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 CO \(_{2}\) in the expired air at its peak, assuming 1 atm 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 \(37^{\circ} \mathrm{C} ?\)

Short answer

The partial pressure of \(\mathrm{CO}_2\) in the expired air at its peak is 0.046 atm, and the molarity of the \(\mathrm{CO}_2\) in the expired air at its peak is 0.0018 mol/L.

Step-by-step solution

Calculate the partial pressure of CO₂

We know the peak concentration of \(\mathrm{CO}_2\) is 4.6% by volume. Since the total pressure of the air is given as 1 atm, we can calculate the partial pressure of \(\mathrm{CO}_2\) using Dalton's law of partial pressures: Partial pressure of \(\mathrm{CO}_2 = \text{Fraction of }$$\mathrm{CO}_2 \times \text{Total pressure}\) Partial pressure of \(\mathrm{CO}_2 = 0.046 \times 1 \: \mathrm{atm} = 0.046 \: \mathrm{atm}\)

Convert the temperature to Kelvin

The body temperature is given as \(37^{\circ} \mathrm{C}\). We need to convert this value to Kelvin since gas laws use Kelvin for temperature. The formula to convert Celsius to Kelvin is: \(K = ^{\circ}\mathrm{C} + 273.15\) \(T = 37^{\circ} \mathrm{C} + 273.15 = 310.15 \: \mathrm{K}\)

Calculate the molarity of CO₂

We will use the ideal gas law to calculate the molarity of \(\mathrm{CO}_2\). The ideal gas law is: \(PV = nRT\) We need to solve for n, the number of moles of \(\mathrm{CO}_2\). We have the partial pressure of \(\mathrm{CO}_2\) from step 1 (P = 0.046 atm) and the temperature from step 2 (T = 310.15 K). The ideal gas constant, R, is equal to 0.0821 L atm/mol K. The volume of the expelled air is not given and cannot be canceled out, so we will solve for the molar concentration (n/V) instead: \((n/V) = \frac{P}{RT}\) Molarity of \(\mathrm{CO}_2 = \frac{0.046 \: \text{atm}}{(0.0821 \: \mathrm{L \: atm/mol \: K})(310.15 \: \mathrm{K})} = 0.0018 \: \text{mol/L}\) The partial pressure of \(\mathrm{CO}_2\) in the expired air at its peak is 0.046 atm, and the molarity of the \(\mathrm{CO}_2\) in the expired air at its peak is 0.0018 mol/L.

Key concepts

Partial Pressure

When we talk about partial pressure, we're referring to the pressure that a single gas in a mixture would exert if it occupied the entire volume on its own.
The idea is simple: every gas in a mixture contributes to the total pressure based on its proportion in the entire mixture.
This is particularly useful for understanding how gases behave in different scenarios, such as in human lungs during breathing.To put it into context, let's consider our exercise where we have carbon dioxide (\(\text{CO}_2\)) in expired air.
We know that its volume concentration at its peak is 4.6%, and the total air pressure is 1 atm.**Using Dalton's Law**

• The formula for partial pressure is:\[\text{Partial Pressure of } \text{CO}_2 = \text{Fraction of } \text{CO}_2 \times \text{Total Pressure}\]
• In this example:\[\text{Partial Pressure of } \text{CO}_2 = 0.046 \times 1 \: \text{atm} = 0.046 \: \text{atm}\]

This means that in the expired air, at its concentration peak, \(\text{CO}_2\) exerts a partial pressure of 0.046 atm.

Ideal Gas Law

The Ideal Gas Law is a fundamental equation in chemistry that describes the behavior of an ideal gas in terms of pressure, volume, temperature, and amount of gas.
This law is succinctly expressed as:\[PV = nRT\]Where:

• \( P \) is the pressure of the gas.
• \( V \) is the volume the gas occupies.
• \( n \) is the moles of gas.
• \( R \) is the universal gas constant, \(0.0821 \: \text{L atm/mol K}\).
• \( T \) is the temperature in Kelvin.

In the example we are solving, we want to find the molarity of the \(\text{CO}_2\) gas, which is essentially the moles of \(\text{CO}_2\) per liter of air.**Calculating Molarity**

• We rearrange the Ideal Gas Law to solve for \( n/V \), the concentration:\[\left( \frac{n}{V} \right) = \frac{P}{RT}\]
• Substituting the values:\[\frac{n}{V} = \frac{0.046 \: \text{atm}}{0.0821 \: \text{L atm/mol K} \times 310.15 \: \text{K}} \approx 0.0018 \: \text{mol/L}\]

Thus, the molarity of \(\text{CO}_2\) in the expired air is approximately 0.0018 mol/L.
This calculation is crucial for understanding the concentration of gases in respiratory processes.

Dalton's Law

Dalton's Law of Partial Pressures is an easy way to understand and calculate how different gases in a mixture, each with their own pressure, combine to give a total pressure.
It tells us that the total pressure of a gas mixture is the sum of the partial pressures of each individual gas.This is particularly handy when dealing with real-world situations like breathing, where multiple gases mix together.
In the expired air example, where the air pressure equals 1 atm, Dalton's Law is used to calculate the partial pressure of \(\text{CO}_2\) as:\[\text{Total Pressure} = \text{Partial Pressure of } \text{O}_2 + \text{Partial Pressure of } \text{CO}_2 + \text{Partial Pressure of other gases}\]**Practical Application**

• The exercise shows that the calculated partial pressure of \(\text{CO}_2\) is simply its proportional concentration multiplied by the total pressure (1 atm).
• This gives a direct and straightforward way to measure individual gas contributions in mixtures, crucial for fields like medicine and environmental science.

Dalton’s Law simplifies the complex interactions in gaseous mixtures and provides an analytical tool to predict how gases will behave when mixed together.