Question

An adult inhales approximately 12 times per minute, taking in about 500 mL of air with each inhalation. Oxygen and carbon dioxide are exchanged in the lungs, but there is essentially no exchange of nitrogen. The exhaled air has a mole fraction of nitrogen of 0.75 and is saturated with water vapor at body temperature, 37°C. If ambient conditions Professor Christina Tang, [email protected] 2 are 25°C, 1 atm and 50% relative humidity, determine what volume of liquid water (mL) would have to be consumed over a two hour period to replace the water loss from breathing.

(a) How much water would have to be consumed on an airplane where the temperature, pressure and relative humidity are 25°C, 1atm and 10%, respectively.

Answer 1

SOLUTION:

Volume of air taken in each inhalation = 500 mL = 0.5 L

Relative humidity = 50%

At 25 oC, vapor pressure of water = 23.8 mm Hg

Relative humidity = partial pressure of water vapor / vapor pressure of liquid

Partial pressure of water vapor = Relative humidity * vapor pressure of liquid

= 23.8 * 0.5 = 11.4 mm Hg

Moles of gas exhaled is given as:

n = PV / RT

n = 1 * 0.5 / (0.0821*(37+273.15)

n = 0.0196 moles of gas/12minutes

For two hour period = 2 * 60 = 120 minutes

Moles of air required in 120 minutes = 120 * 0.0196 / 12

= 0.196 moles of air

Mole fraction of nitrogen = 0.75

Moles of nitrogen = Mole fraction * total moles

= 0.75 * 0.196

= 0.147 moles

Partial pressure of water vapor / partial pressure of nitrogen = moles of water vapor / moles of nitrogen

11.4 / (0.75 * 1) = moles of water vapor / 0.147

Moles of water vapor = 11.4 * 0.147 / 0.75

= 2.2344 moles of water

= 2.2344 * 18 gm = 40.22 gm

Density of water = 1 g / mL

Volume of water = 40.22 / 1

= 40.22 mL

When the person is in aeroplane:

Relative humidity = 10%

Partial pressure of water vapor = 23.8 * 0.1

= 2.38 mmHg

2.38 / 0.75 = moles of water vapor / 0.147

Moles of water vapor = 2.38 * 0.147 / 0.75

= 0.47 moles

= 0.47 * 18

= 8.46 gm

Volume of water = 8.46 / 1

= 8.46 mL