4: Vapour Pressure

At the same temperature, adding a non-volatile solute lowers the solvent mole fraction and reduces the number of solvent molecules escaping from the surface. Raoult’s law therefore predicts a solution pressure below the pure-solvent pressure. If the solute were volatile, it could add a partial pressure, and a solvent mole fraction of one would describe pure solvent rather than a solution.

A liquid boils when its vapour pressure equals the external pressure. At the same external pressure, a liquid with higher vapour pressure reaches this equality at a lower temperature. Therefore, higher vapour pressure corresponds to a lower boiling point, while lower vapour pressure generally corresponds to a higher boiling point. The relation is not absent; it follows directly from the boiling condition.

For the solvent in an ideal solution, Raoult’s law is Psolvent = Xsolvent P°solvent. At a fixed temperature, the pure-solvent pressure is constant, so decreasing the solvent mole fraction decreases its partial vapour pressure in the same proportion. The pressure cannot become negative, and the relation does not remove temperature dependence; P° still depends on temperature.

For a volatile component in an ideal solution, Raoult’s law gives its partial vapour pressure as p = x p°, where x is its mole fraction and p° is its vapour pressure in the pure state. Thus, p = 0.75 × 80 = 60 kPa. The value is lower than 80 kPa because the solvent mole fraction is below one. Options B, C, and D do not follow this calculation.

A non-volatile solute does not appreciably enter the vapour phase. Its particles occupy part of the solution and reduce the mole fraction, and hence the escaping tendency, of solvent molecules. By Raoult’s law, the solvent pressure becomes xsolvent p°, which is below p°; the solute is not rapidly vaporising or destroying solvent.

Each volatile component contributes its own partial vapour pressure to the gas phase. Dalton’s law states that the total pressure is the sum of these partial pressures: Ptotal = pA + pB. For an ideal solution, each partial pressure is found from Raoult’s law, pA = xA pA° and pB = xB pB°. Amount or boiling point alone cannot replace this addition.

Vapour pressure measures the tendency of liquid molecules to escape into the vapour phase at a specified temperature. When intermolecular attractions are weak, less energy is needed for escape, so more molecules enter the vapour and the equilibrium pressure is higher. Strong attractions hold molecules back and generally lower vapour pressure; colour is irrelevant.

The relative lowering of vapour pressure is calculated using the expression (P° − P) / P°, where P° is the vapour pressure of the pure solvent and P is the vapour pressure of the solution. Here, the lowering is 100 − 90 = 10 kPa. Therefore, relative lowering = 10 / 100 = 0.10, or 10%. Hence, option A is correct. Option B is the remaining fraction of vapour pressure, while option C is the absolute lowering, not the relative value.

For a solution containing a non-volatile solute, only the solvent contributes appreciably to the vapour pressure. Raoult’s law leads to (p° − p)/p° = xsolute for a binary solution, and this relation is especially useful for dilute solutions. Therefore, relative lowering equals the mole fraction of solute. Mass, colour, and container volume do not define this colligative relation.

A colligative property depends primarily on the number of dissolved particles relative to solvent particles, not on the chemical identity of those particles. For a non-volatile solute, lowering of vapour pressure follows the solute mole fraction. Thus equal effective particle numbers produce comparable effects, while colour, container shape and the name of the solvent are not governing factors.

Raoult’s law states that the partial vapour pressure of the solvent is psolvent = xsolvent p°solvent. Substitution gives psolvent = 0.9 × 50 = 45 kPa. Since the solvent mole fraction is less than one, its pressure must be lower than the pure-solvent pressure of 50 kPa. Thus 55 and 90 kPa are impossible, while 5 kPa comes from an incorrect operation.

Boiling begins when the vapour pressure of a liquid becomes equal to the pressure exerted on its surface. At that point, vapour bubbles can form throughout the liquid instead of only at the surface. Evaporation does not stop; it continues along with boiling. Freezing and a change in mass are unrelated to this defining pressure condition.

