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BIO A person at rest inhales 0.50 L of air with each
Chapter 18, Problem 66P(choose chapter or problem)
BIO A person at rest inhales 0.50 L of air with each breath at a pressure of 1.00 atm and a temperature of 20.0o C. The inhaled air is 21.0% oxygen. (a) How many oxygen molecules does this person inhale with each breath? (b) Suppose this person is now resting at an elevation of 2000 m but the temperature is still 20.0o C. Assuming that the oxygen percentage and volume per inhalation are the same as stated above, how many oxygen molecules does this person now inhale with each breath? (c) Given that the body still requires the same number of oxygen molecules per second as at sea level to maintain its functions, explain why some people report “shortness of breath” at high elevations.
Questions & Answers
QUESTION:
BIO A person at rest inhales 0.50 L of air with each breath at a pressure of 1.00 atm and a temperature of 20.0o C. The inhaled air is 21.0% oxygen. (a) How many oxygen molecules does this person inhale with each breath? (b) Suppose this person is now resting at an elevation of 2000 m but the temperature is still 20.0o C. Assuming that the oxygen percentage and volume per inhalation are the same as stated above, how many oxygen molecules does this person now inhale with each breath? (c) Given that the body still requires the same number of oxygen molecules per second as at sea level to maintain its functions, explain why some people report “shortness of breath” at high elevations.
ANSWER:Solution 66P Here, we shall have to use the ideal gas equation PV = nRT and the equation for change RTy in pressure with height P = P e 0 . Here, P= pressure, V = volume, R = gas constant, T= absolute temperature, n = number of moles, M = molar mass of the gas, g = acceleration due to gravity and y = altitude Given, volume V = 0.5 L = 0.5 × 10 3m = 5.0 × 10 4 m 3 5 P = 1.00 atm = 1.01 × 10 Pa = 101000 Pa 0 T = 20 C = 293 K R = 8.314 J/mol.K (a) ow, substituting these values in the ideal gas equation, PV = nRT 4 3 101000 Pa × 5.0 × 10 m = n × 8.314 × 293 K n = 0.02 moles Therefore, the number of molecu