2
$\begingroup$

Suppose that I have some substance with a critical temperature of 20°C. That means above that temperature, the substance exists neither as a gas nor a liquid, but instead as a super-critical fluid. Does it mean that no matter how high a pressure I apply on the fluid, it is not going to turn into a liquid above 20°C? Why is that so?

A substance is a liquid if the particles are close enough to interact, i.e. the space between particles is similar across different liquids. It seems that if we increase the pressure (i.e. decrease the available volume), particles should get close enough to interact.

I know that the higher the temperature of a gas, the higher the speed of the particles. How does this affect the phase transition?

What is the role of intermolecular forces? They are part of the explanation why at lower temperatures, gases condense into liquids. It seems that we are ignoring them in the case of super-critical fluids. Are we now saying I just need to apply pressure to decrease the distance between particles? If it weren't for intermolecular forces, I would have to apply much higher pressure to turn a gas into a liquid (at temperatures below the critical temperature).

So what happens when a super-critical fluid is compressed, and how is the interplay of particle speed, intermolecular forces and distance of particles different below and above the critical temperature?

Improve this question
$\endgroup$
17
  • $\begingroup$ If you are worried about density alone, then of course you can put your particles as close as you wish by applying enough pressure. There is nothing special about 20°C in this regard. $\endgroup$
    – Ivan Neretin
    Commented Sep 12, 2019 at 14:01
  • $\begingroup$ Then when the density of a given gas becomes too high, won't it turn into a liquid?? $\endgroup$
    – Jatin Gakhar
    Commented Sep 12, 2019 at 14:17
  • $\begingroup$ so, is it only the closeness of particles of a substance that tells whether it is a liquid or a gas, or it is affected by the speed of these particles too? $\endgroup$
    – Jatin Gakhar
    Commented Sep 12, 2019 at 14:45
  • $\begingroup$ It is complicated. First of all, here is you compound compressed to such-and-such density; how do you tell if it is a liquid or a gas? $\endgroup$
    – Ivan Neretin
    Commented Sep 12, 2019 at 15:51
  • 1
    $\begingroup$ You can see it as you see both the gas and the liquid at the same time, as they have identical properties, so there is no difference between them. Below critical temperature, their properties converge toward the critical point. Imagine a beach (liquid), the cliff(gas) and the cliff edge(phase transition). You go along the beach, which climbs and becomes more like cliff, the cliff descends and become more like a beach. At some point the cliff edge disappear and you cannot say what is what, all is the same, with no sudden change with altitude. Single phase with gradual change with pressure. $\endgroup$
    – Poutnik
    Commented Sep 13, 2019 at 4:11

1 Answer 1

Reset to default
4
$\begingroup$

When you heat up a liquid at constant volume (leaving sufficient space for the gas phase), the density of the liquid will decrease and the intermolecular interactions will weaken. Some of the liquid will transition to vapor, so the vapor above the liquid will get denser and the frequency of intermolecular interactions will increase (it behaves less and less as an ideal gas as the number of collisions increases with increasing density). At the point where the density of the liquid and the gas are the same, there is no more phase boundary (with gravity, we picture liquid on the bottom and gas on the top, but if the density is identical, that is no longer the case).

This is all pretty outlandish for most of us because we are used to constant pressure, not constant volume. Here are two videos showing the process:

Liquid chlorine: Here, it looks like the liquid level stays constant - the expansion of the liquid and the evaporation balance each other out. If you look carefully, though, you can see how the vapor phase color gets more intense, indicating more molecules in the vapor phase. At the same time, the color of the liquid gets less intense, indicating a decrease in density. https://chem.libretexts.org/Courses/University_of_North_Texas/UNT%3A_CHEM_1410_-_General_Chemistry_for_Science_Majors_I/Text/10%3A_Solids%2C_Liquids_and_Solutions/10.13%3A_Critical_Temperature_and_Pressure

Carbon dioxide: When approaching the critical point, the system gets turbid because the difference in density between liquid and gas is no longer sufficient to keep the liquid down and the vapor up. The system is a heterogeneous mixture of liquid and vapor for a while, and then a single pure phase.

https://www.youtube.com/watch?v=GEr3NxsPTOA

Making of video: https://www.youtube.com/watch?v=-gCTKteN5Y4

Improve this answer
$\endgroup$
4
  • 1
    $\begingroup$ I understand that this is meant for beginners, but " the intermolecular interactions will get stronger" is not true. The pair potential need not change, but yes, with increasing density the frequency of interactions increases. $\endgroup$
    – Buck Thorn ♦
    Commented Sep 13, 2019 at 7:10
  • $\begingroup$ @BuckThorn I replaced my incorrect statement with your correct description - it is not more difficult to understand for beginners. (I also edited the question in hopes to get it re-opened because I think the underlying conceptual question is valid and insightful). $\endgroup$
    – Karsten ♦
    Commented Sep 13, 2019 at 11:28
  • 1
    $\begingroup$ There is still room for confusion. For instance water density increases with heating in a small range of T (although this can be considered anomalous) :-). This is a difficult topic to describe in layman's terms. For instance, near the CP the liquid-like phase is sometimes referred to as a "compressible liquid". Imho the substance is more like a liquid than a gas near the CP. Our liquid/gas binary terminology sort of just fails in this region. But your description of the behavior as you move along the coexistence line is reasonable. I was not the downvoter btw. $\endgroup$
    – Buck Thorn ♦
    Commented Sep 13, 2019 at 13:17
  • $\begingroup$ @BuckThorn I appreciate your input! I also appreciate the votes, in either direction. $\endgroup$
    – Karsten ♦
    Commented Sep 13, 2019 at 16:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.