Question

During a typical breathing cycle the \(\mathrm{CO}_{2}\) concentration in the expired air rises to a peak of \(4.6 \%\) by volume. Calculate the partial pressure of the \(\mathrm{CO}_{2}\) at this point, assuming 1 atm pressure. What is the molarity of the \(\mathrm{CO}_{2}\) in air at this point, assuming a body temperature of \(37^{\circ} \mathrm{C} ?\)

Short answer

At the peak of the breathing cycle with 1 atm pressure and a body temperature of 37°C, the partial pressure of CO2 in the air is 0.046 atm, and the molarity of CO2 is 0.00190 mol/L.

Step-by-step solution

Calculate Partial Pressure of CO2

Calculate the partial pressure of CO2 using the given concentration and total pressure. The concentration is 4.6% by volume, which can be converted into a mole fraction by dividing it by 100. Mole Fraction of CO2 = 4.6% / 100 = 0.046 Total Pressure = 1 atm Using the formula, Partial Pressure of CO2 = Total Pressure × Mole Fraction Partial Pressure of CO2 = 1 atm × 0.046 = 0.046 atm

Calculate the Molarity of CO2

To calculate the molarity of CO2 in air at this point, we will use the ideal gas equation: PV = nRT Where P is pressure, V is volume, n is the amount of substance in moles, R is the gas constant, and T is the temperature in Kelvin. We want to find the molarity, which is defined as moles per liter (n/V). Rearrange the equation to solve for n/V: Molarity (n/V) = P / (RT) First, we need to convert the given body temperature from Celsius to Kelvin: Temperature = 37°C + 273.15 = 310.15 K Now we can use the ideal gas equation to find the molarity of CO2: Molarity = (0.046 atm) / ((0.0821 L atm/mol K) × 310.15 K) Molarity = 0.00190 mol/L Therefore, at the peak of the breathing cycle with 1 atm pressure and a body temperature of 37°C, the partial pressure of CO2 in the air is 0.046 atm, and the molarity of CO2 is 0.00190 mol/L.

Key concepts

Partial Pressure

In gas mixtures, the partial pressure is a term used to describe the contribution each gas in a mixture makes to the total pressure. It's an important concept when considering gases in contexts such as breathing or chemical reactions. Each gas exerts a pressure proportionate to its amount or concentration in the mixture. This is called its partial pressure.
In the problem, the expired air contains 4.6% carbon dioxide (\( \mathrm{CO}_2 \)) by volume. This percentage can be directly translated into a mole fraction, which is useful for calculating partial pressure. Since we're assuming a total atmospheric pressure of 1 atm, we multiply the mole fraction by this total pressure to find the partial pressure:

• Partial Pressure of \( \mathrm{CO}_2 \) = Total Pressure × Mole Fraction
• Mole Fraction = 0.046
• Partial Pressure of \( \mathrm{CO}_2 \) = 1 atm × 0.046 = 0.046 atm

This tells us that, even though several gases are present, \( \mathrm{CO}_2 \) contributes 0.046 atm of the total pressure.

Mole Fraction

The mole fraction is a way to express the concentration of a component in a mixture. It's a ratio of the moles of a particular substance to the total moles present in the mixture.
To calculate the partial pressure of a component like \( \mathrm{CO}_2 \), you first need to convert its concentration percentage into a mole fraction. This is done by dividing the percentage by 100. In the given problem:

• Mole Fraction of \( \mathrm{CO}_2 \) = 4.6% / 100 = 0.046

This mole fraction indicates that \( \mathrm{CO}_2 \) represents 4.6% of the total moles of gas in the air. Once you have the mole fraction, it's straightforward to calculate the partial pressure using the total pressure.

Gas Constant

The gas constant, denoted as \( R \), is an important constant when working with the Ideal Gas Law. It provides the link between moles of gas, the physical volume that they occupy, and the temperature they are at, under a certain pressure.
In the given exercise, the Ideal Gas Law equation is used: \( PV = nRT \), where \( R = 0.0821 \, \text{L atm/mol K} \) is the gas constant. This constant is vital for converting units like liters, atmospheres, and degrees Kelvin so we can relate pressure, volume, and temperature in gas calculations.
By using \( R \) in the molarity calculation, we make sure that all units align, resulting in accurate measurements and conversions. It's essential to always check your units with \( R \), to avoid potential mistakes in calculations.

Body Temperature

Body temperature influences the behavior of gases. In the human body, \( 37^{\circ} \mathrm{C} \) is the normal internal temperature, equivalent to 310.15 Kelvin.
At this temperature, the volume and pressure of gases will behave accordingly under the Ideal Gas Law. In our exercise, the body temperature is crucial for calculating the molarity of \( \mathrm{CO}_2 \) because the gas constant \( R \) requires the temperature to be in Kelvin.

• Convert Celsius to Kelvin for calculations: \[ 37^{\circ} \mathrm{C} + 273.15 = 310.15 \, \text{K} \]
• Use this temperature in the equation for accurate results

This conversion ensures that the calculation of molarity, which depends on \( PV = nRT \), is precise and reflects the conditions within the body.