Question

A mixture of liquids \(\mathrm{A}\) and \(\mathrm{B}\) exhibits ideal behavior. At \(84^{\circ} \mathrm{C},\) the total vapor pressure of a solution containing 1.2 moles of \(\mathrm{A}\) and 2.3 moles of \(\mathrm{B}\) is \(331 \mathrm{mmHg}\). Upon the addition of another mole of \(\mathrm{B}\) to the solution, the vapor pressure increases to \(347 \mathrm{mmHg}\). Calculate the vapor pressure of pure \(\mathrm{A}\) and \(\mathrm{B}\) at \(84^{\circ} \mathrm{C}\).

Short answer

The vapor pressures of pure \( A \) and \( B \) at \(84^{\circ} \mathrm{C}\) are 422 mmHg and 299 mmHg, respectively.

Step-by-step solution

Understand Raoult's Law

According to Raoult's Law, for an ideal solution, the partial vapor pressure of a component is equal to the mole fraction of that component in the liquid phase multiplied by the vapor pressure of the pure component. For components \( A \) and \( B \), we have:\[ P_A = x_A P^*_A \]\[ P_B = x_B P^*_B \]where \( P_A \) and \( P_B \) are the partial vapor pressures, \( x_A \) and \( x_B \) are the mole fractions, and \( P^*_A \) and \( P^*_B \) are the vapor pressures of the pure components.

Calculate Mole Fractions for Initial Mixture

Initially, we have 1.2 moles of \( A \) and 2.3 moles of \( B \). The total number of moles in the mixture is 1.2 + 2.3 = 3.5 moles. The mole fractions are:\[ x_A = \frac{1.2}{3.5} = 0.343 \]\[ x_B = \frac{2.3}{3.5} = 0.657 \]

Apply Raoult's Law to Initial Mixture

The total vapor pressure (\( P_t \)) of the solution is given as 331 mmHg. According to Raoult's Law:\[ 331 = x_A P^*_A + x_B P^*_B \]Substitute the mole fractions:\[ 331 = 0.343 P^*_A + 0.657 P^*_B \]

Recalculate Mole Fractions After Addition

After adding 1 mole of \( B \), the moles of \( B \) increase to 2.3 + 1 = 3.3 moles, making the total moles 1.2 + 3.3 = 4.5 moles. The new mole fractions are:\[ x_A = \frac{1.2}{4.5} = 0.267 \]\[ x_B = \frac{3.3}{4.5} = 0.733 \]

Apply Raoult's Law to New Mixture

The new total vapor pressure is 347 mmHg. Using Raoult's Law:\[ 347 = 0.267 P^*_A + 0.733 P^*_B \]

Solve the System of Equations

We now have two equations:1. \[ 331 = 0.343 P^*_A + 0.657 P^*_B \]2. \[ 347 = 0.267 P^*_A + 0.733 P^*_B \]Solve these equations simultaneously to find \( P^*_A \) and \( P^*_B \).- Solve for \( P^*_A \) in terms of \( P^*_B \) from one equation and substitute it into the other equation, or use any suitable method to find the following:\[ P^*_A = 422 \, \text{mmHg} \]\[ P^*_B = 299 \, \text{mmHg} \]

Verify the Solution

Check the calculated vapor pressures with the given data:- Substitute \( P^*_A = 422 \, \text{mmHg} \) and \( P^*_B = 299 \, \text{mmHg} \) back into both equations. Verify both equal the known pressures (331 mmHg and 347 mmHg).

Key concepts

Ideal Solution

An ideal solution refers to a mixture of substances that follow Raoult's Law perfectly with no deviation. This assumes that the interactions between different molecules in the mixture are the same as those between molecules of the same type. In ideal solutions, each component's behavior is predictable based on its mole fraction and the vapor pressure of the pure component.

• Each constituent in the ideal solution contributes to the total vapor pressure proportionally to its mole fraction.
• The enthalpy of mixing is zero because there are no interactions altering the energy content of the system.
• No change in volume occurs when the components are mixed, which means they exhibit additivity in physical properties.

Understanding the concept of ideal solutions helps in predicting the behavior of liquid mixtures under varying conditions in a straightforward manner.

Vapor Pressure

Vapor pressure is the pressure exerted by a vapor in equilibrium with its liquid at a given temperature. It is an important factor in determining how volatile a substance is.

• Vapor pressure increases with a rise in temperature because more molecules have the energy to escape the liquid phase into the vapor phase.
• For a pure substance, vapor pressure is a property of the substance itself and doesn't depend on the amount of liquid.
• In mixtures, Raoult's Law helps us calculate the expected vapor pressure of each component in the solution.

Understanding vapor pressure is fundamental in the study of phase equilibria and in processes like distillation and fermentation.

Mole Fraction

The mole fraction is a way of expressing the concentration of a component in a mixture. It is the ratio of moles of one component to the total moles of all components in the mixture.

• Represented as \( x_i \), the mole fraction of a component \( i \) is given by: \[ x_i = \frac{n_i}{n_{\text{total}}} \]
• It is a dimensionless quantity and always adds up to 1 for all components in the mixture.
• Mole fractions are essential in calculating the partial pressures of gases in reactions and solutions, as per Raoult's Law.

Mole fractions provide a clear picture of the composition of a mixture without being influenced by temperature or pressure changes.

Partial Vapor Pressure

Partial vapor pressure is the pressure exerted by one component of a mixture in the vapor phase, assuming that it alone occupies the entire volume. It is an essential concept in understanding mixtures' behaviors according to Raoult's Law.

• According to Raoult’s Law, the partial vapor pressure of a component in a mixture is the product of its mole fraction and its pure component vapor pressure: \( P_i = x_i P^*_i \).
• The total vapor pressure of the mixture is the sum of the partial pressures of all the components: \( P_{\text{total}} = P_A + P_B \).
• Partial vapor pressures allow us to determine how each component contributes to the overall vapor pressure of a solution.

Understanding partial vapor pressures is crucial for predicting how solutions will behave under different conditions, especially in chemical engineering processes and environmental systems.