Vapor Pressure Calculator

Solve vapor pressure, temperature, or enthalpy of vaporization in either direction.

Four temperature scales, twelve pressure units, and per-mole energy menus; Raoult’s law runs in the same page.

Updated October 13, 2026

CreatorFrank Zhao

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Introduction

Every liquid is slowly throwing molecules into the air above it. The pressure those escaped molecules exert — the vapor pressure — climbs steeply with temperature, and the temperature at which it reaches the surrounding pressure is what we call the boiling point. This calculator works that relationship in both directions, so you can move between a known boiling point and a new pressure, or between two measured points and the heat of vaporization behind them.

It holds two tools that share a page because they answer the two halves of the same question. The first is the Clausius-Clapeyron relation, which needs two pressures and two temperatures and solves for whichever one you leave out — the practical way to ask "what does water do at 0.6 atm?" or "how much energy does this liquid need to evaporate?". The second is Raoult's law, which covers the case where the liquid is not pure: dissolving something non-volatile in a solvent lowers its vapor pressure in proportion to how much of the mixture is solvent.

It is aimed at anyone who needs a working number rather than a derivation: a cook adjusting for altitude or a pressure cooker, a distiller or lab worker choosing a vacuum, an engineer sizing a drying step, or a student checking a homework result against a second method.

4 temperature scales

Celsius, Fahrenheit, kelvin and Rankine

12 pressure units

Pa through GPa, psi, bar, atm, torr, inHg, mmHg

5 energy-per-mole menus

J, kJ, MJ or ft-lb per mol, mmol, µmol, nmol or pmol

Two equations, five solvable unknowns, no hidden constants

Nothing is tabulated behind the scenes. The calculator supplies one fixed value — the gas constant $R = 8.3145\ \mathrm{J/(mol \cdot K)}$ — and otherwise works with exactly the numbers you give it. That is why it can return the heat of vaporization from two boiling points instead of asking you to look one up, and why the answer is only as good as the values you type in.

Quick start

The first group opens on Celsius and pascals, which is the most common way to state a boiling point. Here is a full pass through it, using water's normal boiling point as the reference state.

Describe the state you already know

Water boils at 100 °C under exactly 1 atm. Put 100 in Initial temperature and 101325 in Initial pressure, both in their default units. Every temperature is converted to the absolute scale for you, so you never have to add 273.15 by hand.

Add the heat of vaporization

Type 40660 in Molar enthalpy of vaporization — the energy one mole of water needs to escape at its boiling point. The two menus next to that box say joules and mole, so this is 40 660 J/mol; switching them to kilojoules or millimoles changes how the number is written, never what it is worth.

Give the second state one known value

To find a boiling point, enter the pressure you care about in Final pressure — for a pressure cooker that might be 212782.5 Pa, roughly 2.1 atm. To find a pressure instead, enter the second temperature in Final temperature. Three knowns are enough for the fourth; a row of inputs with only two of them filled in stays quiet until it has what it needs.

Read the answer — and know which box it landed in

The remaining box fills in as soon as the inputs are complete. With the numbers above the calculator returns a final temperature of 122.39 °C: at 2.1 atm, water does not boil until well past its usual 100 °C. That is the whole calculation — three inputs, one result, and the same row can be re-used for any other liquid once you change the heat of vaporization.

The same pass, written out

Rearranged for the temperature, the working is short enough to follow by hand:

T_2

=

\frac{T_1}{1 - k\,T_1}

=

\frac{373.15}{1 - 0.00015172 \times 373.15}

=

395.54\ \mathrm{K}

395.54\ \mathrm{K} - 273.15

=

122.39\ ^\circ\mathrm{C}

That is the number the calculator shows

The factor $k$ is just the logarithm of the pressure ratio scaled by $R/\Delta H_{\mathrm{vap}}$ ; the next section defines it properly. Notice what the result is telling you: a pressure about twice atmospheric raises the boiling point by 22 °C. Pressure cookers exploit exactly this, and the same arithmetic in reverse explains why water boils below 100 °C on a mountain.

