Assume that an exhaled breath of air consists of \(74.8 \% \mathrm{N}_{2}, 15.3 \% \mathrm{O}_{2}, 3.7 \% \mathrm{CO}_{2},\) and 6.2\(\%\) water vapor. (a) If the total pressure of the gases is 0.985 atm, calculate the partial pressure of each component of the mixture. (b) If the volume of the exhaled gas is 455 \(\mathrm{mL}\) and its temperature is \(37^{\circ} \mathrm{C}\) , calculate the number of moles of \(\mathrm{CO}_{2}\) exhaled. (c) How many grams of glucose \(\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right)\) would need to be metabolized to produce this quantity of \(\mathrm{CO}_{2} ?\) (The chemical reaction is the same as that for combustion of \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6} .\) See Section 3.2 and Problem 10.57 )

In this step, use the given percentages of each component and the total pressure to find the partial pressures of each gas. We will use the formula \(P_{gas} = P_{total} \cdot \%_{gas}\), where \(P_{gas}\) is the partial pressure of the gas, \(P_{total}\) is the total pressure, and \%_{gas} is the percentage of the gas in decimal form. a) For N2: - \(P_{N2} = (0.985 \, \text{atm}) \cdot 0.748 = 0.737 \, \text{atm}\) For O2: - \(P_{O2} = (0.985 \, \text{atm}) \cdot 0.153 = 0.151 \, \text{atm}\) For CO2: - \(P_{CO2} = (0.985 \, \text{atm}) \cdot 0.037 = 0.036 \, \text{atm}\) For water vapor: - \(P_{H2O} = (0.985 \, \text{atm}) \cdot 0.062 = 0.061 \, \text{atm}\) The partial pressures are: \(P_{N2} = 0.737 \, \text{atm}\), \(P_{O2} = 0.151 \, \text{atm}\), \(P_{CO2} = 0.036 \, \text{atm}\), and \(P_{H2O} = 0.061 \, \text{atm}\).

In this step, use the Ideal Gas Law, \(PV = nRT\), to find the number of moles of CO2 by rearranging the formula: \(n = \frac{PV}{RT}\). We have the partial pressure of CO2 from step 1, the volume of the exhaled gas (455 mL), and the temperature (37°C). The gas constant R is \(0.0821\, \frac{\text{L atm}}{\text{mol K}}\). First, convert the volume to L and the temperature to K: - Volume: \(455 \, \text{mL} = 0.455 \, \text{L}\) - Temperature: \(37^{\circ}\text{C} + 273.15 = 310.15 \, \text{K}\) Now, use the Ideal Gas Law to find the number of moles of CO2: - \(n_{CO2} = \frac{(0.036 \, \text{atm})(0.455 \, \text{L})}{(0.0821 \, \frac{\text{L atm}}{\text{mol K}})(310.15 \, \text{K})} = 0.00566 \, \text{mol CO2}\)

In this step, use the stoichiometry of the chemical reaction of glucose combustion to find the grams of glucose needed to produce the moles of CO2 calculated in step 2. For the combustion of glucose, the balanced chemical reaction is: \[C_6H_{12}O_6 + 6O_2 \rightarrow 6CO_2 + 6H_2O\] From the balanced chemical equation, 1 mol of glucose produces 6 mol of CO2. Calculate the moles of glucose needed to produce 0.00566 mol of CO2: - \(\text{mol glucose} = \frac{0.00566 \, \text{mol CO2}}{6} = 9.43 \times 10^{-4} \, \text{mol glucose}\) Now, convert moles of glucose to grams using the molar mass of glucose (\(180.16 \, \frac{\text{g}}{\text{mol}}\)): - \(9.43 \times 10^{-4} \, \text{mol glucose} \cdot \frac{180.16 \, \text{g}}{\text{mol}} = 0.170 \, \text{g glucose}\) Thus, to produce the given quantity of CO2, \(0.170 \, \text{g}\) of glucose would need to be metabolized.