10.4: The Clausius-Clapeyron Equation
- Page ID
- 202935
( \newcommand{\kernel}{\mathrm{null}\,}\)
Evaporation
In Section 23.3, the Clapeyron Equation was derived for melting points.
dPdT=ΔHmolarTΔVmolar
However, our argument is actually quite general and should hold for vapor equilibria as well. The only problem is that the molar volume of gases are by no means so nicely constant as they are for condensed phases. (i. e., for condenses phases, both α and κ are pretty small).
We can write:
dPdT=ΔHmolarTΔVmolar=ΔHmolarT[Vgasmolar−Vliquidmolar]
as
Vgasmolar≫Vliquidmolar
we can approximate
Vgasmolar−Vliquidmolar
by just taking Vgasmolar. Further more if the vapor is considered an ideal gas, then
Vgasmolar=RTP
We get
1P.dPdT=dlnPdT=ΔHvapmolarRT2
Equation ??? is known as the Clausius-Clapeyron equation. We can further work our the integration and find the how the equilibrium vapor pressure changes with temperature:
ln(P2P1)=−ΔHvapmolarR[1T2−1T1]
Thus if we know the molar enthalpy of vaporization we can predict the vapor lines in the diagram. Of course the approximations made are likely to lead to deviations if the vapor is not ideal or very dense (e.g., approaching the critical point).
The Clapeyron Equation
The Clapeyron attempts to answer the question of what the shape of a two-phase coexistence line is. In the P−T plane, we see the a function P(T), which gives us the dependence of P on T along a coexistence curve.
Consider two phases, denoted α and β, in equilibrium with each other. These could be solid and liquid, liquid and gas, solid and gas, two solid phases, et. Let μα(P,T) and μβ(P,T) be the chemical potentials of the two phases. We have just seen that
μα(P,T)=μβ(P,T)
Next, suppose that the pressure and temperature are changed by dP and dT. The changes in the chemical potentials of each phase are
dμα(P,T)=dμβ(P,T)
(∂μα∂P)TdP+(∂μα∂T)PdT=(∂μβ∂P)TdP+(∂μβ∂T)PdT
However, since G(n,P,T)=nμ(P,T), the molar free energy ˉG(P,T), which is G(n,P,T)/n, is also just equal to the chemical potential
ˉG(P,T)=G(n,P,T)n=μ(P,T)
Moreover, the derivatives of ˉG are
(∂ˉG∂P)T=ˉV,(∂ˉG∂T)P=−ˉS
Applying these results to the chemical potential condition in Equation ???, we obtain
(∂ˉGα∂P)TdP+(∂ˉGα∂T)PdT=(∂ˉGβ∂P)TdP+(∂ˉGβ∂T)PdTˉVαdP−ˉSαdT=ˉVβdP−ˉSβdT
Dividing through by dT, we obtain
ˉVα∂P∂T−ˉSα=ˉVβ∂P∂T−ˉSβ(ˉVα−ˉVβ)∂P∂T=ˉSα−ˉSβdPdT=ˉSα−ˉSβˉVα−ˉVβ
The importance of the quantity dP/dT is that is represents the slope of the coexistence curve on the phase diagram between the two phases. Now, in equilibrium dG=0, and since G=H−TS, it follows that dH=TdS at fixed T. In the narrow temperature range in which the two phases are in equilibrium, we can assume that H is independent of T, hence, we can write S=H/T. Consequently, we can write the molar entropy difference as
ˉSα−ˉSβ=ˉHα−ˉHβT
and the pressure derivative dP/dT becomes
dPdT=ˉHα−ˉHβT(ˉVα−ˉVβ)=ΔαβˉHTΔαβˉV
a result known as the Clapeyron equation, which tells us that the slope of the coexistence curve is related to the ratio of the molar enthalpy between the phases to the change in the molar volume between the phases. If the phase equilibrium is between the solid and liquid phases, then ΔαβˉH and ΔαβˉV are ΔˉHfus and ΔˉVfus, respectively. If the phase equilibrium is between the liquid and gas phases, then ΔαβˉH and ΔαβˉV are ΔˉHvap and ΔˉVvap, respectively.
For the liquid-gas equilibrium, some interesting approximations can be made in the use of the Clapeyron equation. For this equilibrium, Equation ??? becomes
dPdT=ΔˉHvapT(ˉVg−ˉVl)
In this case, ˉVg≫ˉVl, and we can approximate Equation ??? as
dPdT≈ΔˉHvapTˉVg
Suppose that we can treat the vapor phase as an ideal gas. Certainly, this is not a good approximation so close to the vaporization point, but it leads to an example we can integrate. Since PVg=nRT, PˉVg=RT, Equation ??? becomes
dPdT=ΔˉHvapPRT21PdPdT=ΔˉHvapRT2dlnPdT=ΔˉHvapRT2
which is called the Clausius-Clapeyron equation. We now integrate both sides, which yields
lnP=−ΔˉHvapRT+C
where C is a constant of integration. Exponentiating both sides, we find
P(T)=C′e−ΔˉHvap/RT
which actually has the wrong curvature for large T, but since the liquid-vapor coexistence line terminates in a critical point, as long as T is not too large, the approximation leading to the above expression is not that bad.
