Every liquid constantly loses molecules to the gas above it. In a closed container the vapor builds up until molecules return to the liquid as fast as they leave; the pressure of the vapor at that balance is the vapor pressure. It depends only on the substance and the temperature – not on how much liquid there is – and it rises steeply with temperature, roughly doubling for every 10–15 °C for many organic liquids.
Volatile liquids such as diethyl ether or acetone have high vapor pressures at room temperature because their molecules are held by weak forces; water, with its hydrogen bonds, has only about 3.2 kPa at 25 °C. When the vapor pressure reaches the surrounding pressure the liquid boils, which is why vapor pressure data also tell you boiling points at any pressure.
The Clausius–Clapeyron equation connects two points on the vapor-pressure curve through the molar enthalpy of vaporization. Built-in data for 14 substances supply the normal boiling point (where the vapor pressure is exactly 1 atm) as a reference. Adding a non-volatile solute lowers the vapor pressure in proportion to the solvent’s mole fraction – Raoult’s law, the fourth mode.
How to use the Vapor Pressure Calculator
- Calculate
- Substance
- Enthalpy of vaporization ΔHvap
- Known temperature T₁
- Vapor pressure at T₁
- New temperature T₂
- Target vapor pressure P₂
- Temperature of measurement 1
- Vapor pressure at temperature 1
- Temperature of measurement 2
- Vapor pressure at temperature 2
- Composition given as
- Mole fraction of the solvent x
- Moles of solvent
- Moles of solute particles (× van ‘t Hoff factor)
- Vapor pressure of the pure solvent
Results update as you type, every field has its own unit menu, and everything is calculated in your browser.
How it works
In Clausius–Clapeyron modes the temperatures are converted to kelvin; the vapor pressure at T₂ is P₁ · exp[(ΔHvap/R)(1/T₁ − 1/T₂)], and the other modes rearrange the same equation. For Raoult’s law the solvent’s mole fraction is multiplied by its pure vapor pressure; the difference is the vapor-pressure lowering.
Worked example
Ethanol at 25 °C, from its normal boiling point (351.44 K, 1 atm) and ΔHvap = 38,560 J/mol: (ΔHvap/R)(1/T₁ − 1/T₂) = 4,637.7 × (0.0028454 − 0.0033540) = −2.3588, so P = 1 atm × e^(−2.3588) = 0.0945 atm = 9.58 kPa. Measured: about 7.9 kPa – the equation overestimates because ΔHvap at 25 °C is higher than at 78 °C. For better accuracy, use a reference point closer to the target temperature in Custom mode.
ΔHvap mode: water at 25 °C (3.17 kPa) and 100 °C (101.325 kPa) gives ΔHvap = 8.314 × ln(101.325/3.17) / (1/298.15 − 1/373.15) ≈ 42.7 kJ/mol.
Reference data used for the presets
| Substance | Normal boiling point (1 atm) | ΔHvap at the boiling point |
|---|---|---|
| Water | 99.97 °C | 40.65 kJ/mol |
| Ethanol | 78.29 °C | 38.56 kJ/mol |
| Methanol | 64.7 °C | 35.21 kJ/mol |
| Acetone | 56.05 °C | 29.10 kJ/mol |
| Benzene | 80.09 °C | 30.72 kJ/mol |
| Toluene | 110.63 °C | 33.18 kJ/mol |
| n-Hexane | 68.73 °C | 28.85 kJ/mol |
| Diethyl ether | 34.5 °C | 26.52 kJ/mol |
| Chloroform | 61.17 °C | 29.24 kJ/mol |
| Ammonia | −33.33 °C | 23.33 kJ/mol |
| Methane | −161.48 °C | 8.19 kJ/mol |
| Propane | −42.1 °C | 19.04 kJ/mol |
| n-Butane | −0.5 °C | 22.44 kJ/mol |
| Hydrogen | −252.87 °C | 0.90 kJ/mol |
How to interpret the result
Clausius–Clapeyron with a constant ΔHvap is a straight line of ln P against 1/T. It is very good over a few tens of degrees but drifts over wide ranges and fails near the critical point. Raoult’s law holds for ideal solutions and is a good approximation for dilute solutions of non-volatile solutes.
Tips
- Use kelvin for every temperature in thermodynamic formulas – the unit menus convert for you.
- Watch the units: ΔH is usually in kJ/mol and ΔS in J/(mol·K).
- Tabulated values are for standard conditions (1 bar, usually 25 °C).
- Everything is calculated in your browser.
Common mistakes to avoid
- Mixing kJ and J between ΔH and ΔS.
- Assuming ΔH and ΔS do not change with temperature over large ranges.
- Using Celsius in ln(T₂/T₁).
