Question

An ideal solution is formed by mixing two volatile liquids \(A\) and \(B . X_{A}\) and \(X_{B}\) are the mole fractions of \(A\) and \(B\) respectively in the solution and \(Y_{A}\) and \(Y_{B}\) are the mole fractions of \(A\) and \(B\) respectively in the vapour phase. A plot of \(\frac{1}{Y_{A}}\) along \(y\) -axis against \(\frac{1}{X_{1}}\) along \(x\) -axis gives a straight line. What is the slope of the straight line? (a) \(\frac{P_{B}^{\circ}}{P_{A}^{\circ}}\) (b) \(\frac{P_{A}^{\circ}}{P_{B}^{\circ}}\) (c) \(P_{B}^{\circ}-P_{A}^{\circ}\) (d) \(P_{A}^{\circ}-P_{B}^{\circ}\)

Short answer

\(\frac{P_{B}^{\circ}}{P_{A}^{\circ}}\)

Step-by-step solution

Understand the Relationship between the Mole Fractions and the Partial Pressures

According to Raoult's Law for an ideal solution, the partial vapor pressure of each component is directly proportional to the vapor phase mole fraction of that component times its pure component vapor pressure at the same temperature. The equations are as follows: For compound A: \( P_A = Y_A \times P_A^{\text{o}} \) And for compound B: \( P_B = Y_B \times P_B^{\text{o}} \) Where: \( P_A \) and \( P_B \) are the partial pressures of A and B, respectively, \( Y_A \) and \( Y_B \) are the mole fractions of A and B in the vapor phase, \( P_A^{\text{o}} \) and \( P_B^{\text{o}} \) are the pure component vapor pressures of A and B.

Express the Mole Fractions in Terms of Partial Pressures

From Raoult's Law, we can express \( Y_A \) and \( X_A \) in terms of partial pressures: \( Y_A = \frac{P_A}{P_A^{\text{o}}} \) Since the total pressure is the sum of partial pressures of A and B, and the mole fractions sum up to 1 (\( X_A + X_B = 1 \) and \( Y_A + Y_B = 1 \)), we can write: \( Y_A = \frac{P_A}{P_A + P_B} \)

Relate the Mole Fraction in Liquid Phase to the Vapor Phase by Dalton's Law

Using Dalton's Law and the fact that \( P_{total} = P_A + P_A \), we substitute: \( Y_A = \frac{X_A \times P_A^{\text{o}}}{X_A \times P_A^{\text{o}} + (1-X_A) \times P_B^{\text{o}}} \)

Rearrange the Equation to Get the Desired Ratio

We rearrange the expression for \( Y_A \) to isolate it: \( \frac{1}{Y_A} = \frac{X_A \times P_A^{\text{o}} + (1-X_A) \times P_B^{\text{o}}}{X_A \times P_A^{\text{o}}} \) We want to express this in terms of \( \frac{1}{X_A} \), so we continue the manipulation: \( \frac{1}{Y_A} = \frac{P_A^{\text{o}}}{X_A \times P_A^{\text{o}}} + \frac{(1-X_A) \times P_B^{\text{o}}}{X_A \times P_A^{\text{o}}} \)

Simplify the Equation

Simplifying the above expression, we get: \( \frac{1}{Y_A} = \frac{1}{X_A} + \frac{P_B^{\text{o}} - X_A \times P_B^{\text{o}}}{X_A \times P_A^{\text{o}}} \) \( \frac{1}{Y_A} = \frac{1}{X_A} + \frac{P_B^{\text{o}}}{X_A \times P_A^{\text{o}}} - \frac{P_B^{\text{o}}}{P_A^{\text{o}}} \) Notice that the plot of \( \frac{1}{Y_A} \) against \( \frac{1}{X_A} \) is a straight line, so the coefficient of \( \frac{1}{X_A} \) in the equation will represent the slope.

Find the Slope of the Straight Line

The coefficient of \( \frac{1}{X_A} \) is the slope of the line. From the simplified equation, we can see that the slope is: \( \ Slope = \frac{P_B^{\text{o}}}{P_A^{\text{o}}} \)

Key concepts

Understanding Ideal Solutions

An ideal solution is a mix of two or more substances that exhibits certain properties. These solutions follow Raoult's Law strictly, meaning each component’s vapor pressure contributes to the total vapor pressure based on its mole fraction in the mixture. The intermolecular forces between different components in an ideal solution are equal to the forces amongst the like molecules. Because of this, the enthalpy change when mixing is zero, indicating that no heat is absorbed or released, making the process thermodynamically favorable.

These unique characteristics of ideal solutions also suggest that the volume and boiling point follow a linear relation with the mole fractions of constituents. In the context of homework problems, understanding ideal solutions often involves calculations using these linear relationships, ratios, and properties, which simplifies the problem-solving process.

The Role of Vapor Pressure in Raoult's Law

The concept of vapor pressure is critical when discussing solutions and their behaviors. Vapor pressure is the pressure exerted by the vapor in equilibrium with its liquid at a given temperature. In an ideal solution, the vapor pressure of each component is proportional to its mole fraction, which is elegantly represented by Raoult's Law.

To break it down, Raoult's Law can be expressed as: \( P_i = X_i \times P_i^{\circ} \), where \( P_i \) is the partial vapor pressure of component \(i\), \(X_i\) is the mole fraction of \(i\) in the liquid phase, and \(P_i^{\circ}\) is the pure component's vapor pressure. Problems involving vapor pressure calculations require an understanding of these factors and the ability to relate them through algebraic manipulations for predictions about boiling points, compositions, and even the nature of solutions.

Mole Fraction and Its Impact

The mole fraction is a way of expressing the concentration of a component in a mixture. Defined as the ratio of the number of moles of that component to the total number of moles of all components of the mixture, it is denoted as \(X_i\) for a component \(i\). The mole fraction is unitless and always takes a value between 0 and 1. It plays a pivotal role in Raoult’s Law as it directly affects the partial vapor pressure of each component in a solution.

For a binary solution comprised of two volatile liquids, \(A\) and \(B\), if one knows the mole fraction of one component, the other can be easily calculated since the sum of mole fractions is always equal to 1 (\(X_A + X_B = 1\)). This relationship is essential for problem-solving when dealing with colligative properties or when determining the composition of a vapor in equilibrium with a liquid mixture.