The Clausius-Clapeyron Relation

Why it matters

Every phase diagram you have ever seen — the P–T plane split into solid, liquid, and gas regions by three curves meeting at a triple point — is quantitatively controlled by a single equation. The Clausius-Clapeyron relation is the derivative of those boundary curves. It answers the practical questions that thermodynamics is built to answer: at what temperature does this liquid boil under this pressure? How much does the melting point move if I squeeze the sample? How steep is the vapor-pressure curve, and what does its steepness tell me about the energy locked up in the phase change?

The relation is the reason a pressure cooker works (raise the pressure, raise the boiling point, cook hotter), the reason the ice-water melting line leans backwards (local pressure nudges it toward lower temperature), and the backbone of meteorology, where the exponential rise of saturation vapor pressure with temperature — a direct consequence of the ln P versus 1/T law — governs cloud formation and how much moisture a warming atmosphere can hold (roughly 7% more water vapor per degree Celsius). It is also the standard laboratory method for measuring a latent heat: you never have to run a calorimeter, you just measure vapor pressure at a few temperatures and read the slope.

How it works — step by step

The starting point is phase equilibrium. Two phases coexist along a line in the P–T plane precisely when their molar Gibbs free energies (equivalently, their chemical potentials) are equal:

g_1(T, P) = g_2(T, P)     along the coexistence line

Step 1 — Move along the line. Take a small step (dT, dP) that stays on the boundary. Both Gibbs energies must change by the same amount to keep them equal, so dg₁ = dg₂.

Step 2 — Use the Gibbs differential. For a single component, dg = −s·dT + v·dP, where s is molar entropy and v is molar volume. Applying it to each phase and setting the changes equal:

−s_1·dT + v_1·dP = −s_2·dT + v_2·dP

Step 3 — Solve for the slope. Collecting terms gives (s₂ − s₁)·dT = (v₂ − v₁)·dP, so

dP/dT = Δs / Δv

Step 4 — Replace Δs with latent heat. The transition is isothermal and reversible, so the entropy jump is the latent heat divided by temperature, Δs = L/T. Substituting gives the exact Clapeyron equation:

dP/dT = L / (T · ΔV)

where every symbol is defined in the table below. This form is exact for any first-order transition — melting, vaporization, sublimation, or a solid–solid structural change.

Step 5 — Specialize to a vapor. For a liquid–vapor or solid–vapor boundary, the gas is enormously more voluminous than the condensed phase, so ΔV ≈ V_gas. Treat the vapor as ideal, V_gas ≈ RT/P. Then

dP/dT = L·P / (R·T²)   →   d(ln P)/dT = L / (R·T²)

Step 6 — Integrate. Assuming L is roughly constant over the temperature range, integrate to obtain the celebrated Clausius-Clapeyron equation:

ln(P_2 / P_1) = −(L/R) · (1/T_2 − 1/T_1)

equivalently   ln P = −L/(R·T) + C

A plot of ln P against 1/T is therefore a straight line of slope −L/R. This is where the "just read the slope" latent-heat measurement comes from.

Symbols and units

Symbol	Meaning	SI unit
P	Pressure along the coexistence line (vapor pressure for a liquid–gas line)	Pa
T	Absolute temperature (always Kelvin)	K
dP/dT	Slope of the phase-boundary curve	Pa/K
L	Molar latent heat (enthalpy) of the transition, L = T·ΔS	J/mol
ΔV	Molar volume of the higher-T phase minus the lower-T phase	m³/mol
ΔS	Molar entropy change of the transition	J/(mol·K)
R	Universal gas constant = 8.314	J/(mol·K)

Sign convention: L and T are always positive. ΔV is positive for a normal transition where the higher-temperature phase occupies more volume (liquid→gas, most solid→liquid), which makes dP/dT positive. When the higher-temperature phase is denser — famously ice→water — ΔV is negative and the boundary slopes backwards.

The three coexistence lines

Near the triple point, three curves radiate outward, one for each pairwise transition. Their relative steepness is set entirely by their latent heats and volume changes, per the Clapeyron equation.

Boundary	Transition	Latent heat (water)	Slope character
Sublimation line	solid ⇌ vapor	L_sub ≈ 51.1 kJ/mol	Steepest of the three (largest L, huge ΔV)
Vaporization line	liquid ⇌ vapor	L_vap ≈ 40.7 kJ/mol	Steep, ends at the critical point
Fusion (melting) line	solid ⇌ liquid	L_fus ≈ 6.01 kJ/mol	Nearly vertical; tiny ΔV → enormous dP/dT

Because ΔV for melting is thousands of times smaller than for vaporization, the fusion line is almost vertical: it takes a huge pressure change to move the melting point even a little. For water the fusion line is not just steep but leans slightly left, because ice is less dense than liquid water (ΔV_melt ≈ −1.6 cm³/mol). Note that L_sub ≈ L_fus + L_vap when all three are evaluated at the same temperature (energy is a state function, so subliming equals melting then boiling); near the triple point this gives L_sub ≈ 6.0 + 45 ≈ 51 kJ/mol.

