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Relative Humidity and Vaporization

  1. Oct 3, 2009 #1
    1. The problem statement, all variables and given/known data

    If a person breathes 10 liters per minute of air at 68 degrees F and 50% relative humidity, how much water per day must the internal membranes supply to saturate the air at 98.6 degrees F. (assume all the moisture is exhaled) If each gram of water extracts 580 calories as it is vaporized, how much daily heat loss in kilocalories (food calories) does this represent? (Saturation vapor pressure at 20 degrees C is 16.3 g/m cubed and at 37 degrees C is 44.0 g/m (grams per cubic meter)


    2. Relevant equations

    What I have done....

    l0 liters / min = 600 l / hr = 14,400 l /day

    68 F = 20 C & 98.6 F = 37 C

    580 cal / g @ 20 C = 25520 cal/m or 25.520 kcal / m

    580 cal / g @ 37 C = 10034 cal / m or 10.034 kcal / m

    *water has about 1000 grams in a liter*

    Inhale 50% of 17.3 = 8.65 g/m

    Exhale 100% of 44 g/m

    Relative Humidity = content of moisture / capacity

    3. The attempt at a solution

    I converted most of the equations above so I can use the right units without converting later on.

    I know the volume needed to so call humidify the air is 44 - 8.65 since you half the inhaled air which results in 35.35.

    I think I use molecular mass * partial pressure ( not sure what it is ) / ( R * Temp in Kelvins )

    I find R with

    PV / nT = R

    Do I only have 1 atm of pressure and 1 mol?

    Any guidance is great - thanks.
     
  2. jcsd
  3. Oct 3, 2009 #2
    The saturation var pressure @ 20 degrees is 16.3 and not 17.3.
    the rest of your calculations are OK.

    Because you heat the air it expands. I think it expands after saturating it, so you
    don't have to calculate the volume increase.
     
  4. Oct 4, 2009 #3
    The problem gives it to be 17.3 g / m

    =/

    But interesting idea for the Volume thanks.
     
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