# Phase diagram of water in real life

• fog37
In summary: So, if I have H2O(g), CO2(g), and N2(g) in a container at 25 degrees Celsius and atmospheric pressure, the water will freeze at p=1 atm and T=0 C, right?Yes, water will freeze at p=1 atm and T=0 C.
Borek said:
OK, I see. Seems like English nomenclature is quite convoluted. There are three things:

1. Equilibrium (saturated vapor pressure).
2. Non-equilibrium vapor pressure.
3. Partial pressure.

Calling the first one "vapor pressure" (when it already has two unambiguous names - "equilibrium vp" and "saturated vp"), and the second "partial" (when "partial pressure" typically means "pressure exerted by one of the components of the mixture") seems to be designed to make things confusing. So, when humidity is 50% do we refer to it as "partial partial pressure of water vapor?"

Partial pressure:
"In a mixture of gases, each gas has a partial pressure which is the hypothetical pressure of that gas if it alone occupied the entire volume of the original mixture at the same temperature."
https://en.wikipedia.org/wiki/Partial_pressure

Here the partial is not in relation to the saturated vapor pressure, but to the total pressure of all gasses in the mixture. For example, you could have a (unstable, non-equilibrium) situation where the partial pressure of a gas is higher than its saturated vapor pressure. When you have a gas-liquid or solid-gas equilibrium, the partial pressure of the gas at equilibrium is defined as the vapor pressure. As an analogy, think partial pressure is to vapor pressure as reaction quotient is to equilibrium constant.

Perhaps for this thread, we should take the convention of referring to the partial pressure of a gas when it is at equilibrium with its condensed phase as the equilibrium vapor pressure or saturated vapor pressure.

fog37 said:
I believe I understand, thanks to your help, the concepts you are mentioning: evaporation always occurs when the vapor pressure is less than the equilibrium vapor pressure and boiling when the equilibrium vapor pressure is larger than atmospheric pressure.

So, in the graph below, the coordinates P=1 atm and T=100 C indicate the boiling point (the normal boiling point since any other point sitting along that same line is also a boiling point) and that means the water vapor pressure P has to equal to 1 atm for boiling to occur, i.e. the vapor pressure on its own must be equal to the the same pressure that the entire mixture of gases composing air exerts on the free surface of the water and inside the liquid by pascal principle. It remains that the vertical axis is always and only the equilibrium, saturated, vapor pressure of water no matter if we are talking about the blue (vapor), orange(solid) or green(liquid) area in the diagram...

View attachment 215990

The y-axis is the partial pressure of water vapor not the equilibrium vapor pressure.

Let's take the T = 100°C line for example. When the partial pressure of water vapor is less than 1 atm (i.e. PH2O < 1 atm), the gas phase is the thermodynamically favored phase and net evaporation of liquid into gas will occur. When PH2O > 1 atm, water is the thermodynamically favored phase and you will have net condensation of vapor into liquid. Only when PH2O = 1 atm can you have liquid and gaseous water coexisting at equilibrium. This is consistent with our observations in real life. If you leave a boiling pot of water on the stove (T = 100°C, PH2O ~ 8torr), the water will all eventually boil away. You could only get an equilibrium between the boiling liquid and vapor if you were to close the system to allow PH2O to build up to ~ 1 atm (this is what occurs in a pressure cooker).

The black line that defines the boundary between the vapor phase and the condensed phases (solid or liquid) defines the equilibrium vapor pressure of water at each temperature.

Last edited:
Ygggdrasil said:
Here the partial is not in relation to the saturated vapor pressure, but to the total pressure of all gasses in the mixture.

Ygg, all I meant is that this convention forces misuse of words - "partial pressure" has a known meaning, but if we treat "vapor pressure" as being that of saturated vapor we need to differentiate between "saturated vapor pressure" and "existing vapor pressure" - you suggested the latter is a "partial pressure of water vapor", which uses the same adjective that is already used to mark something else related to pressures. In some context you will need to use it in different meanings and it will be confusing, there would be no such problem if "vapor pressure" would mean exactly what it says.

It is a moot, we are not going to change the convention, but it is a bit off to me.

Bystander
Thank you everyone. I think we can put this topic to sleep. But one more clarification, if possible, before we do that :)

The temperature T in the diagram is of easy interpretation: it represents the internal temperature of the substance itself, no matter its current state. I am also now clear that the y-axis of the diagram represents the partial pressure of water when the water is in vapor form . It makes sense that it must be the pressure of the substance itself specifically when it is in its vapor state.

