The problem can be covered from different methods for development. I will approximate it by the proximity method. We know that the person breathes about 6 liters per minute, that is [tex]6 * 10 - 3 m ^ 3 / min[/tex] (Recall that [tex]1L = 1 * 10 - 3 m ^ 3[/tex])
Given the conditions, we have that at atmospheric pressure with a temperature of 20 ° C the air density is [tex]1.24kgm ^ 3[/tex]
Therefore, from the density ratio, the mass would be
[tex]\rho = \frac{m}{V}\rightarrow m = \rho \dot{V}[/tex]
Here,
m = mass per time unit
V = Volume per time unit
[tex]\rho[/tex] = Density
We have
[tex]m = (6*10^{-3}m^3 / min )(1.24kg/m^3 )[/tex]
[tex]m= 7.44*10^{-3} kg/ min[/tex]
Using the conversion factor from minutes to days,
[tex]m= 7.44*10^{-3} kg/ min(\frac{60min}{1hour})(\frac{24 hours}{1day })[/tex]
[tex]m = 10.7136kg/day[/tex]
Therefore he mass of air in kilograms that a person breathes in per day is 10.7136kg
