4: Vapour Pressure

Raoult’s law gives the solvent partial pressure as Psolvent = xsolventP°solvent. At constant temperature, the pure-solvent pressure is fixed, so decreasing xsolvent directly decreases Psolvent. The pressure need not become zero unless the solvent mole fraction becomes zero, and it does not increase because the relationship is direct, not inverse.

A non-volatile solute does not enter the vapour phase appreciably, but it decreases the mole fraction of the solvent. By Raoult’s law, the solvent vapour pressure becomes p = x solvent × p° solvent, which is lower than p° solvent when x solvent is less than one. The other options contradict the meaning of mole fraction or the existence of a solution.

For a solution containing a non-volatile solute, Raoult’s law gives the relative lowering of vapour pressure as the solute mole fraction: x_solute = (P° − P)/P°. Here, the lowering is 64 − 48 = 16 kPa, so x_solute = 16/64 = 0.25. Therefore, option A is correct. The value 0.75 is the solvent mole fraction, not the solute mole fraction.

For an ideal solution, the partial pressure of each component is pᵢ = xᵢpᵢ°. At equal liquid mole fractions, the component with the larger pure vapour pressure has the larger partial pressure. Vapour-phase mole fraction is proportional to partial pressure, so component A, being more volatile, will be richer in the vapour. B is not absent; it is simply less abundant.

Raoult’s law gives the ideal vapour pressure for a specified composition. If the observed pressure is higher than this ideal value, the solution shows positive deviation. This generally means unlike-molecule attractions are weaker than the corresponding like-molecule attractions, so molecules escape more readily. Negative deviation would instead give a pressure below the ideal value.

Negative deviation means that the measured vapour pressure is lower than the Raoult-law prediction. A usual molecular reason is that unlike molecules attract one another more strongly than the original like molecules did. More energy is then needed for molecules to escape from the liquid, so the escaping tendency and vapour pressure decrease. Weak attraction would cause positive deviation.

For a non-volatile solute, Raoult’s law gives the solution vapour pressure as p = x solvent p° solvent. Substituting the data gives p = 0.85 × 40 = 34 kPa. The value 6 kPa is the lowering, 40 kPa is the pressure of pure solvent, and 85 is a mole fraction expressed incorrectly as a pressure.

Equilibrium vapour pressure is defined for a liquid and its vapour in a closed container at a fixed temperature. At equilibrium, the rate of vaporisation equals the rate of condensation, so the vapour pressure remains constant. In an open vessel vapour escapes continuously and such equilibrium is not established. Freezing and absence of condensation are unrelated to the definition.

For a pure liquid in a closed vessel, equilibrium vapour pressure at a fixed temperature is determined mainly by the liquid’s nature and temperature, not by the amount present, provided some liquid remains. If the amount is reduced, some vapour condenses or evaporates until the same equilibrium pressure is restored. The amount affects how long equilibrium lasts, not its final pressure.

For an ideal solution, Raoult’s law for component A is p A = x A p° A. Thus p A = 0.20 × 180 = 36 kPa. The pure vapour pressure, 180 kPa, would apply when A is pure, while 20 is only the mole fraction and 90 does not follow from the given relation. The liquid mole fraction, not vapour mole fraction, is used.

For a gaseous mixture, Dalton’s law relates the mole fraction of a component to its partial pressure: y_A = p_A/P_total. The total pressure is 20 + 80 = 100 kPa, and therefore y_A = 20/100 = 0.20. Thus option A is correct. The value 0.80 corresponds to component B, while 4.00 is not a possible mole fraction because mole fractions lie between 0 and 1.

Positive deviation occurs when unlike molecules attract each other less strongly than expected from the original liquid interactions. The molecules are then held less firmly in the liquid and escape more easily, increasing the partial pressures and hence the total vapour pressure above the Raoult-law value. Stronger unlike attraction would lower the pressure and produce negative deviation.

