Vapour Pressure: The Starting Point
A liquid kept in a closed vessel evaporates until an equilibrium is reached between the liquid phase and the vapour phase. The pressure exerted by the vapours of the liquid over the liquid surface under these equilibrium conditions is called the vapour pressure. This single quantity is the thread that runs through the entire Solutions unit: relative lowering of vapour pressure, boiling-point elevation, and the distinction between ideal and non-ideal mixtures all trace back to it.
NCERT restricts the discussion to binary solutions — two components only. The solvent is generally the volatile liquid; the solute may be volatile (another liquid) or non-volatile (a dissolved solid such as sucrose, urea, glucose, or sodium chloride). Raoult's law treats both situations from a single principle, so the two cases below are best learned together rather than as separate formulae.
| Quantity | Symbol | Meaning |
|---|---|---|
| Pure vapour pressure | p1°, p2° | Vapour pressure of each pure component at the given temperature |
| Partial vapour pressure | p1, p2 | Contribution of each component to the pressure over the solution |
| Liquid mole fraction | x1, x2 | Composition of the liquid solution, with $x_1 + x_2 = 1$ |
| Vapour mole fraction | y1, y2 | Composition of the vapour in equilibrium with the liquid |
| Total vapour pressure | ptotal | Sum of all partial pressures over the solution |
Raoult's Law for Two Volatile Liquids
Consider a binary solution of two volatile liquids, labelled component 1 and component 2. In a closed vessel both evaporate and an equilibrium is established. The French chemist François-Marie Raoult (1886) gave the quantitative relationship between the partial pressures and the composition. Raoult's law states that for a solution of volatile liquids, the partial vapour pressure of each component is directly proportional to its mole fraction present in the solution.
For component 1, $p_1 \propto x_1$, which on introducing the proportionality constant becomes the pure vapour pressure:
$$ p_1 = p_1^{\circ}\, x_1 $$and similarly for component 2:
$$ p_2 = p_2^{\circ}\, x_2 $$Here $p_1^{\circ}$ and $p_2^{\circ}$ are the vapour pressures of the pure components 1 and 2 at the same temperature. The law contains no adjustable constant beyond the pure-component vapour pressures, which is precisely why a solution that obeys it exactly is called an ideal solution.
"Mole fraction" means the liquid-phase mole fraction
In $p_i = p_i^{\circ}\, x_i$, the $x_i$ is the mole fraction in the solution (liquid), not in the vapour. The vapour composition ($y_i$) is a separate, derived quantity. Mixing these up is the single most common error in vapour-pressure problems.
Use $x_i$ to build partial pressures; use $y_i = p_i / p_\text{total}$ only afterwards.
Total Vapour Pressure and the Linear Plot
According to Dalton's law of partial pressures, the total pressure over the solution is the sum of the partial pressures:
$$ p_\text{total} = p_1 + p_2 = p_1^{\circ}\, x_1 + p_2^{\circ}\, x_2 $$Because $x_1 = 1 - x_2$, this can be written in terms of a single mole fraction:
$$ p_\text{total} = p_1^{\circ} + (p_2^{\circ} - p_1^{\circ})\, x_2 $$Three conclusions follow from this expression. The total pressure can be expressed in terms of the mole fraction of any one component; it varies linearly with $x_2$; and depending on which pure component has the higher vapour pressure, the total pressure rises or falls as $x_2$ increases. Taking component 1 as the less volatile one ($p_1^{\circ} < p_2^{\circ}$), the minimum value of $p_\text{total}$ is $p_1^{\circ}$ (at $x_2 = 0$) and the maximum is $p_2^{\circ}$ (at $x_2 = 1$).
