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Vapor pressure in open container

  1. Feb 1, 2015 #1
    Vapor pressure in closed container is well understood by me.But in closed vessel ,it is a nightmare for me.However, in an open container (e.g. pot on a stove) equilibrium cannot be reached due to the vapor being dispersed into the atmosphere .The definition of boiling is when vapor pressure reaches and/or exceeds the atmospheric pressure. If equilibrium cannot be reached in an open container,how can we define boiling point?This is discussed previously in this forum,I have gone through that still I am not getting.
     
  2. jcsd
  3. Feb 1, 2015 #2

    russ_watters

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    Why does boiling imply equilibrium to you? The liquid (boiling or not) has a vapor pressure based on its temperature, even if that vapor pressure isn't equal to the pressure of the vapor in the surrounding room. Does that help?
     
  4. Feb 1, 2015 #3
    In open container how can we measure vapor pressure?I mean in closed system ,we can look at pressure on walls of the container and that will be vapor pressure.We can measure vapor pressure in open system on?
     
  5. Feb 2, 2015 #4
    The water vapor in the rising bubbles is virtually at equilibrium with the liquid water that surrounds it. This is essentially the same situation that exists in a closed system containing only water. If the liquid water temperature is 100 C, the water vapor pressure inside the bubbles is 1 atm.

    The air above the liquid in the open container is not at thermodynamic equilibrium with the mixture of liquid water and vapor bubbles. The only thing that the air does is control the overall pressure acting on the mixture (typically, 1 atm.). In order for the water to boil (i.e., for bubbles to form below the surface), the temperature has to be high enough for the vapor pressure inside the bubbles to push back the atmosphere so that the bubbles can grow. This happens when the equilibrium vapor pressure matches the atmospheric pressure above the mixture.

    Chet
     
  6. Feb 2, 2015 #5

    russ_watters

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    You don't have to measure it: it is a known thermodynamic property of water. You can read it off a table if you know the temperature.

    Consider the concept of relative humidity. Let's say the relative humidity is 50%. 50% of what?
     
  7. Feb 2, 2015 #6
    Why boiling is indicated by bubbles formed below the surface? Kochendes_wasser02.jpg
    I have been thinking that boiling should be indicated by vapors quite above the vessel/in atmosphere,because boiling occurs when vapor pressure becomes equal to atmospheric pressure and vapor pushes atmospheric air to escape from boundaries of vessel into the atmosphere.
     
  8. Feb 2, 2015 #7
    No. Your description is not what boiling is all about. Do you not see the bubbles of water vapor within the water in the container in your picture. That's what I'm talking about. As I said in my previous response, in order for those bubbles to form under the liquid, the vapor pressure of the water has to equal the pressure of the atmosphere above the boiling liquid, so the volume of the mixture of liquid and bubbles can expand.

    Chet
     
  9. Feb 2, 2015 #8
    And what about surface bubbles?Don't they indicate boiling?
     
  10. Feb 2, 2015 #9
    When the bubbles reach the surface, they release their water vapor into the air. After that, diffusion takes over.

    Chet
     
  11. Feb 2, 2015 #10
    Before boiling point how vapor pressure is defined in open system?
     
  12. Feb 2, 2015 #11

    russ_watters

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    Bubbles form beneath the surface because the heat is being applied to the bottom of the vessel, making the water at the bottom the hottest.
    I'm not sure what you mean by the first part, but the second part is basically correct, though not exactly relevant since boiling first occurs below the surface where there is no air.
    I explained that in post #5.
     
  13. Feb 2, 2015 #12
    I really don't know,why I am not able to grasp the concept of vapor pressure in open system.
     
  14. Feb 2, 2015 #13

    russ_watters

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    I'm not sure you are reading and trying to process the responses you were given, since you didn't respond to some of them. You seem to be speculating on your own instead of listening to the answers you are being given. Let me try to explain it again:

    You asked how vapor pressure (of water) can be measured if the water and air are not in equilibrium in an open container. The answer is that it can't be measured. Do you recognize that answer? Please acknowledge that you read it and have thought about it.
     
  15. Feb 2, 2015 #14
    Before the temperature of the liquid reaches the boiling point, no bubbles are forming. However, evaporation is occurring at the surface of the liquid. At the interface with the overlying air, the partial pressure of water vapor in the air is equal to the equilibrium vapor pressure of water at the surface temperature of the liquid. So, the liquid and water vapor are in equilibrium at the interface between the liquid and the air. This process also happens when the temperature of the liquid water reaches the boiling point. However, in addition to the evaporation process at the air/water interface, there is also transport of water vapor out of the liquid by the bubbles of water vapor that rise to the surface and burst. So, during boiling, the rate at which water vapor escapes from the liquid is much larger than by just evaporation at the surface.

