4: Vapour Pressure

Atmospheric pressure decreases as altitude increases. A liquid boils when its vapour pressure becomes equal to the external pressure. At high altitude, water needs to develop only the lower surrounding pressure, so it reaches the boiling condition at a lower temperature. Its chemical composition does not change merely because of altitude, and its vapour pressure does not become zero. Therefore, lower external pressure is the correct explanation, option A.

At a fixed temperature, high vapour pressure means a large fraction of molecules can escape from the liquid into the vapour phase. This is associated with relatively weak intermolecular attractions and therefore high volatility. Strong attractions hold molecules in the liquid and usually produce lower vapour pressure; forces are not literally zero in a liquid.

A liquid with low vapour pressure has a small escaping tendency at a given temperature, usually because its intermolecular attractions are relatively strong. It must be heated more before its vapour pressure reaches one atmosphere, the condition for the normal boiling point. Thus its normal boiling point may be high, though pressure and molecular details also matter.

An ideal solution obeys Raoult’s law over the composition range and has approximately zero enthalpy and volume change on mixing. This behaviour occurs when like-like, unlike-unlike, and like-unlike interactions are nearly comparable. A large difference would produce positive or negative deviation; zero attraction is impossible for an ordinary condensed liquid.

Benzene and toluene are chemically similar non-polar liquids with comparable molecular sizes and intermolecular attractions. Replacing a benzene molecule by a toluene molecule therefore changes the interaction environment only slightly, so the mixture is close to ideal. Acetone–chloroform, water–nitric acid, and ethanol–water show stronger specific interactions and appreciable deviation.

Acetone and chloroform can form a specific attractive interaction, commonly described through hydrogen bonding involving the chloroform hydrogen and acetone oxygen. This strengthens unlike interactions, reduces the escaping tendency of molecules, and causes negative deviation from Raoult’s law. The components do not become completely non-volatile or disappear chemically.

Dalton’s law relates vapour-phase composition to partial pressure: y_A = p_A/p_total. Rearranging gives p_A = y_A p_total. Substitution yields p_A = 0.4 × 200 kPa = 80 kPa. The value 40 kPa would result from an incorrect factor, while 120 kPa corresponds to the complementary vapour fraction of 0.6. Since a component’s partial pressure is only part of the total pressure, 200 kPa is not the answer. Option B is correct.

The more volatile component has the higher pure vapour pressure. In p_total = x_A p°_A + x_B p°_B, increasing its mole fraction replaces some of the less volatile component and increases the weighted average pressure. Thus total pressure rises toward the pure pressure of the more volatile component; it does not become independent of composition.

The less volatile component has the lower pure vapour pressure. Increasing its liquid mole fraction replaces some of the more volatile component, whose contribution to total pressure is larger. The weighted-average total pressure therefore decreases toward the lower pure-component pressure. This conclusion assumes the ideal-solution model and a fixed temperature.

At a given temperature, the more volatile component has a larger pure vapour pressure. Raoult’s law therefore gives it a larger partial pressure for a comparable liquid mole fraction. By Dalton’s law, its vapour-phase mole fraction is its partial pressure divided by total pressure, so it is generally enriched in vapour relative to the liquid. It still remains partly in the liquid.

Raising temperature increases the average kinetic energy and broadens the molecular energy distribution. More molecules then have enough energy to overcome cohesive forces and escape from the liquid surface. At the new equilibrium, a larger vapour population is present and the pressure is higher. Temperature does not change the molecular molar mass or necessarily the chemical formula.

A 25% lowering means that 25% of the pure-solvent pressure is subtracted. The amount lowered is 0.25 × 80 = 20 kPa. Therefore, the solution vapour pressure is p = 80 − 20 = 60 kPa, or directly p = 0.75 × 80 = 60 kPa. The value 20 kPa is the decrease itself, not the final pressure. Hence option B is correct, while 80 kPa would indicate no lowering.

