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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


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


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    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


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


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    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?
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