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Vapour pressure

  1. Jan 8, 2014 #1
    I read in my Chemistry workbook that:
    "At higher temperatures, vapour pressure increases as more energy is provided to the liquid molecules, allowing them to escape into the gas phase."

    At the boiling point, all energy is gone into breaking bonds and to convert liquid phase molecules into gaseous. Why does it say at higher temperatures, more molecules escape into the gas phase? Once past the boiling point, wouldn't all the original liquid molecules of the original sample be converted to the gas state by then?

    Also, does Dalton's Law of Partial Pressure only apply if all the initial gas constituents occupy the same volume and are all placed in a container of the same volume?
    Last edited: Jan 8, 2014
  2. jcsd
  3. Jan 8, 2014 #2
    No. This doesn't happen at the boiling point. It happens at the critical point. The boiling point of a liquid is merely the temperature at which its equilibrium vapor is equal to atmospheric pressure (103000 Pa). This allows bubbles to form within the liquid. If the atmospheric pressure changed, the boiling point would change. Thus, at the top of a mountain, the boiling point is less than at sea level.

    This is much more restrictive than necessary. Even if the mole fractions of the various species are varying with spatial position, and even if the total pressure is also varying with spatial position, Dalton's Law still applies at each spatial location.
  4. Jan 9, 2014 #3


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    Staff: Mentor

    Past the boiling point yes, everything is in the gas state. But before the boiling point pressure of the saturated vapor gets higher with the temperature.
  5. Jan 9, 2014 #4
    I think that this needs to be modified slightly to past the (atmospheric) boiling point, all the liquid turns to vapor if you continue to add heat. If the vapor that has left the liquid is kept in contact with the liquid, the pressure is maintained at one atmosphere, and no more heat is added, no more vaporization will occur (and you will have a combination of liquid water and water vapor present). In fact, any time you have pure vapor in contact with the liquid (i.e., no air present) and you raise the temperature (even above the atmospheric boiling point), the vapor pressure will rise, but the gas will stop evaporating if you stop adding heat. At that point the temperature and pressure will continue to remain constant.

  6. Jan 9, 2014 #5


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    Staff: Mentor

    I don't think that's required - you won't get past the boiling point without adding enough heat to boil the liquid away, so it is all covered by my statement.
  7. Jan 9, 2014 #6
    Well... reluctantly OK. But, from the OPs question, I sensed a greater degree of confusion and the need for additional explanation. I hope I was wrong.

  8. Jan 9, 2014 #7
    For example, if you have:

    Ne(g) occupies 2L at 100 kPa
    He(g) occupies 1L at 100 kPa
    Ar(g) occupies 3L at 200 kPa

    And then these gases are all mixed into a container of 6L. Based on Dalton's Law of partial pressure, each individual pressure would be summed up (right? Assuming these gases behave as ideal gases). So if it was a 5L container, how would this change the situation?
  9. Jan 9, 2014 #8
    If you're asking me to solve this problem, I will be glad to do it. Is this what you want?

  10. Jan 9, 2014 #9
    I'm just starting to learn this topic and an explanation of the law would be great as well. But yes, going through how to solve this question, or anything related, would be helpful. Thank you!
  11. Jan 9, 2014 #10
    OK. Rather than my laying out the whole thing for you, lets work on it together. We are going to be using the ideal gas law, PV=nRT, and later, applying Dalton's law of partial pressures.

    We are going to be assuming that all three containers have the same temperature T, and that the final 6L container is also at the temperature T. Now, using the ideal gas law, how many moles of each gas are present in each of the three starting containers (in terms of R and T)? The number of moles of each gas does not change when they are put into the final 6L container. In terms of R and T, what is the total number of moles of gas in the final container? You now know that total moles of gas in the 6L container and the volume of the container. Using the ideal gas law, what is the pressure in the 6L container (this will not depend on R and T)? You know the number of moles of each gas in the 6L container. What is the mole fraction of each gas in the container?


    Let's stop here for now.
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