The equations and what they assume

One relation links a liquid's vapor pressure to its temperature through the energy needed to evaporate it. Everything the first group does is that relation, rearranged.

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

P_1

Vapor pressure at the first temperature — the state you know

P_2

Vapor pressure at the second temperature — the state you are asking about

T_1

First temperature, on the absolute scale (kelvin)

T_2

Second temperature, on the absolute scale (kelvin)

\Delta H_{\mathrm{vap}}

Molar enthalpy of vaporization, in joules per mole

R

Gas constant, 8.3145 J/(mol·K), the only fixed value in the model

Solved for each of the five boxes

The relation has four unknowns besides the constant, and the calculator will hand back any one of them. Rearranged for pressure it reads:

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

Solving for a temperature collects the logarithm into a single factor, $k = \dfrac{R}{\Delta H_{\mathrm{vap}}}\ln\dfrac{P_1}{P_2}$ , and then:

T_1 = \frac{T_2}{1 + k\,T_2}

\qquad

T_2 = \frac{T_1}{1 - k\,T_1}

And solving for the heat of vaporization is the slope of $\ln P$ against $1/T$ between your two points:

\Delta H_{\mathrm{vap}} = \frac{R\,\ln\left(P_2/P_1\right)}{\dfrac{1}{T_1} - \dfrac{1}{T_2}}

One useful property falls out of the pressure form: only the ratio of the two pressures matters, so it makes no difference whether both are in pascals, atmospheres or torr. The same is not true of temperature, which is why the absolute scale is non-negotiable — see below.

Why the temperatures must be absolute

The relation contains $1/T$ , so the zero of the scale has to mean "no thermal energy at all". Celsius and Fahrenheit place their zero somewhere convenient instead, and putting them into the formula produces answers that are not just inaccurate but physically impossible. Enter temperatures in whatever scale your source uses — the calculator converts internally:

T_{\mathrm{K}}

=

T_{\mathrm{{}^\circ C}} + 273.15

=

\tfrac{5}{9}\left(T_{\mathrm{{}^\circ F}} - 32\right) + 273.15

=

\tfrac{5}{9}T_{\mathrm{{}^\circ R}}

What the model assumes

The heat of vaporization is treated as constant between your two temperatures. In reality it drifts downward as the temperature rises, which is the single largest source of error and the reason a narrow temperature range gives a better answer than a wide one.
The vapor behaves as an ideal gas, and the volume of the liquid is ignored next to the volume of the vapor it becomes. Both hold well at ordinary pressures and fail near the critical point.
Only two phases are involved. The relation describes liquid-to-vapor (or solid-to-vapor, sublimation) transitions; it says nothing about a substance that decomposes before it boils.
Nothing here is tabulated. You supply the heat of vaporization, so a value borrowed from the wrong temperature or the wrong substance gives a confident wrong answer.

Raoult's law: vapor pressure over a solution

Dissolve something that does not itself evaporate and the solvent's vapor pressure drops. It drops in exact proportion to the fraction of the mixture that is still solvent.

P_{\mathrm{solution}} = P_{\mathrm{solvent}} \times X_{\mathrm{solvent}}

P_{\mathrm{solvent}}

Vapor pressure of the pure solvent on its own, at the same temperature

X_{\mathrm{solvent}}

Mole fraction of the solvent: moles of solvent ÷ total moles in the mixture

P_{\mathrm{solution}}

Vapor pressure the mixture actually exerts

The mole fraction is a plain number between 0 and 1, and a pure solvent is exactly 1 — which is why the calculator accepts 1 and rejects anything above it. All three boxes are solvable, so a measured vapor pressure over a solution works back to the mole fraction, and from there to the composition of the mixture.