An ideal solution obeys Raoult’s law over the whole composition range. Its unlike and like intermolecular interactions are approximately similar, so the enthalpy change of mixing and volume change on mixing are approximately zero. It therefore shows neither positive nor negative deviation. Statement A is incorrect because positive deviation is a property of a non-ideal solution, not an ideal one.

For a non-volatile solute, the solution vapour pressure is p = x solvent p° solvent. Therefore the relative lowering is (p° − p)/p° = (1 − x solvent) = x solute. This is a colligative relation because it depends on the number of dissolved particles, represented by their mole fraction, rather than their colour or the vessel volume.

For a non-volatile solute, only the solvent contributes to the vapour pressure, so Raoult’s law is P_solution = x_solvent P°_solvent. Since x_solute = 0.30, x_solvent = 1 − 0.30 = 0.70. Hence P_solution = 0.70 × 70 = 49 kPa. Option A is correct; 21 kPa is the pressure lowering, not the final vapour pressure.

Boiling begins when the equilibrium vapour pressure of a liquid becomes equal to the external pressure. Vapour pressure rises with temperature. If external pressure is lowered, this equality is reached at a lower temperature, so the boiling point decreases. The molecular mass and chemical formula do not suddenly change, and the vapour pressure does not become zero.

A non-volatile solute lowers the vapour pressure of the solvent at every given temperature because the solvent mole fraction is less than one. Boiling requires vapour pressure to equal the external pressure. The solution must therefore be heated to a higher temperature before this equality is reached, producing elevation of boiling point. The solute need not vaporise for this effect.

A negative-deviation mixture has a vapour pressure lower than the ideal Raoult-law value because unlike molecules attract strongly. Since boiling requires the vapour pressure to reach the external pressure, a higher temperature is needed. A sufficiently strong negative deviation can create a maximum-boiling azeotrope, whose vapour and liquid compositions are the same at that composition. Positive deviation instead favours minimum boiling.

Positive deviation means the actual vapour pressure is higher than the ideal value, usually because unlike-molecule attractions are weaker. The mixture can therefore reach external pressure at a lower temperature. When the deviation is sufficiently large, a minimum-boiling azeotrope may form, with identical liquid and vapour compositions at the azeotropic point. Negative deviation gives the opposite, maximum-boiling tendency.

In an ideal binary solution, the mole fractions add to one. Therefore, x_B = 1 − x_A = 1 − 0.40 = 0.60. Raoult’s law gives the partial pressure of B as p_B = x_B P°_B. Substituting the values, p_B = 0.60 × 120 = 72 kPa, so option A is correct. Using 0.40 would incorrectly use A’s mole fraction for B.

Equilibrium vapour pressure is the pressure of vapour in a closed liquid–vapour system when evaporation and condensation rates are equal; at a fixed temperature it is a state property. Evaporation rate describes how fast molecules leave a surface and depends on temperature, surface area, humidity and air movement. Thus the two quantities are related but not identical.

At the same temperature, a high vapour pressure means that a comparatively large fraction of molecules can escape from the liquid into the vapour phase. This generally indicates weaker intermolecular attractions, or weaker effective cohesion, than in a liquid with lower vapour pressure. It does not prove that the forces are absent or ionic; all real liquids retain intermolecular interactions.

A liquid boils when its vapour pressure equals the external pressure. If its vapour pressure is relatively low at ordinary temperatures, the temperature must be raised further to reach that external pressure. Such a liquid generally has a higher boiling point, although comparisons should be made at the same pressure and temperature conditions. Low vapour pressure does not mean that boiling is impossible.

Raoult’s law states that, for an ideal solution, the partial vapour pressure of component i is p i = x i p° i, where x i is its mole fraction in the liquid and p° i is its vapour pressure in the pure state at the same temperature. A partial pressure cannot exceed the total pressure, and it is directly related to liquid mole fraction.

Adding non-volatile solute increases the total number of solute particles and decreases the mole fraction of the solvent. According to Raoult’s law, p solvent = x solvent p° solvent, so the solvent’s vapour pressure decreases at a fixed temperature. The solute does not add its own vapour pressure because it is non-volatile. The decrease is not an increase or an infinite value.