4: Vapour Pressure

Boiling begins when the equilibrium vapour pressure of the liquid becomes equal to the external pressure. If vapour pressure is still lower, vapour bubbles inside the liquid cannot be sustained against the external pressure, so the liquid has not reached its boiling condition. Evaporation at the surface may still occur; option A does not say that vapour formation is impossible.

Low vapour pressure means that relatively few molecules escape from the liquid at a given temperature. Strong intermolecular attractions hold the molecules more firmly and reduce their escaping tendency. The liquid must then be heated more to make its vapour pressure equal external pressure, so it generally has a higher boiling point. Thus B is correct.

Boiling occurs when the liquid’s equilibrium vapour pressure equals the pressure exerted externally. At this point, vapour bubbles formed throughout the liquid can withstand the surrounding pressure and grow to the surface. Surface evaporation may occur at lower temperatures, but equality of the two pressures specifically identifies the boiling condition. Hence option A is correct.

Atmospheric pressure decreases with altitude. Boiling begins when water’s vapour pressure equals the surrounding atmospheric pressure. Because the required external pressure is lower on a hill station, water reaches equality at a lower temperature. The molecular mass and intrinsic vapour-pressure relation of water do not suddenly change, and the pressure certainly does not become zero.

The ideal line represents the pressure predicted by Raoult’s law. If the actual graph lies above it, the actual vapour pressure is higher than the ideal value at that composition. This is positive deviation, usually associated with weaker unlike-molecule attractions and easier escape into the vapour phase. A graph below the line would indicate negative deviation.

The ideal line gives the Raoult-law pressure. A graph below it means the actual vapour pressure is lower than the ideal prediction, which is negative deviation. This commonly indicates stronger attractions between unlike molecules, holding particles in the liquid more effectively. It does not imply that the liquid is absent or that complete vaporisation has occurred.

Vapour pressure is a pressure, so its SI unit is the pascal (Pa); kilopascal, bar, or other pressure units may also be used. Mole fraction is dimensionless because it is a ratio of amounts of substance. Mole per kilogram is a molality unit, and gram per mole is a molar-mass unit. These latter quantities describe composition or mass, not pressure.

Raoult’s law for component A is p_A = x_A p_A°, where x_A is its mole fraction in the liquid and p_A° is its pure vapour pressure. Substituting the given values gives p_A = 0.2 × 100 = 20 kPa. The pure pressure of 100 kPa would apply only when A is the pure liquid, x_A = 1. The other values do not follow from the stated fraction.

For a binary vapour mixture, Dalton’s law states that total pressure equals the sum of the two partial pressures: P = p_A + p_B. Rearranging gives p_B = P − p_A = 150 − 60 = 90 kPa. Adding 60 and 90 returns the stated total of 150 kPa. The total pressure itself is not the unknown component pressure, and 210 kPa would incorrectly add the total and partial pressure.

A 25% lowering means that the decrease in pressure is 25% of the pure solvent pressure. The decrease is therefore (25/100) × 200 = 50 kPa. The solution pressure is obtained by subtracting this decrease from the original pressure: P_solution = 200 − 50 = 150 kPa. Thus, option B is correct. The value 50 kPa is only the amount of lowering, not the final vapour pressure.

A non-volatile solute has negligible vapour pressure under the stated conditions, so it does not directly contribute a partial pressure to the vapour. The observed pressure mainly comes from the solvent and is reduced because the solvent mole fraction is less than one. The solvent pressure does not increase, and a mole fraction can never exceed one.

A higher pure vapour pressure at the same temperature means greater volatility and a greater tendency to enter the vapour phase. During distillation, the vapour is therefore enriched in that component, while the liquid residue becomes relatively richer in the less volatile component. The more volatile liquid is not restricted to the residue or a solid phase.

The solvent mole fraction is x_solvent = 1 − 0.05 = 0.95. For a non-volatile solute, Raoult’s law gives the solution pressure as p = x_solvent p° = 0.95 × 60 = 57 kPa. Equivalently, the lowering is 0.05 × 60 = 3 kPa, so p = 60 − 3 = 57 kPa. Thus 3 kPa is the lowering, while 65 kPa would incorrectly increase the pressure.