Worked example: glucose in water at 37 °C

Water at body temperature exerts 47.1 torr on its own. A solution made from 500 g of water (27.70 mol, molar mass 18 g/mol) and 100 g of glucose (0.555 mol, molar mass 180.2 g/mol) leaves a solvent mole fraction of:

X_{\mathrm{solvent}}

=

\frac{27.70}{27.70 + 0.555}

=

0.98

Switch both pressure boxes to torr — they open on pascals — then enter 47.1 as the solvent pressure and 0.98 as the mole fraction:

P_{\mathrm{solution}}

=

47.1 \times 0.98

=

46.158\ \mathrm{torr}

That is the value the calculator shows

Switching the result box to torr displays 46.158. The pressure has fallen by about 2%, which sounds small until you remember that vapor pressure is what drives evaporation: the same drop is why a dissolved solute raises the boiling point and lowers the freezing point of its solvent, and why those two effects are proportional to the number of dissolved particles rather than to their identity.

Where this stops being true

Raoult's law describes ideal mixtures: it assumes every molecule interacts with its neighbours the same way regardless of what they are. Real mixtures deviate, and strongly interacting pairs — water and ethanol being the classic example — can deviate so far that the mixture's vapor pressure is higher than the pure solvent's. It also assumes the solute is non-volatile, so it cannot handle a mixture of two things that both evaporate. For those cases the calculation needs activity coefficients or measured vapor-liquid equilibrium data rather than a single multiplication.

Choosing the heat of vaporization

This is the one number the calculator cannot derive for you, and the one that decides how good the answer is. It is not a constant of nature: it changes with temperature, and quoting it without the temperature it belongs to is the most common way to get a plausible but wrong result.

The values below are the vapor pressure of water at ordinary temperatures, which is a useful sanity check whichever direction you are solving. The middle row is the one this guide uses: from water's boiling point and a heat of vaporization of 40 660 J/mol, the calculator returns 70 629.94 Pa at 90 °C.

Temperature	Saturation pressure of water	Typical use of that row
37 °C (310.15 K)	6.28 kPa / 47.1 torr	Body temperature, the Raoult example above
90 °C (363.15 K)	70.63 kPa	Model output from the 100 °C reference state
100 °C (373.15 K)	101.325 kPa / 1 atm	The normal boiling point, the reference state

The same row is a cross-check on the model itself. Independently fitted vapor pressure correlations put water at about 70.14 kPa at 90 °C, so treating the heat of vaporization as a single constant across that 10 °C step lands roughly 0.7% high. That is the honest accuracy of the method at this spacing, and it is why the NIST Chemistry WebBook fluid properties tool is worth consulting for reference values of water and other common fluids.

Get the value from the temperature you have

Use the value at the temperature of your known state, not at some other reference. Water sits near 40 660 J/mol at its normal boiling point; at room temperature the same liquid needs noticeably more energy per mole, so pairing a room temperature value with a boiling point measurement biases the answer.
Narrow ranges are safer. Two points 10 °C apart are well described by a single value; two points 150 °C apart are not, and the error grows quickly with the span.
If you have two measured points, solve for the value instead of looking it up. Enter both pressures and both temperatures, leave the heat of vaporization blank, and the calculator returns the effective value for exactly that range. It is a better constant for your problem than any table entry.
For acetic acid, a documented spread. NIST lists enthalpy of vaporization values for acetic acid from about 23.7 kJ/mol near its boiling point up to roughly 38–43 kJ/mol at lower temperatures on its NIST Chemistry WebBook phase-change page, a good illustration of why the temperature attached to the number matters.

When to use it

The calculator is at its best wherever a pressure and a boiling point need to be traded for one another, or where two measured points should become a physical constant.

Cooking under pressure or at altitude

A pressure cooker raises the boiling point; a mountain lowers it. Enter your local or cooker pressure as the final pressure and read the temperature your food or your brewing water is actually reaching — the difference between 100 °C and 122 °C is the difference between a simmer and a braise.

Distillation and vacuum work

Lowering the pressure lowers the temperature a liquid boils at, which is how heat-sensitive compounds are distilled without decomposing. Solve for the temperature at your pump's pressure, or for the pressure needed to reach a target temperature.

