The adult blue whale has a lung capacity of 5.0×103 l. calculate the mass of air (assume an average molar mass 28.98 g/mol) contained in an adult blue whale’s lungs at 0.2 ∘c and 1.04 atm, assuming the air behaves ideally.

6.7 kg

Explanation:

V = 5000 L = 5000 x 10^-3 m^3 = 5 m^3

T = 0.2 degree C = 273.2 K

P = 1.04 atm = 1.04 x 1.01 x 10^5 Pa = 1.0504 x 10^5 Pa

R = 8.314 in SI system of units

Use the ideal gas equation. Let n be the moles of air occupied in the lungs of whale.

P V = n R T

n = P V / R T

n = (1.0504 x 10^5 x 5) / (8.314 x 273.2) = 231.22

the mass of one mole of air is 28.98 g

So, the mass of 231.22 moles of air = 231.22 x 28.98 = 6700 .88 g = 6.7 kg

attach the statements so i can answer : )