Raoult’s law is followed most closely when the two liquids have similar size, polarity, and intermolecular attractions. Replacing a molecule of one component with the other then causes little change in the liquid environment. Very weak or very strong unlike attractions create positive or negative deviations, while chemical reaction may produce a different system altogether.

Vapour pressure is a pressure exerted by vapour molecules, so it must be expressed in a pressure unit. The pascal is the SI unit of pressure, and kilopascal is a commonly used multiple. Mole fraction is dimensionless, gram per mole is a molar-mass unit, and mole per kilogram is a molality unit. None of those describes pressure. Therefore, kilopascal is the only appropriate choice, making option A correct.

Equilibrium vapour pressure is established by the balance between evaporation and condensation at a given temperature. The liquid’s molecular attractions and composition determine how readily molecules escape. If some liquid remains, changing its amount usually does not change the equilibrium pressure, whereas adding a non-volatile solute lowers it. Boiling occurs when this pressure equals the external pressure.

A non-volatile solute does not enter the vapour phase. Its presence reduces the mole fraction and escaping tendency of solvent molecules. According to Raoult’s law, the solution vapour pressure is lower than that of the pure solvent. It does not normally become zero, and it does not necessarily rise again.

For a non-volatile solute, relative lowering of vapour pressure equals the solute mole fraction. Adding more effective solute particles generally increases that mole fraction, while reducing the solvent mole fraction. Therefore the solution vapour pressure falls more relative to the pure solvent. The exact relationship assumes the usual dilute, approximately ideal solution conditions.

A non-volatile solute contributes negligibly to the vapour phase, while its dissolved particles lower the mole fraction of the solvent. Raoult’s law therefore gives psolution = xsolvent p⁰solvent, with xsolvent less than one, so the solution vapour pressure is lower than that of the pure solvent. It is not necessarily zero and does not remain unchanged.

Adding a non-volatile solute lowers the solution’s vapour pressure. Boiling requires vapour pressure to equal the external pressure, so the solution must be heated to a higher temperature. This is elevation in boiling point. The other options describe unrelated optical, phase, or acid-base phenomena.

A non-volatile solute has negligible vapour pressure at the temperature considered, so it contributes essentially no vapour to the gas phase. The solvent molecules therefore provide almost all of the solution’s vapour pressure, although their partial pressure is lower than in the pure solvent because their mole fraction is reduced.

A non-volatile solute does not enter the vapour phase appreciably. Its particles occupy part of the liquid surface and reduce the mole fraction of solvent. Consequently, fewer solvent molecules can escape into the vapour phase at a given temperature, so the equilibrium vapour pressure falls. The solvent mass does not become zero, and the solute does not necessarily vaporise.

For a solution containing a non-volatile solute, Raoult's law is p = x_solvent p°_solvent. Here p is the solution vapour pressure, x_solvent is the mole fraction of the solvent, and p° is the vapour pressure of the pure solvent. The solute contributes no appreciable vapour pressure, so colour, smell and container dimensions are irrelevant.

When a non-volatile solute is dissolved in a solvent, the solvent mole fraction decreases and fewer solvent molecules escape into the vapour phase. The resulting solution therefore has a lower vapour pressure and is generally less volatile than the pure solvent at the same temperature. Rusting, candle burning and glass breaking involve different chemical or physical processes and do not directly illustrate this colligative effect.

A liquid boils when its vapour pressure becomes equal to the external pressure. If the solution vapour pressure is lower at every given temperature, it must be heated to a higher temperature before reaching that external pressure. Thus its boiling point increases. This is why a non-volatile solute produces boiling point elevation; colour and container appearance do not determine the boiling condition.

For a non-volatile solute, the solution vapour pressure is p = x_solvent p°_solvent. Since adding solute makes x_solvent less than one, p is lower than p°_solvent at the same temperature. The pure solvent has only solvent molecules contributing to escape, whereas the solute contributes no vapour pressure of its own.