Composition of the Vapour Phase
The vapour above the solution has its own composition, generally different from the liquid. If $y_1$ and $y_2$ are the mole fractions in the vapour phase, Dalton's law gives each partial pressure as the product of the vapour mole fraction and the total pressure:
$$ p_1 = y_1\, p_\text{total}, \qquad p_2 = y_2\, p_\text{total}, \qquad \text{or in general } \; p_i = y_i\, p_\text{total} $$Rearranging, $y_i = p_i / p_\text{total}$. Because the more volatile component (the one with the larger $p^{\circ}$) contributes a disproportionately large partial pressure, the vapour phase is always enriched in it relative to the liquid. This is the principle behind fractional distillation and a recurring NEET stem.
Vapour is always richer in the more volatile component
For an equimolar liquid mixture, the two liquid mole fractions are equal, so $y_i = p_i / p_\text{total}$ depends only on the partial pressures, which scale with the pure vapour pressures. The component with the higher $p^{\circ}$ therefore dominates the vapour. In NCERT Example 1.8, $\ce{CH2Cl2}$ ($p^{\circ} = 415$ mm Hg) gives $y = 0.82$ while $\ce{CHCl3}$ ($p^{\circ} = 200$ mm Hg) gives only $y = 0.18$.
Higher pure vapour pressure → more volatile → larger mole fraction in the vapour.
Worked Examples
The vapour pressures of chloroform $\ce{(CHCl3)}$ and dichloromethane $\ce{(CH2Cl2)}$ at 298 K are 200 mm Hg and 415 mm Hg respectively. Calculate (i) the vapour pressure of a solution prepared by mixing 25.5 g of $\ce{CHCl3}$ and 40 g of $\ce{CH2Cl2}$, and (ii) the mole fractions of each component in the vapour phase.
Step 1 — moles. Molar mass of $\ce{CH2Cl2} = 12 + 2 + 71 = 85\ \text{g mol}^{-1}$ and of $\ce{CHCl3} = 12 + 1 + 106.5 = 119.5\ \text{g mol}^{-1}$.
$$ n_{\ce{CH2Cl2}} = \frac{40}{85} = 0.47\ \text{mol}, \qquad n_{\ce{CHCl3}} = \frac{25.5}{119.5} = 0.213\ \text{mol} $$Step 2 — liquid mole fractions. Total moles $= 0.47 + 0.213 = 0.683$.
$$ x_{\ce{CH2Cl2}} = \frac{0.47}{0.683} = 0.688, \qquad x_{\ce{CHCl3}} = 1.000 - 0.688 = 0.312 $$Step 3 — total vapour pressure. Using $p_\text{total} = p_1^{\circ} + (p_2^{\circ} - p_1^{\circ})\,x_2$ with $\ce{CHCl3}$ as component 1:
$$ p_\text{total} = 200 + (415 - 200)\times 0.688 = 200 + 147.9 = 347.9\ \text{mm Hg} $$Step 4 — vapour-phase composition. First the partial pressures, then $y_i = p_i / p_\text{total}$:
$$ p_{\ce{CH2Cl2}} = 0.688 \times 415 = 285.5\ \text{mm Hg}, \qquad p_{\ce{CHCl3}} = 0.312 \times 200 = 62.4\ \text{mm Hg} $$ $$ y_{\ce{CH2Cl2}} = \frac{285.5}{347.9} = 0.82, \qquad y_{\ce{CHCl3}} = \frac{62.4}{347.9} = 0.18 $$Since $\ce{CH2Cl2}$ is the more volatile component, the vapour is richer in it ($y = 0.82$ versus a liquid fraction of $0.688$) — confirming the volatility rule.
A solution that obeys Raoult's law over the whole range is ideal; deviations define real mixtures. See Ideal & Non-Ideal Solutions.
Benzene $\ce{(C6H6)}$ and toluene $\ce{(C6H5CH3)}$ form a nearly ideal solution. At 25 °C their pure vapour pressures are 12.8 kPa and 3.85 kPa respectively. For an equimolar (1 : 1) liquid mixture, find the total vapour pressure and the vapour-phase composition.
