**The Clapeyron equation**

This equation is exact and applies to equilibrium processes between any two phases, i.e melting, boiling and sublimation. It gives the slope of the coexistence curve at (saturated) vapor pressure P and temperature T.

The original derivation by Clapeyron is a geometrical method in which an expression for the PV work done in a Carnot cycle is equated with Carnot’s principle δW = δQ (δT)/T. This classic proof is given in Appendix I.

*Derivation*

If two phases A and B of the same pure substance are in equilibrium with each other, and G_{A} and G_{B} are the respective molar Gibbs free energies, then the condition of equilibrium is G_{A} = G_{B} (if this were not the case, substance in the phase with higher molar free energy would pass into the phase with lower molar free energy, which is a non-equilibrium condition).

At equilibrium dG = 0, which is also the condition for the thermodynamically reversible change dG_{A }= dG_{B}. Since dG = VdP – SdT, we can equate the differential expressions for the two phases:

Since we are dealing with a thermodynamically reversible process at constant pressure and temperature, ΔS =(Q_{p})/T which we can write ΔS = ΔH/T where ΔH is the molar enthalpy of phase transition. Hence

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**Applications of the Clapeyron equation **

**1. Calculate the enthalpy of phase transition ΔH**

If both dP/dT and ΔV at temperature T are known, ΔH can be calculated. Scroll down to Worked example 1

**2. Estimate the effect of pressure change on the boiling point temperature**

If both ΔV and ΔH are known at temperature T, dP/dT can be calculated to estimate the effect of pressure change on the boiling point temperature. Scroll down to Worked example 2

**3. Estimate the effect of pressure change on the melting point temperature**

If both ΔV and ΔH are known at temperature T, dP/dT can be calculated to estimate the effect of pressure change on the melting point temperature. Scroll down to Worked example 3

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*Worked Example 1*

The vapor pressure of water changes by 27.17 mm Hg from 372.15K to 373.15K, and the corresponding volume change (water>vapor) is 3.0114 x 10^{-2} m^{3}mol^{-1}. Estimate the enthalpy of vaporization ΔH_{vap} of water at 373.15K.

__Strategy__

Use the Clapeyron equation to solve for ΔH_{vap}

The SI units are:

P (vapor) = Nm^{-2} (conversion: 1 atm/760 mm Hg = 10^{5} Nm^{-2})

T = K

V = m^{3}mol^{-1}

H = Jmol^{-1}

__Calculation__

1. At 373.15 K, the vapor pressure of water changes by 27.17 mm Hg per kelvin. This may be taken as the value of dP/dT at 373.15K.

2. Convert dP/dT from mm Hg per kelvin to Nm^{-2}K^{-1}

dP/dT = 27.17 x (10^{5}/760) = 3575 Nm^{-2}K^{-1}

3. Compute ΔH = (dP/dT) . TΔV

ΔH = 3575 x 373.15 x 3.0114 x 10^{-2} = 40172 Jmol^{-1}

The enthalpy of vaporization of water ΔH_{vap} at 373.15 is estimated to be 40.17 kJmol^{-1}.

**– – – –**

*Worked Example 2*

The enthalpy of vaporization ΔH_{vap} of water at 373.15K is 40,657 Jmol^{-1} and the corresponding volume change (water>vapor) is 3.0114 x 10^{-2} m^{3}mol^{-1}. Estimate the boiling point temperature at 770 mm Hg.

__Strategy__

Use the inverted Clapeyron equation to solve for dT/dP

The SI units are:

T = K

P = Nm^{-2} (conversion: 1 atm/760 mm Hg = 10^{5} Nm^{-2})

ΔV = m^{3}mol^{-1}

ΔH = Jmol^{-1}

Note that P is the * vapor pressure*. Since T is the boiling point temperature, the vapor pressure and the externally applied pressure will be the same.

__Calculation__

1. Compute dT/dP

dT/dP = (373.15 x 3.0114 x 10^{-2})/40657 = 2.76 x 10^{-4} K(Nm^{-2})^{-1}

2. Compute the pressure increase in Nm^{-2}

Water boils at 373.15K at a pressure of 760 mm Hg. The pressure increase under consideration is therefore 770-760 = 10 mm Hg.

