Clausius–Clapeyron Equation — Vapour Pressure and Boiling Point

By Aniket Bhardwaj · 6 September 2026 · Calculator/Formula Guide

Why does water boil at a lower temperature on a hill station than in Delhi? Why does a pressure cooker cook faster? Both have the same answer, and the Clausius–Clapeyron equation is the arithmetic behind it. It links a liquid's vapour pressure to its temperature, and it is a standard numerical in Class 12, IIT-JAM and GATE physical chemistry.

The formula

ln (P₂ / P₁) = − (ΔH_vap / R) (1/T₂ − 1/T₁) = (ΔH_vap / R) (1/T₁ − 1/T₂)

Both versions in that box are the same equation. Choose one form and always write it that way — most errors in this topic are sign errors, and a fixed habit removes them. The differential form it comes from is d(ln P)/dT = ΔH_vap / (RT²).

What each symbol means

Symbol	Meaning	Unit
P₁, P₂	Vapour pressures at the two temperatures	any pressure unit — but the same one for both, since only their ratio is used
T₁, T₂	The two absolute temperatures	kelvin (K)
ΔH_vap	Molar enthalpy of vaporisation	J mol⁻¹ when R is in J
R	Gas constant	8.314 J K⁻¹ mol⁻¹

The pressure unit cancels because P₂/P₁ is a ratio — kPa, atm, bar, mmHg all work, as long as you do not mix two of them in one problem. Temperature does not cancel: 1/T is meaningless in °C.

What "vapour pressure" actually is

Put a liquid in a closed container. Molecules escape into the space above it, and some come back. When the two rates balance, the pressure of vapour in that space is the saturation vapour pressure at that temperature. It depends only on the liquid and the temperature — not on how much liquid there is, and not on the size of the container.

A liquid boils when its vapour pressure equals the external pressure pushing down on it. That single sentence explains both puzzles in the opening paragraph:

On a hill, the atmospheric pressure is lower, so the vapour pressure reaches it at a lower temperature — water boils below 100 °C, and food cooks more slowly because the temperature is lower.
In a pressure cooker, the pressure inside is deliberately higher, so a higher temperature is needed before boiling starts — and the hotter water cooks faster.

The normal boiling point is the temperature at which vapour pressure equals 1 atm = 101.325 kPa.

Worked example 1 — vapour pressure at a lower temperature

Question: Water has ΔH_vap ≈ 40.7 kJ mol⁻¹ and a vapour pressure of 101.3 kPa at its normal boiling point, 373.15 K. Estimate its vapour pressure at 353.15 K (80 °C).

Prediction first: lower temperature → lower vapour pressure. The answer must come out below 101.3 kPa.

Step 1 — ΔH in joules: ΔH_vap/R = 40 700 / 8.314 = 4895.4

Step 2 — reciprocal temperatures:
1/T₂ = 1/353.15 = 2.831658 × 10⁻³
1/T₁ = 1/373.15 = 2.679887 × 10⁻³
(1/T₂ − 1/T₁) = 1.51771 × 10⁻⁴

Step 3 — apply the equation:
ln(P₂/P₁) = −4895.4 × 1.51771 × 10⁻⁴ = −0.74297

Step 4 — exponentiate:
P₂/P₁ = e⁻⁰·⁷⁴²⁹⁷ = 0.4757
P₂ = 101.3 × 0.4757 = 48.19 kPa

P₂ ≈ 48.2 kPa

Honest note: the measured vapour pressure of water at 80 °C is close to 47.4 kPa. Our estimate is about 2% high, and the reason is worth knowing: we treated ΔH_vap as constant between 80 °C and 100 °C, and it is not quite. A 2% error over a 20 K range is exactly the accuracy this equation is designed to give.

Worked example 2 — finding ΔH_vap from two measurements

This is what an experimenter actually does, and it is the most common exam version.

Question: A liquid has a vapour pressure of 13.3 kPa at 300 K and 53.3 kPa at 330 K. Calculate its enthalpy of vaporisation.

