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Homework Help: Water Evaporation (Thermodynamics)

  1. May 16, 2012 #1
    1. The problem statement, all variables and given/known data

    If the air is at100% relative humidity and a water surface is at the same temperature as that of the air above it, do any of the water molecules leave their liquid state to become vapour?

    2. Relevant equations


    Clausius-Clapeyron equation:

    [itex]\frac{de_s}{dT}= \frac{1}{T} \frac{L_v}{V_v-V_w}[/itex]

    3. The attempt at a solution

    How can I use the Clausius-Clapeyron relation (or any other method) to determine whether the net evaporation is positive or negative?

    If I rewrite the equation in terms of saturation vapor pressure, I don't know the values for volumes. So how will I be able to solve it?

    There are other versions of this problem where the water surface may be warmer/colder or the relative humidity may be different. So, how can I approach such problems?
    I could not see any worked examples anywhere online. So any help is greatly appreciated.
  2. jcsd
  3. May 16, 2012 #2
    This question is not related to the Clausius-Clapeyron equation. The water at the surface is in equilibrium with the water vapor in the air, so there is no net flow of water molecules between the liquid water and the air. But molecules are continually leaving the water surface into the air, and molecules are continually leaving the air and entering the liquid surface. It is just that, at equilibrium, the rates of these two molecular flows are exactly equal.

  4. May 17, 2012 #3
    Thank you very much for the response. So, at 100% relative humidity the net evaporation is positive if water surface is warmer than the air above, and negative if it is colder, otherwise it is zero?
  5. May 17, 2012 #4
    This question is a little more complicated, and enters into the area of interphase mass transfer. The simple answer is "yes," but, under the conditions you describe, the air temperature and the water vapor pressure in the air will not be uniform in the vicinity of the interface (say on the order of 1 mm from the interface). This will be the mass transfer "boundary layer." Within the BL, the air temperature and the water vapor partial pressure will vary very rapidly, from the bulk value outside the BL to the values at the immediate interface. There may also be a temperature BL on the liquid side of the interface. At the interface itself, the temperatures of the air and water will, of course, be equal, and the relative humidity at the interface will closely approach 100%. The thicknesses of the BLs will depend on the rates of diffusion and forced- and natural convection within the two phases. I know that this may sound like mumbo jumbo, but it captures the basics of how the mass transfer occurs.

  6. May 21, 2012 #5
    I see, thank you very much for the explanation. I have one more question. So, what if the air and water surface are at the same temprature but we have an initial relative humidity of 50% (instead of being 100%)?
  7. May 22, 2012 #6
    The relative humidity is defined as the partial pressure of the water vapor in the air divided by the equilibrium vapor pressure (times 100%). If the humidity is 50%, this means that, in the bulk air phase, the water vapor partial pressure will be less than the equilibrium vapor pressure. However, exactly at the interface, the water vapor in the air will be at equilibrium with the liquid water at the interface, and the relative humidity will be 100%. Therefore, there will be a higher water vapor partial pressure in the air at the interface than in the bulk of the air. This will translate into a gradient of water vapor partial pressure (concentration), and a driving force for water vapor to diffuse away from the interface. All this transport will occur within a thin air boundary layer immediately adjacent to the interface. The net result will be a flux of water vapor away from the interface, through the boundary layer, and into the bulk air phase. We call this phenomenon evaporation.
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