Clapeyron & Clausius-Clapeyron Equations — Clapeyron and Clausius–Clapeyron equations (use and interpretation) - Physical Chemistry 1

SECTION 1

Historical Context & Motivation

Scientists have studied how temperature governs the pressure at which two phases coexist since the earliest quantitative work on steam engines and meteorology. In the early nineteenth century, engineers needed reliable predictions of boiling points at various pressures to design efficient engines, while natural philosophers sought a unified thermodynamic description of phase transitions. The Clapeyron equation and its approximate descendant, the Clausius–Clapeyron equation, grew out of this intersection of practical engineering needs and rigorous thermodynamic reasoning.

1834 • Clapeyron's Memoir

Benoît Paul Émile Clapeyron published a graphical reformulation of Carnot's work, introducing the P–V indicator diagram and deriving the exact relation dP/dT = ΔH / (TΔV) for phase equilibrium lines.

1850 • Clausius Formalizes Entropy

Rudolf Clausius recast thermodynamics on the concept of entropy, providing a rigorous foundation for Clapeyron's equation and deriving the approximate integrated form for liquid–vapor equilibria.

1865 • Clausius–Clapeyron Integration

Clausius showed that by assuming the vapor behaves ideally and that the molar volume of the liquid is negligible compared to that of the vapor, one can integrate the Clapeyron equation to obtain ln(P₂/P₁) = −ΔH_vap/R × (1/T₂ − 1/T₁).

1876 • Gibbs and Chemical Potential

J. Willard Gibbs unified phase equilibria under the concept of chemical potential equality, providing the deeper justification dμ(α) = dμ(β) from which the Clapeyron equation is a direct consequence.

1900s–present • Modern Applications

The Clausius–Clapeyron equation remains indispensable in atmospheric science for relating water-vapor saturation pressure to temperature, in materials science for predicting sublimation pressures, and in chemical engineering for distillation design.

The fundamental question these equations answer is deceptively simple: How does the equilibrium pressure between two coexisting phases change when we adjust the temperature? Answering this rigorously requires combining the first and second laws of thermodynamics with the condition of chemical potential equality across a phase boundary, ultimately yielding one of the most broadly applied results in physical chemistry.

SECTION 2

Core Principles & Definitions

Before deriving the equations, we need the thermodynamic framework that underlies them. At equilibrium along a phase boundary, the chemical potential μ of a substance is identical in both coexisting phases. Any infinitesimal displacement along the boundary must preserve this equality, so dμ^α = dμ^β. For a pure substance, the chemical potential equals the molar Gibbs energy G_m, and dG_m = V_m dP − S_m dT. Combining these two facts gives the Clapeyron equation directly.

Phase Equilibrium Condition

At equilibrium across a phase boundary, the chemical potentials in both phases are equal: μ^α(T, P) = μ^β(T, P). This single condition defines the coexistence curve in P–T space.

Molar Transition Quantities

ΔV_trs = V_m^β − V_m^α and ΔH_trs are the molar volume change and molar enthalpy change upon transition. For vaporization, both are positive.

Exact Clapeyron Equation

dP/dT = ΔS_trs / ΔV_trs = ΔH_trs / (T ΔV_trs). This is exact—no approximations are involved.

Clausius–Clapeyron Approximations

For liquid–vapor or solid–vapor transitions: (1) ΔV_trs ≈ V_m,gas (condensed-phase volume neglected), and (2) the vapor obeys the ideal gas law V_m = RT/P.

Integrated Form

Assuming ΔH_vap is constant over the temperature range, integration gives ln(P₂/P₁) = −ΔH_vap/R × (1/T₂ − 1/T₁). A plot of ln P vs. 1/T is linear with slope −ΔH_vap/R.

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KEY TAKEAWAY

Think of a phase boundary as a tightrope: a system walking along it must keep the Gibbs energies of both phases perfectly balanced. The Clapeyron equation tells you the slope of that tightrope in P–T space, determined entirely by how much energy (ΔH) and volume (ΔV) the system exchanges during the transition. The Clausius–Clapeyron form is like switching from an exact topographic map to a simplified elevation profile—you lose some detail, but you gain an equation you can integrate and use directly for vapor-pressure predictions.