Rearranged: ΔH_vap = R × ln(P₂/P₁) ÷ (1/T₁ − 1/T₂)

Step 1: P₂/P₁ = 53.3 / 13.3 = 4.0075, so ln(P₂/P₁) = 1.38817
Step 2: 1/300 = 3.333333 × 10⁻³, 1/330 = 3.030303 × 10⁻³, difference = 3.03030 × 10⁻⁴
Step 3: numerator = 8.314 × 1.38817 = 11.5413
ΔH_vap = 11.5413 ÷ (3.03030 × 10⁻⁴) = 38 086 J mol⁻¹

ΔH_vap ≈ 38.1 kJ mol⁻¹

Check the sign: pressure rose with temperature, so ΔH_vap must be positive. Vaporisation always absorbs heat, so a negative answer here would always be wrong ✓

Worked example 3 — estimating a normal boiling point

Question: A liquid has ΔH_vap = 30.0 kJ mol⁻¹ and a vapour pressure of 40.0 kPa at 320 K. Estimate its normal boiling point (the temperature at which its vapour pressure reaches 101.325 kPa).

Rearranged to give the unknown temperature:

1/T₂ = 1/T₁ − (R / ΔH_vap) × ln(P₂ / P₁)

Step 1: P₂/P₁ = 101.325 / 40.0 = 2.53313, ln = 0.92945
Step 2: R/ΔH_vap = 8.314 / 30 000 = 2.77133 × 10⁻⁴
Step 3: (R/ΔH) × ln = 2.77133 × 10⁻⁴ × 0.92945 = 2.57583 × 10⁻⁴
Step 4: 1/T₂ = 3.125000 × 10⁻³ − 0.257583 × 10⁻³ = 2.867417 × 10⁻³
Step 5: T₂ = 1 / (2.867417 × 10⁻³) = 348.75 K

Normal boiling point ≈ 348.7 K = 75.6 °C

Sanity check: we needed a higher pressure than 40.0 kPa, so we must need a higher temperature than 320 K — and 348.7 K is higher ✓ Note the trap in Step 4: subtracting a small number from a small number, so keep six significant figures until the very end. Rounding 1/T₂ to three figures here would move the answer by several kelvin.

The graph: ln P against 1/T

ln P = − (ΔH_vap / R) · (1/T) + constant

Plot ln P on the y-axis and 1/T on the x-axis and you get a straight line with slope −ΔH_vap/R. Since ΔH_vap is always positive for vaporisation, the line always slopes downward. Multiply the slope by −R to recover ΔH_vap. If the plotted points curve noticeably, ΔH_vap is changing over that range — which is a real result, not a mistake in your plotting.

Sublimation and fusion

The same equation works for a solid turning directly into vapour if you use ΔH_sub in place of ΔH_vap. It does not work in this form for melting, because the derivation assumes the vapour behaves as an ideal gas and that the volume of the condensed phase is negligible compared with the vapour. Neither assumption holds for a solid–liquid transition; that case needs the full Clapeyron equation dP/dT = ΔH / (T ΔV) instead.

Common mistakes that cost marks

Temperature in °C. Every T here is in kelvin. Using 80 instead of 353.15 does not give a slightly wrong answer; it gives a nonsense one.
Mixing pressure units. P₁ in mmHg and P₂ in kPa destroys the ratio. Both in the same unit, always.
ΔH in kJ with R in J. 40.7 must be written 40 700 J mol⁻¹.
Wrong bracket order. (1/T₂ − 1/T₁) carries the minus sign; (1/T₁ − 1/T₂) does not. Write it the same way every time.
Stopping at ln(P₂/P₁). That is a logarithm, not a pressure ratio. Take the exponential, then multiply by P₁.
Rounding the reciprocals early. You are subtracting two numbers that differ in the fourth significant figure. Keep at least six figures until the subtraction is done.
Claiming exact answers. This equation assumes ideal vapour and constant ΔH_vap. Over 100 K or more, treat the answer as an estimate and say so.

Where this appears in exams

Exam	Typical use
CBSE/ICSE Class 12	Vapour pressure, boiling point and the effect of external pressure
IIT-JAM / CUET-PG	Two-point numericals; finding ΔH_vap from two pressures
GATE Chemistry	Slope of a ln P vs 1/T plot; Trouton's rule comparisons
CSIR-NET	Full Clapeyron equation and phase diagrams; solid–liquid and solid–vapour lines

Get the reciprocals right every time. The Clausius–Clapeyron calculator takes any three of P₁, P₂, T₁, T₂ and ΔH_vap and returns the missing one — including the boiling-point version, where the delicate small-difference subtraction is done at full precision.

Open the Clausius–Clapeyron Calculator →

Physical chemistry becomes easy once the pattern behind these two-point equations clicks — the same four steps solve van't Hoff, Arrhenius and this one. ABC Chemistry runs Class 11–12 coaching at the Gurugram centre plus online classes across India: abcchemistry.in.