Hoping to clarify a few questions I have that the textbook doesn't address.

Lets just say I have a piston (at equilibrium) with 100 moles of gaseous compound that looks like this:

enter image description here

And here is the phase diagram for this compound:enter image description here

Now lets say I start increasing the pressure on the piston so that the gas starts to approach the gas-liquid boundary.

1.) What would start to happen as you approach the boundary? For example: Would a few moles of gas begin to condense to liquid? Is this process exponential as you approach boundary?

2.) What is happening at the boundary? For example: Are the gas and liquid in perfect equilibrium (50 moles of each)? By that logic, at the triple point, are there 33 moles of each?

3.) Now let's say we approach the critical point along the liquid-vapor boundary. I read online that its when the gas and liquid become "indistinguishable." If so, that means it has a uniform density. Does that make it another state of matter? If not, why is it special?

  • $\begingroup$ Why the downvote? Let me know what I can do to clarify! $\endgroup$ – Nova Jul 24 at 22:02
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    $\begingroup$ (1) No, the process is not exponential, but abrupt. (2) Yes they are in perfect equilibrium, and that's not what you think it is. The number of moles is irrelevant and can be anything. (3) Call it whatever you like. $\endgroup$ – Ivan Neretin Jul 24 at 22:04
  • $\begingroup$ Thanks for reply! Can you specifically describe what would be happening inside the piston at the boundaries then? $\endgroup$ – Nova Jul 24 at 22:07
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    $\begingroup$ The gas will start condensing into a liquid. As you move the piston further, more gas will condense, until there is none of it left. Then you will be just compressing the liquid. $\endgroup$ – Ivan Neretin Jul 24 at 22:10
  • $\begingroup$ Why is saying that half the molecules would be in one phase (and the other half in the other) erroneous? I'm imaging a layer of liquid with a layer of gas on top. $\endgroup$ – Nova Jul 24 at 22:15

I got that you are trying to visualize what is actually happening. I have not done the compression myself, but this is what I think will be observed. It will be good if someone with real experience can confirm the answer.

  1. At the beginning you are in the region marked 'vapour'. When you are approaching the liquid-vapour boundary from below, before you touch the boundary, you will still observe only vapour, no liquid at all. You are still in the region marked 'vapour'. Nothing will happen.

  2. When you just reached the boundary, you will see some liquid droplets formed. It should look just like some dew formed at the wall. Regarding the relative amount of liquid and vapour at equilibrium, the pressure of vapour present must be equal to pressure exerted by you (or else the piston will move!) and equal to pressure on the boundary (determined by temperature). Say if you try to push the piston further inside, the volume occupied by sample will decrease, and more liquid formed, but you will still stay on the boundary. You will notice that the pressure needed to push piston inside further does not change, because the pressure of vapour does not change (push in, vapour pressure should increase, but some vapour condensed, lower back the pressure). Only when all vapour has been converted into liquid, you can depart from the boundary, move into the region marked 'liquid'. However, before all vapour turned into liquid, if you stop pushing in and hold your piston still, the liquid and vapour amount won't change. So there is no unique relative amount of vapour and liquid at equilibrium as written in your question. Above liquid-vapour boundary you have only liquid and the piston surface is resting on the liquid surface.

  1. That state is known as supercritical fluid.
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  • $\begingroup$ Interesting, so at the boundary, if I was to lock the piston in place. The relative ratio of vapour and liquid would become volume dependent? $\endgroup$ – Nova Aug 7 at 15:23
  • $\begingroup$ Yes, that conforms to Le Chatelier principle. $\endgroup$ – TheLearner Aug 8 at 4:13

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