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
(Recall that
)
Given the conditions, we have that at atmospheric pressure with a temperature of 20 ° C the air density is

Therefore, from the density ratio, the mass would be

Here,
m = mass per time unit
V = Volume per time unit
= Density
We have


Using the conversion factor from minutes to days,


Therefore he mass of air in kilograms that a person breathes in per day is 10.7136kg