SECTION 3

Phase Diagram & the Coexistence Curve

A pressure–temperature phase diagram provides the natural geometric context for the Clapeyron equation. Each phase boundary—solid–liquid, liquid–vapor, and solid–vapor—is a curve along which dP/dT is given by the Clapeyron relation. The following diagram illustrates a generic one-component P–T phase diagram with the three coexistence curves meeting at the triple point and the liquid–vapor boundary terminating at the critical point. The slope of each boundary is dictated by the sign and magnitude of ΔH_trs/ΔV_trs.

A generic one-component P–T phase diagram. The solid–liquid boundary (violet) is nearly vertical because ΔV_fus is small. The liquid–vapor boundary (cyan) has a moderate positive slope, and the solid–vapor boundary (amber) is the steepest in terms of dT/dP. The red annotation shows that the local slope is given everywhere by the Clapeyron equation.

Notice that the solid–liquid boundary is nearly vertical because the volume change upon melting is very small compared to the volume change upon vaporization. For water, ΔV_fus is actually negative (ice is less dense than liquid water), giving the solid–liquid line a negative slope—an anomaly with important consequences for glaciology and aquatic life. Each of these slopes is quantified exactly by the Clapeyron equation, making the diagram not just qualitative but a direct graphical representation of thermodynamic data.

SECTION 4

Mathematical Framework

Derivation of the Clapeyron Equation

Consider two phases α and β of a pure substance in equilibrium at temperature T and pressure P. Along the coexistence curve, the chemical potentials remain equal: μ^α = μ^β. Taking the total differential of both sides and using the fundamental relation dμ = V_m dP − S_m dT, we set V_m^α dP − S_m^α dT = V_m^β dP − S_m^β dT. Rearranging yields the exact Clapeyron equation.

CLAPEYRON EQUATION (EXACT)

dP / dT = ΔS_trs / ΔV_trs = ΔH_trs / (T × ΔV_trs)

where ΔS_trs = S_m^β − S_m^α, ΔV_trs = V_m^β − V_m^α, and ΔH_trs = T ΔS_trs at the equilibrium temperature T. This equation is exact and applies to any first-order phase transition (solid–liquid, liquid–vapor, solid–vapor, or polymorphic transitions).

Derivation of the Clausius–Clapeyron Equation

For transitions involving a vapor phase (liquid–vapor or solid–vapor), two simplifying assumptions are physically reasonable away from the critical point. First, V_m,liquid ≪ V_m,vapor, so ΔV_vap ≈ V_m,vapor. Second, the vapor obeys the ideal gas equation V_m,vapor = RT/P. Substituting into the Clapeyron equation gives the differential form of the Clausius–Clapeyron equation.

CLAUSIUS–CLAPEYRON (DIFFERENTIAL)

d(ln P) / dT = ΔH_vap / (R T²)

This form is equivalent to dP/dT = PΔH_vap / (RT²). It relates the rate of change of the natural logarithm of vapor pressure to the reciprocal of T², weighted by the enthalpy of vaporization and the gas constant R.

CLAUSIUS–CLAPEYRON (INTEGRATED, TWO-POINT FORM)

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

Assumes ΔH_vap is constant over the temperature range [T₁, T₂]. P₁ and P₂ are the vapor pressures at temperatures T₁ and T₂ (in Kelvin). R = 8.314 J mol⁻¹ K⁻¹.

CLAUSIUS–CLAPEYRON (SINGLE-POINT FORM)

ln P = −ΔH_vap / (RT) + C

The constant C subsumes the integration constant and entropy of vaporization. A plot of ln P vs. 1/T yields a straight line with slope −ΔH_vap/R.

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When Do the Approximations Fail?

The Clausius–Clapeyron approximations break down near the critical point, where the liquid and vapor densities converge (V_m,liquid is no longer negligible relative to V_m,vapor), the vapor departs significantly from ideality, and ΔH_vap itself approaches zero. In such cases, one must revert to the exact Clapeyron equation or use an equation of state that captures non-ideal behavior.

SECTION 5

The Clausius–Clapeyron Plot: ln P vs. 1/T

One of the most powerful experimental applications of the Clausius–Clapeyron equation is the ln P versus 1/T plot. Because the integrated form predicts a linear relationship between ln P and 1/T (with slope −ΔH_vap/R), measuring vapor pressures at several temperatures and plotting the data allows determination of the enthalpy of vaporization from a simple linear regression. Any deviation from linearity signals that ΔH_vap varies with temperature, prompting more sophisticated treatments.

