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"Boiling does not occur when liquid is heated in a closed vessel. On heating continuously vapour pressure increases. At first a boundary is visible between the liquid and vapour phases because liquid is more denser than vapour. As temperature increases more and more molecules go to vapour phase and density of vapours rises. At the same time liquid becomes less dense. It expands because molecules move apart. When the density of liquid and vapours becomes the same; the clear boundary between the two phases disappears. This temperature is called critical temperature"

That is EXACTLY what my textbook says. And that is where I am having doubts. It's because - a gas might still be found in its gaseous phase below the critical temperature if the pressure is considerably below its critical pressure ($P_c$) and it has a volume sufficiently larger than its critical volume ($V_c$).

isotherm

Just consider any point in the Pressure-Volume isotherms of $\ce{CO2}$ given above, which is in the region of gas-liquid equilibrium (within the shaded dome shaped curve) but has a volume greater than the critical one. For e.g a point that lies

  1. Within the shaded region. and
  2. On a line passing through B and parallel to the pressure axis.

Let that point represent the state of liquid inside the vessel. From there if we increase the temperature, we would arrive at a point on the boundary of the dome shaped curve, (viz. B) but that point would be on a lower isotherm than the critical isotherm. And it looks like B is also a point where the two densities becomes equal.(isn't it?) Based on that premise, we can't say that the temperature arrived at in the above (heating liquid in a clossed vessel) situation would necessarily be the Critical temperature.

What I want to ask is if my reasoning is correct or am I missing anything?

My question can be rephrased in the following manner also
Is it true that critical temperature is the only temperature at which the density of the gaseous phase of a gas becomes equal to its liquid density?

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  • $\begingroup$ Providing that there is enough liquid in the vessel to begin with, your book is correct. $\endgroup$
    – Jon Custer
    Commented Mar 15, 2016 at 20:51
  • $\begingroup$ "Providing" of course is his quibble, so he is right. However if the liquid were evaporating then you'd see the volume of liquid getting less and less as the temperature is raised until it was all gone. The book is proposing that a significant volume of liquid just all of a sudden disappears. For example a container half full of liquid just below Tc. $\endgroup$
    – MaxW
    Commented Mar 15, 2016 at 21:00
  • $\begingroup$ What i am saying is that i found the quoted paragraph in my textbook and i am not sure if the information provided is sufficient to conclude that it would surely be the critical temperature whenever the phases become indistinguishable, because gas below the critical temperature is also not always in the liquid phase and requires sufficient pressure to reach the equilibrium phase. Note: Right now i am not in a position to upload a diagram to make my point clearer to others. I would try my best to describe it with words. $\endgroup$
    – user25714
    Commented Mar 16, 2016 at 1:47
  • $\begingroup$ Could you please insert the citation to the textbook you use for proper attribution and transparency. $\endgroup$
    – Martin - マーチン ♦
    Commented Mar 16, 2016 at 10:09
  • $\begingroup$ Here. But, frankly it is not just my book that says so. I've found similar paragraphs in several places too. Which makes me believe that i might have missed some points. $\endgroup$
    – user25714
    Commented Mar 16, 2016 at 12:40

5 Answers 5

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Ok, I added a point Z to the diagram which I've added below, and I'm going to quote you text in italics below the image...

enter image description here

Just consider any point in the Pressure-Volume isotherms of $\ce{CO2}$ given above, which is in the region of gas-liquid equilibrium (within the shaded dome shaped curve) but has a volume greater than the critical one. For e.g a point that lies:

  1. Within the shaded region.
    and
  2. On a line passing through B and parallel to the pressure axis.

Let that point represent the state of liquid inside the vessel.

Question 1 - My point Z meets your two conditions doesn't it?

... From there if we increase the temperature, we would arrive at a point on the boundary of the dome shaped curve, (viz. B) but that point would be on a lower isotherm than the critical isotherm.

Assuming you agree with point Z, I agree with the idea above. The next two sentences is where you're wrong.

And it looks like B is also a point where the two densities becomes equal.(isn't it?) Based on that premise, we can't say that the temperature arrived at in the above (heating liquid in a closed vessel) situation would necessarily be the Critical temperature.

Let's start at point A, and follow the purple isotherm. At point A all the carbon dioxide is gas. At point B all of the carbon dioxide is still gas. But as you start to goto point C from point B the carbon dioxide vapor becomes supersaturated and liquid carbon dioxide forms. Now when you get to C there is so little volume that all the carbon dioxide is liquid. Thus only in the blue shaded area under the curve do you have two phases gas and liquid.

Liquid carbon dioxide is denser than gaseous carbon dioxide. So the gas will be on top and the liquid below.

