Clausius-Clapeyron Equation

Equivalent forms

\ln\frac{P_2}{P_1} = -\frac{L}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right)

P(T) = P_0 \exp\left(-\frac{L}{R T}\right)

A single ODE governs every phase boundary on every P-T diagram — from ice melting to neutron stars.

Symbol	Name	SI unit	Dimension	Range
$P$	Vapor pressureoutput Equilibrium vapor pressure at temperature T	Pa	M·L⁻¹·T⁻²	1000 – 200000
$T$	Temperature Absolute temperature	K	Θ	250 – 500
$L$	Latent heat Molar latent heat of vaporization	J/mol	M·L²·T⁻²·N⁻¹	10000 – 100000

Unit systems

Symbol	SI	CGS	Imperial
$P$	Pa	dyn/cm²	psi
$T$	K	K	°R
$L$	J/mol	erg/mol	BTU/mol

Where it holds

Valid when (1) vapor behaves as ideal gas, (2) liquid volume is negligible vs vapor, (3) L is approximately constant. Fails near the critical point.

Dimensional analysis

$[dP/dT] = (M\cdot L^{-1}\cdot T^{-2}\cdot \Theta ^{-1})$ , $[L/(T\cdot \Delta v)] = (M\cdot L^{2}\cdot T^{-2}\cdot N^{-1})/(\Theta \cdot L^{3}\cdot N^{-1}) = M\cdot L^{-1}\cdot T^{-2}\cdot \Theta ^{-1} \checkmark$

Discovery

Rudolf Clausius & Émile Clapeyron · 1834

Clapeyron derived the original form from Carnot's cycle; Clausius gave it a rigorous thermodynamic basis using the second law in 1850.

Try this

Why does water boil at 70°C on Mount Everest?

Water has latent heat 2.26e6 J/kg. Estimate the boiling point at 30 kPa (Everest summit).

Research status: stable

Real-world applications

Pressure cookers (raise boiling point to cook faster)
Refrigeration cycle design
Atmospheric science — cloud condensation, humidity
High-altitude cooking adjustments

Common misconceptions

The integrated form $\ln (P) = -L/RT$ assumes constant L; over wide T ranges L itself varies with T
The equation works for sublimation and melting too, with appropriate $L and \Delta v$
For solid-liquid lines, $\Delta v$ can be tiny and even negative (e.g., water-ice), making dP/dT extremely steep or inverted

Experimental verification

Vapor-pressure tables for water, ammonia,

CO_{2}

confirm the equation to within 5% from triple point

to \sim 80

% of critical temperature.

Derivation

At phase equilibrium,

d\mu

_liquid

= d\mu

_vapor.

Using

d\mu = -s dT + v dP

, set the two equal: -

s_l dT + v_l dP = -s_v dT + v_v dP

.

Solving:

dP/dT = (s_v - s_l)/(v_v - v_l) = L/(T\cdot \Delta v)

since

L = T\cdot \Delta s

.

Approximating v_v ≫ v_l and

v_v = RT/P

gives

dP/dT = LP/(RT^{2})

, which integrates to

\ln (P) = -L/(RT)

+ const.

Limiting cases

T \to \infty

⟶

P \to P_0

At very high T the exponential saturates and curve flattens.

L \to 0

⟶

P =

constNo latent heat means no T-dependence of vapor pressure.

T \to T

_critical⟶ Equation breaks downLatent heat vanishes at the critical point; phase distinction disappears.

What if…

What if you double the latent heat?

Vapor-pressure curve becomes much steeper; small T changes produce huge P shifts — typical of refractory liquids.

What if the substance has zero latent heat?

There's no phase transition — solid, liquid, gas merge (supercritical fluid regime).

What

if \Delta v

is negative (e.g., ice-water)?

dP/dT becomes negative — pressure lowers the melting point. Why ice skates work.

1

Boiling point of water at Mt. Everest

Given ·

P 1:: 101325
T 1:: 373.15
L:: 40700
P 2:: 30000

Find · T_2

Steps

$\ln (P_2/P_1) = \ln (30000/101325) = -1.217$
$Set -1.217 = -(40700/8.314)\cdot (1/T_2 - 1/373.15)$
$1/T_2 - 1/373.15 = 1.217 \times 8.314/40700 = 2.486e-4$
$1/T_2 = 1/373.15 + 2.486e-4 = 2.929e-3 \to T_2 \approx 343\,\mathrm{K} (70^{\circ}C)$

Result ·

\ln (30000/101325) = -(40700/8.314)\cdot (1/T_2 - 1/373.15) \to T_2 \approx 343\,\mathrm{K} = 70^{\circ}C

2

Vapor pressure of water at 80°C

Given ·

P 1:: 101325
T 1:: 373.15
T 2:: 353.15
L:: 40700

Find · P_2

Steps

Compute $1/353.15 - 1/373.15 = 1.520e-4\,\mathrm{K}^{-1}$
Multiply by L/R: $40700/8.314 \times 1.520e-4 = 0.744$
$P_2 = 101325 \times e^(-0.744) \approx 48.2\,\mathrm{kPa}$ (measured: 47.4 kPa)

Result ·

\ln (P_2/101325) = -(40700/8.314)\cdot (1/353.15 - 1/373.15) = -0.703 \to P_2 \approx 50.2\,\mathrm{kPa}

Ideal Gas Law→First Law of Thermodynamics→Entropy Change→