Expired air is saturated with water vapor and generally contains more of it than does inspired air. There is usually a net loss of water from the body due to breathing. Water vapor pressure in the at saturation in the lungs (at 37°C) is 47 mmHg, while in expired air at 33°C it is 37.7 mmHg. Consider the case of someone who is breathing dry air in and moist air out at a rate of 15,000 L/day. This corresponds to a mean respiratory minute volume of 10.4 L/min, and to a fairly typical daily metabolic rate. The atmospheric pressure is 760 mmHg (i.e., standard pressure). Since 1 mol (18 g) of water vapor occupies 22.4 L at standard temperature and pressure, 1 L has a mass of 18/22.4 = 0.80 g. Ignoring air volume corrections, what is the daily rate of water loss (in g) if the air is inspired completely dry and expired saturated with water vapor at (a) 37°C and (b) 33°C? Note that solving this problem involves several steps. The professor and TA will help with this if you have difficulty with it. Recall from chapter 21 that you can determine the fractional concentration of a gas if you know its partial pressure and the total pressure (Px = Fx × Ptotal). So you can determine the fractional concentration of water vapor in exhaled air at each temperature. Next, because you now the total volume of air exhaled in a day, you can determine the amount of water vapor exhaled per day. Finally, because you know the mass of water vapor per liter, you can determine the mass of exhaled water vapor. As always, be careful that the units work out properly.
Expired air is saturated with water vapor and generally contains more of it than does inspired air. There is usually a net loss of water from the body due to breathing. Water vapor pressure in the at saturation in the lungs (at 37°C) is 47 mmHg, while in expired air at 33°C it is 37.7 mmHg. Consider the case of someone who is breathing dry air in and moist air out at a rate of 15,000 L/day. This corresponds to a mean respiratory minute volume of 10.4 L/min, and to a fairly typical daily metabolic rate. The atmospheric pressure is 760 mmHg (i.e., standard pressure). Since 1 mol (18 g) of water vapor occupies 22.4 L at standard temperature and pressure, 1 L has a mass of 18/22.4 = 0.80 g. Ignoring air volume corrections, what is the daily rate of water loss (in g) if the air is inspired completely dry and expired saturated with water vapor at (a) 37°C and (b) 33°C? Note that solving this problem involves several steps. The professor and TA will help with this if you have difficulty with it. Recall from chapter 21 that you can determine the fractional concentration of a gas if you know its partial pressure and the total pressure (Px = Fx × Ptotal). So you can determine the fractional concentration of water vapor in exhaled air at each temperature. Next, because you now the total volume of air exhaled in a day, you can determine the amount of water vapor exhaled per day. Finally, because you know the mass of water vapor per liter, you can determine the mass of exhaled water vapor. As always, be careful that the units work out properly.
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Question
Transcribed Image Text:The next three questions go together.
21. Expired air is saturated with water vapor and generally contains more of it than does inspired air.
There is usually a net loss of water from the body due to breathing. Water vapor pressure in the
at saturation in the lungs (at 37°C) is 47 mmHg, while in expired air at 33°℃ it is 37.7 mmHg.
Consider the case of someone who is breathing dry air in and moist air out at a rate of 15,000
L/day. This corresponds to a mean respiratory minute volume of 10.4 L/min, and to a fairly
typical daily metabolic rate. The atmospheric pressure is 760 mmHg (i.e., standard pressure).
Since 1 mol (18 g) of water vapor occupies 22.4 L at standard temperature and pressure, 1 L has
a mass of 18 / 22.4 = 0.80 g. Ignoring air volume corrections, what is the daily rate of water loss
(in g) if the air is inspired completely dry and expired saturated with water vapor at (a) 37°℃ and
(b) 33°C? Note that solving this problem involves several steps. The professor and TA will help
with this if you have difficulty with it.
Recall from chapter 21 that you can determine the fractional concentration of a gas if you
know its partial pressure and the total pressure (Px = Fx × Ptotal). So you can determine the
fractional concentration of water vapor in exhaled air at each temperature. Next, because you
now the total volume of air exhaled in a day, you can determine the amount of water vapor
exhaled per day. Finally, because you know the mass of water vapor per liter, you can determine
the mass of exhaled water vapor. As always, be careful that the units work out properly.
22. From these two answers, how much water is saved daily, when the inspired air is dry, by
breathing out air at 33°C rather than at 37°C?
23. In fact, water vapor pressure of atmospheric air is not generally zero, as in the two previous
calculations, and it varies considerably. For the next calculation, a representative value of 13
mmHg can be used, which corresponds to saturation near 15°C, or under-saturation at higher
temperatures. What is the daily rate of water loss if the inspired air has a water vapor pressure of
13 mmHg and the expired air, saturated at 33°C, has a water vapor pressure of 37.7 mmHg?
Because air is rarely completely dry, the answer to Question 21 is not completely realistic.
Hint: If air has some water vapor in it, then the gradient of water vapor is lessened between
exhaled air and ambient air (how it is lessened is really all that you have to figure out in order to
answer this question; the rest of the calculation is the same as for Q21) and water loss will be
reduced.
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