The Clausius–Clapeyron plot for a hypothetical liquid. Data points (cyan dots) of measured vapor pressures fall along a straight line when plotted as ln P vs. 1/T. The slope of the best-fit line equals −ΔH_vap/R, from which the molar enthalpy of vaporization can be extracted.

The linearity of this plot is a direct test of the assumption that ΔH_vap is temperature-independent. In reality, ΔH_vap generally decreases as temperature rises because the liquid expands and the intermolecular interactions weaken. This curvature becomes noticeable over large temperature ranges or as one approaches the critical temperature, where ΔH_vap → 0. For modest temperature intervals far from T_c, the linear approximation is excellent.

Substance	$T_b$ (K)	ΔH_vap (kJ/mol)	Slope = −ΔH_vap/R (K)
Water	373	40.7	−4893
Ethanol	351	38.6	−4641
Benzene	353	30.7	−3693
Diethyl ether	308	26.5	−3188
Mercury	630	59.1	−7109

Representative enthalpies of vaporization and corresponding Clausius–Clapeyron slopes.

SECTION 6

Worked Example: Vapor Pressure of Water

Let us apply the Clausius–Clapeyron equation to predict the boiling point of water at a reduced pressure—a classic problem encountered in altitude-dependent cooking and vacuum distillation.

At what temperature does water boil at 0.70 atm?

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Step 1 — Identify Given Values

We know water boils at T₁ = 373.15 K at P₁ = 1.00 atm. We want T₂ when P₂ = 0.70 atm. The molar enthalpy of vaporization is ΔH_vap = 40.7 kJ mol⁻¹ = 40 700 J mol⁻¹. R = 8.314 J mol⁻¹ K⁻¹.

T₁ = 373.15 K, P₁ = 1.00 atm, P₂ = 0.70 atm, ΔH_vap = 40 700 J mol⁻¹
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Step 2 — Write and Rearrange the Clausius–Clapeyron Equation

ln(P₂/P₁) = −(ΔH_vap/R) × (1/T₂ − 1/T₁). Rearranging to isolate 1/T₂: 1/T₂ = 1/T₁ − [ln(P₂/P₁) × R] / ΔH_vap
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Step 3 — Compute ln(P₂/P₁)

ln(0.70 / 1.00) = ln(0.70) = −0.3567. Note that because P₂ < P₁, this quantity is negative, consistent with a lower boiling temperature at reduced pressure.

ln(P₂/P₁) = −0.3567
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Step 4 — Solve for 1/T₂

1/T₂ = 1/373.15 − [(−0.3567) × 8.314] / 40 700 = 0.002680 − (−2.966 / 40 700) = 0.002680 − (−7.289 × 10⁻⁵) = 0.002680 + 7.289 × 10⁻⁵ = 0.002753 K⁻¹.

1/T₂ = 0.002753 K⁻¹
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Step 5 — Calculate T₂

T₂ = 1 / 0.002753 = 363.2 K.

T₂ ≈ 363 K ≈ 90 °C
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Step 6 — Interpret the Result

At 0.70 atm—roughly the atmospheric pressure at an altitude of about 3 000 m—water boils at approximately 90 °C rather than 100 °C. This is consistent with the well-known observation that cooking times increase at high altitude because water cannot reach as high a temperature before boiling.

SECTION 7

Clapeyron vs. Clausius–Clapeyron: Strengths & Limitations

Understanding when to use the exact Clapeyron equation versus the approximate Clausius–Clapeyron form is crucial for applying these results correctly. The table below compares the two across several important dimensions.

