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Relative humidity changes

  1. Aug 22, 2014 #1
    1. The problem statement, all variables and given/known data
    A room with volume 2000 cubic metres has air at T=25degrees Celsius with relative humidity 80%. Density of water vapor in saturated air at 25deg Celsius is 22.8g/m^3. Temperature and pressure of room remains constant. What mass of water vapor must be removed from this air to reduce relative humidity to 50%?

    2. Relevant equations

    relative humidity/100 = water vapor pressure/saturated water pressure

    3. The attempt at a solution
    using above formula and density definition we get (18.3x2000)-(11.4x2000) = 6.9kg.
    However the textbook answer is double this? Where have I gone wrong?
    Last edited: Aug 22, 2014
  2. jcsd
  3. Aug 22, 2014 #2


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    Science Advisor

    Try that computation again.
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