Liquid A and B form an ideal solution. The vapour pressure of pure liquids A and B are 350 and 750 mm Hg respectively at the same temperature. If $ x_A $ and $ x_B $ are the mole fraction of A and B in solution while $ y_A $ and $ y_B $ are the mole fraction of A and B in vapour phase then :
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In an ideal solution, the vapour phase is richer in the more volatile component (the one with the higher vapour pressure). Here, B has a higher vapour pressure than A, so the vapour phase will have a higher mole fraction of B compared to the solution. This implies that the ratio of mole fractions of A to B will be smaller in the vapour phase than in the solution.
To solve this problem, we need to understand the behavior of ideal solutions and the concepts of Raoult's law regarding vapour pressure in a solution.
Understanding Raoult's Law:
For an ideal solution, the partial vapour pressure of each component is given by Raoult's Law:
\(P_A = x_A \cdot P_{A}^{\circ}\) and \(P_B = x_B \cdot P_{B}^{\circ}\), where \(P_{A}^{\circ}\) and \(P_{B}^{\circ}\) are the vapour pressures of pure components A and B, respectively.
Given Data:
For pure A, \(P_{A}^{\circ} = 350 \, \text{mm Hg}\)
For pure B, \(P_{B}^{\circ} = 750 \, \text{mm Hg}\)
Total Vapour Pressure:
The total vapour pressure \(P_{\text{total}}\) is the sum of the partial pressures:
Since \(\frac{7}{15} < 1\), it follows that \(\frac{x_A}{x_B} > \frac{y_A}{y_B}\).
Thus, the correct answer is \(\frac{x_A}{x_B} > \frac{y_A}{y_B}\).
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Approach Solution -2
To determine the relationship between the mole fractions of components A and B in both the liquid and vapor phases, we can use Raoult's Law for ideal solutions:
Raoult's Law: For a two-component system, the partial vapor pressure of each component is given by: \( P_{A} = P_{A}^0 \times x_A \) \( P_{B} = P_{B}^0 \times x_B \) where \( P_{A}^0 \) and \( P_{B}^0 \) are the vapor pressures of the pure components A and B, respectively.
Total Vapor Pressure: The total vapor pressure of the solution is: \( P_{total} = P_{A} + P_{B} = P_{A}^0 \times x_A + P_{B}^0 \times x_B \)
Mole Fraction in Vapor Phase: The mole fractions in the vapor phase can be calculated using Dalton's Law: \( y_A = \frac{P_{A}}{P_{total}} = \frac{P_{A}^0 \times x_A}{P_{total}} \) \( y_B = \frac{P_{B}}{P_{total}} = \frac{P_{B}^0 \times x_B}{P_{total}} \)
Comparison: We need to compare \( \frac{x_A}{x_B} \) with \( \frac{y_A}{y_B} \). Given that \( P_{A}^0 = 350 \) mm Hg and \( P_{B}^0 = 750 \) mm Hg, and knowing that \( P_{B}^0 > P_{A}^0 \), it follows that component B is more volatile than component A. Therefore, in the vapor phase, the mole fraction of B will be greater as compared to A, leading to: \( \frac{x_A}{x_B} > \frac{y_A}{y_B} \)
Thus, the correct answer is: \( \frac{x_A}{x_B}>\frac{y_A}{y_B} \)
