Clapeyron and Clausius Clapeyron Equations

Clapeyron and Clausius-Clapeyron Equations

The Clapeyron Equation

In a one-component phase diagram the lines separating the phases (for example, solid from liquid, liquid from vapor, and so on) are not arbitrary, they are determined by the principles of thermodynamics and equilibrium. Thermodynamics provides an equation for each line which gives pressure, p, as a function of temperature T, in other words an equation of the form p = p(T) . It is now our task to find the form of the function p(T). The diagram below is a portion of a hypothetical phase diagram and the phase line on this diagram separates the phases α and β. In other words, the phases α and β are in equilibrium with each other at any and all points on the line.

Phase transitions occur at constant temperature and pressure at the points on this line. Consider the phase transition,

α → β. (1)

Since the α and β phases are in equilibrium with each other at any point on the line we know that the change in the Gibbs free energy for the transition given by Equation 1 is zero everywhere on the α-β phase line. That is,

ΔG = 0. (2)

We already know that for any process at constant temperature it must be true that

ΔG = Δ H − TΔS, (3)

so that

(4)

for a phase transition.

The various quantities appearing in these equations are defined as usual by,

ΔG = G_β − G_α,                (5)
ΔS = S_β − S_α.                (6)
ΔH = H_β − H_α.                (7)

We will also define ΔV the same way,

ΔV = V_β − V_α. (8)

As we have already said, ΔG = 0 along the α-β phase line. That is, ΔG is constant along the line and it is being held constant at the particular value of zero. We want the slope of the line. Therefore, the derivative we want is,

. We can write this derivative in terms of quantities we know using the cyclic rule (Euler's chain relation) as,

. (9)

We know from dG = − SdT + Vdp, or dΔG = − ΔSdT + ΔVdp , that

(10)

and

. (11)

Using Equations10 and 11 in Equation 9 (the two minus signs cancel) we get,

. (12)

Equation 12 is the Clapeyron equation. The Clapeyron equation is thermodynamically exact. It contains no approximations. There is another version of the Clapeyron equation which we obtain by inserting Equation 4 for ΔS into Equation 12,

. (13)

Equation 13 is useful when we want to integrate dp to find p as a function of T. The integration of Equation 12 or 13 is easiest if we can make the approximation that either ΔS or ΔH is reasonably constant over the temperature range. However, ΔH usually varies more slowly with temperature than does ΔS, so that the approximation is better when we integrate 13. Prepare Equation 13 for integration,

. (14)

To integrate Equation 14 we must know how ΔH and ΔV depend on temperature. We have already said that ΔH can be regarded as approximately constant, but what about ΔV? ΔV is certainly not even near constant if one of the components of the phase transition is a gas (see the next section), but if the transition is solid → solid or solid → liquid we can approximate ΔV as a constant also. Then Equation 14 integrates to,

. (15)

If we let T₂ and p₂ range over the phase line and call them T and p, respectively, we can write pressure as a function of temperature for a solid → solid or solid → liquid phase transition line as follows,

(16)

The Clausius-Clapeyron Equation

Equations 12 and 13 are exact and are valid for all types of phase transitions, solid → liquid, solid → gas, liquid → gas, and so on. However, for solid → gas and liquid → gas phase transitions we can make two approximations which will give us a very useful equation called the Calusius-Clapeyron equation. The first approximation comes from noting that the volume of a given amount of gas is much larger that the volume of an equivalent amount of solid or liquid. This being true, we can approximate ΔV by the volume of the gas. That is,

ΔV = V_gas − V_liquid ≈ V_gas (17)

ΔV = V_gas − V_solid ≈ V_gas . (18)

In the second approximation we replace V_gas by the volume the gas would have if it were ideal. That is, we set V_gas = nRT/p. With these two approximations Equation 13 becomes

, (19)

where the bar over the enthalpy, H, means that it is the enthalpy per mole. Equation 19 is the Clausius-Clapeyron equation. We should emphasize that Equation 19 is not exact (because it contains two approximations) and it only applies when one of the phases in the phase transition is a gas. Equation 19 can be prepared for integration is follows,

. (20)

Equation 20 can be integrated if we know how ΔH changes with temperature or, as we will now do, if we regard ΔH as approximately constant. With the ΔH regarded as constant Equation 20 integrates to

(21)

Equation 21 is the integrated form of the Clausius-Clapeyron equation. If we want pressure as a function of temperature we can let the point (p₂, T₂) range over the phase line and call it simply (p, T) and then take the antilog of both sides,

. (22)

Notice that if we know two points on the curve we can solve for ΔH, or if we know one point on the curve and ΔH we can solve for any other point or even get the entire curve. Equation 22 is very useful for estimating vapor pressures at various temperatures when ΔH and the vapor pressure at one temperature are known.

Other details and interesting stuff

1. You can get fancy with Equations 19 and 20 if you want. For example, we know that ΔH for a phase transition is not really constant. If we wanted to take into account the temperature dependence of ΔH we need to know the heat capacities of the two phases, or rather the difference in the heat capacity for the two phases,

ΔC_p = C_pβ − C_pα. (23)

Then

. (24)

Initially we regarded ΔH as constant. The next level of approximation would be to regard ΔC_p as constant so that,

. (25)

Using this expression for ΔH in Equation 20 we get

, (26)

or,

. (27)

Equation 27 integrates to,

(28)

You can get even fancier. The next level of approximation would be to include the temperature dependence of the heat capacities, but we won't deal with that here.

2. Equation 12 does not depend on ΔG being zero. It only assumes that ΔG is constant for the α → β transition. Of course, if ΔG is not zero then equation 4 is not true which makes Equation 13 invalid. Further, if ΔG is not zero then either the α phase or the β phase is thermodynamically unstable. (We usually use the word metastable to describe this situation. Examples are supercooled liquids, superheated liquids, supercooled vapors, and some metastable solids - like diamond.) Nevertheless, one can visualize a family of curves running along side the phase diagram curve with ΔG constant at some value other than zero.

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Copyright 2004, W. R. Salzman
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Last updated 1 Dec 05
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