Vapour Pressure and Boiling

Water boils at 100 °C only if the surrounding pressure equals 1 atm. Climb to the summit of Denali, where the air pressure drops to about 0.47 atm, and water boils near 79 °C. Understanding why requires connecting a liquid's vapour pressure curve to the ambient pressure it must match. This relationship governs everything from pressure-cooker cooking to vacuum distillation in a chemistry lab. It also explains why high-altitude cake recipes fail and why boiling point is not a fixed property of a substance but a moving target set by pressure. This article builds directly on the Clausius–Clapeyron equation. You will see how vapour pressure curves define boiling points. Then you will see how normal and reduced boiling points let us compare substances fairly. Finally, two worked examples will show the numbers in action.

Key takeaways

Boiling occurs at the temperature where a liquid's vapour pressure equals the external pressure.
The normal boiling point $T_b$ is defined at exactly 1 atm (101.325 kPa).
The standard boiling point uses 1 bar (100 kPa) instead, giving slightly lower values.
Vapour pressure curves are steep and exponential, so small pressure changes shift boiling points noticeably.
Reduced variables ( $T_r = T/T_c$ ) let chemists compare boiling behavior across dissimilar substances.
Key idea: boiling is not an intrinsic property of temperature alone — it is the crossing point between a substance's vapour pressure curve and the applied external pressure.

The core insight

A liquid boils when its equilibrium vapour pressure, which rises steeply and nonlinearly with temperature, becomes equal to the pressure pressing down on its surface; because that vapour pressure curve is fixed by the substance's enthalpy of vaporization while the external pressure is set by the environment, the boiling point is really the temperature coordinate of an intersection, not a fundamental constant.

The physical picture

Every pure liquid held in a closed container develops an equilibrium vapour pressure, $P^*(T)$ , as fast-moving molecules escape the liquid surface and slow ones return. Raise the temperature, and more molecules acquire enough kinetic energy to escape, so $P^*$ rises. This dependence is not linear — it is close to exponential, as captured by the Clausius–Clapeyron equation, because escaping requires overcoming the enthalpy of vaporization, $\Delta H_{vap}$ , an energy barrier that appears in an exponent. Boiling is fundamentally different from evaporation at a free surface. Evaporation happens at any temperature below boiling, restricted to the liquid–air interface. Boiling happens throughout the bulk liquid: bubbles of pure vapour nucleate internally and grow, which is only mechanically possible once the internal vapour pressure inside a bubble can push back against the total external pressure squeezing it. That condition is

P^*(T_b) = P_{ext}

The normal boiling point is the special case where $P_{ext}$ is fixed at 1 atm (101.325 kPa) — the traditional reference used in most older tables, engineering handbooks, and cooking altitude charts. IUPAC's current standard boiling point instead fixes $P_{ext}$ at 1 bar (100 kPa), a slightly lower pressure, which nudges the boiling point down by a small but nonzero amount (a fraction of a kelvin for most liquids). Because the two conventions differ, a chemist should always specify which one a quoted " $T_b$ " refers to.

Figure

Vapour pressure curve crossing the external pressure line defines the boiling point

The steepness of that blue curve is the whole story. Because $P^*$ depends exponentially on $-\Delta H_{vap}/RT$ , a modest shift in external pressure — say, from sea level to a mountain town at 0.8 atm — produces a real, sometimes several-degree, shift in the crossing temperature. Substances with small $\Delta H_{vap}$ (weak intermolecular forces, like diethyl ether) have flatter curves in absolute-pressure terms and are comparatively more sensitive to altitude in relative percentage terms; substances with strong hydrogen bonding, like water, still shift noticeably but require larger pressure changes to shift the same number of degrees.

Figure

Reduced boiling points cluster many substances onto a common scale

The core relations

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

where: $P_1^*, P_2^*$ are equilibrium vapour pressures at temperatures $T_1, T_2$ ; $\Delta H_{vap}$ is the (assumed constant) molar enthalpy of vaporization; $R = 8.314\ \text{J mol}^{-1}\text{K}^{-1}$ .

P^*(T_b) = P_{ext} \quad\Longrightarrow\quad T_b \text{ is the normal (1 atm) or standard (1 bar) boiling point}

where: $T_b$ is defined implicitly as the root of this equation for a chosen $P_{ext}$ .

T_r = \frac{T}{T_c}, \qquad P_r = \frac{P}{P_c}

where: $T_c, P_c$ are the critical temperature and pressure of the substance; reduced boiling points $T_{b,r} = T_b/T_c$ typically fall between about 0.5 and 0.7 for many liquids (Guldberg's/Trouton-type rule of thumb), letting chemists compare volatility across chemically unrelated substances on one universal scale.

Derivation APredicting the boiling point at a new pressure from one known vapour pressure datum

Goal. Given a liquid's normal boiling point (where $P^* = 1$ atm) and its enthalpy of vaporization, find the temperature at which it boils under a different external pressure $P_2$ — the calculation behind altitude-adjusted cooking times and vacuum distillation setpoints.

