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Vapour Pressure

General Chemistry • Liquids and Solids

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Liquids and solids

Vapour pressure connects particle motion to boiling behavior

Vapour pressure is the pressure produced by particles that escape from a liquid into the gas phase when evaporation and condensation reach dynamic equilibrium.

Learning target

  • Explain evaporation and condensation at the particle level.
  • Predict how temperature and intermolecular forces affect vapour pressure.
  • Compare volatility and connect vapour pressure to boiling point.
  • Use the Clausius-Clapeyron relationship as a temperature-pressure model.
Closed container with liquid and vapour particles A liquid at the bottom of a sealed container has particles escaping upward and returning downward as gas particles. Sealed container Liquid phase Gas phase evaporation condensation

Why it matters

Vapour pressure explains everyday observations and laboratory techniques

A liquid’s vapour pressure tells us how easily its particles enter the gas phase. That single idea helps explain boiling, distillation, smell, evaporation rates, and why some liquids must be stored in tightly sealed containers.

Boiling

Boiling begins when pressure matches outside pressure

A liquid boils when its vapour pressure equals the external pressure. Lower external pressure means a lower boiling temperature.

Distillation

Different liquids separate because they vaporize differently

A more volatile liquid usually has a higher vapour pressure and tends to enter the gas phase more readily.

Storage

Sealed systems matter

In a closed container, escaping particles build pressure. In an open container, many vapour particles leave the system.

Microscopic cause
Particles with enough kinetic energy overcome attractions and escape the liquid surface.

causes

Macroscopic result
The gas above the liquid exerts pressure on the container walls.

Core concept

Vapour pressure is an equilibrium pressure

In a closed container, evaporation does not continue forever. As more vapour particles collect above the liquid, some collide with the surface and condense. At equilibrium, the rate of evaporation equals the rate of condensation.

Dynamic equilibrium model Two equal arrows represent equal evaporation and condensation rates in a closed container. Closed system vapour particles liquid particles evaporation rate condensation rate dynamic equilibrium

What “dynamic” means

Particles are still moving. Some continue to evaporate, and some continue to condense. The system looks stable macroscopically because the two rates are equal.

\[ \text{rate of evaporation} = \text{rate of condensation} \]

At a fixed temperature, this balance produces a characteristic vapour pressure for that liquid.

Vocabulary and variables

The important words describe particles, pressure, and escaping tendency

A strong vapour-pressure explanation uses both molecular language and measurable quantities.

Term or variable Meaning Common units or comparison Particle-level interpretation
Vapour pressure, \(P_{\text{vap}}\) Pressure exerted by vapour above a liquid at equilibrium. atm, kPa, mmHg, torr More gas particles above the liquid means more collisions with the container walls.
Temperature, \(T\) Measure of average kinetic energy. K for gas-law and Clausius-Clapeyron calculations Higher temperature increases the fraction of particles energetic enough to escape.
Volatility Tendency of a substance to vaporize. Compared qualitatively as low or high High volatility usually means weaker attractions and higher vapour pressure.
\(\Delta H_{\text{vap}}\) Enthalpy required to vaporize one mole of liquid. kJ/mol Larger values mean stronger attractions must be overcome.

Key habit

For temperature equations, convert Celsius to Kelvin before calculating: \(T(\text{K}) = T(^{\circ}\text{C}) + 273.15\).

Main relationships

Temperature raises vapour pressure, while stronger attractions lower it

Vapour pressure depends mainly on how much kinetic energy particles have and how strongly they attract one another.

Clausius-Clapeyron connection

For many introductory problems, the temperature dependence of vapour pressure is modeled by:

\[ \ln\left(\frac{P_2}{P_1}\right) = -\frac{\Delta H_{\text{vap}}}{R} \left(\frac{1}{T_2}-\frac{1}{T_1}\right) \]

Here \(P_1\) and \(P_2\) are vapour pressures at Kelvin temperatures \(T_1\) and \(T_2\), and \(R = 8.314\ \text{J mol}^{-1}\text{K}^{-1}\).

Increase \(T\)

More particles can escape

The high-energy tail of the particle energy distribution becomes larger.

Increase attractions

Fewer particles can escape

Stronger intermolecular forces hold particles in the liquid more effectively.

Reach boiling

\(P_{\text{vap}} = P_{\text{external}}\)

Normal boiling point occurs when \(P_{\text{vap}} = 1\ \text{atm}\).

Interactive simulation

Change temperature and attraction strength to model vapour pressure

Move the sliders. Higher temperature increases escape, while stronger intermolecular forces reduce escape. The index below is a teaching model, not an exact experimental pressure.

Relative \(P_{\text{vap}}\) 42
Escaping tendency moderate
Volatility medium

At moderate temperature and moderate attractions, evaporation and condensation can balance with a moderate vapour pressure.

