Problem 69 - Chemistry The Central Science In Si Units

Question

A rigid vessel containing a \(3: 1 \mathrm{~mol}\) ratio of carbon dioxide and water vapor is held at \(200^{\circ} \mathrm{C}\) where it has a total pressure of \(202.7 \mathrm{kPa}\). If the vessel is cooled to \(10^{\circ} \mathrm{C}\) so that all of the water vapor condenses, what is the pressure of carbon dioxide? Neglect the volume of the liquid water that forms on cooling.

Step-by-Step Solution

Answer

The pressure of carbon dioxide after cooling is 152.025 kPa.

Step 1: Initial Conditions and Dalton's Law Application

Initially, the gas mixture has a total pressure of 202.7 kPa at 200°C with carbon dioxide and water vapor in a 3:1 mole ratio. According to Dalton’s Law of Partial Pressures, the total pressure is the sum of the partial pressures of the gases. Let \( P_{\text{CO}_2} \) be the partial pressure of carbon dioxide and \( P_{\text{H}_2O} \) be the partial pressure of water vapor. Since there is a 3:1 mol ratio, \( P_{\text{CO}_2} = 3P_{\text{H}_2O} \). Hence, the total pressure can be expressed as \( P_{\text{total}} = P_{\text{CO}_2} + P_{\text{H}_2O} = 4P_{\text{H}_2O} \). Therefore, \( 4P_{\text{H}_2O} = 202.7 \, \text{kPa} \).

Step 2: Calculate Initial Partial Pressure of Water Vapor

From the equation \( 4P_{\text{H}_2O} = 202.7 \, \text{kPa} \), solve for \( P_{\text{H}_2O} \):\( P_{\text{H}_2O} = \frac{202.7}{4} = 50.675 \, \text{kPa} \).

Step 3: Calculate Initial Partial Pressure of Carbon Dioxide

Using the relation \( P_{\text{CO}_2} = 3P_{\text{H}_2O} \), substitute the value of \( P_{\text{H}_2O} \):\( P_{\text{CO}_2} = 3 \times 50.675 = 152.025 \, \text{kPa} \).

Step 4: Cooling the Vessel and Condensation of Water Vapor

Upon cooling to 10°C, all the water vapor condenses into liquid water, meaning the only gas remaining in the vessel is carbon dioxide. Since the vessel is rigid and the liquid water formed has negligible volume, the number of moles of carbon dioxide and the container's volume do not change.

Step 5: Calculate Final Pressure of Carbon Dioxide after Cooling

The pressure of carbon dioxide remains unchanged with just a temperature change in a rigid vessel because the volume for gases only reduces by the volume of water vapor. Thus, the pressure of carbon dioxide after cooling stays as its partial pressure before cooling:\( P_{\text{CO}_2} = 152.025 \, \text{kPa} \).

Key Concepts

Partial Pressure

Partial pressure is a fundamental concept in the study of gases. It describes the pressure a single type of gas in a mixture would exert if it occupied the entire volume by itself. In our exercise, Dalton's Law of Partial Pressures is utilized. It states that the total pressure of a gas mixture is the sum of the partial pressures of each individual gas in the mixture.

For a mixture of gases, if you know the total pressure and the mole ratio of the gases, you can determine the partial pressure of each gas. In the given exercise, the total pressure is 202.7 kPa, and by knowing the 3:1 mole ratio of CO2 to H2O, we can find that the partial pressure of CO2 is three times that of H2O. This is crucial for solving problems related to gas mixtures, as it allows us to isolate the pressure contributions from individual gases.

Gas Laws

Gas laws are essential for understanding the behavior of gases under various conditions. They provide relationships between pressure, volume, temperature, and the number of moles. In our exercise, the rigid vessel scenario is an application of these principles, particularly focusing on pressure and temperature changes.

When cooling occurs at constant volume, as in a rigid vessel, the key gas law concept to understand is that pressure is directly proportional to temperature (in Kelvin) according to Gay-Lussac's Law. However, since the water vapor condenses to liquid, this reduces the total volume occupied by the gases, leaving carbon dioxide as the primary gas. This aspect simplifies to Dalton’s Law because the total number of moles changes due to the water condensing.

Phase Change

Phase changes are physical changes of a substance from one state of matter to another, such as from gas to liquid or solid. In the context of our problem, the water vapor undergoes a phase change from vapor to liquid when the vessel is cooled to 10°C. This is because water vapor condenses at much lower temperatures.

During this condensation, the water vapor exerts no pressure in its liquid state, which means the partial pressure of water vapor effectively drops to zero once it becomes liquid. This results in carbon dioxide being the only remaining gas, which still follows the gas laws without any such phase change, maintaining a pressure of 152.025 kPa.

Mole Ratio

Mole ratio is the ratio between the amounts in moles of any two compounds involved in a chemical reaction. This concept helps to determine relationships between substances in chemical reactions and gaseous mixtures. In the exercise, the given 3:1 mole ratio of carbon dioxide to water vapor is crucial for solving the problem.

Knowing the mole ratio helps in calculating partial pressures using Dalton’s Law. For every one mole of water vapor, we have three moles of carbon dioxide. Therefore, the partial pressure of carbon dioxide is three times greater than that of the water vapor. Understanding these relationships through mole ratios facilitates solving complex problems involving mixtures of gases by simplifying the mathematical relationships.