Step 1. Start from the two-point Clausius–Clapeyron relation. Treating $\Delta H_{vap}$ as constant over the modest temperature range involved (a good approximation far from the critical point),

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

Here state 1 is the known normal boiling point, so $P_1 = 101.325\ \text{kPa}$ and $T_1 = T_b$ .

Step 2. Isolate the unknown reciprocal temperature. Rearranging for $1/T_2$ ,

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

Notice the sign: if $P_2 < P_1$ (lower external pressure, as at altitude), the logarithm is negative, so the term subtracted is negative, meaning $1/T_2$ increases — and therefore $T_2$ decreases. Boiling point drops with pressure, matching physical intuition.

Step 3. Invert to solve for $T_2$ .

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

Step 4. Interpret as the new crossing point. This $T_2$ is exactly the temperature at which the same exponential vapour pressure curve, evaluated at the new pressure $P_2$ , reproduces the equilibrium condition $P^*(T_2) = P_2$ . It is the horizontal line in the earlier figure sliding down (lower $P_{ext}$ ) or up (higher $P_{ext}$ ), and reading off wherever it now crosses the blue curve.

\boxed{T_2 = \left(\dfrac{1}{T_b} - \dfrac{R}{\Delta H_{vap}}\ln\dfrac{P_2}{P_1}\right)^{-1}}

Worked examples

Example 1 — Foundational: water boiling on a mountain. Water has a normal boiling point $T_b = 373.15\ \text{K}$ at $P_1 = 101.325\ \text{kPa}$ , and its molar enthalpy of vaporization near this temperature is $\Delta H_{vap} = 40.7\ \text{kJ mol}^{-1}$ . Find the boiling point at Denali base camp, where atmospheric pressure is $P_2 = 68.0\ \text{kPa}$ . Step 1 — compute the pressure ratio.

\frac{P_2}{P_1} = \frac{68.0}{101.325} = 0.6711

\ln(0.6711) = -0.3990

Step 2 — compute $1/T_1$ .

\frac{1}{T_1} = \frac{1}{373.15\ \text{K}} = 2.6799\times10^{-3}\ \text{K}^{-1}

Step 3 — compute the correction term.

\frac{R}{\Delta H_{vap}} = \frac{8.314}{40{,}700} = 2.0428\times10^{-4}\ \text{K}^{-1}\ (\text{per unit of } \ln)

\frac{R}{\Delta H_{vap}}\ln\!\left(\frac{P_2}{P_1}\right) = (2.0428\times10^{-4})(-0.3990) = -8.151\times10^{-5}\ \text{K}^{-1}

Step 4 — assemble $1/T_2$ and invert.

\frac{1}{T_2} = 2.6799\times10^{-3} - (-8.151\times10^{-5}) = 2.7614\times10^{-3}\ \text{K}^{-1}

T_2 = \frac{1}{2.7614\times10^{-3}} = 362.2\ \text{K} = 89.1\ ^\circ\text{C}

Sanity check. Real-world tables list water's boiling point near Denali base camp (~2200 m, ~76 kPa) around 92 °C, and near the ~68 kPa used here (roughly 3600 m elevation) the accepted value is close to 89–90 °C. Our answer of 89.1 °C is consistent — the drop of about 11 °C from sea level matches the well-known mountaineering rule that water boils noticeably cooler at altitude, sometimes requiring pressure cookers for safe food preparation.

Example 2 — Applied: vacuum distillation setpoint and reduced boiling point of an organic liquid. A chemist wants to distill ethanol under reduced pressure to avoid decomposition of a co-dissolved thermally sensitive product. Ethanol's normal boiling point is $T_b = 351.4\ \text{K}$ (78.25 °C) with $\Delta H_{vap} = 38.6\ \text{kJ mol}^{-1}$ . At what pressure will ethanol boil at a gentler 40 °C (313.15 K), and how does ethanol's reduced boiling point compare to water's? Part A — find required vacuum pressure. Use the same two-point relation, now solving for $P_2$ given $T_2 = 313.15$ K:

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

Compute the reciprocals:

\frac{1}{T_1} = \frac{1}{351.4} = 2.8458\times10^{-3}\ \text{K}^{-1}, \qquad \frac{1}{T_2} = \frac{1}{313.15} = 3.1934\times10^{-3}\ \text{K}^{-1}

\frac{1}{T_2} - \frac{1}{T_1} = 3.1934\times10^{-3} - 2.8458\times10^{-3} = 3.476\times10^{-4}\ \text{K}^{-1}

Compute $\Delta H_{vap}/R$ :

\frac{\Delta H_{vap}}{R} = \frac{38{,}600}{8.314} = 4643\ \text{K}

Multiply:

\ln\!\left(\frac{P_2}{P_1}\right) = -(4643)(3.476\times10^{-4}) = -1.614

Exponentiate:

\frac{P_2}{P_1} = e^{-1.614} = 0.1991

P_2 = (0.1991)(101.325\ \text{kPa}) = 20.2\ \text{kPa}

So the rotary evaporator or vacuum line must be pulled down to roughly 20 kPa (about 150 torr) to make ethanol boil at 40 °C — a realistic, commonly used vacuum distillation setpoint, confirming the calculation is physically sensible. Part B — reduced boiling point comparison. Ethanol's critical temperature is $T_c = 513.9\ \text{K}$ . Its reduced normal boiling point is