10 mm Hg = 10 x (10^{5}/760) Nm^{-2} = 1316 Nm^{-2}

3. Put the two facts together

dT/dP tells us that the boiling point is raised by 2.76 x 10^{-4} K for 1 Nm^{-2} increase in vapor pressure.

So a 1316 Nm^{-2} increase in vapor pressure will raise the boiling point temperature by 1316 x 2.76 x 10^{-4} = 0.36K

The boiling point of water at 770 mm Hg is estimated at 373.15 + 0.36 = 373.51K

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*Worked Example 3*

At 273.15K the molar enthalpy of fusion of water ΔH_{fusion} is 6 x 10^{3} Jmol^{-1} and the corresponding volume change (ice>water) is –1.6 x 10-6 m^{3}mol^{-1}. Estimate the melting point of ice at a pressure of 150 atmospheres.

__Strategy__

Use the inverted Clapeyron equation to solve for dT/dP

The SI units are:

T = K

P = Nm^{-2} (conversion: 1 atm/760 mm Hg = 10^{5} Nm^{-2})

ΔV = m^{3}mol^{-1}

ΔH = Jmol^{-1}

Note that P is the * vapor pressure*. Since T is the melting point temperature, the vapor pressure and the externally applied pressure will be the same.

__Calculation__

1. Compute dT/dP

dT/dP = (273.15 x –1.6 x 10^{-6})/(6 x 10^{3}) = –7.28 x 10^{-8} K(Nm^{-2})^{-1}

2. Compute the pressure increase in Nm^{-2}

Ice melts at 273.15K at a pressure of 1 atmosphere. The pressure increase under consideration is therefore 150-1 = 149 atmospheres.

149 atmospheres = 1.49×10^{7} Nm^{-2}

3. Put the two facts together

dT/dP tells us that the melting point is lowered by –7.28 x 10^{-8} K for 1 Nm^{-2} increase in vapor pressure.

So a 1.49×10^{7} Nm^{-2} increase in vapor pressure will lower the melting point by

(1.49 x 10^{7}) x (–7.28 x 10^{-8}) = –1.08K

The melting point of ice at a pressure of 150 atmospheres is estimated at 273.15 – 1.08 = 272.07K

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**The Clausius-Clapeyron equation**

The Clausius-Clapeyron equation is not exact, since approximations are used in its derivation. Its use is confined to processes involving vapor phase equilibrium, but nevertheless the equation is very useful.

*Derivation*

Beginning with the Clapeyron equation

use is made of the fact that for sublimation and vaporization processes, there is generally a large difference in molar volume between the two phases. In this case, ΔV can be approximated as the molar volume of the vapor V_{vap}.

It is then assumed that the behavior of the vapor approximates that of an ideal gas, so that the molar volume V_{vap} can in turn be replaced with RT/P where P is the vapor pressure, giving

This is the Clausius-Clapeyron equation, which approximates the slope of the liquid-vapor coexistence curve at vapor pressure P and boiling point temperature T. Using the mathematical identity

the Clausius-Clapeyron equation can also be written in the form

If we further assume that ΔH_{vap} is independent of temperature, integration of the above equation can be performed

This is the integrated form of the Clausius-Clapeyron equation. If the vapor pressure P_{1} is known at boiling point temperature T_{1}, this equation can be used to estimate the boiling point temperature T_{2} at another pressure P_{2}. Alternatively, if the boiling point temperatures are known at vapor pressures P_{1} and P_{2}, the enthalpy of vaporization ΔH_{vap} can be estimated.

**– – – –**

**Applications of the Clausius-Clapeyron equation**

or

and its integrated form

**1. Estimate the enthalpy of vaporization ΔH _{vap}**

If both dP/dT and the vapor pressure P are known at boiling point temperature T, the enthalpy of vaporization ΔH

_{vap}can be estimated. The value of R needs to be known. Scroll down to Worked example 4

**2. Estimate the effect of pressure change on the boiling point temperature**

If the vapor pressure P_{1} is known at boiling point temperature T_{1}, the boiling point temperature T_{2} at another pressure P_{2} can be estimated. The values of R and ΔH_{vap} need to be known. Scroll down to Worked example 5

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*Worked Example 4*

dP/dT is 27.17 mm Hg per kelvin for water at 373.15K. Estimate the enthalpy of vaporization ΔH_{vap} of water at this temperature. The vapor pressure of water at 373.15K is 10^{5} Nm^{-2}, and the gas constant R is 8.3145 JK^{-1}mol^{-1}.