Step 1 — liquid mole fractions. A 1 : 1 mixture gives $x_{\text{benzene}} = x_{\text{toluene}} = 0.5$.
Step 2 — partial and total pressures.
$$ p_{\text{benzene}} = 0.5 \times 12.8 = 6.40\ \text{kPa}, \qquad p_{\text{toluene}} = 0.5 \times 3.85 = 1.925\ \text{kPa} $$ $$ p_\text{total} = 6.40 + 1.925 = 8.325\ \text{kPa} $$Step 3 — vapour-phase composition.
$$ y_{\text{benzene}} = \frac{6.40}{8.325} = 0.77, \qquad y_{\text{toluene}} = \frac{1.925}{8.325} = 0.23 $$Although the liquid is exactly equimolar, the vapour contains roughly 77 % benzene because benzene is the more volatile component. This is the reasoning the NEET 2016 question below tests directly.
Raoult's Law for a Non-Volatile Solute
When the dissolved substance is a non-volatile solid — sucrose, urea, glucose, sodium chloride — only the solvent molecules enter the vapour phase. In a pure liquid the entire surface is occupied by solvent molecules; once solute is added, the solute particles occupy part of the surface, so fewer solvent molecules escape and the vapour pressure of the solution falls below that of the pure solvent.
Letting subscript 1 denote the solvent, the general form of Raoult's law gives the vapour pressure of the solution as the solvent's partial pressure alone:
$$ p_\text{total} = p_1 = p_1^{\circ}\, x_1 $$The lowering depends only on the quantity of non-volatile solute, not its chemical identity: 1.0 mol of sucrose and 1.0 mol of urea in the same mass of water lower the vapour pressure of water by nearly the same amount. This nature-independence is the basis of the colligative property treated in the sibling note on relative lowering of vapour pressure.
| Situation | Vapour-phase species | Working equation |
|---|---|---|
| Two volatile liquids | Both components | $p_\text{total} = p_1^{\circ}x_1 + p_2^{\circ}x_2$ |
| Non-volatile solute in volatile solvent | Solvent only | $p_\text{total} = p_1^{\circ}x_1$ |
| Gas dissolved in liquid (Henry) | The gas | $p = K_H\, x$ |
Raoult's Law as a Special Case of Henry's Law
According to Raoult's law, the partial pressure of a volatile component is $p_i = p_i^{\circ}\, x_i$. For a gas dissolved in a liquid, the solubility is governed instead by Henry's law, $p = K_H\, x$, where $K_H$ is the Henry's law constant. Comparing the two expressions shows that both make the partial pressure directly proportional to the mole fraction; only the proportionality constant differs — $K_H$ in one case, $p^{\circ}$ in the other.
Raoult's law is therefore a special case of Henry's law in which the constant $K_H$ becomes equal to the pure vapour pressure $p_1^{\circ}$. This unification is worth committing to memory: it ties the two proportional laws of the Solutions unit into a single statement and is a favourite single-line NEET assertion.
For the gas-in-liquid form and the meaning of $K_H$, read Henry's Law.
Raoult's Law in one screen
- Volatile–volatile: $p_1 = p_1^{\circ}x_1$, $p_2 = p_2^{\circ}x_2$; partial pressure $\propto$ liquid mole fraction.
- Total pressure: $p_\text{total} = p_1^{\circ}x_1 + p_2^{\circ}x_2 = p_1^{\circ} + (p_2^{\circ}-p_1^{\circ})x_2$ — linear in $x_2$.
- Vapour composition: $y_i = p_i / p_\text{total}$; the vapour is always richer in the more volatile component.
- Non-volatile solute: only the solvent contributes, $p_\text{total} = p_1^{\circ}x_1$; lowering is nature-independent.
- Raoult's law = special case of Henry's law with $K_H = p^{\circ}$.
- A solution obeying Raoult's law over the full range is ideal; minimum $p_\text{total} = p_1^{\circ}$, maximum $= p_2^{\circ}$.