Boiling Point Calculator

Find where a liquid boils at a new pressure, or work back from two measured boiling points.

Nine tabulated substances, twelve pressure scales, and reverse solves from a target boiling temperature to the pressure that produces it.

Updated October 3, 2026

CreatorFrank Zhao

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Introduction

Every liquid has a boiling point, and almost every liquid has more than one. Water boils at 100 °C under a standard atmosphere, at 93 °C on a 2 000 m mountain, and at 81 °C once you pull a third of the atmosphere away. This calculator moves a liquid between those states using the Clausius–Clapeyron relation, in either direction: give it a pressure and it returns the boiling temperature, give it a temperature and it returns the pressure that produces it.

Nine common liquids are built in — water, acetone, ammonia, butane, ethanol, hydrogen, methane, methanol and propane — each with its tabulated heat of vaporization and normal boiling point. If you have measured your own reference point, you can type over it and work from that instead.

The idea worth taking away first

A boiling point is not a property of the liquid. It is a property of the liquid at a given pressure. Only the heat of vaporization belongs to the substance on its own, which is why that box is locked to the Substance menu while the two pressure and two temperature boxes stay open. Locking all four would quietly bake in "one standard atmosphere" as a hidden assumption — which is fine for a lookup and wrong for anything measured.

Quick start

Pick the substance

The menu fills the reference state with that liquid's tabulated normal boiling point at one standard atmosphere (1 013.25 hPa). Its heat of vaporization appears in the read-only box above. If you uncheck Show heat of vaporization the box hides, but the value still does its job in the calculation.

Set the State 1 pair if your experiment differs

Leave both alone when you are quoting a published normal boiling point. Change them when you actually measured a boiling point at some other pressure — State 1 is the anchor everything else is measured from.

Enter the State 2 pressure and read the temperature

Any of twelve pressure scales works, and the unit menu only changes how the number is written, never the number itself. The solved boiling point turns blue to show the engine produced it.

Or go the other way

Fill in a State 2 boiling point instead of a pressure and the pressure that produces it is solved for you — the usual move when you want to hold a vessel at a chosen temperature and need to know how hard to pressurise it.

Worked example: water at half an atmosphere

Water boils at 373.15 K under 101 325 Pa. What does it do at 50 000 Pa? Rearranging the relation for the unknown temperature gives

\frac{1}{T_{2}}

=

\frac{1}{T_{1}} + \frac{R}{\Delta H}\,\ln\!\left(\frac{P_{1}}{P_{2}}\right)

and substituting the State 1 values and the gas constant:

\frac{1}{T_{2}}

=

\frac{1}{373.15} + \frac{8.31446}{40657}\,\ln\!\left(\frac{101325}{50000}\right)

=

2.67989\times10^{-3} + 2.04503\times10^{-4}\times 0.706310

=

2.82433\times10^{-3}

T_{2}

=

354.07\ \mathrm{K}

=

80.92\ ^\circ\mathrm{C}

That is the value the calculator shows

Steam tables put the true figure nearer 81.3 °C — the gap is the model's approximation, and it is quantified in how far to trust it. Lower the pressure and the boiling point falls; raise it and the boiling point rises. There is no pressure at which the answer stops being a liquid, which is why you can run this backwards.

The relation and what it assumes

The calculator works along one equation, the integrated form of the Clausius–Clapeyron relation:

\ln\!\left(\frac{P_{2}}{P_{1}}\right) = -\,\frac{\Delta H}{R}\left(\frac{1}{T_{2}} - \frac{1}{T_{1}}\right)

Read it as: take the logarithm of how much the pressure changed, and it equals the molar heat of vaporization divided by the gas constant, times how much the inverse absolute temperature changed. On a graph of $\ln P$ against $1/T$ , a liquid that obeyed this exactly would be a straight line, and the slope would be the heat of vaporization.

