How this instrument works
A liquid boils when its vapor pressure — the pressure of vapor it throws off as it evaporates — climbs to match the pressure pressing down on its surface. Turn down the surrounding pressure and the liquid needs less vapor pressure to win that contest, so it boils at a lower temperature. That's the whole reason water boils at a cooler 90°C on a mountain and a hotter 121°C inside a pressure cooker: the surrounding pressure, not the water itself, has changed.
The Clausius-Clapeyron equation is the quantitative version of that idea. It says the natural log of the ratio of two vapor pressures is proportional to the difference in 1/T between the two temperatures, scaled by the substance's molar heat of vaporization and the gas constant R: ln(P2/P1) = −(ΔHvap/R)(1/T2 − 1/T1). Rearranged to solve for the unknown temperature, it becomes the formula this instrument runs — hand it a known boiling point at a known pressure, plus the substance's heat of vaporization, and it returns the boiling point at any new pressure.
The heat of vaporization, ΔHvap, is what makes the equation substance-specific: it's the energy needed to tear one mole of liquid molecules apart into vapor, and it varies enormously across substances depending on how strongly their molecules attract each other. Water's hydrogen bonds give it an unusually high ΔHvap (~40.7 kJ/mol) — one reason water resists boiling away compared to most other liquids of similar molecular weight.
- Enter the reference pressure P1 — a pressure at which you already know the boiling point (101.325 kPa for standard atmospheric pressure).
- Enter the reference boiling point T1 at that pressure, in degrees Celsius.
- Enter the substance's molar heat of vaporization, ΔHvap, in J/mol — this value is substance-specific and comes from a reference table.
- Enter the new pressure P2 you want the boiling point for.
- Read the predicted boiling point T2 at that new pressure.
Worked example — water's boiling point at half atmospheric pressure
Water boils at 100°C at standard sea-level pressure, 101.325 kPa, with a molar heat of vaporization of about 40,700 J/mol. Drop the pressure to exactly half — 50 kPa, roughly the pressure at the summit of a very high mountain — and the Clausius-Clapeyron equation predicts a boiling point of about 80.9°C. Textbooks and reference calculators round this to 'approximately 82°C,' which lands in the same neighborhood; the small gap comes down to how many decimal places are carried through the logarithm and the exact ΔHvap figure used.
The relationship isn't linear — cutting pressure in half doesn't cut the boiling point (in Kelvin) in half, or even by a fixed number of degrees. It's the logarithm of the pressure ratio that matters, which is why the same equation, run at 70 kPa (roughly the pressure around 3,000 m elevation), predicts water boiling at about 89.8°C rather than something evenly spaced between the two endpoints.
Questions
Why does water boil at a lower temperature at high altitude?
Because atmospheric pressure drops with altitude, and a liquid boils as soon as its vapor pressure reaches whatever pressure is pushing down on it. At lower atmospheric pressure, water's vapor pressure doesn't have to climb as high to win that contest, so it reaches boiling at a lower temperature. At the roughly 70 kPa pressure found around 3,000 m elevation, water boils near 90°C instead of 100°C — a well-known headache for high-altitude cooking, since food actually cooks at a lower temperature even though the water is still 'boiling.'
What is the Clausius-Clapeyron equation used for?
It relates a pure substance's vapor pressure to its temperature along the liquid-vapor equilibrium line, and it's the standard tool for two closely related jobs: predicting the boiling point at a new pressure if you know it at one reference pressure (what this instrument does), or predicting the vapor pressure at a new temperature if you know it at one reference temperature. Both directions use the exact same underlying relationship, just solved for a different unknown.
Do I need the heat of vaporization for the specific substance, or can I use water's value for anything?
You need the value for whatever substance you're actually working with — ΔHvap is substance-specific and reflects how strongly that liquid's molecules attract each other. Using water's ~40,700 J/mol for, say, ethanol (~38,600 J/mol at its boiling point) will be close but not exact; using it for something with very different intermolecular forces, like a nonpolar hydrocarbon, will give a meaningfully wrong answer. Look up the correct ΔHvap for your substance in a chemistry reference table or the NIST WebBook.
Is the Clausius-Clapeyron equation exact, or an approximation?
It's derived with a few simplifying assumptions: that the vapor behaves as an ideal gas, that the liquid's volume is negligible compared to the vapor's, and that ΔHvap doesn't change with temperature over the range being considered. Those assumptions hold well for moderate pressure ranges well below a substance's critical point, which covers most everyday boiling-point and altitude calculations, but the equation drifts from measured values as conditions get more extreme.
What happens if I enter the same pressure for P1 and P2?
You get back exactly the reference temperature you put in, unchanged. When P2 equals P1, the ratio P2/P1 is 1, and the natural log of 1 is exactly 0 — so the entire pressure-dependent term in the equation vanishes and T2 = T1. It's a useful sanity check: this instrument's own internal correctness test relies on exactly that identity.