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Can anyone answer this?

  1. Mar 28, 2005 #1
    An air mass is at 90 degrees F. It contains 20 grams of vapor/m^3. What is the relative humidity of this air mass?

    What would the relative humidity of this air mass be if its temperature were increased to 100 degrees F?

    What is the dew point of this air mass?

    The temperature is now expected to drop to 50 degrees F. How much water vapor (g/m^3) do you expect the air mass to lose?
  2. jcsd
  3. Mar 28, 2005 #2
    guess not.
  4. Mar 28, 2005 #3

    Andrew Mason

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    What is the saturated vapour density of water at 90 degrees F in g/m^3? Divide 20 by that number and you will have the relative humidity. To work the density out from the vapour pressure, if you can't find it on a table, you will have to use the ideal gas equation.

    The dew point is the temperature at which 20 g/m^3 is the saturated vapour density of water. Work out what this is in terms of pressure and just go down the graph of vapour pressure vs. temperature to find this temperature.

    Find the saturated vapour density of water at 50F and subtract that from 20 g/m^3. That is the amount of mass/m^3 of water that will precipitate out.

    Last edited: Mar 28, 2005
  5. Mar 28, 2005 #4
  6. Mar 31, 2005 #5
    thanks a lot Gamma and Mason.
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