4 ways in

Any one of the four state boxes can be the unknown

R = 8.31446

J/(mol·K), exact since the 2019 SI redefinition

ΔH locked

A property of the liquid, never solved backwards

The five boxes and who fills them

P_{1},\,T_{1}

State 1, your reference point. Editable. Normally the tabulated normal boiling point at one standard atmosphere.

P_{2},\,T_{2}

State 2, the condition you are asking about. Editable. Whichever of the pair you leave blank is the one the engine solves.

\Delta H

The heat of vaporization. Locked to the Substance menu, and deliberately never solved for.

The assumption that decides your accuracy

The integration treats $\Delta H$ as a constant. For a real liquid it is not — it falls as the liquid warms. The NIST reference data for acetone, for instance, tabulate 33.8 kJ/mol at 254 K and 9.2 kJ/mol at 498 K. NIST Chemistry WebBook, acetone So the narrower the temperature span you ask about, the better the answer. That is the single most useful thing to know when reading a result.

Where this form stops working

Set the State 2 denominator to zero and you find the pressure at which the solved temperature diverges:

P^{*} = P_{1}\,\exp\!\left(\frac{\Delta H}{R\,T_{1}}\right)

For water at one atmosphere that is about 4.98 × 10¹⁰ Pa, roughly 490 000 atmospheres. Nothing you will ever boil at is anywhere near it — but past that pressure the line runs through zero and the "temperature" comes out negative. The calculator watches for it and says so before you get there: a caution when you are within a factor of two of $P^{*}$ , and a firm warning naming the pressure once you pass it.

Worked examples

1. Cooking at altitude

The standard atmosphere thins with height. Work the State 2 pressure out from the atmospheric lapse rate first:

p = p_{0}\left(1 - \frac{L\,h}{T_{0}}\right)^{5.25588}, \quad T_{0} = 288.15\ \mathrm{K}, \quad L = 0.0065\ \mathrm{K\,m^{-1}}

with $p_{0} = 101\,325\ \mathrm{Pa}$ at sea level. Feeding those pressures into State 2 with water as the substance gives the familiar slower boil at altitude:

Height	State 2 pressure	Boiling point
sea level	1 013.25 hPa	100.0 °C
1 000 m	898.75 hPa	96.6 °C
2 000 m	794.95 hPa	93.2 °C
3 000 m	701.09 hPa	89.8 °C

Three degrees by three thousand metres is enough to matter for anything that needs a set temperature. These are the calculator's own figures; how close they are to measured behaviour is in how far to trust it.

2. Holding a vessel at 80 °C

Water boils at 80 °C at about 473 hPa. This calculator returns 482.4 hPa.

A useful working figure, and about 2 % above the tabulated pressure. Use it for sizing a vacuum line or a pump curve, not for setting a relief valve. The reason it reads high is the same constant-heat assumption: quoted at the boiling point, $\Delta H$ is at its smallest over the range below, so the solved temperature lands low and the solved pressure lands high.

3. A cryogenic liquid

Hydrogen is included because it shows the scale of the equation.

Its normal boiling point is 20.27 K — minus 252.88 °C — with a heat of vaporization of 0.904 kJ/mol, the smallest of the nine. Drop it to 500 hPa and the calculator returns −255.24 °C, i.e. 17.91 K. Nothing breaks: the units menu handles Celsius, Fahrenheit and kelvin, and a sub-zero Celsius boiling point is a perfectly ordinary answer for a substance that has none in Fahrenheit worth quoting.

Tips and pitfalls

Five habits that keep the arithmetic honest. None of them need a calculator to check — they are all things to do before you type.

Keep the temperature span small

Asking about a pressure 5 K from where the liquid already boils is where this relation is most trustworthy. A 30 K span is fine for estimation; a 200 K span is an extrapolation dressed up as arithmetic.

Use your own State 1 when you have one

If your apparatus already boils at a known pressure and temperature, type both in. The answer then reflects your substance under your conditions rather than a handbook average.

Watch which box is blue

Blue means the engine solved it from the others. If a result is blue and disagrees with a table, the disagreement is usually in the span, not the arithmetic.

