area of (P,T) space where they separate again. $\endgroup$ – Jon Custer Jan 11 '19 at 15:06
  • $\begingroup$ @Jon Custer. Yes is right at the point that I have visualisation troubles. Is that sentences such the two phases coexist and the two phases are indistinguishable are found in many places and even the same paragraph. For years I've skipped this dilemma by placing me infinitesimally out centered from the very CP :) $\endgroup$ – Alchimista Jan 11 '19 at 15:52
  • 1
    $\begingroup$ They aren't two phases - they have merged into one. At some simple conceptual level, the difference between a gas and a liquid is how much space is between the atoms/molecules in the two phases. At some density, they pretty much look the same. Well, from a thermodynamic perspective, the two indeed merge into one, with a single Gibbs free energy term. Change the (P,T) situation, and they separate back out. But you can move from one to the other without undergoing a first order phase transition by going through (P,T) space appropriately. $\endgroup$ – Jon Custer Jan 11 '19 at 15:56

To illustrate what is happening near the critical point for a fluid, I will look at carbon dioxide, a fairly well known system showing a critical point. It is of some use technologically, particularly in MEMS (microelectromechanical systems) technology for critical-point drying.

Lets model CO2 as a van der Waals fluid, with

$(p + {a \over v^{2}})(v-b) = RT$

The $a$ and $b$ coefficients can be found on Wikipedia.

One rewrites the van der Waals equation to get $p$ as a function of $v$ at constant temperature $T$. For a series of temperatures up to and just past the critical point temperatures one obtains a plot like:

enter image description here

Take the lower line (270K). If you are holding the system at 270K and 40bar, there are two distinct volumes that are stable solutions to the van der Waals equation. These represent, respectively, what we would call the liquid (lower volume ~0.07L) and gas (higher volume, ~0.4L) phases. The two phases have distinct properties (the volume occupied at a given pressure and temperature). They will be in equilibrium with each other, and the balance of phases will be dictated by other considerations (adding/subtracting enthalpy to convert one phase to the other).

As the temperature increases, one continues to see that there are multiple volumes at a given pressure until near 305K. There it looks like the local minima is going away, and indeed at 310K it is clear that the pressure is now a single-valued function of volume. (The critical point of CO2 is officially 304.25K at 73.9 bar.)

Now, can you go from the lower volume (liquid) phase at 270K to the larger volume (gas) phase at 270K without going through a first order phase transition? Sure thing, through suitable variations of pressure and temperature. Increase the pressure to, say, 100bar at 270K. Now increase the temperature to 310K, holding at 100bar. You are now on the top yellow curve. Decrease the applied pressure to about 40bar, giving a volume way over on the right hand side of the plot. Now drop the temperature to 270K. Congratulations, you are now over on the low volume (~0.4 L/mol at 270K) side of the plot having never gone through a first order phase transition.


what does it mean that there's no distinction between the two?

Given a closed system that consists of a pure liquid in equilibrium with its vapor, as the liquid temperature increases, its density decreases. Because the vapor pressure of the liquid rises as temperature increases, some of that liquid goes into the vapor phase, so the vapor density increases as the liquid temperature increases. At some point, the temperature is high enough for the liquid density to equal the vapor density. This is the critical temperature, and above this temperature, there is only a super-critical fluid with one density.

Does that mean that the specific enthalpy gap between the liquid and gaseous states is also zero?

Since there is only one phase above the critical temperature, there is no "enthalpy gap" because there is no liquid to vaporize.


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