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Why is the Vapour pressure of Water Constant ?

  1. Oct 8, 2010 #1
    Now if i have the vapour pressure of water to be 200mm and the water occupies 4 L out of 20 L closed vessel.Which implies that the Vapour occupies 16 L

    Now if i shift the water to a 30 L closed vessel.The vapour pressure of water vapour remains the same even though it occupies a greater volume of 26 L.

    So how is this possible ?
  2. jcsd
  3. Oct 8, 2010 #2
    Assuming STP, changing the volume above the liquid water will result in more water vaporizing to get to the same pressure. Think PV=nRT.
  4. Oct 10, 2010 #3
    [OK, second attempt to answer this. My first attempt was thrown away when I clicked "submit", and discovered that the forum software had silently logged me out for my convenience and safety. I will refrain from transcribing my verbal comments at that time. An hour's work, while I figured this out, pfft!]

    Consider what's going on at the molecular level, and look closely at the surface of the liquid, at the liquid/gas interface.

    The vapor pressure of a liquid arises from the balance of probabilities that a liquid-phase molecule will have enough energy to escape the surface ("boiling"), versus a gas-phase molecule not having enough energy to avoid being reabsorbed ("condensing") if it strikes the surface. For a given compound, these probabilities depend solely on the temperature, which is why the vapor pressure varies with temperature.

    Suppose we are at equilibrium. A few extra high energy liquid molecules happen to escape, and the partial pressure of our compound increases. With the increased pressure, the probability that a given molecule will strike the surface increases, as does the average force with which the molecules bounce off each other (which is exactly what "increased pressure" means), and a corresponding number of low-energy molecules will be knocked back into the pool and reabsorbed, dropping the pressure back down again. Thus is equilibrium maintained.

    Even if there is another gas compound present above the liquid, all that matters is the molecules of the liquid compound of interest.

    Further, the vapor pressure is an absolute, not a percentage of the total pressure. So, if our container is pressurized to 1000 torr, and the vapor pressure of our liquid is 10 torr, the other gasses in the container will make up the remaining 990 torr. Or, if our container is pre-pressurized to 1000 torr, and we inject a small enough amount of our liquid so as not to change the total volume significantly, the total pressure will rise to 1010 torr.

    Suppose now that our gas volume is so large that even if all the liquid converts to gas, it will not raise the partial pressure of our compound to the vapor pressure. Thus, with the kitchen window open to the world, our tea water boils away.

    I eagerly await correction on those points where I am wrong.
    Last edited: Oct 10, 2010
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