Drying, degassing and storage

How readily a solvent leaves a surface is set by its vapor pressure at that temperature. Comparing two candidate solvents, or two operating temperatures, is a single run of the first group each time.

Colligative properties and solution work

The second group answers "how much does dissolving this lower the vapor pressure?", which is the first step behind boiling point elevation, freezing point depression and osmotic behavior. The boiling point elevation calculator takes that pressure drop through to the new boiling point, and the molar mass calculator converts the masses in a recipe into the moles the mole fraction needs.

Where it is the wrong tool

Two cases are worth naming. First, near the critical point, where the liquid and vapor become indistinguishable: the ideal-gas assumption behind the relation fails completely, and no single heat of vaporization describes the transition. Second, mixtures of two or more volatile components, where the vapor above the liquid is a different composition from the liquid itself; Raoult's law cannot capture that, and you need measured vapor-liquid equilibrium data or activity coefficients. A pure component over a moderate temperature range is where this calculator is reliable.

Questions

I have two boiling points — can the calculator find the heat of vaporization for me?

Yes, and it is the most reliable way to use the page. Enter both pressures and both temperatures and leave the heat of vaporization box empty. The result is the effective value across exactly your range, which is more trustworthy than a table entry measured somewhere else. It works the other way too: if you have one pressure-temperature pair and a value from a handbook, the calculator gives you the second temperature or pressure.

Why does the row stay blank until I fill in three boxes?

The relation connects four quantities: two pressures, two temperatures and the heat of vaporization — five boxes in total, with the constant supplied for you. To solve for one unknown you need the others, so a row with only two entries has nothing to work with and stays quiet rather than guessing. A blank box is not an error; it is the calculator waiting for the last input. If a box turns red instead, the value is present but breaks a real boundary — a temperature at or below absolute zero, a pressure of zero (the logarithm of a zero ratio is undefined), or two identical temperatures that make the ratio meaningless.

Does it matter which point I call initial and which final?

No — the relation is symmetric in that sense, and the labels only decide which box is which. The one thing to keep consistent is the pairing: each pressure belongs to the temperature beside it. Mixing a pressure from one state with a temperature from another is the easiest mistake to make and produces an answer that looks entirely reasonable.

Can I mix units — torr for one pressure and kilopascals for the other?

Yes. Each pressure box has its own menu, and both values are normalised before they meet, so torr against kilopascals is fine. The same is true of temperature, where you can state one boiling point in Celsius and the other in Fahrenheit. What you cannot do is mix a temperature level with a temperature difference, but the calculator never asks you for a difference — the only temperature-difference quantities here are the heat of vaporization and the gas constant, and both are handled internally.

Why is my answer a few degrees off a published table?

Two reasons, usually. The heat of vaporization is treated as constant between your points, while in reality it falls as the temperature rises — over a wide range that alone can shift a boiling point by several degrees. And your value may belong to a different temperature than the state you entered. If the gap matters for your work, narrow the range: solve for the heat of vaporization from two measured points close to the state you care about, then use that value for the final step.

How far to trust it

The model is an approximation, and the heat of vaporization is the weak point. It is assumed constant across your two temperatures. The wider the gap between them, the less that holds — the guide's own 10 °C water step already runs about 0.7% high against an independently fitted vapor pressure curve.
Raoult's law covers ideal, dilute solutions of a non-volatile solute only. Real mixtures deviate, sometimes strongly, and a mixture of two volatile components needs measured equilibrium data instead of one multiplication.
Physical reality beyond the model is yours to supply. Non-ideal gas behavior, dissolved air, superheating, a substance that reacts or decomposes before it boils, and pressure gradients inside a vessel are all outside what a two-point relation can describe.
Treat results as estimates, not as specifications. For pressure-vessel safety, process design, pharmaceutical or food-safety work, or anything where being a few degrees out carries a real cost, verify against measured data or an appropriate reference before acting — and consult a qualified engineer or chemist for the decision itself.

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