Worked example — boiling point on Everest

How hot must water get to boil at the summit of Mount Everest, where atmospheric pressure is about P₂ = 0.333 atm? Water boils where its vapor pressure equals the ambient pressure. At sea level, P₁ = 1 atm at T₁ = 373.15 K (100 °C). Use the integrated Clausius-Clapeyron equation with L_vap = 40.7 × 10³ J/mol and R = 8.314 J/(mol·K):

ln(P_2/P_1) = −(L/R)·(1/T_2 − 1/T_1)

ln(0.333)   = −(40700/8.314)·(1/T_2 − 1/373.15)

−1.0996     = −4895.4·(1/T_2 − 0.0026799)

1/T_2 − 0.0026799 = 2.246 × 10⁻⁴

1/T_2 = 0.0029045   →   T_2 ≈ 344.3 K ≈ 71 °C

So water boils at roughly 71 °C on Everest — cool enough that a boiled egg never fully sets and pressure cookers become essential. The measured value is about 68–71 °C depending on the exact summit pressure, so a one-line calculation with a constant latent heat nails it to a couple of degrees. The small discrepancy comes from L_vap actually rising a little at lower temperatures, which the constant-L assumption ignores.

History

Benoît Paul Émile Clapeyron, a French engineer, published the exact form dP/dT = L/(T·ΔV) in 1834 in his memoir reworking Sadi Carnot's ideas — it predates the formal concept of entropy. Rudolf Clausius, having introduced entropy in the 1850s, recast the equation on a rigorous thermodynamic basis and derived the integrated vapor-pressure form that now carries both names. The relation is one of the earliest quantitative triumphs of thermodynamics, connecting a purely mechanical measurement (how pressure shifts a transition temperature) to a caloric quantity (latent heat) without any microscopic model.

Common misconceptions

• "It's an approximation." The Clapeyron equation dP/dT = L/(T·ΔV) is exact for any first-order transition. Only the integrated ln P versus 1/T Clausius-Clapeyron form is approximate, because it assumes an ideal vapor, a negligible liquid volume, and a constant L.
• "You can use Celsius." Never. T appears as an absolute temperature (and squared in the differential form), so it must be in Kelvin. A Celsius substitution gives nonsense, especially near 0 °C.
• "Vapor pressure depends on how much air is above the liquid." Vapor pressure is a property of the liquid–vapor equilibrium at a given temperature, independent of any inert gas present. Boiling occurs when this vapor pressure equals the total ambient pressure.
• "The straight line works all the way to the critical point." It breaks down there: ΔV → 0 and L → 0, the ideal-gas approximation fails badly, and the line simply terminates. The exact Clapeyron slope stays finite (0/0 resolved), but the log-linear plot curves.
• "Latent heat is constant." L_vap of water falls from 45.0 kJ/mol at 0 °C to 40.7 kJ/mol at 100 °C to zero at the 374 °C critical point. Over a narrow range the constant-L integral is fine; over a wide one you must keep L(T) inside the integral.
• "ΔV is the volume change of the liquid." ΔV is the difference between the two phases' molar volumes. For vaporization it is dominated by the gas (≈ 30,000 cm³/mol for steam versus ≈ 19 cm³/mol for water), which is exactly why the V_gas ≈ RT/P shortcut is so good.

JavaScript — Clausius-Clapeyron calculations

const R = 8.314;  // J/(mol·K)

// Exact Clapeyron slope of a coexistence line, in Pa/K
// L in J/mol, T in K, dV (V_high - V_low) in m^3/mol
function clapeyronSlope(L, T, dV) {
  return L / (T * dV);
}

// Vapor pressure at T2 given a reference point (P1, T1), integrated form
// Assumes constant latent heat L (J/mol)
function vaporPressure(P1, T1, T2, L) {
  return P1 * Math.exp(-(L / R) * (1 / T2 - 1 / T1));
}

// Boiling temperature at a given ambient pressure (invert the integrated form)
function boilingTemp(Pambient, P1, T1, L) {
  // ln(P/P1) = -(L/R)(1/T - 1/T1)  ->  solve for T
  const invT = 1 / T1 - (R / L) * Math.log(Pambient / P1);
  return 1 / invT;
}

// Latent heat from the slope of a ln P vs 1/T fit
function latentHeatFromSlope(slope) {
  return -slope * R;  // slope is d(ln P)/d(1/T)
}

const L_vap = 40700;               // J/mol, water at 100 C
const P0 = 101325, T0 = 373.15;    // 1 atm, 100 C

console.log(vaporPressure(P0, T0, 343.15, L_vap).toFixed(0), 'Pa at 70 C');   // ~32,000 Pa
console.log((boilingTemp(0.333 * P0, P0, T0, L_vap) - 273.15).toFixed(1), 'C on Everest'); // ~71 C
console.log(latentHeatFromSlope(-4895).toFixed(0), 'J/mol from slope -4895 K'); // ~40,700

Where it shows up

• Meteorology. Saturation vapor pressure of water follows Clausius-Clapeyron; a warmer atmosphere holds ~7%/°C more moisture, intensifying rainfall.
• Cooking and altitude. Pressure cookers raise P to raise the boiling point (≈121 °C at 2 atm); high-altitude kitchens fight the opposite effect.
• Vacuum and freeze-drying. Lyophilization uses the sublimation line — low pressure lets ice go straight to vapor without melting.
• Meteorology of ice / glaciology. The backward-sloping fusion line means high local pressure can lower the melting point of ice.
• Materials science. Solid–solid transition lines (e.g. graphite–diamond) obey the exact Clapeyron form, guiding high-pressure synthesis.
• Refrigeration. The vapor-pressure curve of a refrigerant sets the pressures at which it evaporates and condenses in the cycle.