This seems to imply that in the phase diagram, in order to determine the thermodynamically favored and stable state (liquid, solid, vapor or gas) of a sample of water at a specific T and P, we must always imply that water in its vapor form is present. At a specific temperature T, it is the water vapor and its pressure that dictates the state of the water sample. But it is not the pressure of the water sample. It is the pressure of the water vapor existing above the water sample... If P=0 it would mean that the partial pressure is zero and that no water vapor is present above the water sample. That seems to be an impossible situation in real life or in the lab...

fog37 said:
This seems to imply that in the phase diagram, in order to determine the thermodynamically favored and stable state (liquid, solid, vapor or gas) of a sample of water at a specific T and P, we must always imply that water in its vapor form is present. At a specific temperature T, it is the water vapor and its pressure that dictates the state of the water sample. But it is not the pressure of the water sample. It is the pressure of the water vapor existing above the water sample...
What about when the water (or steam) is alone in a pipe or tank? As said above, only when you are on the line between states can you have two states in equilibrium. All the rest of the time you only have one.
If P=0 it would mean that the partial pressure is zero and that no water vapor is present above the water sample. That seems to be an impossible situation in real life or in the lab...
I would agree with that -- P=0 is not really achievable.

Please be aware that we're talking *perfect* liquids / gasses etc, ignoring nucleation issues.
The comment about 'boiling granules' hit the spot: Pure liquids in a smooth container have a nasty habit of super-heating then 'bumping' violently.
I've seen a dozy (*) colleague's 2-litre beaker eject most of its contents in one spew...

( That drowned the bunsen burners playing on its diffuser so, even though the mess was under a fume-hood, we shut off the gas before fetching a mop... )

If you cannot have porous chips to provide nucleation, you must use glass beads or equivalent. Plan_B is to use a rotary evaporator, where evaporation from the thin film on flask up-side dominates the process. Even so, anti-bump precautions are essential lest the main volume super-heat...
==
*) He worked 'Continental' shifts, had just switched from 'Lates' to 'Earlies' and was in full-on, jet-lagged zombie mode...

Asymptotic
Borek said:
Ygg, all I meant is that this convention forces misuse of words - "partial pressure" has a known meaning, but if we treat "vapor pressure" as being that of saturated vapor we need to differentiate between "saturated vapor pressure" and "existing vapor pressure" - you suggested the latter is a "partial pressure of water vapor", which uses the same adjective that is already used to mark something else related to pressures. In some context you will need to use it in different meanings and it will be confusing, there would be no such problem if "vapor pressure" would mean exactly what it says.

It is a moot, we are not going to change the convention, but it is a bit off to me.
I agree that the existing terminology is confusing but I don't quite understand your confusion here.

Let's take a glass of water in a normal room (25°C, 8 torr water vapor). I would say that partial pressure of water vapor is 8 torr (PH2O = 8 torr). Is this what you mean by the existing vapor pressure? Am I using the term "partial pressure" incorrectly here?

@Russ watters,

Well, water alone in a tank soon develop, by evaporation, vapor in the tank itself (if there is room for the vapor to fill) and equilibrium is eventually reached once the partial pressure of water vapor because the saturated vapor pressure.

I guess you mean liquid water inside a container with volume equal to the volume of liquid water with no room for the vapor to occupy...

Last edited:
Ygggdrasil said:
Let's take a glass of water in a normal room (25°C, 8 torr water vapor). I would say that partial pressure of water vapor is 8 torr (PH2O = 8 torr). Is this what you mean by the existing vapor pressure? Am I using the term "partial pressure" incorrectly here?

What I was referring to is that in exactly this context "partial pressure of water is 8 torr" is ambiguous, as it is not clear whether "partial" refers to the fact there are other gases present (but the vapor itself is saturated), or to the fact saturated vapor pressure is 77 torr and the existing one (real, observed, measured one) is only 8 torr (so is a part of 77). It is not that I am confused, it is more like English convention makes the wording unnecessarily ambiguous. Or at least that's how I see it.

Just to confirm another bit of information: are the water vapor in the atmosphere or liquid water inside a container affected by the other gases in the atmosphere and their pressure?

When talking about boiling (assuming no nucleation) the water vapor inside the forming bubble must have a saturated vapor pressure equal to 1 atm (760 torr) to overcome the pressure of 760 torr inside the liquid water. This internal liquid pressure is transmitted, by Stevin principle, from the free interface to any other point inside the liquid. So the saturated vapor pressure inside the bubble must be 760 torr to overcome the pressure from all the added partial pressures of the gases outside the container...

fog37 said:
Just to confirm another bit of information: are the water vapor in the atmosphere or liquid water inside a container affected by the other gases in the atmosphere and their pressure?

