How Does Surface Curvature Affect Saturation Vapour Pressure?

Click For Summary
SUMMARY

The discussion centers on the relationship between surface curvature and saturation vapor pressure, emphasizing that vapor pressure is influenced by both temperature and surface tension. Surface tension, a property of liquids, affects the evaporation process at the liquid-vapor interface, where molecules transition between phases. The Kelvin equation is referenced to explain how internal and external pressures, along with surface tension, impact vapor pressure in curved surfaces, such as droplets. The Clausius-Clapeyron equation is also mentioned, providing a mathematical framework for understanding these relationships.

PREREQUISITES
  • Understanding of surface tension in liquids
  • Familiarity with the Kelvin equation
  • Knowledge of the Clausius-Clapeyron equation
  • Basic principles of phase transitions
NEXT STEPS
  • Study the Kelvin equation and its applications in thermodynamics
  • Explore the Clausius-Clapeyron equation in detail
  • Investigate the effects of surface tension on vapor pressure in different liquids
  • Learn about phase equilibrium and its implications in physical chemistry
USEFUL FOR

Students and professionals in physical chemistry, chemical engineering, and materials science who are interested in the thermodynamic properties of liquids and phase transitions.

EIRE2003
Messages
107
Reaction score
0
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?
 
Physics news on Phys.org
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
p=p*e^{ \frac{Vm \Delta P}{RT} } where the \Delta P 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?
 

Similar threads

  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 22 ·
Replies
22
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 10 ·
Replies
10
Views
4K
  • · Replies 24 ·
Replies
24
Views
6K
  • · Replies 18 ·
Replies
18
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 1 ·
Replies
1
Views
10K