    Chet
     
  16. Feb 2, 2015 #15
    I think I am getting the concept.But I can't interpret only one line i.e
     
  17. Feb 2, 2015 #16
    Well, you are aware that, if you had a closed container with both air and water vapor in the head space, then, at equilibrium, the partial pressure of the water vapor in the head space would be equal to the equilibrium vapor pressure of water at the temperature in the container, correct? At the interface of an evaporating liquid in an open system, the vapor in the gas phase virtually instantaneously comes to equilibrium with the liquid at the temperature of the interface (at the interface only). This has been verified experimentally over the ages by showing that the observed rate of evaporation in all cases is consistent with the corresponding rate of diffusion of vapor from the interface.

    Chet
     
  18. Feb 2, 2015 #17
    Please bear with me
    Here does vapor pressure of water at the surface temperature of the liquid mean vapor pressure of water vapor on the liquid water surface?
     
  19. Feb 2, 2015 #18
    I don't quite get the distinction. Pressure is isotropic, so it acts equally in all directions at a given spatial location. Is that what you are asking?

    Chet
     
  20. Feb 2, 2015 #19
    Can you please complete this sentence -"rate of diffusion of vapor from the interface" TO ...?
     
  21. Feb 2, 2015 #20
    Rate of diffusion of vapor from the interface into the bulk air in the room. The partial pressure of water vapor in the air decreases with distance from the interface. This gradient in water vapor partial pressure provides the driving force for diffusive transport of water vapor away from the interface (Fick's law). What do you think the partial pressure of water vapor is in the bulk room air containing a pot of boiling water, if the bulk room air temperature is about 20 C, and the relative humidity in the bulk of the room air is 50%? (The bulk room air refers to the majority of the air that is not right next to the pot of boiling water).

    Chet
     
  22. May 23, 2016 #21
    How would you define vapor pressure even before bubbles are not formed?

    You stated somewhere that vapor pressure i.e defined in the closed system as the equilibrium pressure of water-vapor interface; That is similar to the case where bubbles are in thermodynamic equilibrium with water in the open system.

    I understood partial pressure of vapor present above the liquid surface.
     
  23. May 23, 2016 #22
    If you have a glass of liquid water sitting open to the atmosphere (and no boiling is occurring), the partial pressure of water vapor away from the glass is still only about 0.01 atm. However, at the liquid water interface with the air, the partial pressure of the water vapor is equal to the equilibrium vapor pressure at the liquid water temperature. But, overall, the system is not at equilibrium. There is a driving force for water vapor to diffuse away from the interface. There is a higher partial pressure near the interface than out in the bulk of the air. This is how the water vapor that evaporates is transported away into the surrounding atmosphere. Typically, the region over which the water vapor partial pressure varies from the equilibrium vapor pressure to the bulk partial pressure in the air is only a cm or 2 away from the interface. What we are dealing with here is a Transport Process, rather than an equilibrium situation. To learn more about transport processes, see Mass Transfer Operations by Treybel.
     
  24. May 24, 2016 #23
    I referred some textbook and found that vapor pressure is the property of liquid which is also called saturation pressure at particular temperature. Correct me if I am wrong.

    Either raise the temperature to increase the vapor pressure of liquid or reduce the surrounding pressure so that vapor pressure would win the race against partial pressure of water vapor above it. That's how boiling occurs.

    The whole process is not in thermodynamic equilibrium; we have to just worry about liquid vapor interface partial pressure of vapor; rest all is mass transfer away from liquid as per Fick's law.
     
  25. May 24, 2016 #24

    jbriggs444

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    Rather that is how evaporation occurs. If the vapor pressure of the liquid is greater than the partial pressure already present in the adjacent atmosphere then the rate of evaporation is greater than the rate of condensation. As you say, the two are not in equilibrium. The race is not even.

    Boiling occurs when the vapor pressure of the liquid at some point is greater than the pressure in the liquid at that point. In this condition, a bubble which formed at that point would grow as liquid evaporates into it. The gas in the bubble is pure vapor. The partial pressure of the vapor in the bubble is equal to the total ambient pressure. In order to "win the race" and have a positive net rate of evaporation into the bubble, one needs the vapor pressure of the liquid to exceed the total ambient pressure, not just the ambient partial pressure of the vapor in question.
     
  26. May 24, 2016 #25
    We say phase change process from liquid to vapor (below 1 atm say); also a boiling phenomenon without bubbles. That means there we have to consider only vapor pressure of fluid at that temperature>partial pressure of vapor above the liquid.
     
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