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At room temperature and 1 atm pressure liquid water should be a compressed liquid. Now if that is the case then why do we have water vapor at these conditions? I can't make sense of this based on just a single component phase diagram. Should I be thinking about water and air as a system and then think about a gas phase component of the air+water mixture from a 2 component T-(x,y) diagram (or P-(x,y))?

Or another possibility is that my interpretation of the P in the P-T diagram is incorrect. and P corresponds to the partial pressure of water?

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    $\begingroup$ The phase diagram describes the one component system, pure water. P is the vapor pressure of water [not a partial pressure], there is no other component present. You state that in your question then contradict yourself. At RT and one atm there can be no head space and no water vapor present. Addition of an inert gas [air] headspace allows water to vaporize into the headspace [there is empty space between the air molecules]. It is no longer a one component system $\endgroup$
    – jimchmst
    Commented May 5 at 23:20
  • $\begingroup$ That is what I was getting at. We cannot use the P-T diagram of water to understand its phases under air then right? Also I was a bit confused because it says that P in a P-T diagram is the partial pressure on en.wikipedia.org/wiki/Phase_diagram#/media/…. $\endgroup$ Commented May 5 at 23:43
  • $\begingroup$ @Poutnik but according to the phase diagram at 1 atm and 25 deg C water should not be on the liquid-solid or liquid-vapor lines. it should be a compressed liquid and hence should have 0 vapor pressure. $\endgroup$ Commented May 6 at 15:38
  • $\begingroup$ You have not still got what was posted in comments and the answer. These phase diagrams are for a single component, in this case water. there is no air there. // At 25 deg C, (saturated) vapor pressure is about 0.02 atm, as you can see on the diagram. If vapor partial pressure is higher that (saturated) vapor pressure 0.02 atm, vapor condenses. If lower, water evaporates. // If it happened there was 1 atm of vapor at 25 deg C, it would condense until it reaches p-T curve. // The total pressure 1 atm is relevant only for the boiling point, when vapor pressure equals total pressure. $\endgroup$
    – Poutnik
    Commented May 6 at 16:16
  • $\begingroup$ By other words, no matter how big the total gas pressure is (1), water vapor condenses only if its partial pressure is bigger than the (saturated) vapor pressure for the current temperature. // (1) - For really big pressures, water has counter-intuitively even higher vapor pressure than for low pressures. $\endgroup$
    – Poutnik
    Commented May 6 at 16:48

3 Answers 3

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Think of molecules of water liquid escaping into vapor, and molecules of water vapor condensing into liquid. Both processes occur simultaneously, and an equilibrium is reached in a closed system. The molecules in vapor form create a specific vapor pressure at a given temperature, as explained by the kinetic theory of gases.

For some materials, vapor is exceedingly low. For example, gallium has a vapor pressure of ~1 Pa at 1310 K! At room temperature, it's quite low, perhaps 10-40Pa at it's melting point, ~303 K, just below body temperature (This formula might give more accurate results). In that case, there is, indeed, a sharp line demarcating liquid and vapor.

Other materials have vapor pressure so high that even the solid sublimes to vapor, such as $\ce{CO2}$ and iodine. And as ice, does, as well, so that even well below 0°C, ice cubes in the 'fridge slowly disappear. For these, your question and comments do point out that a simple flat phase diagram is insufficient to express the relationship -- and you might consider the work Josiah Willard Gibbs and James Clark Maxwell did on a three-dimensional phase model. (BTW, in Europe, Gibbs was considered the first "American chemist".) That model shows, more clearly, that solid, liquid and gas phases are not mutually exclusive, explains phase coexistence, and the triple point, where all three phases coexist in equilibrium at a specific temperature and pressure.

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    $\begingroup$ Thanks for the answer. But my confusion comes from thinking about it from a equilibrium thermodynamic pov. I understand how the rate of evaporation and condensation will become equal to establish equilibrum and that will give us a vapor pressure. But if we just think about it from the pov of a phase diagram of water, then at 1 atm and 25 deg C it should be a compressed liquid and hence there should be no vapor. Why can't we explain it from the phase diagram of H2O? $\endgroup$ Commented May 6 at 2:52
  • $\begingroup$ @chem_101_102 I suspect your problem is to think of thermodynamic equilibrium as a static equilibrium. It isn't. The equilibrium is established as a dynamic exchange between vapor and liquid. The phase diagram does not invalidate this view. $\endgroup$
    – matt_black
    Commented May 6 at 11:29
  • $\begingroup$ @matt_black I get that equilibrium is not static. There is a dynamic exchange of particles between the 2 phases if we are on the Liquid-Vapor line. The amount going from Liquid to Vapor is equal to the amount going from Vapor to liquid. But that is not the source of my confusion. At 1 atm pressure and 25 deg C the phase diagram says that water should be a compressed liquid with no vapor (it shouldn't be on the liquid vapor equilibrium line). But we still have vapor at these conditions. Why is that? $\endgroup$ Commented May 6 at 14:49
  • $\begingroup$ @chem_101_102, The answer has been enlarged a bit to address your question; I hope it helps answer it, $\endgroup$ Commented May 6 at 15:32
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I think that it is a good question: if you have ice in liquid water, even at 1°C (just above the melting point), all the ice will disappear and you will have 100% liquid. Then why do you have always some gas in equilibrium with a liquid (or solid as DrMoishe Pippik pointed out) ?

