Vapour Pressure & Boiling Point (Clausius–Clapeyron)

Set the pressure above the water. The page works out where it boils from one number, the latent heat, checks that against the measured steam tables, and shows how far the simple law carries before it drifts.

Worked example — Vapour Pressure & Boiling Point (Clausius–Clapeyron)

Boiling point, from one latent heat	71.10 °C
Boiling point, measured (IAPWS-95)	71.81 °C
How far the model is out	−0.71 K
Latent heat at that boiling point, measured	41.95 kJ/mol

These are the values the calculator shows when the page opens. Change any input above to recompute them.

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Lesson

The theory — Vapour Pressure & Boiling Point (Clausius–Clapeyron)

A liquid's vapour pressure is the pressure of its own vapour in equilibrium with it, and it rises steeply with temperature. The liquid boils at the temperature where that pressure reaches the pressure above it. The Clausius–Clapeyron equation gives the rise: ln(P₂/P₁) = −(ΔH_vap/R)(1/T₂ − 1/T₁), so ln P against 1/T is a straight line of slope −ΔH_vap/R. Read one way it turns a pressure into a boiling point; read the other, it turns two boiling points into a latent heat.

What each symbol means

P: the pressure above the liquid, absolute, in bar: 1.01325 at sea level
T_b: the boiling point, in kelvin inside the formula
ΔH_vap: the molar latent heat of vaporisation: 40.65 kJ/mol for water at 100 °C
R: the gas constant, 8.314 J/(mol·K)

Assumes: Three things, each false somewhere. The latent heat is the same at every temperature, when it is 45.05 kJ/mol at the triple point and zero at the critical point. The vapour is an ideal gas. The liquid takes up no room beside it. Over this page's middle range the errors largely cancel, which is the second insight. The anchor is 100 °C at 1 atm, as on the colligative properties page, which takes its boiling-point constant from the same 40.65 kJ/mol; on ITS-90 the measured normal boiling point is 99.97 °C. And the pressure is absolute: a canner's gauge reads the excess over the air.
Breaks when: Near the critical point, 373.95 °C and 220.64 bar, where the latent heat goes to zero and liquid and vapour stop being different things; the model has no critical point and keeps going. Below the triple point there is no liquid at all, and ice has its own curve, set by the latent heat of sublimation. And for a mixture, where the vapour pressure is shared between the components and the boiling point climbs as the more volatile one leaves, one liquid's curve is not enough.

Water boils at 71.8 °C on Everest and 115.8 °C in a canner 🖖

Press Summit of Everest, then Canner at 10.5 psi. Nothing about the water changed, and its boiling point moved by 44 degrees. A liquid boils when its vapour pressure reaches the pressure above it, so a boiling point belongs to a pressure as much as to a liquid, and 100 °C belongs to one pressure only. Clausius–Clapeyron with the latent heat water has at 100 °C, 40.65 kJ/mol, gets both ends from the sea-level point alone: 71.10 and 116.01 °C, against IAPWS-95's 71.81 and 115.81. The USDA puts that canner at 240 °F, 115.56 °C, and insists on one for vegetables and meat because, in its own words, botulinum spores are very hard to destroy at boiling-water temperatures.

The latent heat falls by a quarter and the line barely bends 🖖

Read the last two columns of the table. From 1 atm to 40 bar the latent heat at the boiling point falls from 40.65 to 30.87 kJ/mol, 24%. The slope of ln P against 1/T, which is the only thing the model uses, falls from 41.33 to 38.42, 7%. Clapeyron's exact slope is the latent heat divided by how much PV/RT grows when the liquid boils, and at high pressure the steam is less of an ideal gas and the liquid has swollen, so that divisor shrinks almost as fast as the latent heat does. That is why one number carries so far: in the table the model is within a kelvin from the summit to 10 bar. The error panel draws the whole sweep, which crosses zero twice and falls away at both ends. At the triple point the model is 4.71 K low, because the latent heat there is 45.05 kJ/mol, 11% above what it assumes; at 200 bar it is 13.46 K low.

