4: Vapour Pressure

Stronger attractions between unlike molecules stabilize the liquid mixture more than expected. Fewer molecules can escape into the vapour phase, so the observed vapour pressure falls below the Raoult-law value. This is negative deviation. Strong attraction does not automatically mean no deviation or compulsory boiling; the actual result depends on comparison with the ideal prediction.

Raoult’s law gives the ideal vapour pressure expected from the composition. If the measured or actual vapour pressure is higher than this value, the solution shows a positive deviation. This usually indicates that unlike-molecule attractions are weaker than the corresponding like-molecule attractions, allowing particles to escape more readily. Negative deviation would mean an actual pressure lower than the ideal value.

The value calculated from Raoult's law is the ideal expectation. When the experimental vapour pressure is lower than that expectation, the difference is negative and the solution shows negative deviation. Stronger unlike-molecule attractions often cause this effect. It should not be confused with the rate of evaporation or with the separate event of boiling.

At a fixed temperature, the equilibrium vapour pressure of a pure liquid depends mainly on the liquid's nature and temperature, not on how much liquid is present, provided some liquid remains. Increasing the amount may increase the time or total vapour needed to reach equilibrium, but it does not double the final pressure. If all liquid disappears, the equilibrium condition changes.

At the same temperature, adding a non-volatile solute lowers the mole fraction of water and therefore lowers its partial vapour pressure. A larger effective amount of such solute produces a greater lowering, assuming a comparable solution model. Pure water has the highest vapour pressure among these choices, and heating pure water raises rather than lowers its vapour pressure.

In an ideal solution, the solvent's partial vapour pressure is P_solvent = X_solvent P°_solvent. When X_solvent decreases, fewer solvent molecules are statistically available at the surface and the partial pressure falls proportionally. Their energy does not become zero, the solvent is not destroyed, and external pressure does not enter the liquid as a substance.

A pressure below the Raoult-law value is negative deviation. Strong attraction between unlike molecules holds them more firmly in the liquid phase, so fewer molecules escape into the vapour. Weak unlike attraction would favour positive deviation and a higher pressure. The other choices are physically incorrect or contradict the definition of a solution.

An ideal solution obeys Raoult's law for each component throughout the composition range: PA = XA PA° and PB = XB PB°. Its unlike-molecule interactions are comparable to like-molecule interactions, not absent. Ideal solutions are not restricted to solid solutes, and colour has no role in identifying ideality.

In a closed vessel, liquid molecules continuously evaporate while vapour molecules continuously condense back into the liquid. At equilibrium, these two opposing rates become equal, so the macroscopic vapour pressure remains constant even though molecular movement continues. Equilibrium does not mean that particles stop or that one of the two processes disappears.

Adding solute increases the total number of moles while the moles of solvent remain unchanged. Therefore Xsolvent = nsolvent/ntotal decreases. Raoult's law then gives P = Xsolvent P°, so the solvent vapour pressure decreases as well. The solvent need not gain mass, and a non-volatile solute does not make it more volatile.

Dissociation changes one formula unit into several solute particles. Colligative effects depend on the effective number of particles, so dissociation increases the particle count and therefore increases the lowering of vapour pressure. In calculations this is represented by a van’t Hoff factor greater than one for substantial dissociation. It does not make the solution pressure exceed that of the pure solvent.

Lowering of vapour pressure is a colligative property, so it depends on the effective number of solute particles rather than simply on the original number of formula units. During association, several solute particles combine and the effective particle count decreases. The mole fraction of solute therefore becomes smaller than expected, and the vapour-pressure lowering decreases. Dissociation would have the opposite effect by increasing the number of particles.

An ideal solution is defined as one in which each component obeys Raoult’s law throughout the complete composition range. The unlike interactions are approximately comparable to the like interactions, so mixing produces no significant enthalpy or volume change. Ideal does not mean that all intermolecular forces vanish, and its vapour pressure is not generally zero.

