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.