Liquid-Vapor Equilibrium and Vapor Pressure

  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. Correction: "Above the pistons we have weights that do a initial pressure of 300 mm Hg"

    Nobody knows the answer?
     
  4. etudiant

    etudiant 905
    Gold Member

    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. Chestermiller

    Staff: Mentor

    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. 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?
     
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