Raoult's Law — Vapour Pressure of Solutions
Slide the composition and watch two volatile liquids build the total vapour pressure.
About this simulation
In an ideal solution each volatile component contributes a partial pressure equal to its pure vapour pressure times its mole fraction (pᴬ = xᴬ·p°ᴬ), so the total is a straight line between the two pure liquids. Students vary the mole fraction and the two pure vapour pressures, and toggle ideal / positive / negative deviation to see p_total bulge above or dip below the ideal line as A–B attractions differ from A–A and B–B. Dropping p°_B to near zero recovers the non-volatile-solute case — the relative lowering that underlies every colligative property — and the vapour composition y_A shows the vapour is always richer in the more volatile liquid.
How to use it
- Drag 'Mole fraction of A' from 0 to 1 and watch the purple total-pressure line run straight from pure B on the left to pure A on the right.
- Change 'Pure V.P. of A (volatile)' and 'Pure V.P. of B' and see each partial-pressure line pivot about zero at the end where that liquid is absent.
- Switch 'Behaviour' from Ideal to Positive or Negative and compare the total curve with the dashed ideal line that appears.
- Try the 'Non-volatile solute' preset, where B barely evaporates, and read how far the total pressure falls below p°A.
- At the defaults, read pA = 40.0 kPa and p = 55.0 kPa at the top of the stage: yA = 40 ÷ 55 ≈ 0.73, printed there as 'vapour yA = 0.73', richer in A than the 0.50 you set.
What to notice
- For an ideal solution the partial pressures pA and pB are straight lines, so the total vapour pressure is also a straight line between the two pure values.
- Positive deviation makes the total curve bulge above the dashed ideal line; negative deviation makes it sag below, while both ends stay fixed at the pure vapour pressures.
- The vapour is richer than the liquid in the more volatile component: with p°A larger than p°B, yA comes out bigger than xA.
- When B is almost non-volatile, the vapour pressure is nearly all pA, and the drop below p°A grows with the amount of B added.
The formulas
- pA = xA · p°A, pB = xB · p°B
- Partial pressure of each component (p) equals its mole fraction in the liquid (x) times its vapour pressure when pure (p°).
- p(total) = xA · p°A + xB · p°B = p°B + (p°A − p°B) · xA
- Total vapour pressure of an ideal solution, a straight line in the mole fraction of A.
- yA = pA / p(total)
- Mole fraction of A in the vapour above the liquid, from Dalton's law of partial pressures.
Questions
What is Raoult's law for a solution of two volatile liquids?
Raoult's law says that the partial vapour pressure of each volatile component equals its mole fraction in the solution multiplied by the vapour pressure of the pure component. The total pressure is the sum of the two partial pressures. In the simulation, keep Behaviour on Ideal and move the mole-fraction slider: both partial-pressure lines and the total line stay perfectly straight.
What is the difference between positive and negative deviation from Raoult's law?
In positive deviation the A–B attractions are weaker than A–A and B–B, so molecules escape more easily and the vapour pressure is higher than Raoult's law predicts; ethanol and acetone is a common example. In negative deviation A–B attractions are stronger, so vapour pressure is lower, as in acetone and chloroform. Switch Behaviour to Positive or Negative and compare the curve with the dashed ideal line.
Why does adding a non-volatile solute lower the vapour pressure of a solvent?
Only the solvent can escape into the vapour, and its partial pressure is its mole fraction times its pure vapour pressure. Adding solute lowers the solvent's mole fraction, so the vapour pressure falls. The relative lowering, (p° − p)/p°, equals the mole fraction of solute. Load the 'Non-volatile solute' preset and lower the mole fraction of A to see the total pressure drop.
Why is the vapour richer in the more volatile component than the liquid?
The more volatile liquid has the larger pure vapour pressure, so it contributes a bigger share of the total pressure than its share of the liquid. By Dalton's law its vapour mole fraction is its partial pressure divided by the total, which is larger than its liquid mole fraction. Set p°A above p°B and divide pA at the blue dot by the total beside the purple dot: the result is larger than the mole fraction of A you set.
What are the conditions for an ideal solution?
An ideal solution obeys Raoult's law at every composition. This happens when the attractions between unlike molecules are about the same as between like molecules, so mixing involves no heat change (ΔmixH = 0) and no volume change (ΔmixV = 0). Mixtures of very similar liquids, such as benzene and toluene or n-hexane and n-heptane, come close to ideal.