Feature	Clapeyron (Exact)	Clausius–Clapeyron (Approximate)
Applicable transitions	All first-order transitions: solid–liquid, liquid–vapor, solid–vapor, polymorphic	Only transitions involving a vapor phase (liquid–vapor, solid–vapor)
Approximations	None — thermodynamically exact	$V_m$ ,condensed ≈ 0; vapor is ideal gas; ΔH_vap constant
Data requirements	Requires ΔH_trs and ΔV_trs (including molar volumes of both phases)	Requires only ΔH_vap
Integrability	Not easily integrated because ΔH and ΔV are T- and P-dependent	Readily integrated to give ln P vs. 1/T
Accuracy near $T_c$	Exact everywhere, given accurate input data	Fails as T → $T_c$ because all three assumptions break down
Solid–liquid use	The primary tool for solid–liquid boundaries (e.g., effect of pressure on melting point)	Not applicable — both phases are condensed

Comparison of the exact Clapeyron and approximate Clausius–Clapeyron equations.

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KEY TAKEAWAY

The Clapeyron equation is the "parent," and the Clausius–Clapeyron equation is its "child," produced by introducing three simplifying assumptions that work well for vapor-phase transitions far from the critical point. The parent equation must be consulted whenever those assumptions fail—for solid–liquid transitions, at high pressures, or near critical conditions. This relationship resembles using Ohm's law (V = IR) for ideal resistors while reverting to the full constitutive equations for nonlinear circuit elements.

SECTION 8

Connections to Advanced Theory

The Clapeyron and Clausius–Clapeyron equations are starting points for more refined treatments. In advanced thermodynamics and statistical mechanics, several extensions and generalizations are commonly encountered.

This Lesson	Advanced Extension
ΔH_vap assumed constant over [T₁, T₂]	Kirchhoff's equation: ΔH_vap(T) = ΔH_vap(T_ref) + ∫Δ $C_p$ dT, leading to ln P as a polynomial in 1/T (Antoine equation, Wagner equation)
Ideal gas approximation for vapor	Real-gas corrections via fugacity f replacing P: d(ln f) / dT = ΔH_vap / (RT²). Equations of state (van der Waals, Peng–Robinson) provide $V_m$ ,vapor.
Pure one-component system	Multicomponent extensions: Clausius–Clapeyron combined with Raoult's law or activity coefficients for mixtures; Gibbs–Duhem constraints on coexistence surfaces
First-order transitions only	Ehrenfest equations for second-order transitions (continuous ΔV and ΔS, but discontinuous heat capacity and compressibility)
Classical thermodynamics derivation	Statistical mechanical derivation via partition functions; lattice models; renormalization group theory near the critical point

How this lesson's results connect to advanced thermodynamic theory.

The Ehrenfest equations deserve special mention: they extend the Clapeyron logic to second-order phase transitions (e.g., superconducting transitions, some structural transitions in solids) where ΔS_trs = 0 and ΔV_trs = 0 but the heat capacity and compressibility are discontinuous. In modern critical-phenomena theory, even the Ehrenfest classification has been superseded by a more nuanced understanding rooted in the renormalization group, but the Clapeyron equation remains the indispensable starting point for all first-order transitions.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

The solid–liquid coexistence curve for water has a negative slope (dP/dT < 0). Using the Clapeyron equation, explain what this tells you about the sign of ΔV_fus for water, and discuss the physical origin of this anomaly.

Show answer

From the Clapeyron equation dP/dT = ΔH_fus/(TΔV_fus), since ΔH_fus > 0 (melting is endothermic) and T > 0, a negative dP/dT requires ΔV_fus < 0. This means ice has a larger molar volume than liquid water—ice is less dense than liquid water. Physically, this arises because the hydrogen-bonding network in ice adopts an open hexagonal structure with significant void space, which partially collapses upon melting. This is anomalous; for most substances ΔV_fus > 0 and the solid–liquid line has a positive slope.

Question 2: BASIC CALCULATION

The vapor pressure of ethanol is 1.00 atm at 351 K and ΔH_vap = 38.6 kJ mol⁻¹. Use the Clausius–Clapeyron equation to estimate the vapor pressure of ethanol at 330 K.

Show answer

Using ln(P₂/P₁) = −(ΔH_vap/R)(1/T₂ − 1/T₁) with P₁ = 1.00 atm, T₁ = 351 K, T₂ = 330 K: ln(P₂/1.00) = −(38 600/8.314)(1/330 − 1/351) = −4642 × (0.003030 − 0.002849) = −4642 × 1.810 × 10⁻⁴ = −0.8402. Therefore P₂ = exp(−0.8402) = 0.432 atm ≈ 0.43 atm. This makes physical sense—reducing the temperature below the normal boiling point should yield a vapor pressure below 1 atm.

