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What factors determine the rate of evaporation?

  1. Sep 5, 2008 #1
    And if I'm heating a liquid that contains things that evaporate at different temperatures, will I have an increasing amount of trouble separating these things as I increase the amount of heat applied?
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  3. Sep 5, 2008 #2


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    Evaporation is the tendency for an atom or molecule to (1) leave and (2) not come back. So anything that affects these two processes will affect the rate.

    Yes, fractional distillation could be less convenient at a higher input power, in that the material might pass too quickly between the temperature at which component A evaporates and the temperature at which component B evaporates.
  4. Sep 5, 2008 #3
    If I mix an equal volume of three materials, A, B and C. Each having a different evaporation temperature. Let's say the evaporation temperature of B is equally above the evaporation temp of A as it is below the evaporation temp of C.

    Would the evaporation temperature of the whole now be equal to the evaporation temperature of material B?

    Or would material A, having the lowest evaporation temperature, begin to leave the mixture as soon as it's individual evaporation temperature is reached?

    In other words would there be any advantage to maintaining the mixture at the evaporation temperature of material A?

    And to explore my original intention a little more let's compare two cylindrical vessels used to evaporate liquids. One twice the diameter of the other, but each holding the same amount of an identical liquid. We'll say the goal is to reduce the volume of liquid by half.

    Assuming I can apply heat perfectly without losses, would the power requirements to reach the goal be the same? And more importantly, would the evaporation rate be the same? Do I save time with the larger diameter vessel? If so, is it achieved at the cost of an increased amount of power? Is there a simple mathematical relationship in this?
  5. Sep 5, 2008 #4


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    Actually, "evaporation temperature," on its own, doesn't really mean anything, so I was being imprecise in my answer above. All materials evaporate at a non-zero rate above 0K. Are we talking about the boiling temperature? Or perhaps we're talking about a temperature where the partial pressure is a certain amount, e.g., 10 kPa. In any case, we should be a little more specific.

    If A, B, and C are totally immiscible (that is, the heat or enthalpy of mixing is zero, with oil and water being a good example), then the evaporation processes will be uncoupled, and A, then B, then C will leave the mixture (assuming they are all present on the surface). However, the situation will be different if they form a compound, for example.

    In your thought experiment, the cylinder with the larger surface area will be more efficient in evaporation at a given temperature. If the heating process if fast compared to the evaporation time, then the same power is required in each case (because the volume of liquid is equal). However, if the heating process is slow (that is, significant evaporation occurs as the temperature increases to its final value), then the cylinder with the larger surface area will require less power.
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