Problem 28 - Chemistry: Principles and Reactions

Question

Exhaled air contains \(74.5 \% \mathrm{~N}_{2}, 15.7 \% \mathrm{O}_{2}, 3.6 \% \mathrm{CO}_{2}\), and \(6.2 \% \mathrm{H}_{2} \mathrm{O}\) (mole percent). (a) Calculate the molar mass of exhaled air. (b) Calculate the density of exhaled air at \(37^{\circ} \mathrm{C}\) and \(757 \mathrm{~mm} \mathrm{Hg}\), and compare the value obtained with that for ordinary air (MM = \(29.0 \mathrm{~g} / \mathrm{mol}\) ).

Step-by-Step Solution

Answer

Question: Calculate the molar mass and density of exhaled air and compare it with the density of ordinary air. Answer: The molar mass of exhaled air is 29.09 g/mol. The density of exhaled air is 1.146 g/L, which is slightly higher than the density of ordinary air, which is 1.142 g/L.

Step 1: Get the molar masses of each component

We need to know the molar masses (in g/mol) of each component present in the exhaled air. They are as follows: - \(\mathrm{N}_{2}: 28.02 ~\mathrm{g/mol}\) - \(\mathrm{O}_{2}: 32.00 ~\mathrm{g/mol}\) - \(\mathrm{CO}_{2}: 44.01 ~\mathrm{g/mol}\) - \(\mathrm{H}_{2}\mathrm{O}: 18.02 ~\mathrm{g/mol}\)

Step 2: Calculate the molar mass of exhaled air

We will now use the mole percentage of each component to find the molar mass of exhaled air. The calculation is as follows: Molar mass of exhaled air = \((74.5 \% \times 28.02) + (15.7 \% \times 32.00) + (3.6 \% \times 44.01) + (6.2 \% \times 18.02)\) Molar mass of exhaled air = \((0.745 \times 28.02) + (0.157 \times 32.00) + (0.036 \times 44.01) + (0.062 \times 18.02)\) Molar mass of exhaled air = \(29.09 ~\mathrm{g/mol}\)

Step 3: Use the ideal gas law to find the density of exhaled air

We will apply the ideal gas law represented as \(PV=nRT\), where: - P is the pressure in atm - V is the volume in liters - n is the number of moles - R is the ideal gas constant (\(0.0821 ~\mathrm{L \cdot atm} / (\mathrm{K} \cdot\mathrm{mol})\)) - T is the temperature in Kelvin First, we need to convert the temperature to Kelvin and the pressure to atm: - Temperature: \((37 + 273.15) = 310.15 ~\mathrm{K}\) - Pressure: \(757~\mathrm{mmHg} \times \dfrac{1~\mathrm{atm}}{760 ~\mathrm{mmHg}} = 0.996 ~\mathrm{atm}\) Now, let's find the density. The density of a gas is given by the formula \(\rho = \dfrac{n \cdot MM}{V}\). Replacing n from the ideal gas law and modifying, we get: \(\rho = \dfrac{P \cdot MM}{R \cdot T}\) Now, we can plug the values and calculate the density of exhaled air: \(\rho = \dfrac{0.996 ~\mathrm{atm} \times 29.09 ~\mathrm{g/mol}}{0.0821 ~\mathrm{L \cdot atm / (K \cdot mol)} \times 310.15 ~\mathrm{K}} = 1.146 ~\mathrm{g/L}\)

Step 4: Compare the density of exhaled air with ordinary air

We have calculated the density of exhaled air and now we will compare it with the density of ordinary air (MM = \(29.0 ~\mathrm{g/mol}\)). Using the same formula as in Step 3, we can find the density of ordinary air: \(\rho_{ordinary} = \dfrac{0.996 ~\mathrm{atm} \times 29.0 ~\mathrm{g/mol}}{0.0821 ~\mathrm{L \cdot atm / (K \cdot mol)} \times 310.15 ~\mathrm{K}} = 1.142 ~\mathrm{g/L}\) Thus, the density of exhaled air (\(1.146 ~\mathrm{g/L}\)) is slightly higher than the density of ordinary air (\(1.142 ~\mathrm{g/L}\)).

Key Concepts

Ideal Gas Law

The Ideal Gas Law is an essential equation in chemistry and physics that relates the pressure, volume, temperature, and number of moles of a gas. Its formula is given by \[ PV = nRT \]

• \( P \) stands for pressure, measured in atmospheres (atm).
• \( V \) is the volume in liters.
• \( n \) denotes the number of moles of gas.
• \( R \) is the ideal gas constant, which is approximately \( 0.0821 \, \text{L} \, \text{atm} / (\text{K} \, \text{mol}) \).
• \( T \) is the temperature in Kelvin.

To use this equation, make sure to convert all your measurements to the correct units: temperature to Kelvin and pressure to atmospheres. It's a versatile tool that helps to calculate the unknown parameters if other three are known, making it particularly useful for finding gas densities.

Gas Density

Gas density is a measure of how much mass a gas contains in a given volume. It is usually expressed in grams per liter (g/L). The density of a gas can be determined using the Ideal Gas Law by modifying it slightly to obtain the formula: \[ \rho = \dfrac{P \cdot MM}{R \cdot T} \] where \( \rho \) denotes density, \( MM \) is the molar mass of the gas, \( P \) is the pressure in atm, \( R \) is the ideal gas constant, and \( T \) is the temperature in Kelvin. This formula helps in determining how dense a particular gas is under specified conditions of temperature and pressure, allowing for comparisons between different gases, such as exhaled air versus ordinary air.

Molecular Composition

The molecular composition of a gas mixture indicates the types and mole percentages of molecules that make up the gas. In the case of exhaled air, it includes:

• Nitrogen \((\mathrm{N}_2)\): 74.5%
• Oxygen \((\mathrm{O}_2)\): 15.7%
• Carbon Dioxide \((\mathrm{CO}_2)\): 3.6%
• Water Vapor \((\mathrm{H}_2\mathrm{O})\): 6.2%

Knowing the molecular composition is essential as it allows the calculation of the molar mass of the mixture. To find the molar mass of this composition, multiply the molar mass of each component by its mole fraction (the percentage divided by 100) and sum the results. This gives insight into the overall characteristics of the exhaled air.

Exhaled Air Composition

Exhaled air is a complex mixture of gases. It has different properties compared to the inhaled air due to the various physiological processes inside the human body. The most prevalent components include nitrogen, oxygen, carbon dioxide, and water vapor, each contributing to the unique composition of the exhaled breath. The changes in composition from inhaled to exhaled air primarily reflect the exchange of oxygen and carbon dioxide in the lungs during respiration. This exchange is crucial for maintaining oxygen levels in the bloodstream and expelling carbon dioxide, a byproduct of metabolism. By understanding the composition of exhaled air, we gain insights into our body's efficiency in gas exchange and its overall metabolic state.