Interactive vapour pressure simulation A liquid in a sealed container with gas particles, evaporation arrows, condensation arrows, and a vapour pressure gauge. closed container liquid particles held by attractions vapour particles create pressure escape return vapour pressure high low 42 / 100

Graph reasoning

Vapour pressure increases nonlinearly with temperature

The same temperature change has a larger pressure effect at higher temperatures. Liquids with weaker attractions sit higher on the vapour-pressure curve.

Highlight a liquid type

Medium-attraction liquids have moderate vapour pressure. Their boiling point is higher than a weak-attraction liquid but lower than a strong-attraction liquid.

When \(P_{\text{vap}}\) reaches the external pressure, bubbles can form throughout the liquid and boiling occurs.

Vapour pressure curves versus temperature Three upward-curving vapour pressure curves show weak, medium, and strong intermolecular attractions. external pressure high low temperature vapour pressure weak attractions medium attractions strong attractions lower boiling point higher boiling point

Worked example

Estimate vapour pressure at a new temperature

A liquid has \(P_1 = 44.6\ \text{mmHg}\) at \(20.0^{\circ}\text{C}\). Its \(\Delta H_{\text{vap}}\) is \(38.6\ \text{kJ/mol}\). Estimate \(P_2\) at \(35.0^{\circ}\text{C}\).

  1. Convert temperatures to Kelvin.

    \(T_1 = 293.15\ \text{K}\) and \(T_2 = 308.15\ \text{K}\).

  2. Use joules for \(\Delta H_{\text{vap}}\).

    \(38.6\ \text{kJ/mol} = 38600\ \text{J/mol}\), so the units match \(R = 8.314\ \text{J mol}^{-1}\text{K}^{-1}\).

  3. Substitute into Clausius-Clapeyron.

    \(\ln(P_2/44.6) = -(38600/8.314)(1/308.15 - 1/293.15)\).

  4. Solve for the pressure ratio.

    \(\ln(P_2/44.6) \approx 0.771\), so \(P_2/44.6 \approx e^{0.771} \approx 2.16\).

  5. Calculate the final pressure.

    \(P_2 \approx 44.6 \times 2.16 = 96.3\ \text{mmHg}\).

Reasonableness check

The temperature increased, so the vapour pressure should increase. The estimated value is higher than \(44.6\ \text{mmHg}\), so the direction makes sense.

Common misconception

High vapour pressure does not mean strong intermolecular forces

A common mistake is to think that high pressure above a liquid means the liquid particles strongly attract one another. For vapour pressure, the relationship is usually the opposite.

Mistake

“This liquid has high vapour pressure, so its particles must be strongly attracted.”

Correction

High vapour pressure usually means particles escape easily, so the intermolecular attractions are relatively weak.

Comparison of weak and strong attractions Weak attractions lead to many vapor particles and higher pressure. Strong attractions lead to few vapor particles and lower pressure. Weak attractions many escaped particles, high vapour pressure Strong attractions few escaped particles, low vapour pressure

Practice check

Predict vapour pressure from particle-level evidence

Two liquids are compared at the same temperature in identical closed containers.

Liquid A

Particles have relatively weak intermolecular attractions. Many particles are observed in the gas phase at equilibrium.

Liquid B

Particles have strong intermolecular attractions. Only a few particles are observed in the gas phase at equilibrium.

Question

Which liquid has the higher vapour pressure, and which liquid is expected to have the higher normal boiling point?

Show answer

Liquid A has the higher vapour pressure because more particles escape into the gas phase at the same temperature. Liquid B is expected to have the higher normal boiling point because stronger attractions require a higher temperature before \(P_{\text{vap}}\) reaches \(1\ \text{atm}\).

Apply the topic

Use vapour pressure reasoning in calculations and conceptual questions

A good solution connects three levels: particle attractions, energy distribution, and the measured pressure above the liquid.

How to apply this topic in future problems

First identify whether the problem is asking about temperature, intermolecular forces, volatility, boiling point, or a numerical pressure change. Then choose particle reasoning, graph reasoning, or the Clausius-Clapeyron equation.

Final summary

The essential takeaways

Vapour pressure is an equilibrium pressure.

It forms in a closed container when evaporation and condensation occur at equal rates.

Temperature increases vapour pressure.

Higher temperature gives more particles enough energy to escape from the liquid surface.

Stronger attractions lower vapour pressure.

Particles are held more tightly, so fewer enter the gas phase at the same temperature.

Boiling occurs when pressures match.

A liquid boils when \(P_{\text{vap}}\) equals the external pressure.

Volatility follows escaping tendency.

More volatile liquids usually have higher vapour pressure and lower boiling points.

Use Kelvin in calculations.

The Clausius-Clapeyron relationship requires absolute temperature values.