Do not use this for mixtures

Salt water and alcohol mixtures boil at temperatures that depend on composition, and the heat of vaporization of the mixture is not the heat of vaporization of either pure component. Adding salt raises the boiling point; the number this calculator returns will not know that.

A boiling point is not a pressure rating

A liquid boiling at 120 °C is not thereby safe in a vessel at that pressure — the vessel also has to hold the vapour, and anything else dissolved in the liquid changes the answer.

How far to trust it

This is an estimating tool built on a linearised relation, not a source of design data. The error is not random — it grows predictably as the temperature span widens, and it always leans the same way for a substance quoted at its boiling point.

Measured against water vapour-pressure data

State 2 pressure	This calculator	Vapour-pressure data	Difference
89 875 Pa	96.62 °C	96.67 °C	−0.06 K
79 495 Pa	93.22 °C	93.34 °C	−0.12 K
70 109 Pa	89.80 °C	90.00 °C	−0.20 K
50 000 Pa	80.92 °C	81.34 °C	−0.42 K

The reference column is the boiling point implied by the vapour-pressure coefficients published for water on the NIST Chemistry WebBook, which also lists the normal boiling point as 373.17 ± 0.04 K — the 100.0 °C everyone uses. Read the table as: within about 25 K of the normal boiling point, expect agreement to a few tenths of a degree. Go further down and the temperature error roughly doubles per doubling of the distance from State 1, and the pressure error — when you solve for pressure instead — grows faster still, because the same temperature error is amplified by the exponential.

The other boundary: this is for pure liquids

Every built-in substance is a pure liquid with a tabulated heat of vaporization. For a solution, use the Raoult's law calculator instead — it works from the mole fraction of the volatile component and its pure-component vapour pressure, which is what a boiling-point elevation or depression actually depends on.

Pressures must be positive and no greater than 1 × 10¹² Pa; boiling points must be above absolute zero and below 6 000 K. Anything outside those ranges is flagged rather than quietly accepted.
The nine tabulated values are rounded handbook figures, printed to three or four significant figures. The calculator keeps each one exactly as published rather than implying a precision the source does not have.
Results are for education, estimation and planning. Anything involving a pressure vessel, a distillation column, a cryogenic system or a safety margin belongs against real vapour-pressure data and a qualified engineer.

Questions

Why can I not edit the heat of vaporization?

Because it is a property of the liquid rather than of your experiment, and the Substance menu is the honest source for it. It is also never solved backwards: if the engine were allowed to derive it from four inputs, the grey box would silently change to whatever those inputs implied, and you would have no way to correct it. If you need a value from a specific table, pick the matching substance and compare — the tabulated figures are quoted with their evaluation temperatures stated.

Why does the same substance give a different boiling point than a textbook?

Almost always the reference state. A textbook number is almost always quoted at one standard atmosphere; if you enter a State 1 pressure of anything else without also changing State 1 temperature, you are anchoring to a different point on the line. If the reference state is right, the remaining difference is the constant-heat approximation described above, and the error table gives its size.

Does adding salt raise or lower the boiling point?

It raises it — salt is a non-volatile solute, so it lowers the vapour pressure of the water and the solution has to be heated further to reach the surrounding pressure. That is a composition effect this calculator does not model, because it needs the mole fraction and the pure component's vapour pressure rather than a single tabulated constant. See the note on mixtures for the tool that does.

How far can I take the pressure before the answer stops meaning anything?

The calculator warns you at two points. Once you are within a factor of two of the divergence pressure it cautions you that accuracy is degrading; past it the linearised form has given up and the temperature has run through zero. For water at one atmosphere that ceiling is around 4.98 × 10¹⁰ Pa:

P^{*} = P_{1}\,\exp\!\left(\frac{\Delta H}{R\,T_{1}}\right)

In practice the honest limit is far lower — if you are extrapolating more than a few tens of kelvin, use real data.

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