Answer
6464.9 grams of air
Work Step by Step
We have the Volume of the lungs along with the pressure and temperature. First we convert the temperature from 0C to Kelvin by adding 273 so we have 273K. Then we can use ideal gas law:
PV = nRT or in this case
n = PV/RT = (1atm) * (5.0 x 10^3L) / (0.0821 * 273K) =223.08 moles
Then since the molar mass is 28.98 g/mol we have
223.08 moles * (28.98 g/ 1mol) = 6464.9 grams of air
