4: Vapour Pressure

Negative deviation means that the actual vapour pressure is lower than the ideal Raoult-law prediction. This occurs when A–B attractions are stronger than the attractions in the pure liquids, holding molecules more firmly in the liquid phase. Fewer molecules escape, so pressure falls. Very weak attraction would instead cause positive deviation.

For an ideal binary solution, Raoult’s law gives pA = xA p°A and pB = xB p°B. Thus pA = 0.5 × 100 = 50 kPa and pB = 0.5 × 60 = 30 kPa. Dalton’s law then gives the total pressure as ptotal = pA + pB = 50 + 30 = 80 kPa. Adding pure pressures directly would incorrectly ignore the mole fractions.

When a non-volatile solute lowers the solvent’s vapour pressure, the solution must be heated to a higher temperature before its vapour pressure can reach the external pressure required for boiling. Thus the boiling point is elevated. The effect does not mean that mass or molecules disappear, and colour change is not a colligative consequence.

Lowering of vapour pressure is a colligative property. For comparable solutions, its magnitude is governed by the effective number or mole fraction of solute particles; electrolytes may produce more particles through dissociation. Thus particle count and concentration determine the order. Colour, container height and the length of a chemical name have no physical role in this comparison.

Heating increases molecular kinetic energy and generally raises vapour pressure for every liquid. The increase is especially evident for a volatile liquid whose molecules can readily enter the vapour phase as energy rises. A liquid whose molecules never vaporize is physically inconsistent under ordinary conditions, and a system with no liquid or no temperature change cannot be used for this comparison.

A pressure above the Raoult-law value is positive deviation. It can arise when unlike-molecule attractions are weaker than the attractions in the pure liquids, so molecules escape from the mixed liquid more easily. Stronger unlike attraction would hold molecules back and produce negative deviation. The solution must contain molecules and a vapour phase for the pressure comparison.

A temperature increase raises the kinetic energy of liquid molecules. Immediately after heating, more molecules have enough energy to escape from the surface, so the evaporation rate increases. The vapour pressure also tends to rise, and condensation may later increase until a new dynamic equilibrium is reached. Evaporation does not become zero or remain unchanged.

For a volatile component, p_i = x_i p_i° in an ideal solution. A large pure vapour pressure p_i° can therefore produce a large partial pressure, depending also on its mole fraction. Since total pressure is the sum of partial pressures, this component may substantially raise it. It does not erase the other mole fraction or make the mixture non-volatile.

Adding a non-volatile solute lowers the solvent mole fraction and hence lowers the solution vapour pressure at a given temperature. Boiling still requires vapour pressure to equal external pressure, so the solution must be heated more than the pure solvent to reach that condition. The solute does not boil quickly, and neither mass loss nor infinite pressure explains the elevation.

Before calculating, identify which component contributes to the vapour phase and determine its mole fraction. For an ideal solution, Raoult’s law then gives its partial pressure as p_i = x_i p_i°. If more than one component is volatile, calculate each partial pressure and add them. Container colour, wording length and name length have no relevance.

At equilibrium, vapour pressure is determined by the temperature and the nature of the liquid, because these control molecular escape and condensation tendencies. If sufficient liquid remains, changing its amount or the vessel size does not change the equilibrium pressure at fixed temperature; it changes only how much vapour is present. Surface colour is not a governing thermodynamic property.

A larger surface area can increase the initial rate of evaporation, so equilibrium may be reached sooner. However, at fixed temperature, the equilibrium vapour pressure is determined by the liquid's nature and temperature, provided liquid remains present. Surface area affects rate, not the final equilibrium pressure.

For an ideal binary solution, Raoult’s law gives pA = xA pA° and pB = xB pB°. Since xB = 1 − xA, the total pressure is ptotal = xA pA° + (1 − xA)pB°, which is a linear expression in xA. Therefore the total-pressure versus composition graph is a straight line joining the pure-component pressures, not a parabola or discontinuous curve.

