How Does Surface Curvature Affect Saturation Vapour Pressure?

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Discussion Overview

The discussion centers on the relationship between surface curvature and saturation vapor pressure, exploring the underlying physical principles and equations that govern this relationship. Participants examine the role of surface tension, evaporation processes, and relevant equations such as the Kelvin equation and the Clausius-Clapeyron equation.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants note that the pressure exerted by a saturated vapor depends on temperature and surface curvature, questioning why curvature affects this relationship.
  • One participant suggests that surface tension is a key factor, as it involves cohesive forces among molecules at the liquid-vapor interface.
  • Another participant discusses how pressure applied to a liquid can increase vapor pressure, linking this to surface curvature and surface tension effects on droplet behavior.
  • A participant references the Kelvin equation as relevant to understanding the relationship between curvature and vapor pressure.
  • One participant presents a mathematical derivation involving the Clausius-Clapeyron equation, expressing uncertainty about determining a constant in the equation.

Areas of Agreement / Disagreement

Participants express various viewpoints on the relationship between surface curvature and saturation vapor pressure, with no consensus reached on the specifics of the mechanisms involved or the mathematical interpretations presented.

Contextual Notes

Some limitations include unresolved assumptions regarding the definitions of terms like surface tension and curvature, as well as the dependence on specific conditions for the equations discussed.

EIRE2003
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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?
 
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where does the evaporation occur?
 
EIRE2003 said:
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?

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.
 
See "Kelvin equation."
 
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.
 
From Clausius Clapeyron Equation ,
dlnPs/dT=H/RT2,
we can get
lnPs=-H/RT+C,ie,
Ps=exp(-H/RT+C)=Aexp(-H/RT), A=exp(C).

But how can we get A? I don't know.
My answer is right?
 

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