If a woman breathes in air at 22.0 C with a total barometric pressure of 0.97 atm and warms to her core temp. of 37.6 C, what is the partial pressure of the 3 components of air in her lungs?

Question

Air is approximately 78% nitrogen, 21% oxygen, and 1% argon (note that unless stated explicitly, gas concentrations are always expressed in terms of partial pressures (or mole fractions), so a gas "concentration" of 78% implies 78% of the total pressure is due to that gas).

That average volume of breath a healthy adult woman takes is 400. mL

Answer 1

At 22.0ºC (295K), and 0.97 atm, find the moles of each gas, so that you can then find the mole fraction of each gas. Use the ideal gas law (PV = nRT) to do this

For N₂: P = 0.97 atm x 0.78 = 0.76 atm. n = (0.76)(0.400L) / (0.0821)(295) = 0.013 mols

For O₂: P = 0.97 atm x 0.21 = 0.20 atm. n = (0.2)(0.400) / (0.0821)(295) = 0.0033 mols

For Ar: P = 0.97 atm x 0.01 = 0.0097 atm. n = (0.0097)(0.400) / (0.0821)(295) = 0.00016 mols

TOTAL moles of gas = 0.013 + 0.0033 + 0.00016 = 0.016 mols

Mol fraction N₂ = 0.013 mol / 0.016 mol = 0.81

Mol fraction O₂ = 0.0033 mol / 0.016 mol = 0.21

Mol fraction Ar = 0.00016 mol / 0.016 mol = 0.01

Total pressure @ 37.6ºC (310.8K) = P = nRT/V = (0.016 mol)(0.0821 Latm/Kmol)(310.8K) / 0.400 L = 1.02 atm

Partial pressure of N₂ = 0.81 x 1.02 atm = 0.83 atm

Partial pressure of O₂ = 0.21 x 1.02 atm = 0.21 atm

Partial pressure of Ar = 0.01 x 1.02 atm = 0.010 atm