1))))A person’s lungs might hold 6.0 x 10-3 m3 of air at body temperature (37 0C) and atmospheric pressure (101 kPa). Given that air is 21% oxygen (O2), find the number of O2 molecules in the lungs
ans: 2.97 x 1022
2))
If the person takes a particularly deep breath, so that the lungs hold a total of 1.5 x 1023 molecules, what is the new volume of the lungs?
[Answer: 6.35 x 10-3 m3 ]
1)V = 6*10^-3 m^3, T =37 oC =37+273 = 310 K, P =101 kPa
From ideal ga equation
PV = nRT
101*1000*6*10^-3 = n*8.314*310
n = 0.235 moles
only 21% oxygen
n = N/Na
N = 0.21*0.235*6.023*10^23
N = 2.97*10^22
(2) N = 1.5*10^23
PV = (N/Na)RT
101*1000*V = (1.5*10^23/(6.023*10^23))*8.314*310
V = 6.35*10^-3 m^3
1))))A person’s lungs might hold 6.0 x 10-3 m3 of air at body temperature (37 0C)...
A young male adult takes in about 5.80 x 10-4 m3 of fresh air during a normal breath. Fresh air contains approximately 21% oxygen. Assuming that the pressure in the lungs is 1.03 x 105 Pa and air is an ideal gas at a temperature of 310 K, find the number of oxygen molecules in a normal breath.