Inputs explained
| Field | What to enter |
|---|---|
| Calculate | Choose one: Vapor pressure at a new temperature, Temperature for a given vapor pressure, Enthalpy of vaporization from two measurements, Vapor pressure of a solution (Raoult’s law) |
| Substance | Choose one: Custom – enter your own values, Water (bp 99.97 °C, ΔHvap 40.65 kJ/mol), Ethanol (bp 78.29 °C, ΔHvap 38.56 kJ/mol), Methanol (bp 64.7 °C, ΔHvap 35.21 kJ/mol) … |
| Enthalpy of vaporization ΔHvap | Units: kJ/mol, J/mol, kcal/mol, cal/mol, eV (per particle) |
| Known temperature T₁ | Units: °C, K, °F |
| Vapor pressure at T₁ | Units: kPa, atm, bar, hPa (mbar), mmHg (Torr), psi … |
| New temperature T₂ | Units: °C, K, °F |
| Target vapor pressure P₂ | Units: kPa, atm, bar, hPa (mbar), mmHg (Torr), psi … |
| Temperature of measurement 1 | Units: °C, K, °F |
| Vapor pressure at temperature 1 | Units: kPa, atm, bar, hPa (mbar), mmHg (Torr), psi … |
| Temperature of measurement 2 | Units: °C, K, °F |
| Vapor pressure at temperature 2 | Units: kPa, atm, bar, hPa (mbar), mmHg (Torr), psi … |
| Composition given as | Choose one: Mole fraction of the solvent, Moles of solvent and of solute particles |
| Mole fraction of the solvent x | A number |
| Moles of solvent | Units: mol, mmol, µmol, nmol, kmol |
| Moles of solute particles (× van ‘t Hoff factor) | Units: mol, mmol, µmol, nmol, kmol |
| Vapor pressure of the pure solvent | Units: kPa, atm, bar, hPa (mbar), mmHg (Torr), psi … |
What each result means
- Vapor pressure at T₂ = [pref]×exp([Hv]×1000÷8.314462618×(1÷[tref] – 1÷([New temperature T₂] + 273.15)))
- Temperature at which the vapor pressure is P₂ (boiling point at P₂) = 1÷(1÷[tref] – 8.314462618×ln([Target vapor pressure P₂]÷[pref])÷([Hv]×1000)) – 273.15
- Enthalpy of vaporization ΔHvap = 8.314462618×ln([Vapor pressure at temperature 2]÷[Vapor pressure at temperature 1])÷(1÷([Temperature of measurement 1] + 273.15) – 1÷([Temperature of measurement 2] + 273.15))÷1000
- Entropy of vaporization ΔHvap/T₂ = 8.314462618×ln([Vapor pressure at temperature 2]÷[Vapor pressure at temperature 1])÷(1÷([Temperature of measurement 1] + 273.15) – 1÷([Temperature of measurement 2] + 273.15))÷([Temperature of measurement 2] + 273.15)
- Reference point used: temperature = [tref] – 273.15
- Reference point used: vapor pressure = [pref]
- Enthalpy of vaporization used = [Hv]
Formula
ln(P₂/P₁) = (ΔHvap/R) · (1/T₁ − 1/T₂) · ΔHvap = R · ln(P₂/P₁) / (1/T₁ − 1/T₂) · Raoult: P_solution = x_solvent · P°_solventWorked example
Inputs — Calculate: Vapor pressure at a new temperature; Substance: Ethanol (bp 78.29 °C, ΔHvap 38.56 kJ/mol); New temperature T₂: 25 °C.
Results — Vapor pressure at T₂ = 9.5801 kPa; Reference point used: temperature = 78.29 °C; Reference point used: vapor pressure = 101.325 kPa; Enthalpy of vaporization used = 38.56 kJ/mol; At T₂ the liquid = stays below its normal boiling point at 1 atm.
Key terms
- Vapor pressure
- The pressure of a vapor in equilibrium with its liquid (or solid) at a given temperature.
- Volatility
- The tendency of a substance to vaporise; high vapor pressure means high volatility.
- Raoult's law
- P_solvent = x_solvent · P°_solvent for ideal solutions.
- Clausius–Clapeyron equation
- ln(P₂/P₁) = (ΔHvap/R)(1/T₁ − 1/T₂), relating vapor pressure to temperature.
Frequently asked questions
What is the vapor pressure of ethanol at 25 °C?
Clausius–Clapeyron from the normal boiling point (78.29 °C, ΔHvap = 38.56 kJ/mol) gives about 9.6 kPa; the measured value is 7.9 kPa. The gap shows that ΔHvap rises as temperature falls (42.3 kJ/mol at 25 °C).
What is the vapor pressure of water at 25 °C?
3.17 kPa (23.8 mmHg). See the vapour pressure of water calculator for dedicated formulas.
Why does vapor pressure increase with temperature?
More molecules have enough kinetic energy to escape the liquid. The fraction grows exponentially, following e^(−ΔHvap/RT).
How do I find ΔHvap from two vapor pressures?
ΔHvap = R · ln(P₂/P₁) / (1/T₁ − 1/T₂). Water at 25 °C (3.17 kPa) and 100 °C (101.3 kPa) gives about 42.7 kJ/mol – the average over that range.
What is Raoult's law?
The vapor pressure of the solvent above an ideal solution equals its mole fraction times the vapor pressure of the pure solvent: P = x·P°.
How much does sugar lower the vapor pressure of water?
0.5 mol of sucrose in 5.55 mol (100 g) of water gives x = 0.917, so at 25 °C P drops from 3.17 to 2.91 kPa.
Does the amount of liquid affect vapor pressure?
No. Vapor pressure is an intensive property – a drop and a lake of water at the same temperature have the same vapor pressure.
What formula does the Vapor Pressure Calculator use?
ln(P₂/P₁) = (ΔHvap/R) · (1/T₁ − 1/T₂) · ΔHvap = R · ln(P₂/P₁) / (1/T₁ − 1/T₂) · Raoult: P_solution = x_solvent · P°_solvent.