T_{b,r} = \frac{351.4}{513.9} = 0.684

Water's critical temperature is $T_c = 647.1\ \text{K}$ , giving

T_{b,r} = \frac{373.15}{647.1} = 0.577

Both values sit in the typical 0.5–0.7 range for ordinary liquids, but ethanol's reduced boiling point is noticeably higher than water's. This reflects the fact that ethanol's normal boiling point sits relatively closer to its own critical point than water's does — consistent with ethanol's weaker, more easily disrupted hydrogen-bond network compared to water's exceptionally strong and extensive one relative to its critical temperature.

Limiting cases & common traps

Situation	What actually happens	Common error
Pressure below the triple-point pressure	Liquid cannot exist; solid sublimes directly to vapour	Assuming a "boiling point" still exists at any pressure
Very low external pressure (deep vacuum)	Boiling point can drop far below room temperature	Confusing this with the liquid simply evaporating away
Near the critical point	Liquid and vapour become indistinguishable, $\Delta H_{vap} \to 0$	Applying Clausius–Clapeyron with constant $\Delta H_{vap}$ far above $T_r \approx 0.8$
Normal vs standard boiling point confusion	1 atm (101.325 kPa) vs 1 bar (100 kPa) give slightly different $T_b$	Treating tabulated " $T_b$ " values as interchangeable without checking the reference pressure
Superheating in very smooth containers	Liquid exceeds $T_b$ without bubbling because no nucleation sites exist	Assuming boiling always occurs exactly at the equilibrium crossing temperature in practice

The chemist behind it

The vapour-pressure–temperature relationship traces back to Benoît Paul Émile Clapeyron's 1834 analysis of Carnot's ideal heat engine, later refined by Rudolf Clausius into the approximate exponential form used throughout this article. But the practical tabulation of boiling points at reduced pressure — essential for the vacuum distillation techniques used in Example 2 — owes enormous debt to the systematic experimental work of chemists cataloguing vapour pressure curves across the nineteenth and twentieth centuries, work that made modern fractional and vacuum distillation routine laboratory operations rather than guesswork.

Origins · 1 minRudolf Clausius and how this was discovered

Rudolf Clausius

1822–1888 · German physicist who reformulated Clapeyron's relation into the modern exponential vapour-pressure law and helped found classical thermodynamics's second law

Ask the tutor: How does adding a nonvolatile solute shift a liquid's vapour pressure curve and its boiling point, as described by Raoult's Law?

Why does boiling point change with altitude but melting point barely does?

Boiling directly involves the vapour phase, whose pressure is exquisitely sensitive to temperature through the exponential Clausius–Clapeyron relation, so matching a lower external pressure requires a substantially lower temperature. Melting, by contrast, is an equilibrium between two condensed phases (solid and liquid), both nearly incompressible. The Clapeyron slope $dP/dT$ for melting is enormous (steep), meaning huge pressure changes produce only tiny shifts in melting point — typically a fraction of a degree even over kilometers of altitude change.

Is the normal boiling point the same as the "standard" boiling point quoted in databases?

Not necessarily. The normal boiling point uses exactly 1 atm (101.325 kPa) as the reference pressure, a convention from before SI units were standardized. IUPAC's modern standard boiling point uses 1 bar (100 kPa) instead. Because 1 bar is slightly less than 1 atm, the standard boiling point is always a little lower than the normal boiling point — often by a few tenths of a kelvin, though the exact gap depends on the substance's $\Delta H_{vap}$ .

Why does superheated liquid sometimes fail to boil exactly at the predicted temperature?

The Clausius–Clapeyron prediction describes thermodynamic equilibrium between bulk liquid and vapour, but boiling in practice requires bubble nucleation, which needs microscopic defects, dissolved gas pockets, or rough surfaces to get started. In a very smooth, clean vessel a liquid can be heated past its true boiling point without bubbling — a metastable state called superheating — until a disturbance triggers sudden, sometimes violent, boiling ("bumping"). This is why boiling chips are added in lab distillations.

How does reduced boiling point help compare unrelated substances?

Raw boiling points in kelvin depend on arbitrary absolute temperature scales and don't reveal how "close" a liquid's normal boiling behavior sits relative to its own critical point. Dividing by the critical temperature, $T_r = T_b/T_c$ , produces a dimensionless number directly comparable across chemically unrelated substances — helium, water, benzene — revealing patterns in intermolecular forces and volatility that raw kelvin values obscure.

Does the enthalpy of vaporization used in these calculations really stay constant?

Only approximately, and only away from the critical point. $\Delta H_{vap}$ actually decreases as temperature rises, falling to exactly zero at the critical point where liquid and vapour become identical phases. Over modest temperature ranges — tens of kelvin, as in both worked examples here — treating it as constant introduces only small errors, typically a few percent, which is why the two-point Clausius–Clapeyron equation remains a reliable practical tool.