Question

This question is from CBE 2200 Process Fundamentals - Mass and Energy Balances.

. An adult takes roughly 12 breaths each minute, inhaling approximately 500 mL with each breath. Oxygen and carbon dioxide are exchanged in the lungs. The amount of nitrogen exhaled equals the amount inhaled, and the molt traction of nitrogen in the exhaled air is 0.75. The exhaled air is saturated with water vapor at body temperature, 37 degree C. Estimate the increase in the rate of water loss (g/day) when a person breathing air at 23degree C and a relative humidity of 50% enters an airplane in which the temperature is also 23 degree C but the relative humidity is M.

Any help would be greatly appreciated, thanks!

Answer 1

A relative humidity of 50% means partial pressure of water vapour = 0.5*equilibrium water vapour pressure and at 10%, it is 0.1*equilibrium vapor pressure.

V = 500mL (since the exhaled air is saturated with water vapour, the difference in humidity is enough to calculated the water loss), T = 296C. From litereature, equilibrium water vapor pressure at 23 is taken and is equal to 2809.3 Pa.

Therefore, n = PV/RT = (0.5-0.1)*2809.3*0.0005/(8.314*296) = 2.283x10^-4 moles of water.

mass of water = n*18 = 4.11 mg

Loss of water per day = 4.11*12*60*24 = 71013.41 mg or 71 g of water.