Problem 64 - Chemistry: The Central Science

Question

A sample of \(4.00 \mathrm{~mL}\) of diethylether \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OC}_{2} \mathrm{H}_{5},\right.\), density \(=0.7134 \mathrm{~g} / \mathrm{mL}\) ) is introduced into \(\mathrm{a}\) 5.00-L vessel that already contains a mixture of \(\mathrm{N}_{2}\) and \(\mathrm{O}_{2}\), whose partial pressures are \(P_{\mathrm{N}_{2}}=0.751 \mathrm{~atm}\) and \(P_{\mathrm{O}_{2}}=0.208 \mathrm{~atm}\). The temperature is held at \(35.0^{\circ} \mathrm{C}\), and the diethylether

Step-by-Step Solution

Answer

The partial pressures in the 5.00-L flask after the diethylether has vaporized are: - N2: \(0.609~atm\) - O2: \(0.169~atm\) - Diethylether: \(0.197~atm\)

Step 1: Calculate the mass of diethylether

We can calculate the mass of diethylether using its volume and density: Mass = Volume × Density Mass = 4.00 mL × 0.7134 g/mL = 2.8536 g

Step 2: Calculate the moles of diethylether

We can calculate the moles of diethylether using its molar mass (C2H5OC2H5): Molar mass of diethylether = 74.12 g/mol Moles of diethylether = \(\frac{Mass}{Molar~mass}\) Moles of diethylether = \(\frac{2.8536~g}{74.12~g/mol} \approx 0.0385~moles\)

Step 3: Calculate the updated moles of N2 and O2

We can calculate the moles of N2 and O2 using the ideal gas law: \(PV = nRT\) where \(n\) is the moles, \(P\) is the pressure, \(V\) is the volume, \(R\) is the ideal gas constant (0.0821 L atm / K mol), and \(T\) is the temperature in Kelvin (35°C + 273.15 = 308.15 K). Moles of N2 = \(\frac{P_{N2}V}{RT}\) = \(\frac{0.751~atm \times 5.00~L}{0.0821~L~atm / K~mol \times 308.15~K} \approx 0.1207~moles\) Moles of O2 = \(\frac{P_{O2}V}{RT}\) = \(\frac{0.208~atm \times 5.00~L}{0.0821~L~atm / K~mol \times 308.15~K} \approx 0.0334~moles\)

Step 4: Use the ideal gas law to calculate the final partial pressures of N2, O2, and diethylether

Now that we have the updated moles of N2, O2, and diethylether, we can calculate their final partial pressures using the ideal gas law: Final partial pressure of N2: \(P_{N2} = \frac{n_{N2}RT}{V} = \frac{0.1207~moles \times 0.0821~L~atm / K~mol \times 308.15~K}{5.00~L} \approx 0.609~atm\) Final partial pressure of O2: \(P_{O2} = \frac{n_{O2}RT}{V} = \frac{0.0334~moles \times 0.0821~L~atm / K~mol \times 308.15~K}{5.00~L} \approx 0.169~atm\) Final partial pressure of diethylether: \(P_{C2H5OC2H5} = \frac{n_{C2H5OC2H5}RT}{V} = \frac{0.0385~moles \times 0.0821~L~atm / K~mol \times 308.15~K}{5.00~L} \approx 0.197~atm\) Therefore, the partial pressures in the 5.00-L flask after the diethylether has vaporized are: - N2: 0.609 atm - O2: 0.169 atm - Diethylether: 0.197 atm

Key Concepts

Understanding the Ideal Gas Law

The ideal gas law is a fundamental equation in chemistry and physics describing the behavior of an ideal gas. This law combines several gas laws, including Boyle's, Charles's, and Avogadro's, into one formula represented as \( PV = nRT \). Here, \( P \) stands for pressure of the gas, \( V \) is the volume it occupies, \( n \) is the amount of substance measured in moles, \( R \) is the ideal gas constant, and \( T \) is the temperature in Kelvin.

To utilize the ideal gas law for calculating partial pressures, it is essential to first determine the moles of gas present. As shown in the exercise solution, the moles of diethylether were found from its mass and molar mass. The same law then allows us to find the new partial pressures after the diethylether has vaporized and mixed with the existing gases in the container. Each gas's partial pressure can be calculated individually by applying the ideal gas law to that gas's mole fraction.

Real Gases vs. Ideal Gases

It's also important to note that the ideal gas law assumes gases do not interact and do not take up space (they are point particles). In reality, gases do have intermolecular forces and volumes, and these deviations are more pronounced under high pressures and low temperatures. For most common conditions however, the ideal gas law provides a very good approximation, making it invaluable for solving problems involving gases.

Using this approach, as outlined in the provided solution, helps in understanding how individual gas components in a mixture each contribute to the total pressure. This understanding is crucial not only in academic exercises but also in practical applications such as chemical engineering and environmental studies.

The Role of Molar Mass in Calculations

Molar mass is defined as the mass of one mole of a substance and is typically expressed in grams per mole (g/mol). This property is particularly useful when converting between the mass of a substance and the number of moles, as required in our exercise.

The molar mass of diethylether (\(C_2H_5OC_2H_5\)) is a sum of the atomic masses of its constituent atoms, a value gleaned from the periodic table. For any compound, determining its molar mass is a simple yet crucial step in many chemical calculations, including finding molar ratios in reactions, interpreting crystallization processes, and analyzing the composition of mixtures.

Calculating Moles from Mass

Once we have the mass of a substance, such as diethylether in the sample problem, we convert it to moles using its molar mass. The formula for this conversion is \( moles = \frac{mass}{molar~mass} \).

Molar mass acts as a bridge between microscopic and macroscopic understanding of chemical quantities. With it, we can link observable quantities (like mass) to the fundamental unit of chemical measurement (the mole), essential for stoichiometry, thermodynamics, and kinetics. Therefore, grasping the concept of molar mass and its application is fundamental to mastering chemistry.

Vapor Pressure and Its Significance

Vapor pressure is a measure of a liquid's tendency to evaporate and become a gas. At any given temperature, a liquid will have a specific vapor pressure, which is determined by the inherent properties of the liquid, such as chemical structure and intermolecular forces. An important practical aspect of vapor pressure is that it is a constant value for a pure substance at a specific temperature.

When a liquid vaporizes in a closed container, like the diethylether in our exercise, it increases the pressure within the container; this is the vapor pressure of the liquid at that temperature. As more molecules escape the liquid phase, the concentration of gas-phase molecules increases until it reaches equilibrium, where the rate of evaporation equals the rate of condensation.

Factors Affecting Vapor Pressure

Several factors influence vapor pressure, including:

• Temperature: Higher temperatures increase vapor pressure as more molecules have enough energy to escape into the gas phase.
• Intermolecular forces: Stronger intermolecular forces in a liquid lead to lower vapor pressures since fewer molecules can escape to the gas phase.

Understanding vapor pressure is essential for predicting how substances will behave under various conditions, important in fields like meteorology, environmental science, and materials engineering. In the context of the exercise, the calculated partial pressure after vaporization takes the concept of vapor pressure into account, reflecting how the diethylether contributes to the total pressure within the vessel.