__Strategy__

Use the Clausius-Clapeyron equation to solve for ΔH_{vap}

The SI units are:

T = K

P (vapor) = Nm^{-2} (conversion: 1 atm/760 mm Hg = 10^{5} Nm^{-2})

R = JK^{-1}mol^{-1}

ΔH = Jmol^{-1}

__Calculation__

1. Convert dP/dT from mm Hg per kelvin to Nm^{-2}K^{-1}

dP/dT = 27.17 x (10^{5}/760) = 3575 Nm^{-2}K^{-1}

2. Compute ΔH_{vap} = 1/P x dP/dT x RT^{2}

ΔH_{vap} = 10^{-5} x 3575 x 8.3145 x (373.15)^{2} = 41388 Jmol^{-1}

The enthalpy of vaporization of water ΔH_{vap} at 373.15K is estimated to be 41.39 kJmol^{-1} (which is about 2% higher than the experimental value of 40.66 kJmol^{-1}).

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*Worked Example 5*

Water boils at 373.15K at a pressure of 1 atmosphere. At what temperature will water boil in a pressure cooker operating at a pressure of 2 atmospheres? The enthalpy of vaporization ΔH_{vap} of water at 373.15K is 40657 Jmol^{-1} and the gas constant R is 8.3145 JK^{-1}mol^{-1}.

__Strategy__

Use the integrated Clausius-Clapeyron equation to solve for T_{2}

The SI units are:

T = K

R = JK^{-1}mol^{-1}

ΔH = Jmol^{-1}

The units of P_{1} and P_{2} are immaterial, so long as they are the same.

Note that P_{1} and P_{2} are * vapor pressures*. Since T

_{1}and T

_{2}refer to boiling point temperatures, the vapor pressures P

_{1}and P

_{2}will be the same as the externally applied pressures.

__Calculation__

Water in a pressure cooker operating at a pressure of two atmospheres boils at 394K, or 121°C. This explains why vegetables cook so fast in these devices.

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**Appendix 1**

*The Clapeyron equation – Clapeyron’s proof *

Émile Clapeyron’s 1834 paper *Mémoire sur la Puissance Motrice de la Chaleur* (Memoire on the Motive Power of Heat) contains the first analytical representation of the Carnot cycle, depicted above as the quadrilateral ABCD. By assigning infinitely small values to the variations of volume V and vapor pressure P during the four successive operations of the cycle, Clapeyron renders the quadrilateral a parallelogram.

He then demonstrates that the parallelogram ABCD, representing the work done in a complete cycle, is equal in area to BCEF since both stand on the same base BC and lie between the same parallels T and T–δT.

The base-altitude product BF.PQ therefore equates to the work done in a complete cycle.

Now BF=δP is the increase in vapor pressure per δT rise of temperature at constant volume, while PQ=δV equates to δQ/L_{v} where δQ is the heat absorbed and L_{v} is the latent heat of expansion per unit increase in volume

Since by Carnot’s principle the work δW done in a complete cycle is δQ (δT)/T

This is the famous Clapeyron equation.

In modern notation L_{v} is ΔH/ΔV and since vapor pressure is independent of volume the subscript can be omitted and the equation written

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Hello Akash and thank you for your question. Every chemical substance, whether it is water or a hydrocarbon or even a metal, exerts a vapor pressure which always increases with temperature T. Whether the vapor pressure at a given temperature results in the substance being a solid, liquid or vapor depends on the external pressure P acting on it. The phase diagram represents this P,T relationship and can be constructed for any chemical substance.

Below is a link to a page showing the phase diagram for the cyclical hydrocarbon benzene (C6H6) whose melting and boiling points are actually quite similar to those of water:

https://www.engineeringtoolbox.com/benzene-benzol-properties-d_2053.html

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What will be the difference in this phase diagram if the substance I a hydrocarbon in place of water?

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