Two boiling points weigh a latent heat 🖖

The line works backwards. Two boiling points fix its slope, the slope is −ΔH_vap/R, and so ΔH_vap = R ln(P/P₀) / (1/T₀ − 1/T). Press Summit of Everest and read the row that does this with IAPWS-95's 71.81 and 99.97 °C: 41.80 kJ/mol. Water's measured latent heat midway between them, at 85.34 °C, is 41.33, so a barometer and a thermometer have weighed a heat of vaporisation to 1.1%. That 1.1% is the divisor from the insight above, because the line measures the latent heat divided by the growth in PV/RT, and steam at 85 °C is not quite an ideal gas. Type 41.80 into the latent-heat box and the model lands on the summit's boiling point to 0.02 K.

Problem solved in full

A pressure canner at Everest base camp 4 steps
A canner's weight holds the steam at a set pressure above the air around it. At sea level 10.5 psi on the gauge gives the USDA's 240 °F. At Everest base camp, where West's expedition measured 400.4 Torr, what gauge pressure does 240 °F take, and how hot does a canner with a 15 psi weight get? Press Canner at 10.5 psi for the sea-level case.

Open this problem in the tool
1. A gauge reads the pressure above the air, and the water boils at the absolute pressure. At sea level 10.5 psi is 0.7239 bar on top of 1.01325, 1.7372 bar in all, and the page puts the boiling point there at 116.01 °C.
  
  $P = 1.01325 + 10.5 \times 0.0689476 = 1.7372 bar \Rightarrow T_{b} = 116.01^{\circ} C$
2. At base camp the air supplies 400.4 Torr, 0.5338 bar. The same 10.5 psi now makes only 1.2578 bar, and the water boils at 106.26 °C.
  
  $P = 0.5338 + 0.7239 = 1.2578 bar \Rightarrow T_{b} = 106.26^{\circ} C$
3. Run the model backwards: which absolute pressure boils water at 240 °F, which is 115.56 °C?
  
  $P = 1.01325 exp [- \frac{40 , 650}{8 . 3145} (\frac{1}{388 . 71} - \frac{1}{373 . 15})] = 1.7117 bar$
4. Take away what the air already supplies, and turn the rest into the gauge's units.
  
  $1.7117 - 0.5338 = 1.1779 bar = \frac{1 . 1779}{0 . 0689476} = 17.08 psi$
Answer

17.08 psi on the gauge, 1.1779 bar above the air. A 15 psi weight at base camp holds 1.5680 bar, and the water boils at 112.86 °C by the model and 112.69 °C by IAPWS-95: 2.69 K short of 240 °F by one and 2.87 K by the other. The weight fixes how far the steam may exceed the air, and the air is what the mountain took away, which is why the USDA tells anyone canning above 1,000 feet to add pressure.

$0.5338 + 15 \times 0.0689476 = 1.5680 bar \Rightarrow T_{b} = 112.86^{\circ} C 115.56 - 112.86 = 2.69 K$

Learning path

Mixtures and solutions

References (3)

The measured curve the page holds its model against: water's saturation table, from the triple point to the critical point, as the NIST WebBook evaluates IAPWS-95: NIST Chemistry WebBook, NIST Standard Reference Database 69: "Thermophysical Properties of Fluid Systems", water, saturation properties. Source equation: W. Wagner and A. Pruß, J. Phys. Chem. Ref. Data 31(2), 2002, pp. 387-535.
The canner: 240 °F at 10.5 psi on the gauge at sea level, why low-acid food needs it, and the instruction to add pressure above 1,000 feet: U.S. Department of Agriculture, Complete Guide to Home Canning. Agriculture Information Bulletin No. 539, revised 2015.
The two Everest pressures, 253.0 Torr on the summit and 400.4 Torr at base camp, measured on the American Medical Research Expedition to Everest: J. B. West, S. Lahiri, K. H. Maret, R. M. Peters Jr and C. J. Pizzo, "Barometric pressures at extreme altitudes on Mt. Everest: physiological significance." Journal of Applied Physiology 54(5), 1983, pp. 1188-1194. DOI 10.1152/jappl.1983.54.5.1188

By the 42Math team · Reviewed against published sources. Corrections welcome.