Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Saturation Vapour Pressure

  1. Oct 21, 2005 #1
    The pressure exerted by a saturated vapour depends on the temperature and the curvature of the liquids surface.

    Why does it depend on the curvature of the liquids surface?
  2. jcsd
  3. Oct 21, 2005 #2


    User Avatar
    Science Advisor
    Homework Helper

    where does the evaporation occur?
  4. Oct 21, 2005 #3


    User Avatar

    Staff: Mentor

    I believe one is referring to 'surface tension', which is a liquid property - the molecules are still in continuous contact. In a two phase system, liquid and vapor, the molcules are evaporating at the liquid vapor interface.

    A vapor bubble must expand against a liquid, and it the tension in the liquid (surface tension) which is providing resistance to the bubble expansion. The surface tension is dependent on the cohesive forces among the molecules.
  5. Oct 21, 2005 #4


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    See "Kelvin equation."
  6. Oct 24, 2005 #5


    User Avatar
    Science Advisor
    Homework Helper

    my first reply was "a drunken post", just to clear up this query further
    When pressure is exerted on a liquid, whether through the introduction of an inert gas or by direct porous piston, the vapor pressure increases. Pressure is applied to the liquid, more gases escape.
    The relation to surface curvature is in which one considers the variation of the pressure due to surface tension with the curvature of a particular liquid. A droplet of water for instance, experiences an internal pressure as well as an external pressure, along with surface tension, which acts to shrink the droplet, and can be equated with the external pressure. This additional "external pressure" serves to increase the vapor pressure (compared to that of a bulk liquid).
    the equation to remember is
    [tex]p=p*e^{ \frac{Vm \Delta P}{RT} } [/tex] where the [tex] \Delta P [/tex] pertains to the change in the total pressure experienced by a liquid.
  7. May 11, 2008 #6
    From Clausius Clapeyron Equation ,
    we can get
    Ps=exp(-H/RT+C)=Aexp(-H/RT), A=exp(C).

    But how can we get A? I dont know.
    My answer is right?
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?