For an ideal solution, Raoult’s law gives the solvent’s partial vapour pressure as p₁ = x₁p₁⁰, where x₁ is the mole fraction of the solvent and p₁⁰ is the vapour pressure of the pure solvent at the same temperature. Thus, with temperature fixed, the partial pressure changes directly with x₁. The solute’s colour, container shape, and solution volume are not the governing quantities in this relation.

A non-volatile solute has negligible vapour pressure at the stated temperature, so it does not add appreciably to the vapour above the solution. Its particles occupy part of the liquid phase and reduce the mole fraction, and therefore the escaping tendency, of solvent molecules. By Raoult’s law, p_solvent = x_solvent p⁰_solvent, and x_solvent is less than 1 after adding solute; hence the solution vapour pressure is lower.

Both volatile components contribute molecules to the vapour phase. Dalton’s law states that the total pressure of a gas mixture is the sum of the partial pressures. Therefore, for two components, Ptotal = pA + pB; in an ideal solution these are obtained from Raoult’s law as xA pA° and xB pB°. Amount, boiling point, or density difference alone is not the formula.

An ideal solution obeys Raoult’s law throughout the complete composition range, not merely at one dilution. Its intermolecular attractions between unlike molecules are approximately comparable with those between like molecules, so the heat and volume changes of mixing are nearly zero. Ion formation is not required, and its vapour pressure is certainly not always zero.

Positive deviation means that the observed partial or total vapour pressure is greater than the value predicted by Raoult’s law. This commonly occurs when unlike intermolecular attractions are weaker than like attractions, allowing molecules to escape more easily. Such mixtures may show higher volatility and can form minimum-boiling azeotropes; the pressure is not independent of composition.

Negative deviation occurs when unlike molecules attract one another more strongly than the corresponding like molecules. The stronger attraction holds particles in the liquid and makes escape into the vapour phase more difficult. Consequently, the observed vapour pressure is lower than the Raoult-law value. Complete evaporation, precipitation, or zero mole fraction is not the defining cause.

For a solution containing a non-volatile solute, the solvent pressure is p = x_solvent p⁰_solvent. If only a very small amount of solute is present, x_solvent is close to 1, so p is close to p⁰. A large solute amount or a very low solvent mole fraction produces a larger lowering. “Beyond saturation” is not a normal condition that makes the vapour pressure closer to that of the pure solvent.

When unlike molecules attract weakly, mixing does not hold them as strongly as the original like-molecule interactions did. More molecules can escape from the liquid, so the actual vapour pressure exceeds the Raoult-law prediction. This is positive deviation. Negative deviation would require stronger unlike attraction, while ideal behaviour requires nearly equal attractions.

Vapour pressure measures the tendency of liquid molecules to enter the vapour phase at a specified temperature. A high value means that a relatively large fraction of molecules can overcome intermolecular attractions and escape. Such a liquid is called more volatile. Motionless molecules, infinite attraction, and exclusive existence as a solid would contradict the meaning of high vapour pressure.

In a closed container, liquid molecules continuously vaporise while vapour molecules continuously condense. At dynamic equilibrium, the two rates are equal, so the number of vapour molecules and the pressure remain constant, although molecular motion continues. At a fixed temperature, equilibrium vapour pressure is therefore steady with time; it does not continuously rise, fall, or become zero.

Raising temperature increases the average kinetic energy of liquid molecules. A larger number of molecules then possess enough energy to overcome cohesive forces and enter the vapour phase. The equilibrium concentration of vapour rises, so vapour pressure increases. Molecular mass and chemical formula do not change merely because temperature rises, and the liquid does not become non-volatile.

The boiling point is the temperature at which the vapour pressure of a liquid becomes equal to the pressure exerted on its surface. At this condition, vapour bubbles can form throughout the liquid and continue to rise, so the liquid boils rather than merely evaporating from its surface. Melting and freezing involve solid–liquid equilibrium, while evaporation can occur below the boiling point; therefore option A is correct.