Assume that an exhaled breath of air consists of 74.8% N2, 15.3% O2, 3.7% CO2, and 6.2% water vapour.
a) If the total pressure of the gases is 0.980 atm, calculate the partial pressure of each component of the mixture.
b) If the volume of the exhaled gas is 455 mL and its temperature is 37°C, calculate the number of moles of CO2 exhaled.
c) How many grams of glucose (C6H12O6) would need to be metabolized to produce this quantity
of CO2? (The quantity you calculated in b)). Hint: The chemical reaction for the metabol-ization process is the same as that for the combustion of C6H12O6.

a) By the Dalton's Law, in a gas mixture, the total pressure is the sum of the partial pressures, and the partial pressure is the molar fraction of the gas multiplied by the total pressure.

PN₂ = 0.748*0.980 =0.733 atm

PO₂ = 0.153*0.980 = 0.150 atm

PCO₂ = 0.037*0.980 = 0.036 atm

Pwater = 0.062*0.980 = 0.061 atm

b) The number of moles of CO₂ can be calculated by the ideal gas law:

PV = nRT, where P is the pressure, V is the volume (0.455 L), n is the number of moles, R is the gas constant (0.082 atm.L/mol.K), and T is the temperature (37°C + 273 = 310 K).

0.036*0.455 = 0.082*310*n

25.42n = 0.01638

n = 6.44x10⁻⁴ mol

c) For the combustion reaction of glucose:

C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O

So, the stoichiometry is:

1 mol of glucose ------- 6 moles of CO₂

x ------- 6.44x10⁻⁴ mol of CO₂

By a simple direct three rule:

6x = 6.44x10⁻⁴

x = 1.073x10⁻⁴mol of glucose

Glucose has a molar mass equal to 180 g/mol, and its mass is the molar mass multiplied by the number of moles:

m = 180x1.073x10⁻⁴

m = 0.02 g