Vapour Liquid Equilibria Questions and Answers

Answer: A

Correct answer: False

Vapour–liquid equilibrium depends on temperature through both the vapour pressures and the activity coefficients, so data taken at one temperature do not transfer to another. What can be done is to fit an activity coefficient model to the data and use the model, with temperature-dependent parameters, to predict at other conditions - which is why correlation rather than interpolation is the standard practice.

Answer: B

Its formulation as a single monotonic function means a simple root-finding method cannot jump between spurious solutions. That numerical robustness is why every flash routine uses it. A bisection method converges on it without difficulty.

Answer: C

Summing two partial pressures each proportional to mole fraction gives a straight line on a Pxy plot. Departures from that straight line are the visible signature of non-ideality. A bulge above the line indicates positive deviations.

Answer: D

Correct answer: 55 kilopascals

Raoult’s law for an ideal solution gives \[ P = x_1P_1^{sat} + x_2P_2^{sat} = 0.5\times80 + 0.5\times30 = 40 + 15 = 55\ \mathrm{kPa}. \] The total pressure is the mole-fraction-weighted average of the pure vapour pressures, so for an equimolar mixture it is simply their mean - and adding them, which gives the 110 offered, would describe two separate pure liquids rather than a solution.

Answer: A

Both are true and the reason gives the arithmetic: the vapour mole fraction is the partial pressure over the total, and a larger vapour pressure raises that ratio above the liquid mole fraction. The effect is what every distillation column exploits.

Answer: B

The first two are correct and describe standard practice. The third is wrong, since a model needs data across the composition range, and the dilute ends are the most demanding to measure accurately. Analytical error grows as a component becomes dilute.

Answer: C

Correct answer: the excess Gibbs energy curve has a common tangent

Two liquid phases coexist when a single tangent line touches the Gibbs energy curve at two compositions, since at those points the chemical potential of each species is the same in both phases. That construction requires the curve to be non-convex, which in turn requires a sufficiently positive excess Gibbs energy - so strongly positive deviations are the condition under which a model predicts a split.