Question

Problem 15.64: The effect of altitude on the lungs.

Part A

Calculate the change in air pressure you will experience if you climb a 1250 m mountain, assuming that the temperature and air density do not change over this distance and that they were 22.0 ∘C and 1.20 kg/m3respectively, at the bottom of the mountain. Pa

Part B

If you took a 0.500 L breath at the foot of the mountain and managed to hold it until you reached the top, what would be the volume of this breath when you exhaled it there? L

Answer 1

The constant R of the air is R = 287 J/kg*K.
R = Rg/M with Rg = gas constant 8.314 J/mole*K end M = molecular mass of the air 28.968 g/mole.
The equation of the ideal gas is P*v = R*T.
v is specific volume and the density ρ = 1/v
P/ρ = R*T.
ρ = 1.2 kg/m³
T = 22 C° = 295,15 K.
R = 287 J/kg*K
=> P1 = ρ*R*T = 1.2*295.15*287 = 101649.66 Pa (Pascal).
If the density (ore specific volume) and the temperature don't change the pressure change only for the altitude. The Stevino law P = ρ*g*h says that the air column over the ground is
h = P/ρ*g = 101649.66/(9,81*1,2) = 8634.86 m

P2 = ρ*g*(8634.8 - 1250) = 86845.248 Pa.

$\Delta P=P_{1}-P_{2}$

=101649.66-86845.248 =14803.752 Pa

B)
Is the Boyle law: P*V = cost.
=> P1*V1 = P2*V2
=> 101649,66*V1 = 86845.248*V2
=> V2 = 1.17 V1 with V1 = 0.500L

=0.585L