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Chapter 10: Problem 107

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 \(99.8 \mathrm{kPa}\), 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 \(10.57 .\) )

Short Answer

Expert verified
The partial pressures of the gas components in the exhaled breath are 74.7 kPa for N鈧�, 15.3 kPa for O鈧�, 3.69 kPa for CO鈧�, and 6.19 kPa for H鈧侽. The number of moles of CO鈧� exhaled is 0.00217 moles, and approximately 0.0652 grams of glucose need to be metabolized to produce this amount of CO鈧�.

Step by step solution

01

Calculate Partial Pressures

We are given the percentage composition of the exhaled breath of air and its total pressure (99.8 kPa). We will use this information to calculate the partial pressure of each gas component. The formula for partial pressure is: Partial Pressure = (% of gas component) 脳 (Total Pressure) So, for each gas component, we will calculate the partial pressure as follows: 1) N鈧�: Partial Pressure (N鈧�) = \(0.748 \times 99.8 \mathrm{kPa} = 74.7 \mathrm{kPa}\) 2) O鈧�: Partial Pressure (O鈧�) = \(0.153 \times 99.8 \mathrm{kPa} = 15.3 \mathrm{kPa}\) 3) CO鈧�: Partial Pressure (CO鈧�) = \(0.037 \times 99.8 \mathrm{kPa} = 3.69 \mathrm{kPa}\) 4) H鈧侽: Partial Pressure (H鈧侽) = \(0.062 \times 99.8 \mathrm{kPa} = 6.19 \mathrm{kPa}\)
02

Calculate the Number of Moles of CO鈧� Exhaled

We are given the volume of the exhaled gas (455 mL) and its temperature (37掳C). We can use this information along with the partial pressure of CO鈧� (calculated in Step 1) to find the number of moles of CO鈧� exhaled by using the ideal gas law: PV = nRT Where: P = partial pressure of CO鈧� (3.69 kPa) V = volume (0.455 L) n = number of moles R = gas constant (8.314 J mol鈦宦筀鈦宦� or 8.314 kPa L mol鈦宦筀鈦宦�) T = temperature in Kelvin (37掳C + 273.15 = 310.15 K) Rearranging the formula to find the number of moles (n): n = PV / RT n = \((3.69 \mathrm{kPa})(0.455 \mathrm{L})/((8.314 \mathrm{kPa\:L\:mol}^{-1} \mathrm{K}^{-1})(310.15\: \mathrm{K}))=0.00217\: \mathrm{mol}\) The number of moles of CO鈧� exhaled is 0.00217 moles.
03

