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Liquid-Vapor Equilibrium and Vapor Pressure

  1. Mar 25, 2013 #1
    I'm studying vapor pressure and I'm having trouble to understand some concepts.

    Vapor pressure depends only on the temperature of the liquid in question right? For example, if we have a container with water at 30° C, it doesn't matter the experiment, would it ALWAYS be a vapor phase with pressure = vapor pressure of water at 30°C = 31.8 mmHg?

    So consider the following experiment: In a container we have water at 30°C. One end is closed while in the other we have a piston. Above the pistons we have weights that do a initial pressure of 200mm Hg on the water. Is there water in the vapor state? I would say yes, but if so the vapor should exert a pressure of 31.8mmHg, and where would be the other 268.2mm Hg? The only explanation I thought is that a few bubbles of gas above the liquid makes the 31.8mmHg but the liquid still stays in contact with the piston making the other 268.2 mmHg. But I haven't read this in anywhere, is it right?
  2. jcsd
  3. Mar 26, 2013 #2
    Correction: "Above the pistons we have weights that do a initial pressure of 300 mm Hg"

    Nobody knows the answer?
  4. Mar 26, 2013 #3


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    I do not see the need to subtract the vapor pressure anywhere. There is the same pressure beneath the piston as beneath the closed end, so the full effect of the piston weight is transmitted.
  5. Mar 26, 2013 #4
    There wouldn't be any vapor present. Assuming there is a vacuum above the piston, you would have to raise the temperature enough for the vapor pressure of water to equal the pressure applied by the weights (200 mm Hg). That would be about 66 C. Until then, you would only have liquid water.
  6. Mar 27, 2013 #5
    I can't understand why!
    When we look at the Maxwell-Boltzmann distribution of molecular energy we see that any liquid has some molecules with energy enough to pass to the gaseous state. On the surface of the example I gave, shouldn't be liquid water turning into water vapor?
  7. Jan 1, 2016 #6
    Hi, sorry for digging out this old thread, but I have the exact same question and I can't get a firm grasp of the concepts... as I understand it, any container that is filled to any appreciable level will have an outward pushing pressure of whatever gas phase is already inside it (i.e air) plus the vapors from the liquid phase as a result of it's vapor pressure - and I understand that this happens at any temperature (but only when the vapor pressure surpasses the pressure of the gas above the liquid do we have boiling).

    So by this reasoning, in the example above, I would expect the pressure of a container semi-filled with water at equilibrium at 30°C to be 300mmHg in total; 31.8mmHg from the vapor pressure of the water, and 268.2mmHg from the air above the water. Isn't that the case?

    What I can't seem to understand is if the container was filled EXCLUSIVELY with water, what would happen?
  8. Jan 1, 2016 #7
    Yes, if there is air present in the head space. The OP was assuming that there is no air present in the heat space, so, if a total pressure of 300 mmHg were applied, there would only be liquid water present.
    Are you assuming that there is no air present in the head space? If so, if you apply a pressure of 300 mmHg with a piston and hold the temperature constant at 30 C, there will be no vapor present at equilibrium, and the piston will be pressing down solely on liquid water.
  9. Jan 2, 2016 #8
    Thanks for the response!
    Yes, I'm assuming no air present in the head space. But then how would the piston be balanced? Would the water exert 31.8mmHg on the piston wall from it's fugacity/vapor pressure and another 268.2mmHg in reaction to the piston "pressing" on it? I.e. would we treat it like a reaction force from the incompressible water onto the piston weight, just as we would treat one solid body atop another?
  10. Jan 2, 2016 #9
    Sure. The water would exert a force of 300 mg Hg on the piston. This would be required for force equilibrium. And we wouldn't have to break down how this comes about, since it will be all liquid water. And, the water doesn't have to be considered incompressible for this to apply.

    Have you ever seen a phase diagram for water? If not, google "water phase diagram." Also, what does the phase rule tell you if you have a liquid phase with no vapor? The number of phases is 1. What is the number of degrees of freedom (T and P)?

  11. Jan 2, 2016 #10
    I'm not sure what info the phase diagram provides with regards to this situation. What am I supposed to see?
    As far as the phase rule, phase p=1, components c=1, so degrees of freedom f=c-p+2=1-1+2=2 right? Does this mean that we can independently vary temperature and pressure?
  12. Jan 2, 2016 #11
    Jaumzaum said:

    "Vapor pressure depends only on the temperature of the liquid in question right? For example, if we have a container with water at 30° C, it doesn't matter the experiment, would it ALWAYS be a vapor phase with pressure = vapor pressure of water at 30°C = 31.8 mmHg?"

    I believe that you are confusing equilibrium vapor pressure with actual vapor pressure. In the free atmosphere, you can have a variety of vapor pressures over a 30°C surface of water, depending upon environmental conditions. I know that many laboratory physicists carelessly use the phrase "vapor pressure" to mean equilibrium vapor pressure, but the atmospheric scientist who reads that phrase gets an entirely different message.
  13. Jan 2, 2016 #12
    Notice on the phase diagram the area labelled liquid. Throughout this region, for all the temperatures and pressures indicated, there is only one phase present at equilibrium. That is liquid water.

    Concerning the phase rule, you are correct that there are 2 degrees of freedom. That means that you can vary both temperature and pressure independently when you have only a liquid phase present. So the phase rule tells you the same thing as the phase diagram in this regard.

    Last edited: Jan 2, 2016
  14. Jan 9, 2016 #13
    Thanks for the reply Chestermiller, sorry I haven't gotten back to the thread until now.

    But how do we interpret the water phase diagram when we have the other case of a liquid with room above it for a vapor phase? I mean where is the "clue" that there should be two phases present if we rely solely on the phase diagram?
  15. Jan 9, 2016 #14
    The clue is that the temperature and pressure are on the saturation line if two phases are present. If the pressure is above the saturation line or, equivalently, the temperature is below the saturation line, then there is only liquid water present.
  16. Jan 9, 2016 #15
    Hmm I suspect I may have some misconceptions about what it means to have a vapor phase present...
    I mean, in a closed vessel half filled with water, even before the boiling point, aren't there both liquid and a saturated vapor phase present? That's why we have evaporation, right? The liquid always has a vapor pressure/fugacity, i.e. there will always be some molecules escaping from the liquid into the vapor phase, so in an open atmosphere the liquid will evaporate, but in a closed vessel the rate of molecules leaving the liquid phase reaches an equilibrium with the rate that molecules re-enter it, so we have the notion of saturated vapor pressure, am I mistaken? Do we only say that we have a vapor phase when we are above the boiling point?
  17. Jan 9, 2016 #16
    No, you're not mistaken. As you have correctly done, we need to make the distinction between when we have equilibrium, and when we don't have equilibrium.
    No, of course not. As you said, you have a vapor phase when you have both liquid and vapor in the container. The phases may or may not be at equilibrium. If the pressure of the species in the vapor phase is less than the equilibrium vapor pressure at the liquid temperature, the liquid can be boiling. If there is also air present, then to get boiling, the equilibrium vapor pressure at the liquid temperature must be at least equal to the total pressure of the air and water vapor in the gas phase.

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