"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 point might still be found below the critical temperature where the density of both the phases become equal 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 $CO_2$ 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.
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?

NOTE: 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?


Actually, if you are starting out at some point along the line BC and are increasing the temperature and pressure within your closed container, unless your initial average specific volume of vapor and liquid on BC is exactly equal to the critical specific volume, you will not reach the critical point. If you are to the left of the critical specific volume on BC, then you will reach a point of all liquid, and, if you are to the right of the critical specific volume on BC, you will reach a point of all vapor.

Either way, the original quote is not correct. The phase diagram clearly shows this.

  • $\begingroup$ I get that. But i am talking about a line passing through B and parallel to the pressure axis not the volume axis. What happens then? $\endgroup$ – Phill2 Mar 16 '16 at 12:23
  • $\begingroup$ And I'd have to add that the line is not in the diagram. I want you to suppose that line. $\endgroup$ – Phill2 Mar 16 '16 at 12:28
  • $\begingroup$ At point B, you will have all saturated vapor present, and, if you raise the temperature, you will move into the superheated region where there will be no liquid. In a closed vessel, you can only reach the critical conditions if there is enough liquid present so that it does not all evaporate before you reach the critical. $\endgroup$ – Chet Miller Mar 16 '16 at 12:33
  • $\begingroup$ Ok. We assume that we have sufficient volume. And the initial state of our system is the point on the line as i supposed in the question. Now if we heat the system continuously untill we reach B, we'd reach the state when the gas is in its saturated vapour phase with not a single molecule condensed. That would mean we have turned every molecule into vapour so there's no question of some 'meniscus' between the phases to disappear right? $\endgroup$ – Phill2 Mar 16 '16 at 12:59
  • $\begingroup$ I am asking about meniscus because I was having a tough time distinguishing between the critical point and point B (or any other point on the dome shaped curve facing the gaseous/vapour side). $\endgroup$ – Phill2 Mar 16 '16 at 13:04

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