Question

Determine the percent composition of air in the lungs from the following composition in partial pressures: PN2=565mmHg, PO2=108mmHg, PCO2=37mmHg, PH2O=50mmHg; all at 37∘C and 1atm pressure.

A % N2

B % O2

C %CO2

D % H2O

Answer 1

PN2=565mmHg, PO2=108mmHg, PCO2=37mmHg, PH2O=50mmHg; all at 37∘C and 1atm pressure. Since total pressure is 760 torr (sum of partial pressure of all components).

partial pressure of N2 in atm = 565/760 = 0.743 atm

n = PV/RT

we will use the literature value of volume of lungs which is 6 L

n = 0.743 atm x 6L/ 0.08205 L atm K⁻¹ mol⁻¹ x 310

n = 4.458/25.435 = 0.175 moles that is 0.175 x 28 = 4.91 g (28 is MW of N₂)

partial pressure of O2 in atm = 108/760 = 0.142 atm

n = 0.142 atm x 6L/ 0.08205 L atm K⁻¹ mol⁻¹ x 310

n = 0.852/25.435 = 0.0335 moles that is 0.0335 x 32 = 1.072 g (MW of O₂ is 32)

partial pressure of CO₂ in atm = 37/760 = 0.048 atm

n = 0.048 atm x 6L/ 0.08205 L atm K⁻¹ mol⁻¹ x 310

n = 0.292/25.435 = 0.0115 moles that is 0.0115 x 44 = 0.505 g (44 is MW of CO₂)

partial pressure of H₂O in atm = 50/760 = 0.0657 atm

n = 0.0657 atm x 6L/ 0.08205 L atm K⁻¹ mol⁻¹ x 310

n = 0.395/25.435 = 0.0155 moles that is 0.0155 x 18 = 0.279 g (18 is MW of H₂O)

%N2 = wt of N2/total weight = [4.91/(4.91 +1.072 +0.505+0.279)] x 100 = 72.56%

%O2 = wt of O2/total weight = [1.072/(4.91 +1.072 +0.505+0.279)] x 100 = 15.84%

%CO₂ = wt of CO₂/total weight = [0.505/(4.91 +1.072 +0.505+0.279)] x 100 = 7.46%

%H₂O =wt of H₂O/total weight = [0.279/(4.91 +1.072 +0.505+0.279)] x 100 = 4.12%