If we, instead, integrate both sides, the left from P1 to P2, and the right from T1 to T2, we find
∫P2P1dlnP=∫T2T1ΔˉHvapRT2dTln(P2P1)=−ΔˉHvapR(1T2−1T1)=ΔˉHvapR(T1−T1T1T2)
assuming that ΔˉHvap is independent of T. Here P1 is the pressure of the liquid phase, and P2 is the pressure of the vapor phase. Suppose we know P2 at a temperature T2, and we want to know P3 at another temperature T3. The above result can be written as
ln(P3P1)=−ΔˉHvapR(1T3−1T1)
Subtracting the two results, we obtain
ln(P2P3)=−ΔˉHvapR(1T2−1T3)
so that we can determine the vapor pressure at any temperature if it is known as one temperature.
In order to illustrate the use of this result, consider the following example:
Example 23.4.1
At 1bar, the boiling point of water is 373K. At what pressure does water boil at 473K? Take the heat of vaporization of water to be 40.65kJ/mol.
Solution
Let P1=1bar and T1=373K. Take T2=473K, and we need to calculate P2. Substituting in the numbers, we find
lnP2(bar)=−(40.65kJ/mol)(1000J/kJ)8.3145J/mol⋅K(1473K−1373K)=2.77P2(bar)=(1bar)e2.77=16bar
Learning Objectives
- Apply the Clausius-Clapeyron equation to estimate the vapor pressure at any temperature.
- Estimate the heat of phase transition from the vapor pressures measured at two temperatures.
The vaporization curves of most liquids have similar shapes with the vapor pressure steadily increasing as the temperature increases (Figure 10.4.1).

A good approach is to find a mathematical model for the pressure increase as a function of temperature. Experiments showed that the vapor pressure P and temperature T are related,
P∝exp(−ΔHvapRT)
where ΔHvap is the Enthalpy (heat) of Vaporization and R is the gas constant (8.3145 J mol-1 K-1).
A simple relationship can be found by integrating Equation ??? between two pressure-temperature endpoints:
ln(P1P2)=ΔHvapR(1T2−1T1)
where P1 and P2 are the vapor pressures at two temperatures T1 and T2. Equation ??? is known as the Clausius-Clapeyron Equation and allows us to estimate the vapor pressure at another temperature, if the vapor pressure is known at some temperature, and if the enthalpy of vaporization is known.
Alternative Formulation
The order of the temperatures in Equation ??? matters as the Clausius-Clapeyron Equation is sometimes written with a negative sign (and switched order of temperatures):
ln(P1P2)=−ΔHvapR(1T1−1T2)
Example 10.4.1: Vapor Pressure of Water
The vapor pressure of water is 1.0 atm at 373 K, and the enthalpy of vaporization is 40.7 kJ mol-1. Estimate the vapor pressure at temperature 363 and 383 K respectively.
Solution
Using the Clausius-Clapeyron equation (Equation ???), we have:
P363=1.0exp[−(40,7008.3145)(1363K−1373K)]=0.697atm
P383=1.0exp[−(40,7008.3145)(1383K−1373K)]=1.409atm
Note that the increase in vapor pressure from 363 K to 373 K is 0.303 atm, but the increase from 373 to 383 K is 0.409 atm. The increase in vapor pressure is not a linear process.
Discussion
We can use the Clausius-Clapeyron equation to construct the entire vaporization curve. There is a deviation from experimental value, that is because the enthalpy of vaporization varies slightly with temperature.
The Clausius-Clapeyron equation can be also applied to sublimation; the following example shows its application in estimating the heat of sublimation.
Example 10.4.2: Sublimation of Ice
The vapor pressures of ice at 268 K and 273 K are 2.965 and 4.560 torr respectively. Estimate the heat of sublimation of ice.
Solution
The enthalpy of sublimation is ΔHsub. Use a piece of paper and derive the Clausius-Clapeyron equation so that you can get the form:
ΔHsub=Rln(P273P268)1268K−1273K=8.3145ln(4.5602.965)1268K−1273K=52,370Jmol−1
Note that the heat of sublimation is the sum of heat of melting (6,006 J/mol at 0°C and 101 kPa) and the heat of vaporization (45,051 J/mol at 0 °C).
Exercise 10.4.2
Show that the vapor pressure of ice at 274 K is higher than that of water at the same temperature. Note the curve of vaporization is also called the curve of evaporization.
Example 10.4.3: Vaporization of Ethanol
Calculate ΔHvap for ethanol, given vapor pressure at 40 oC = 150 torr. The normal boiling point for ethanol is 78 oC.
Solution
Recognize that we have TWO sets of (P,T) data:
- Set 1: (150 torr at 40+273K)
- Set 2: (760 torr at 78+273K)
We then directly use these data in Equation ???
ln(150760)=−ΔHvap8.314[1313−1351]ln150−ln760=−ΔHvap8.314[1313−1351]−1.623=−ΔHvap8.314[0.0032−0.0028]
Then solving for ΔHvap
ΔHvap=3.90×104 joule/mole=39.0 kJ/mole
Advanced Note
It is important to not use the Clausius-Clapeyron equation for the solid to liquid transition. That requires the use of the more general Clapeyron equation
dPdT=ΔˉHTΔˉV
where ΔˉH and ΔˉV is the molar change in enthalpy (the enthalpy of fusion in this case) and volume respectively between the two phases in the transition.