Problem 59 - Chemistry: The Central Science in SI Units, Global Edition

Question

During a person's typical breathing cycle, the \(\mathrm{CO}_{2}\) concentration in the expired air rises to a peak of \(4.6 \%\) by volume. (a) Calculate the partial pressure of the \(\mathrm{CO}_{2}\) in the expired air at its peak, assuming \(101.3 \mathrm{kPa}\) pressure and a body temperature of \(37^{\circ} \mathrm{C}\). (b) What is the molarity of the \(\mathrm{CO}_{2}\) in the expired air at its peak, assuming a body temperature of \(37^{\circ} \mathrm{C} ?\)

Step-by-Step Solution

Answer

The partial pressure of \(\mathrm{CO}_{2}\) in the expired air at its peak is 4.6598 kPa, and the molarity of the \(\mathrm{CO}_{2}\) in the expired air at its peak is 0.0182 M.

Step 1: Part (a) - Calculation of Partial Pressure

First, let's find the partial pressure of the \(\mathrm{CO}_{2}\) in the expired air at its peak. The mole fraction of the \(\mathrm{CO}_{2}\) in the expired air is 4.6% by volume, which can be converted to a ratio: Mole Fraction of \(\mathrm{CO}_{2} = \frac{4.6}{100} = 0.046\) Since the total pressure is given as 101.3 kPa, we can now find the partial pressure of \(\mathrm{CO}_{2}\) using the formula: Partial Pressure of \(\mathrm{CO}_{2} = 0.046 \times 101.3 \, \mathrm{kPa}\) Partial Pressure of \(\mathrm{CO}_{2} = 4.6598 \, \mathrm{kPa}\)

Step 2: Part (b) - Calculation of Molarity

Now let's find the molarity of the \(\mathrm{CO}_{2}\) in the expired air at its peak. To do this, we will first convert the given temperature of \(37^{\circ} \mathrm{C}\) to Kelvin: Temperature in \(K = 37^{\circ} \mathrm{C} + 273.15 \,K\) Temperature in \(K = 310.15 \,K\) Assuming that \(\mathrm{CO}_{2}\) behaves as an ideal gas, we can use the Ideal Gas Law to find the molarity given partial pressure and temperature as: \(PV = nRT\) Where \(P\) is the partial pressure, \(V\) is the volume, n is the amount of substance in moles, \(R\) is the ideal gas constant (8.314 J/mol⋅K), and \(T\) is the temperature in Kelvin. We are interested in molarity, which is given by: Molarity \(= \frac{n}{V}\) Rearranging the Ideal Gas Law: \(\frac{n}{V} = \frac{P}{RT}\) By plugging the values of the partial pressure of \(\mathrm{CO}_{2}\) and the temperature in Kelvin, we get: Molarity \(= \frac{4.6598 \, \mathrm{kPa}}{(8.314 \times 10^{-3} \, \frac{\mathrm{kPa}\cdot\mathrm{L}}{\mathrm{mol}\cdot\mathrm{K}} )(310.15 \, K)}\) Molarity \(= 0.0182 \, \frac{\mathrm{mol}}{\mathrm{L}}\) So the molarity of the \(\mathrm{CO}_{2}\) in the expired air at its peak is 0.0182 M.

Key Concepts

Partial Pressure

Partial pressure is crucial in understanding gases within a mixture, such as air. In our example, we're interested in the partial pressure of carbon dioxide (CO₂) in expired air. This is calculated using the mole fraction and total pressure. The mole fraction is essentially the proportion of a specific gas present in a mixture.
To find the partial pressure of CO₂, you multiply its mole fraction by the total atmospheric pressure. The mole fraction of CO₂ here is 0.046, or 4.6%, hence:

• Partial Pressure of CO₂ = Mole Fraction × Total Pressure
• Partial Pressure of CO₂ = 0.046 × 101.3 kPa
• Partial Pressure of CO₂ ≈ 4.66 kPa

This tells us that CO₂ contributes a pressure of about 4.66 kPa to the total pressure in breathed-out air.

Ideal Gas Law

The Ideal Gas Law is a fundamental relation between the pressure, volume, temperature, and amount (moles) of gas. It is expressed as:\[ PV = nRT \]where:

• \( P \): Pressure in kilopascals (kPa)
• \( V \): Volume in liters (L)
• \( n \): Number of moles
• \( R \): Ideal gas constant (8.314 J/(mol⋅K) or 0.08314 L⋅bar/(mol⋅K))
• \( T \): Temperature in Kelvin (K)

To find the molarity of CO₂, which is moles per volume (mol/L), we rearrange this equation as:\[ \frac{n}{V} = \frac{P}{RT} \]Using our calculation from before, we plug in the partial pressure and temperature:\[\frac{4.66 \, \mathrm{kPa}}{(8.314 \, \times 10^{-3} \, \mathrm{kPa}\cdot\mathrm{L}/\mathrm{mol}\cdot\mathrm{K})(310.15 \, \mathrm{K})}\approx 0.0182 \, \mathrm{M}\]This result reveals that at body temperature, the molarity of CO₂ is 0.0182 mol/L.

Molarity

Molarity measures the concentration of a solution, stated as the amount of solute (CO₂ here) per liter of solution. It's denoted as M and calculated by dividing the number of moles by the volume in liters. In this exercise, knowing the partial pressure and using the ideal gas constant facilitated finding the molarity of CO₂.
Think of molarity as a way to express how much CO₂ is available within the breath, a critical aspect when considering gaseous exchanges in the lungs. It’s a straightforward formula influenced significantly by temperature and pressure. At standard body conditions or specific experimental setups, like in our example, this helps to simplify calculations of gas exchange processes efficiently.

Body Temperature

Body temperature is a pivotal parameter in determining the behavior of gases within biological systems. It usually averages around 37°C (98.6°F), but for thermodynamic calculations, it's essential to convert it into Kelvin (K). This is because Kelvin avoids negative numbers in gas law equations and aligns with the absolute scale needed in scientific calculations.
The conversion from Celsius to Kelvin is straightforward:

• Add 273.15 to the Celsius temperature. For instance, 37°C + 273.15 = 310.15 K.

In relation to gas laws, the temperature influences how freely gas molecules move and how they occupy space. Thus, if the temperature fluctuates, it can significantly affect gas pressure and volume, necessitating adjustments in equations such as the Ideal Gas Law. Understanding how temperature affects these properties is crucial in explaining physiological phenomena such as breathing.