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Chapter 10: Problem 110

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 \(\left.10.59 .\right)\)

Short Answer

Expert verified
(a) Using Dalton's law of partial pressures: \(P_{N_{2}} = 0.748 × 0.985 \mathrm{~atm} = 0.736 \mathrm{~atm}\), \(P_{O_{2}} = 0.153 × 0.985 \mathrm{~atm} = 0.151 \mathrm{~atm}\), \(P_{CO_{2}} = 0.037 × 0.985 \mathrm{~atm} = 0.0364 \mathrm{~atm}\), and \(P_{H_{2}O} = 0.062 × 0.985 \mathrm{~atm} = 0.061 \mathrm{~atm}\). (b) Using the ideal gas law: \(n = \frac{PV}{RT} = \frac{0.0364 \mathrm{~atm} \times 0.455 \mathrm{~L}}{0.0821 \mathrm{~L·atm/mol·K} \times 310.15 \mathrm{~K}} = 5.40 \times 10^{-3} \ moles \ CO_{2}\). (c) Moles of glucose: \(\frac{5.40 \times 10^{-3} \ moles \ CO_{2}}{6} = 9.00 \times 10^{-4} \ moles \ C_{6}H_{12}O_{6}\), grams of glucose: \(9.00 \times 10^{-4} \ moles \ C_{6}H_{12}O_{6} \times 180.16 \ g/mol = 0.162 \ g \ C_{6}H_{12}O_{6}\).

Step by step solution

01

Calculate partial pressures of each gas

Using the given percentages for each gas and Dalton's law of partial pressures, we can find the partial pressure of each gas component: \(P_{total} = P_{N_{2}} + P_{O_{2}} + P_{CO_{2}} + P_{H_{2}O}\) Given, total pressure: \(P_{total} = 0.985 \mathrm{~atm}\) Calculate the partial pressure of each component by multiplying the total pressure by its percentage: \(P_{N_{2}} = 0.748 × P_{total}\) \(P_{O_{2}} = 0.153 × P_{total}\) \(P_{CO_{2}} = 0.037 × P_{total}\) \(P_{H_{2}O} = 0.062 × P_{total}\)
02

Determine the number of moles of CO2 exhaled

Using the ideal gas law, we will express the number of moles of CO2 as follows: \(PV = nRT\) where \(P\) is the pressure of the gas, \(V\) is the volume, \(n\) is the number of moles, \(R\) is the ideal gas constant and \(T\) is the temperature in Kelvin. We can use the partial pressure of CO2 calculated in Step 1 and the given volume of the exhaled gas (0.455 L) and temperature (37°C = 310.15 K). The ideal gas constant, \(R\), has a value of 0.0821 L·atm/mol·K. Now, rearrange the formula to solve for the number of moles of CO2: \(n = \frac{PV}{RT}\)
03

Calculate the grams of glucose needed to produce the CO2

Using the balanced chemical equation for glucose metabolism (C6H12O6 + 6O2 -> 6CO2 + 6H2O), we can find the amount of glucose needed to produce the moles of CO2 calculated in Step 2. 1 mole of glucose (C6H12O6) produces 6 moles of CO2: \(1 \ mole \ C_{6}H_{12}O_{6} \ → \ 6 \ moles \ CO_{2}\) Now, use the stoichiometry of the balanced chemical equation to find the moles of glucose required to produce the moles of CO2 exhaled: \[Moles \ of \ Glucose = \frac{Moles \ of \ CO_{2} \ exhaled}{6}\] Finally, convert moles of glucose to grams using the molar mass of glucose (\(C_{6}H_{12}O_{6} = 180.16 \ g/mol\)): \[Grams \ of \ glucose = Moles \ of \ glucose × Molar \ mass \ of \ glucose\]

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