Question

when breathing slowly a person inhales air at room temperature and exhales it at body temperature. We may assume that the lungs maintain ordinary atmospheric pressure as they expand. If a resting person inhales 0.6L of air at 24 degrees what volume does the air occupy when it is exhaled at 37 degrees and what mass is that? ans 0.626L and 7.74 * 10^-4 kg. Given density of air to be 1.29 kg/m3.

Answer 1

Use PV=nRT
where
P=pressure
V=volume
T=temperature in Kelvin (Celsius+273.15°)
n=number of moles (quantity of substance)
R=gas constant

In the given case, the quantity of air remain the same, so n is a constant. The pressure is assumed constant, so the equation is reduced to
V1/T1=V2/T2
where 1 and 2 represent initial and final states.
V1=0.6L
T1=273.15+24°
T2=273.15+37°
V2=(V1/T1)*T2
=0.6L*(310.15/297.15)

At 0°C, air weighs 1.29 kg/m².
So the volume of 0.6L at 24° has to be transformed to the volume at 0° for the calculation of mass.

Also,
1.29 kg-m^-3
=1.29 g-l^-1


Volume at 0°
=(V1/T1)*T2
=0.6L*(273.15/297.15)
=0.552L
Mass of air = 0.552*1.29 g
=0.712 g

Note: the given answer (7.74*10^-4 kg) for the mass assumes that air weighs 1.29 kg/m^3 at 24°C.