Raoult’s law describes ideal behaviour, where interactions between unlike molecules are similar to those between like molecules. In a non-ideal solution, mixing may make unlike attractions substantially weaker or stronger. This changes how readily molecules escape and causes positive or negative deviation. The mole-fraction sum and pressure units are mathematical or descriptive facts, not causes of deviation.

For a solution with a non-volatile solute, relative lowering equals the solute mole fraction: (P° − P)/P° = X_solute. Thus X_solute = 0.2. Since solvent and solute mole fractions add to one, X_solvent = 1 − X_solute = 1 − 0.2 = 0.8. Therefore option B is correct. The value 0.2 belongs to the solute, while 1.2 and 2.0 cannot be mole fractions.

For dilute solutions under comparable conditions, lowering of vapour pressure depends on the number of solute particles, not their chemical identity. Equal mole amounts of non-volatile solutes that do not dissociate produce equal numbers of dissolved particles. Therefore, the lowering will be nearly the same, assuming the solvent, temperature, and concentration basis are comparable. Colour does not affect it.

For a pure liquid or solution at equilibrium, vapour pressure is determined mainly by temperature and composition, provided some liquid phase remains. Changing the amount of liquid then changes the time needed to establish equilibrium, but not the equilibrium pressure. If all liquid disappears, or the system is open and vapour escapes, the stated equilibrium condition no longer applies reliably.

A lower-than-ideal vapour pressure means that fewer molecules escape to the vapour phase than Raoult’s law predicts. A common molecular explanation is stronger attraction between unlike components, which holds the particles more tightly in the liquid. This is associated with negative deviation. Easier escape would increase pressure and indicate positive deviation instead.

For an ideal binary solution, apply Raoult’s law separately to each volatile component. For A, P_A = X_A P_A° = 0.5 × 100 = 50 units. For B, P_B = X_B P_B° = 0.5 × 200 = 100 units. Dalton’s law then gives P_total = P_A + P_B = 50 + 100 = 150 units. Therefore option C is correct; 50 and 100 are only partial pressures.

Raoult’s law gives PA = XA P°A and PB = XB P°B. At constant temperature, the pure-component pressures remain fixed, so each partial pressure changes directly with its own liquid-phase mole fraction. Increasing XA therefore raises PA, while decreasing XB lowers PB. The two pressures do not both move in the same direction merely because the solution remains binary.

Raoult’s law uses the liquid-phase mole fraction and the vapour pressure of the corresponding pure component. Therefore, first identify which mole fraction belongs to which component and note the relevant pure pressure. Then calculate the partial pressure, and add partial pressures if both components are volatile. Blind addition, colour, and ignoring temperature can lead to invalid reasoning.

In a binary solution, the mole fractions satisfy xA + xB = 1. Therefore, when xA increases, xB must decrease. For an ideal solution, Raoult’s law gives pB = xB p°B; at fixed temperature, p°B remains constant. A decrease in xB consequently decreases the partial pressure of B. The total pressure may behave differently, but B’s own partial pressure decreases.

Vapour pressure is the pressure exerted by vapour when the liquid and vapour phases are in dynamic equilibrium at a specified temperature. In an open vessel, vapour molecules continuously escape into the surroundings, so the vapour does not accumulate to establish a stable equilibrium pressure. The liquid still evaporates and contains molecules; the difficulty is the absence of a closed system, not zero temperature.

An ideal solution has nearly equal A–A, B–B and A–B intermolecular interactions. Mixing then causes little change in enthalpy or volume, and each component obeys pA = xA pA° and pB = xB pB°. The components must contact each other; a mixture of only solids is not this vapour-pressure situation, and zero temperature is not the defining criterion.

Positive deviation means the observed partial or total vapour pressure is greater than the value predicted by Raoult’s law. This usually results when unlike-molecule attractions are weaker than the corresponding like-molecule attractions, allowing molecules to escape more readily. Therefore the pressure is higher; a lower pressure would represent negative deviation.