Atmospheric pressure decreases with increasing altitude. Water boils when its vapour pressure equals the surrounding pressure, so at high altitude it needs to reach a smaller pressure and does so at a lower temperature. The water does not become non-volatile, and its mole fraction is not the reason. Vapour pressure still increases with temperature.

Increasing temperature raises the average kinetic energy of molecules, so more can escape into the vapour phase. Weak intermolecular attraction also makes escape easier. These two factors act in the same direction and produce high vapour pressure. Low temperature, strong attraction and non-volatility suppress escape; container colour has no direct role.

For a volatile component in an ideal solution, its partial pressure is calculated using p = x p°. Here x = 0.25 and p° = 120 kPa, so p = 0.25 × 120 = 30 kPa. A mole fraction of one-fourth produces one-fourth of the pure-component pressure at the same temperature. The other values either ignore the mole fraction or use an incorrect operation.

A non-volatile solute has negligible vapour pressure under the stated conditions, so it contributes essentially nothing to the vapour phase. The volatile solvent is therefore the principal, usually the only significant, contributor to total pressure. Its pressure is reduced from p° to xsolvent p° by the solute. Container walls and colour do not generate vapour pressure.

At a fixed temperature, a liquid with a greater tendency to escape into the gas phase establishes a higher equilibrium vapour pressure in a closed vessel. Faster evaporation is evidence of greater volatility and generally weaker effective intermolecular attraction. Therefore, the faster-evaporating liquid has the higher vapour pressure. Zero vapour pressure or absence of molecules cannot explain evaporation, and lower pressure would indicate slower escape.

In a pressure cooker, steam raises the pressure above the liquid. Water must then develop a vapour pressure equal to this higher external pressure, which requires a higher temperature. The boiling water is therefore hotter than at ordinary pressure, transferring heat faster to food. Pressure does not make water non-volatile or reduce its boiling point.

Adding more non-volatile solute increases the number of solute particles and lowers the mole fraction of the solvent. Raoult’s law gives psolvent = xsolvent p°, so a lower xsolvent produces a lower solvent vapour pressure at the same temperature. The pressure cannot return to the pure-solvent value unless the solute is removed or its fraction becomes negligible.

The lowering of vapour pressure is the direct difference between the vapour pressure of the pure solvent and that of the solution: Δp = p° − p. Substituting the given values gives Δp = 40 − 36 = 4 kPa. This question asks for absolute lowering, so no division by 40 is required. Dividing would instead calculate relative lowering, which would be 0.10.

For a dilute solution containing a non-volatile solute, the relative lowering of vapour pressure is approximately equal to the solute mole fraction: (p° − p)/p° ≈ xsolute. Since xsolute = 0.2, the relative lowering is approximately 0.2, or 20%. The solvent mole fraction would be 0.8, but it is not the quantity requested; values above one are not possible for a mole fraction.

For an ordinary dilute aqueous salt solution, the dissolved salt is treated as a non-volatile solute, while water is the volatile solvent. Adding salt lowers the mole fraction of water and, by Raoult’s law, lowers the water vapour pressure compared with pure water at the same temperature. Therefore, pure water has the higher vapour pressure. The conclusion assumes the usual ideal or dilute-solution treatment.

A solution containing solvent and one solute is a binary solution, so the mole fractions must add to one: xsolvent + xsolute = 1. Therefore, xsolute = 1 − 0.6 = 0.4. The fact that the solute is non-volatile affects vapour-pressure calculations, but it does not change the mole-fraction sum. A mole fraction of 1.6 is impossible, and 0.6 simply repeats the solvent value.

A 15% lowering means that the relative lowering of vapour pressure is 15/100 = 0.15. For a dilute solution with a non-volatile solute, relative lowering is approximately equal to the solute mole fraction. Hence xsolute ≈ 0.15. The value 0.85 is the corresponding solvent mole fraction, while 15 is still a percentage rather than the dimensionless decimal fraction required here.

Boiling occurs when vapour pressure equals external pressure. A liquid with higher vapour pressure reaches that required pressure at a lower temperature, so it generally has a lower boiling point and is more volatile. The word “generally” allows comparison at the same external pressure; the other statements reverse or deny this pressure-temperature relationship.