Vapor–liquid equilibrium (Raoult’s law): Calculate the total pressure of a liquid mixture (benzene, toluene, o-xylene) at 100 °C given their pure-component vapor pressures.
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A756.2 mmHg
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B780.5 mmHg
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C801.5 mmHg
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D880.5 mmHg
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E900.0 mmHg
Answer
Correct Answer: 801.5 mmHg
Explanation
Introduction / Context:This problem tests application of Raoult’s law for an ideal liquid solution at a fixed temperature. The total pressure above the mixture equals the sum of each component’s partial pressure, which is the product of its liquid mole fraction and its pure-component vapor pressure at that temperature.
Given Data / Assumptions:
- Liquid moles: benzene = 0.45 kmol, toluene = 0.44 kmol, o-xylene = 0.23 kmol.
- Total moles in liquid = 0.45 + 0.44 + 0.23 = 1.12 kmol.
- P^sat at 100 °C (mmHg): benzene = 1340, toluene = 560, o-xylene = 210.
- Assume ideal solution behavior (Raoult’s law) and negligible gas-phase non-ideality.
Concept / Approach:For each component i: p_i = x_i * P_i^sat, where x_i is the liquid mole fraction. The total pressure P_total = Σ p_i across all components.
Step-by-Step Solution:
Compute x_benzene = 0.45 / 1.12 = 0.401786.Compute x_toluene = 0.44 / 1.12 = 0.392857.Compute x_oxylene = 0.23 / 1.12 = 0.205357.Partial pressures: p_B = 0.401786 * 1340 ≈ 538.39 mmHg.p_T = 0.392857 * 560 ≈ 220.00 mmHg.p_X = 0.205357 * 210 ≈ 43.13 mmHg.Total pressure P_total = 538.39 + 220.00 + 43.13 ≈ 801.52 mmHg.Verification / Alternative check:Recalculate with rounded x_i values to two or three decimals; the total remains ≈ 801.5 mmHg, confirming robustness to small rounding differences.
Why Other Options Are Wrong:
- 756.2 / 780.5 mmHg: Underestimates; likely from arithmetic or mole-fraction rounding errors.
- 880.5 / 900.0 mmHg: Overestimates; would require higher P^sat weighting or incorrect x_i.
Common Pitfalls:Forgetting to normalize mole fractions, mixing activity coefficients into an ideal law problem, or confusing mass fractions with mole fractions.
Final Answer:801.5 mmHg