If it wasn't this way, would water boil at lower temperatures at higher elevations (where the air pressure is lower)?

Nik_2213
Well, frost and dew form based on the amount water vapor and it partial pressure present in the air. Other gas species (oxygen, nitrogen, etc.) in the air don't seem to play a role.

In the case of boiling, the pressure of all the gases is transmitted inside the liquid water via Pascal's principle and the bubble forms when it overcomes that 1 atm pressure...

Unfortunately, under the title "practical" we discuss water behaviour in terms of "ideal", "eqiulibria", etc. Water is a very unusual liquid (and in moderate temperatures, vapour) because of autoassociation, caused mainly by hydrogen bonding. Associates are dynamic in nature, exchange molecules and change "n" in the general formula (H2O)n. In the years of my studies, professors equals n to 2, which was easily visible from IR spectra taken in mild conditions. It doesn't fit to statistical mechanics calculations and later it was understood that water dimer has strong prevalence to single molecular geometry, so its IR peaks are stronger than those of another associates, existing in many forms.
In practice, this all means that water behaves as it has molecular weight much higher than 18. Compare boiling points of gases having Mw about 20 and Mw's of liquids having boiling points about 100 deg C. As the degree of association in liquid water should differ from these I am water vapor, their phase equlibria are very problematic and situation is rather as in multicomponent solution. Starting from this point of view it is easier to understand water behavior in biological sstems.
Some discussion parts refers to perfect liquids in ideal equlibria conditions, some to practical evaporation (boiling stones, rotary evaporators) and the title contains "phase diagram". All these appraches have different starting points and mehodologies. In addition, the nomenclature used is (in my view) partially taken from chemical process engineering (because of literature sources). This jargon uses many well defined terms in a strange meanings, resulting from industrial practice tradition. Intersting ponit of view appears in geophysics and climatology, where of importance is water transport through the atmosphere. Most important are flows of liquid water and air, as without the air movement, water (practically heavier than other atmospheric gases) occupies solid and liquid Earth surface. Without oceanic currents, their surfaces should contain much less water and more salts, which result in higher boilimg point and lower vapor pressure.
For physicists it should be much easier to refer to simpler liquids, which vapors behaves as ideal gases. For example the "noble gases".
Cheers,

Nik_2213
Something that was said probably in this thread (but not in my summary):

the pressure P in the phase diagram of water represents the partial pressure of water vapor at that specific temperature (not the saturated vapor pressure) except on the lines where two states of the substance coexist in equilibrium. On those lines, the pressure P represents instead the saturated vapor pressure.

Assume we move horizontally from the very right of the diagram, where the substance is a gas, to the left. That means the temperature is decreasing and the pressure is constant. Once we run into the line of vaporization, the system (water) becomes single component and biphase: it starts existing as liquid and vapor in the point on the curve. At that point, if we continue to reduce the temperature T, the pressure P does not remain constant but decreases and assumes the vapor pressure values exactly along the vaporization curve line.

fog37 said:
Something that was said probably in this thread (but not in my summary):

the pressure P in the phase diagram of water represents the partial pressure of water vapor at that specific temperature (not the saturated vapor pressure) except on the lines where two states of the substance coexist in equilibrium. On those lines, the pressure P represents instead the saturated vapor pressure.

Assume we move horizontally from the very right of the diagram, where the substance is a gas, to the left. That means the temperature is decreasing and the pressure is constant. Once we run into the line of vaporization, the system (water) becomes single component and biphase: it starts existing as liquid and vapor in the point on the curve. At that point, if we continue to reduce the temperature T, the pressure P does not remain constant but decreases and assumes the vapor pressure values exactly along the vaporization curve line.

That is correct. If liquid and vapor are at equilibrium and you lower the temperature, the system will no longer be in equilibrium. Vapor will condense into liquid until the partial pressure of water drops to the new vapor pressure at the lower temperature and equilibrium is re-established.

Nik_2213

• Chemistry
Replies
9
Views
2K
• Chemistry
Replies
10
Views
1K
• Chemistry
Replies
11
Views
1K
• Chemistry
Replies
26
Views
3K
• Chemistry
Replies
8
Views
2K
• Chemistry
Replies
4
Views
3K
• Chemistry
Replies
21
Views
5K
• Mechanics
Replies
5
Views
1K
• Atomic and Condensed Matter
Replies
1
Views
2K
• Thermodynamics
Replies
2
Views
1K