Question

With each breath, a person at rest breathes in about 0.50 L of air, 20.9% of which is O2, and exhales the same volume of air containing 16.3% O2. In the lungs, oxygen diffuses into the blood, and is then transported throughout the body. Severe illness (altitude sickness) and even death can result if the amount of oxygen is too low. At sea level, atmospheric pressure is 1.00 atm, but at 3048 m (10,000 ft) it is reduced to 0.695 atm; the percentage of oxygen remains the same in both cases. Suppose that the temperature is 20 °C at both altitudes. What is the net number of oxygen molecules in each complete breath at the following altitudes?

(a) at sea level = molecules

(b) at an altitude of 3048 m = molecules

Answer 1

a) at see level

T = 20 C = 20 + 273 = 293 k

P = 1 atm = 1.013*10^5 pa

V = 20.9% of 0.5 L

= 0.209*0.5*10^-3

= 1.045*10^-4 m^3

let n is the no of moles of the O2 present.

use, P*V = n*R*T

n = P*V/(R*T)

= 1.013*10^5*1.045*10^-4/(8.314*293)

= 0.004346 moles

no of O2 molecules present, N = n*Na

= 0.004346*6.023*10^23

= 2.62*10^21 molecules

b) at an altitude of 3048 m

T = 20 C = 20 + 273 = 293 k

P = 0.695 atm = 0.695*1.013*10^5 pa

V = 20.9% of 0.5 L

= 0.209*0.5*10^-3

= 1.045*10^-4 m^3

let n is the no of moles of the O2 present.

use, P*V = n*R*T

n = P*V/(R*T)

= 0.695*1.013*10^5*1.045*10^-4/(8.314*293)

= 0.00302 moles

no of O2 molecules present, N = n*Na

= 0.00302*6.023*10^23

= 1.82*10^21 molecules