Calculate the Amount of Glucose Metabolized

To find out how much glucose is metabolized to produce the calculated amount of CO鈧�, we will use the stoichiometry of the chemical reaction: C鈧咹鈧佲倐O鈧� (glucose) + 6 O鈧� 鈫� 6 CO鈧� + 6 H鈧侽 From the balanced equation, we can see that 1 mole of glucose produces 6 moles of CO鈧�. So, we can calculate the moles of glucose needed to produce 0.00217 moles of CO鈧�: Moles of glucose = \((0.00217 \mathrm{mol\: CO_{2}})(1 \mathrm{mol\: C_{6} H_{12} O_{6}}/6\: \mathrm{mol\: CO_{2}})= 3.62 \times 10^{-4}\: \mathrm{mol\: C_{6} H_{12} O_{6}}\) Next, we will convert the moles of glucose to grams using the molar mass of glucose, which is 180.16 g/mol: Mass of glucose = \((3.62 \times 10^{-4}\: \mathrm{mol\: C_{6} H_{12} O_{6}})(180.16 \mathrm{g/mol})=0.0652\: \mathrm{g}\) Therefore, approximately 0.0652 grams of glucose needs to be metabolized to produce the calculated amount of CO鈧� exhaled.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Ideal Gas Law
The Ideal Gas Law is a handy tool when dealing with gases, linking pressure, volume, temperature, and moles together in a neat equation: \(PV = nRT\). This equation helps to understand the behavior of gases under different conditions.
  • **P** stands for pressure, usually measured in Pascal or kPa (1 kPa = 1,000 Pascal).
  • **V** is the volume of the gas, often given in liters (L) or milliliters (mL).
  • **n** represents the number of moles of gas.
  • **R** is the universal gas constant, which is 8.314 J/mol路K or 8.314 kPa路L/mol路K.
  • **T** is temperature, measured in Kelvin (K). Convert degrees Celsius to Kelvin by adding 273.15.
The law is perfect for calculations involving the number of moles of gas, as seen in our example, where it helped determine moles of CO鈧� based on volume, pressure, and temperature. Keep in mind that this applies to ideal conditions, meaning no interactions between gas molecules, which is a useful approximation for many real-life gases.
Stoichiometry
Stoichiometry is the branch of chemistry that quantifies the relationships between the reactants and products in chemical reactions. It is crucial for calculating how much of each substance is involved, like in our example with glucose and CO鈧�.In our reaction:- **Glucose** (C鈧咹鈧佲倐O鈧�) reacts with oxygen to produce carbon dioxide and water.The balanced chemical equation is:\[\text{C}_6\text{H}_{12}\text{O}_6 + 6\text{O}_2 \rightarrow 6\text{CO}_2 + 6\text{H}_2\text{O}\]From this equation:
  • 1 mole of glucose produces 6 moles of CO鈧�.
This ratio is crucial for our calculations. By knowing the moles of one substance, we can deduce the moles of others involved through these stoichiometric coefficients. This allows us to determine how much glucose is necessary to produce the exhaled CO鈧�.
Mole Calculation
Calculating moles is a common task in chemistry that helps connect physical masses and molecules. Moles hint at the number of entities, like atoms or molecules, using Avogadro's number as a bridge between the microscopic and macroscopic levels.In our example, we utilize the Ideal Gas Law to compute the number of moles of CO鈧� exhaled. Given:
  • Volume: 455 mL (which is 0.455 L after conversion)
  • Partial Pressure: 3.69 kPa (specific to CO鈧�)
  • Temperature: 310.15 K (converted from 37掳C)
We rearrange the Ideal Gas Law equation \(PV = nRT\) to isolate **n**:\[n = \frac{PV}{RT}\]Substituting in our numbers:\[n = \frac{(3.69\text{ kPa})(0.455\text{ L})}{(8.314\text{ kPa L/mol K})(310.15\text{ K})} = 0.00217\text{ moles}\]Thus, we find that 0.00217 moles of CO鈧� are exhaled, highlighting the power of the mole concept in connecting theory to practical measurements.
Gas Composition
Gas composition concerns the makeup of a gas mixture, giving us the fraction or percentage of each gas in the mix. This is key for calculating partial pressures, which are the pressures each gas would exert if it alone occupied the entire volume.In our case, the exhaled breath is composed of:
  • 74.8% nitrogen (N鈧�)
  • 15.3% oxygen (O鈧�)
  • 3.7% carbon dioxide (CO鈧�)
  • 6.2% water vapor (H鈧侽)
To find the partial pressure for each gas, we multiply its percentage by the total pressure (99.8 kPa). For example:- Partial pressure of CO鈧�: \(0.037 \times 99.8\text{ kPa} = 3.69\text{ kPa}\)This technique helps in understanding how each gas behaves independently in the mix.
Thus, knowing the gas composition allows us to dissect the behavior of complex mixtures and calculate specifics such as the partial pressures or contributions of each gas to a mixture's properties.

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Most popular questions from this chapter

A piece of dry ice (solid carbon dioxide) with a mass of \(20.0 \mathrm{~g}\) is placed in a 25.0-L vessel that already contains air at \(50.66 \mathrm{kPa}\) and \(25^{\circ} \mathrm{C}\). After the carbon dioxide has totally sublimed, what is the partial pressure of the resultant \(\mathrm{CO}_{2}\) gas, and the total pressure in the container at \(25^{\circ} \mathrm{C} ?\)

Suppose you have a fixed amount of an ideal gas at a constant volume. If the pressure of the gas is doubled while the volume is held constant, what happens to its temperature? [Section 10.4\(]\)

A quantity of \(\mathrm{N}_{2}\) gas originally held at \(531.96 \mathrm{kPa}\) pressure in a 1.00 - \(\mathrm{L}\) container at \(26^{\circ} \mathrm{C}\) is transferred to a \(12.5-\mathrm{L}\) container at \(20^{\circ} \mathrm{C}\). A quantity of \(\mathrm{O}_{2}\) gas originally at \(531.96 \mathrm{kPa}\) and \(26^{\circ} \mathrm{C}\) in a \(5.00-\mathrm{L}\) container is transferred to this same container. What is the total pressure in the new container?

In the Dumas-bulb technique for determining the molar mass of an unknown liquid, you vaporize the sample of a liquid that boils below \(100^{\circ} \mathrm{C}\) in a boiling-water bath and determine the mass of vapor required to fill the bulb. From the following data, calculate the molar mass of the unknown liquid: mass of unknown vapor, \(1.012 \mathrm{~g}\); volume of bulb, \(354 \mathrm{~cm}^{3}\); pressure, \(98.93 \mathrm{kPa} ;\) temperature, \(99{ }^{\circ} \mathrm{C}\).

In the contact process, sulfur dioxide and oxygen gas react to form sulfur trioxide as follows: $$2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{SO}_{3}(g)$$ At a certain temperature and pressure, \(50 \mathrm{~L}\) of \(\mathrm{SO}_{2}\) reacts with \(25 \mathrm{~L}\) of \(\mathrm{O}_{2}\). If all the \(\mathrm{SO}_{2}\) and \(\mathrm{O}_{2}\) are consumed, what volume of \(\mathrm{SO}_{3}\), at the same temperature and pressure, will be produced?
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