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  • $\begingroup$ To your question: yes, Z is the point.. Regarding your answer: I am getting it now. But could you tell me what is supersaturated? Is it same as simply saturated? Assuming it is, if we reach point B, we get a vessel full of saturated vapour. Phase distinction is still not lost at that temp. because we can get liquid if we want by reducing the volume. Now, above B, our vessel is full of unsaturated vapour for each corresponding higher temp. untill we reach the critical isotherm and that is when the two phases literally merge, and their densities become equal. Am I correct now? $\endgroup$
    – user25714
    Commented Mar 17, 2016 at 4:51
  • $\begingroup$ Supersaturated in this case means that there is too much carbon dioxide in the gas phase and some of the carbon dioxide has to go into the liquid state to reach equilibrium between the gas phase and liquid phase. (Like fog forming when temperature drops outside.) // The gas density and the liquid density never will be equal. At point E carbon dioxide is one phase, a "supercritical fluid", which is different than a gas or a liquid. en.wikipedia.org/wiki/Supercritical_fluid $\endgroup$
    – MaxW
    Commented Mar 17, 2016 at 5:00
  • $\begingroup$ Also, if any gas above its critical temperature and pressure is called supercritical fluid, what is it called when its above critical temp. but below critical pressure? $\endgroup$
    – user25714
    Commented Mar 17, 2016 at 5:01
  • $\begingroup$ A "liquid" see phase diagram for CO2 en.wikipedia.org/wiki/Supercritical_fluid#Phase_diagram $\endgroup$
    – MaxW
    Commented Mar 17, 2016 at 5:03
  • $\begingroup$ Its gas not liquid.. :) $\endgroup$
    – user25714
    Commented Mar 17, 2016 at 5:13
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Lets pick up any point in the gas-liquid equilibrium. That would be any point on an isotherm below critical temperature but the point should be in phase equilibrium. Now when we are increasing the temperature while keeping the volume constant, we are increasing pressure. Now as we increase pressure we are moving the point to the isotherm of a little higher temperature. Now if we keep on increasing the temperature, we will reach on a point on the critical isotherm. The point will be below the critical pressure, not the critical temperature. Further heating would make it supercritical.

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  • $\begingroup$ Below the critical pressure, not temperature? Could you elaborate.. There's a dome shaped region in the isotherm that represents the gas liquid equilibrium phase and that has its peak touched by the critical isotherm. Now, isnt it true that the boundary of that dome on the right side of its peak represents the points where the line between gas and liquid phase disapear? Note: Right now i am not in a position to upload a diagram to make my point clearer to others. I would try my best to describe it with words. $\endgroup$
    – user25714
    Commented Mar 16, 2016 at 1:36
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I am unsure about some details on what you exposed. For any two points have the same density in the diagram above, they have to be located in the same vertical line. The right hand side of the dome line correspond to the volume of pure gas in hypotetical equilibrium with the liquid (both have the same chemical potential). Notice that the total amount of matter is fixed. Left side of the curve correspond to liquid volume. Below critic temperature, liquid is denser than vapour. As temperarure increase they densities get closer up the critic point that they turn to be the same.

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Bear with me on this one. We chose a point in the shaded region and just vertically below the point B. Now keeping the volume constant we start heating. As random motion of particles rises, we get an increase in pressure. Pressure keeps increasing until it reaches the point B. Here the liquid is in equilibrium with gas. The distinction between liquid and gas phase is there. But if we keep heating further the equilibrium mixture turned completely into gas and on further heating pressure will reach upto a point on the dark blue isotherm. All points on an isotherm i.e. the dark blue line are on the same temperature, hence our fluid reached the critical temperature. Distinction is lost when we get above that dark blue line. The shaded region is not an isotherm. It represents the region where gas and liquid are in equilibrium and the boundaries of the shaded region represent the point where our substance just started phase change and got into equilibrium. Phases are distinct. Isotherms are the lines on the graph and the critical isotherm ( dark blue line, damn it how many times do I have to say that) is the deciding factor. Above it our substance behave as a supercritical fluid ( has the properties of both liquid i.e. solubility and gases i.e. mobility).

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The "shaded region" is an unstable state and represents a region where vapor an liquid are in equilibrium.

Your question refers to points "in the shaded region". This region does not represent a stable state; in that region $\left(\frac{\partial P}{\partial V}\right)_T$ is positive; this violates principles of thermodynamic stability. Points in the shaded region will result in phase separation into two phases, marked "B" and "C" in the text. Since density is reciprocal volume (as long as n the number of molecules total is constant), your book is correct, at least as far as I can see.

In other words, points on the line "BC" will represent a two-phase system, with variable proportions of the two phases, but with phase densities given by point "C" for the liquid and "B" for the vapor phase.

The Point "Z"

Other answers mention a new point "Z". This is also in the shaded region. Therefore it represents a region where phase separation into liquid and vapor occur.

The impossibility of liquid and vapor densities being equal...

The liquid density is given by the x-coordinate of "C"-like points, i.e. points on the left side of the edge of the shaded region.

The vapor density is given by the x-coordinate of "B"-like points, i.e. points on the right side of the edge of the shaded region.

Since the right side of the shaded region always has a higher x-coordinate (in this case, x represents volume), the vapor density will always be greater than the liquid density.

As temperature rises, the difference in densities of vapor and liquid gets smaller and smaller, as we ascend the right side (vapor phase) and left side (liquid phase) of the shaded region.

...except at the critical point

When we reach the critical point, the difference between liquid and vapor density disappears. Because the shaded region is convex, there is no possibility of this occurring anywhere else.

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