Use Raoult’s law for both volatile components. The mole fraction of B is xB = 1 − 0.40 = 0.60. Thus pA = 0.40 × 200 = 80 kPa and pB = 0.60 × 100 = 60 kPa. The total pressure is the sum of the partial pressures: ptotal = 80 + 60 = 140 kPa. Hence option B is correct.

Raoult's law is followed most closely when replacing a molecule of one component by a molecule of the other produces little change in intermolecular interactions. Similar molecular size, shape and polarity make like-like and unlike attractions comparable. Stronger or weaker unlike attractions cause negative or positive deviations, respectively.

At constant temperature, a liquid-vapour system with some liquid present re-establishes the saturation equilibrium after a volume change. If volume increases, some liquid evaporates; if it decreases, some vapour condenses. The final pressure therefore returns approximately to the temperature-dependent equilibrium vapour pressure, not a value proportional to volume.

Positive deviation means the mixture has a vapour pressure higher than the ideal prediction because unlike attractions are relatively weak. A pressure-composition maximum corresponds, at constant external pressure, to a temperature minimum, producing a minimum-boiling azeotrope. Negative deviation instead favours a maximum-boiling azeotrope.

A negative deviation results from unusually strong unlike-molecule attraction, giving lower vapour pressure than Raoult's law predicts. At a fixed external pressure, a lower vapour pressure requires a higher temperature to boil. Thus the pressure minimum corresponds to a maximum in boiling temperature, characteristic of a maximum-boiling azeotrope.

Acetone and chloroform show negative deviation because specific interactions, including hydrogen bonding involving the chloroform hydrogen and acetone oxygen, make unlike molecules attract strongly. This lowers their escaping tendency and vapour pressure below the Raoult-law value. Hexane-heptane is closer to ideal, while water-oil is largely immiscible.

Benzene and toluene are structurally similar non-polar liquids with comparable molecular sizes and intermolecular forces. Consequently, unlike interactions are close to like interactions and the mixture can approximately obey Raoult's law. Acetone-water and ethanol-water have strong specific interactions, while water-chloroform shows marked non-ideality and limited miscibility.

Vapour pressure is the equilibrium pressure established by evaporation and condensation at a fixed temperature, usually considered in a closed system. Evaporation rate is a kinetic quantity and can change with surface area, air movement, humidity and temperature. Thus a larger surface may speed evaporation without changing the final equilibrium vapour pressure.

For a solution containing a non-volatile solute, Raoult’s law gives Δp/p° = xsolute. Thus, the actual lowering is Δp = xsolute × p° = 0.15 × 80 = 12 kPa. The value 0.15 is the relative lowering, whereas 12 kPa is the lowering expressed as pressure. Therefore, option A is correct; 68 kPa is the solution pressure, not the lowering.

The governing concepts are Raoult’s law and Dalton’s law. For an ideal binary solution, the partial pressure of each component is pᵢ = xᵢpᵢ°. Since xA = 0.25, the liquid mole fraction of B is xB = 1 − 0.25 = 0.75. Thus pA = 0.25 × 150 kPa = 37.5 kPa, and pB = 0.75 × 50 kPa = 37.5 kPa. Dalton’s law states that the total pressure is the sum of the partial pressures, so P = pA + pB = 37.5 + 37.5 = 75 kPa. Therefore option A is correct. The pure pressure of A cannot simply be used as the total pressure because A occupies only one-quarter of the liquid mole fraction. Similarly, adding the pure pressures without weighting would be incorrect.

The lowering is Δp = 72 − 54 = 18 kPa. Its relative value is Δp/p° = 18/72 = 0.25. For a solution containing a non-volatile solute, Raoult’s law states that the relative lowering equals the solute mole fraction, xsolute = Δp/p°. Hence xsolute = 0.25. The other values do not follow from the required ratio, so option A is correct.

Positive deviation usually reflects weaker unlike-molecule interactions than like-molecule interactions. On mixing, molecules may therefore pack less efficiently and the mixture can show a positive excess volume, meaning its volume may increase. This is a general possibility, not an absolute rule for every system. The other options incorrectly deny any volume change or solution formation.