I think that the problem comes from not defining your system precisely:

At room temperature and pressure, if you have a bottle of water completely filled with liquid (no space for a gas above it), there will be no formation of water vapour in the bottle, so water is truly 100% liquid in these conditions.

However, if outside of the liquid water there is a gas (could be vacuum, same idea), which can exchange heat and matter with the liquid water. If there is no water in the gas phase to start with, moving some water from the liquid to the gas will increase the total entropy, as mixing gases is spontaneous.

Below the boiling point, evaporation is a surface process it happens between the two systems. Above the boiling point, the liquid water boils in the volume, it does not depend on exchange with a gas.

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  • $\begingroup$ Thank you for the answer. The first part of your answer makes sense to me. If we put substances in a sealed container then I can see how we can use the phase diagram to predict the state of the substance. And i follow the logic for why we see evaporation when we open the bottle. But am I correct in my understanding that when you have an open system, then I can't really use the phase diagram to predict the state of the system? $\endgroup$ Commented May 6 at 14:53
  • $\begingroup$ Well you can definitely predict the state of a piece of water with the phase diagram even if it is in contact with the atmosphere. The water that did not move to the gas phase through the liquid/atmosphere interface is in the state predicted in the phase diagram. It is just that some of the material excited the system so you lost some of it. $\endgroup$
    – Guillaume
    Commented May 6 at 15:53
  • $\begingroup$ Thank you this cleared it up for me. Just to clarify, I should be thinking about the water that has evaporated in the air as a different phase altogether right? I mean distinct of the vapor phase of water that the PT diagram describes. And this happens by a different mechanism i.e. water molecules escaping into air rather than forming a pure water vapor phase over the liquid (which would be described by the PT diagram). $\endgroup$ Commented May 6 at 16:54
  • $\begingroup$ *I mean distinct that then vapor phase ... $\endgroup$ Commented May 6 at 17:16
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Why do we have water vapor at room temperature and atmospheric pressure when water is a compressed liquid at these conditions?

If we kept vapor partial pressure below saturated vapor pressure, we would have no liquid water at equilibrium. All water would evaporate.


These phase diagrams are for a single component, in this case water. there is no air there.

The p-T curve between liquid and solid region of the water phase diagram describes vapor pressure of water for the given temperature.

It does not mean the phase diagram cannot be used if air is involved. But you have to know how to interpret them. The pressure in the diagram is not the total pressure and the diagram described equilibrium. an equilibrium cannot be reached in the open system.

At $\pu{20 ^{\circ}C}$, (saturated) vapor pressure is about $\ce{0.02 atm}$, as you can see on the diagram.

  • If vapor partial pressure is higher than (saturated) vapor pressure $\ce{0.02 atm}$, vapor condenses.
  • If lower, water evaporates.

If it happened there was $\ce{1 atm}$ of vapor at $\pu{20 ^{\circ}C}$, it would condense until it reaches p-T curve.

The total pressure $\ce{1 atm}$ is relevant only for the boiling point, when vapor pressure equals total pressure.

By other words, no matter how big the total gas pressure is (*), water vapor condenses only if its partial pressure is bigger than the (saturated) vapor pressure for the current temperature.


(*) - For really big pressures, water has counter-intuitively even higher vapor pressure than for low pressures.

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  • $\begingroup$ It does not require "really big" pressures that effect is responsible for the difference in T between the normal freezing point and the triple point. In both cases liquid and solid are in equilibrium with the vapor. the presence of inert gas changes the temperature which changes the vapor pressure[or vice versa]. The equilibrium on the liquid-solid line does not involve vapor there is no head space. $\endgroup$
    – jimchmst
    Commented May 6 at 19:41
  • $\begingroup$ Well, for low pressures, compared to critical pressure, the effect is negligible. $\endgroup$
    – Poutnik
    Commented May 6 at 20:17

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