The Clausius-Clapeyron Equation

In 1834 Benoit Paul Émile Clapeyron (1799 Feb 26 – 1864
Jan 28) published his "Memoir on the Motive Power of Heat" in the* Journal de
l’Ecole Polytechnique *and in 1843 he had it republished in Poggendorf’s *
Annalen der Physik*. In Section IV of that paper he derived a simple equation
that describes a liquid’s latent heat (he called it caloric) of vaporization as
a function of differential changes in the liquid’s vapor pressure in relation to
differential changes in the liquid’s temperature.

In 1850 Rudolf Clausius published in Poggendorf’s *
Annalen der Physik *a long paper "On the Motive Power of Heat, and on the
Laws which can be Deduced from it for the Theory of Heat". He derived something
close to the modern expression of the Clausius-Clapeyron equation in Section II
in devising his Equation 21.

The modern derivation of the equation follows a different route, one based more on the work of Josiah Willard Gibbs and the use of thermodynamic potentials, which neither Clapeyron nor Clausius knew.

We start with the Gibbs-Duhem relation, which we derive from the scaling property of extensive variables. If we alter all of the extensive variables describing a system by the same scale factor, the total energy contained in the system must change by the same factor. We have then

(Eq’n 1)

If we rewrite the scale factor as α=1+β with |β|<<1, that equation becomes

(Eq’n 2)

which gives us the master equation of thermodynamics,

(Eq’n 3)

In those situations in which we want to use temperature and pressure as our independent variables, the factors that we can control, we use the Gibbs free energy as our master equation,

(Eq’n 4)

in accordance with the Legendre transformation of the system’s energy content. That gives us the differential as

(Eq’n 5)

But we also have the differential Gibbs free energy as

(Eq’n 6)

Comparing those two equations gives us

(Eq’n 7)

which expresses the Gibbs-Duhem relation.

We want to apply that relation to a study of a system whose active part consists of a certain amount of a single chemical species. We have the system in contact with reservoirs that keep it at a constant temperature and pressure. Under those circumstances our target substance exists in two phases and we want to discover the conditions of those phases existing in equilibrium with each other.

In that situation the system reaches equilibrium by so evolving that the Gibbs free energy of the system achieves the minimum possible value. If we plot the substance’s phases on a temperature versus pressure grid, then the points representing the temperatures and pressures at which the system reaches equilibrium between the two phases form a curve across the plot: the conditions under which all of the system consists of phase A lie on one side of that curve and the conditions under which all of the substance consists of phase B lie on the other side. In order to have a minimum value at equilibrium the Gibbs free energy must have the same value on both sides of that curve; if N particles change from one phase to the other, crossing one given point on the equilibrium curve, then, in accordance with Equation 4, we must have

(Eq’n 8)

If we impose minuscule changes, dT and dp, in the temperature and pressure upon our system in such a way that the system remains in equilibrium, then Equation 8 becomes

(Eq’n 9)

which we transform via the Gibbs-Duhem relation to

(Eq’n 10)

Rearranging that equation, we get

(Eq’n 11)

Now we assume that the phase change of our chosen substance occurs between the liquid (phase B) and the gas (phase A) phases. In that case

(Eq’n 12)

in which H_{V} represents the heat of vaporization of the substance
(measured in joules per mole of material). We also assume that we have made the
pressure of the system light enough that we can make the molar volume of the gas
very much larger than the molar volume of the liquid, V_{A}>>V_{B},
so that we can write Equation 11 as

(Eq’n 13)

in which I have used the ideal gas law, pV=RT, in its form for one mole of gas, to replace the molar volume (with the universal gas constant, R=8.314 joule per mole per degree Kelvin). That equation rearranges and then integrates readily to

(Eq’n 14)

The first of those equations displays one form of the Clausius-Clapeyron equation.

Because we based that equation on the equilibrium between the liquid and gas phases of a substance, the pressures in that equation correspond to the vapor pressure of the liquid at two different temperatures. Thus, if we know the vapor pressure of a liquid at one temperature, Equation 14 enables us to calculate the vapor pressure at any other temperature. Because a liquid boils when its vapor pressure equals or exceeds the pressure of the atmosphere pressing down on it, we can take the boiling temperature of a liquid at normal atmospheric pressure (14.7 pounds per square inch or 101,325 newtons per square meter) as our known point. As an example let’s calculate the vapor pressure of water at ten-degree intervals between 0 C and 100 C.

For water we have the heat of vaporization at 100 Celsius
(373 Kelvin, which we must use in our calculation) as 539.55 calories per gram,
which corresponds to H_{V}=2,257,477 joules per kilogram. The gas
constant, 8.314 joule per mole per degree Kelvin, applied to water (18 grams per
mole or 55.55... moles per kilogram) takes the value 461.89 joules per kilogram
per degree Kelvin. To calculate the vapor pressure p at the temperature T we use
a two-step process, in the first step of which we calculate an intermediary
number X. First we have Equation 14 in the form

(Eq’n 15)

and then we have

(Eq’n 16)

Applying those formulae gives us Column I in the following table (with vapor pressure expressed in pounds per square inch):

T |
I |
II |
III |

0 C/273 K |
0.1211 |
0.0886 |
0.0912 |

10 C/283 K |
0.2280 |
0.1781 |
0.1781 |

20 C/293 K |
0.4110 |
0.3392 |
0.3325 |

30 C/303 K |
0.7127 |
0.6155 |
0.5957 |

40 C/313 K |
1.1931 |
1.0701 |
1.0282 |

50 C/323 K |
1.9346 |
1.7893 |
1.7156 |

60 C/333 K |
3.0472 |
2.8893 |
2.7760 |

70 C/343 K |
4.6741 |
4.5203 |
4.3674 |

80 C/353 K |
6.9979 |
6.8684 |
6.6970 |

90 C/363 K |
10.2467 |
10.1693 |
10.0301 |

100 C/373 K |
14.7000 |
14.7000 |
14.7000 |

(Table 1)

Column II lays out the measured values of the vapor pressure and we see a serious discrepancy between those values and the calculated values in Column I. However, our calculations come close enough to the measured values that we cannot legitimately dismiss the Clausius-Clapeyron equation as invalid: we can only state that it stands before us flawed and/or incomplete. We know that we left the equation flawed: we dismissed the volume of the liquid phase as negligible and used the ideal gas law rather than the more accurate van der Waals version. And we know that we may have missed some subtle aspect of phase change in our original analysis and therefore failed to take it into account. Correcting those errors provides the subject of a subsequent essay in this series.

Column III shows vapor pressures calculated through a revised version of Equation 15. In this case I modified Equation 15 by dividing 14.7 psi by the measured vapor pressure at 283 Kelvin and working backward through Equations 16 and 15. Thus I got Equation 15 in the form

(Eq’n 17)

That semi-empirical result gives us figures good enough for some preliminary engineering calculations, but we still see discrepancies which indicate the need for a deeper understanding of vapor pressure and phase change and a suitable revision of the Clausius-Clapeyron equation.

gabh