# Saturated vapour pressure - why an equilibrum is reached?

1. Feb 20, 2015

### Greg777

Hello,
I've read about the concept of saturated vapour pressure on some sites but none of them really explain why the equilibrum happens and that's what I want to ask you. Why the number of particles leaving the surface and joining the surface can't change over time and never reach the equilibrum?

2. Feb 20, 2015

### Bystander

How do you think it might change?

3. Feb 20, 2015

### Greg777

For example the act of leaving the surface by the particle is determined by whether it has sufficient "excess" of kinetic energy. But whether the particle will leave the surface depends if it's close to the surface and I think that's completely random. And also rejoining the surface seems random to me. I don't see why an equilibrum should be reached.

4. Feb 20, 2015

### Bystander

The difference between "Avogadro's Number" and "Avogadro's Number + 1?" You might want to look at "fluctuation theory/phenomena" to get a sense for the scales of systems for which "random" exchanges/motions of discrete particles/molecules are at all significant.

5. Feb 20, 2015

### Greg777

Ok, so now I know that randomness doesn't really matter but I'm still confused about the equilibrum - I know that the same amount of particles leave and rejoin the surface but does this number of leaving/joining particles increase over time? So e.g. X particles leave and X particles return and after a while 2X particle leave and 2X particles return, then 3X leave and 3X return and so on?

6. Feb 20, 2015

### Bystander

No.

7. Feb 20, 2015

### Greg777

So the number of particles leaving and joining the surface increases up to the equilibrum and then stays constant?

8. Feb 20, 2015

### Bystander

Or, decreases, depending upon the direction from which the equilibrium is approached. Other thing to examine is the definition of pressure, particularly in respect to vapor pressure; pressure can be considered in terms of d(mv)/dt per unit area. If you increase or decrease the collision rates of molecules, you are changing their partial pressures.

9. Feb 20, 2015

### Greg777

Under what circumstances the number may decrease?
I understood that randomness don't matter but the most important concept I don't get is why the number of leaving/joining particles may increase but then stay constant? Why (even if in isolated container) all the particles won't turn into gas after some time? In other words, why a fixed number of gaseous particles is reached and it stops changing after that?

10. Feb 20, 2015

### Bystander

Supersaturated vapor condenses until it reaches equilibrium.
Liquid evaporates until it saturates a vapor volume.
Amount of vaporization is limited by the volume available to vapor at the vapor pressure of the liquid. The number of molecules in the vapor phase does not keep increasing. The vapor pressure is the upper limit of pressure the liquid can contribute to the pressure in the available volume.

11. Feb 20, 2015

### Staff: Mentor

What factors determine how many particles will rejoin the surface?

I think you are focusing too much on the "leaving".

12. Feb 20, 2015

### Puma

Lets say you have one molecule of water leaving the surface of some water into a vacuum. This one particle of water bounces around the vacuum container for 5 seconds before it hits the surface of the water again and rejoins the body of water. Lets say only one molecule of water has the energy to escape from the surface of the water every five seconds. You then, on average, have reached an equilibrium of one particle in the vacuum above the water at any time.

13. Feb 21, 2015

### lightarrow

At equilibrium, the number dN1/dt of particles leaving the liquid phase for unit time is equal to the number dN2/dt of those joining the liquid phase (or you won't have equilibrium) but this number is proportional to the area A of the separation surface and also depends on the temperature T and on the vapour density rho:

dN1/dt = A*f(T)
dN2/dt = A*g(T)*rho

At the equilibrium: dN1/dt = dN2/dt so :

f(T) = g(T)*rho --> rho = f(T)/g(T) = h(T)

so at constant temperature T the vapour density rho (and so even its pressure) is constant. Conclusion: the number dN1/dt or dN2/dt doesn't vary *if the temperature is constant and the area A doesn't vary.

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lightarrow