Currently, I have been learning about P-T diagrams and vapour pressure and I have a question that seems like it should have a simple explanation yet I am not able to understand it.

Taking the P-T diagram of water for example,

Suppose we choose a point that corresponds to the liquid region in the P-T diagram. From what I know of vapour pressure, every pure liquid in a closed container has a vapour pressure and will eventually reach a liquid-vapour equilibrium. If this is the case, what is the meaning of a "liquid region" in a P-T diagram if there must always exist a vapour-liquid equilibrium? And by extension why is there even a coexistence curve between liquid and vapour in the first place?

  • $\begingroup$ It is normal practice to include a reference to the figure unless it is your own :) A figure may also be copyrighted. $\endgroup$
    – porphyrin
    Mar 28, 2019 at 8:47
  • $\begingroup$ Wow, now that's completely misleading question title... And what you say is incorrect. You can get some water with no contact with air very easily. $\endgroup$
    – Mithoron
    Mar 30, 2019 at 0:44

2 Answers 2


The gas phase is unique in its ability to expand to fill space, so if there is space above or around a solid or liquid, molecules will eventually enter the gas phase in order to fill that space. That's the vapor-liquid equilibrium you described.

The liquid region of the phase diagram is the region where, in a closed space, some of the material will remain indefinitely in the liquid phase, even though some is also in the vapor phase. For example, in a sealed but expandable container (like a piston) maintained at an external pressure of 1 atm, the equilibrium state at 80 C has some water and some water vapor. But at 110 C, it will be all vapor.

The "coexistence curve" is a misnomer for the curve you are referring to. It does not mean that the liquid and vapor simply coexist (which as you pointed out can happen throughout the liquid region). It means that the equilibrium vapor pressure is equal to the external pressure. At that point, you cannot increase the temp of the liquid any further if the pressure remains constant. All heat added will go towards converting liquid to gas. Only after it is all gas will the temp increase. In order to get liquid at a higher temp, you have to increase the pressure. We define this as the boiling point.

  • $\begingroup$ Thank you for the explanation. So it would be correct to say that the P on on the y axis refers to external pressure and not vapour pressure? And it can only be interpreted as vapour pressure along the coexistence curve? If so, then throughout the liquid region is is correct to say that there always exists a vapour-liquid equilibrium but the amount of vapour pressure differs depending on the temperature no? Then what mathematically defines the vapour pressure in the liquid region because I know that the Clausius-Clapyeron defines vapour pressure at the coexistence curve. $\endgroup$ Mar 28, 2019 at 16:09
  • $\begingroup$ @rofldude188 If you draw a vertical line from the point representing current conditions to the liquid/vapor boundary line, the pressure value at that point on the line represents the equilibrium vapor pressure at that temp. So mathematically, it's given by the C-C equation even though the total pressure has moved you off the boundary line. $\endgroup$
    – Andrew
    Mar 28, 2019 at 18:15
  • $\begingroup$ If the external pressure is 1 atm,, doesn't the internal pressure also have to be 1 atm? At 80 C, the equilibrium vapor pressure of water is about 0.5 atm. So, with an external pressure of 1 atm, the piston will not be in equilibrium. The only way it can be in equilibrium is if there is no water vapor, and the water in the cylinder is all liquid at 1 atm. $\endgroup$ Mar 28, 2019 at 22:26
  • $\begingroup$ @ChetMiller - yes, the internal pressure must also be 1 atm. I'm assuming that another gas is present so the piston is at equilibrium before the water vaporizes. As the water equilibrates with water vapor, the volume increases, but the total pressure internal remains 1 atm. The partial pressure of the water vapor will be the 0.5 atm you mentioned. The other gas will contribute the other 0.5 atm. $\endgroup$
    – Andrew
    Mar 28, 2019 at 22:42
  • $\begingroup$ @Andrew If the volume of the closed container is very big (e.g. $1000$L) will all the liquid turn into gas? $\endgroup$
    – Anton
    Mar 8, 2021 at 19:32

The liquid region corresponds to pure liquid at a pressure greater than the equilibrium vapor pressure at the liquid temperature. Imagine liquid in a cylinder, with a piston pressing down on the liquid with a pressure greater than the equilibrium vapor pressure. It is also possible to have the liquid being pressed upon by a non-condensible gas at a total pressure above the equilibrium vapor pressure; of course, in this case, at equilibrium, there is also vapor within the non-condensible gas with a partial pressure equal to the equilibrium vapor pressure. But, as far as the bulk liquid is concerned, this gas phase in this case is essentially the same as a piston.

  • $\begingroup$ So it would be correct to imagine the liquid region as a liquid in a closed container with an external pressure corresponding to the P value on the y-axis of the diagram? If so, then how would one determine the vapour pressure here as it obviously exists but will be significantly less because of the greater external pressure? $\endgroup$ Mar 28, 2019 at 17:55
  • $\begingroup$ If you have a gas phase over the water (including a non-condensible gas), the partial pressure of the water vapor in the gas phase is equal to the equilibrium vapor pressure of water at the system temperature. If there is no gas phase over the liquid water, then the vapor pressure is irrelevant. $\endgroup$ Mar 28, 2019 at 18:26

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.