Question 3: INTERMEDIATE

Vapor pressure data for a certain organic liquid yield a best-fit line of ln(P/Pa) = −4250/T + 22.8 when plotted as ln P vs. 1/T. Determine (a) ΔH_vap in kJ mol⁻¹, and (b) the normal boiling point (where P = 101 325 Pa).

Show answer

(a) The slope is −ΔH_vap/R = −4250 K, so ΔH_vap = 4250 × 8.314 = 35 335 J mol⁻¹ ≈ 35.3 kJ mol⁻¹. (b) At the normal boiling point, ln(101 325) = 11.526. Setting 11.526 = −4250/T_b + 22.8, we get 4250/T_b = 22.8 − 11.526 = 11.274, so T_b = 4250/11.274 ≈ 377 K ≈ 104 °C.

Question 4: APPLIED

An ice skater exerts a pressure of approximately 5.0 MPa on the ice through the blade. The normal melting point of ice is 273.15 K at 0.1013 MPa, ΔH_fus = 6.01 kJ mol⁻¹, and ΔV_fus = −1.63 × 10⁻⁶ m³ mol⁻¹. Using the Clapeyron equation (not Clausius–Clapeyron!), estimate the melting point of ice under the blade.

Show answer

Using dP/dT = ΔH_fus/(TΔV_fus), we approximate ΔT ≈ (TΔV_fus/ΔH_fus) × ΔP. ΔP = 5.0 − 0.1013 ≈ 4.9 MPa = 4.9 × 10⁶ Pa. ΔT = (273.15)(−1.63 × 10⁻⁶)/(6010) × (4.9 × 10⁶) = (273.15 × (−1.63 × 10⁻⁶) × 4.9 × 10⁶)/6010 = (273.15 × (−7.987))/6010 = −2181/6010 ≈ −0.36 K. The melting point is lowered by only about 0.36 K to approximately 272.8 K (−0.4 °C). This shows that pressure-induced melting under a skate blade contributes very little to the lubrication—modern research attributes skating to a premelting surface layer.

Question 5: CRITICAL THINKING

A student measures the vapor pressure of a liquid at five temperatures and obtains a ln P vs. 1/T plot that is distinctly curved rather than linear. Discuss at least three physical reasons why this might occur, and explain what modifications to the Clausius–Clapeyron treatment would address each one.

Show answer

(1) ΔH_vap is temperature-dependent: The difference in heat capacities between liquid and vapor (ΔC_p) means ΔH_vap changes with T via Kirchhoff's equation. Incorporating ΔH_vap(T) into the integral yields a ln P expression that is polynomial in 1/T (e.g., the Antoine equation). (2) The vapor is non-ideal: At higher pressures, intermolecular forces in the vapor cause deviations from PV = nRT. Replacing P with fugacity f or using an equation of state for V_m,vapor corrects this. (3) The molar volume of the liquid is not negligible: As T → T_c, V_m,liquid grows and approaches V_m,vapor; keeping the full ΔV in the Clapeyron equation becomes necessary. A fourth consideration is possible association or dissociation in the vapor phase, which alters the effective molar mass. Addressing all of these leads one to the rigorous Clapeyron equation combined with an accurate equation of state and calorimetric data.

SUMMARY

Summary & Key Concepts

Summary

The Clapeyron equation dP/dT = ΔH_trs / (T ΔV_trs) is an exact thermodynamic result for the slope of any first-order phase boundary in P–T space. It follows directly from the equality of chemical potentials in coexisting phases and requires knowledge of both the molar enthalpy and volume change of the transition. It is the appropriate equation for solid–liquid boundaries and any situation where both phases have comparable densities.

The Clausius–Clapeyron equation d(ln P)/dT = ΔH_vap/(RT²) is an approximate form valid for vapor-phase transitions when the condensed-phase volume is negligible and the vapor behaves ideally. Its integrated two-point form ln(P₂/P₁) = −(ΔH_vap/R)(1/T₂ − 1/T₁) enables direct calculation of unknown vapor pressures or boiling points, and the ln P vs. 1/T plot provides an experimental route to ΔH_vap. These approximations fail near the critical point, where one must return to the exact Clapeyron equation or employ equations of state.