Gibbs Free Energy and Pressure, Chemical Potential, Fugacity

The Gibbs free energy depends on pressure as well as on temperature. The pressure dependence of the Gibbs free energy in a closed system is given by the combined first and second laws and the definition of Gibbs free energy as,

(1)              .

If we hold temperature constant and vary only the pressure we can prepare Equation 1 for integration from pressure p1 to p2 as follows:

(2)              ,




(4)             .

Equation 4 is quite general and applies to all isotropic substance: solids, liquids, ideal gases, and real gases. We will apply it first to isotropic solids and liquids.


Solids and Liquids

First level of approximation

We know that solids and liquids are not very compressible so, to a first approximation, we can regard the volume in Equation 4 as constant (as long as the range of pressure is not too large). Then the V in Equation 4 comes out of the integral and we can integrate easily to get,

(5)              .

Second level of approximation.

We know, however, that solids and liquids are slightly compressible and that we can define the isothermal compressibility as

(6)              .

Our second level of approximation is to regard κ as approximately constant. (In fact, κ is not constant, but the variation with pressure is so small it can be ignored unless enormous pressures - megabars - are involved.) With κ regarded as constant we can rearrange Equation 6 and integrate it to find an expression for V as a function of p (which can then be substituted into Equation 4 and integrated.) Rearrangement of Equation 6 looks like,

(7)              .

Integrate from p1 to p2 (and volume goes from V1 to V2) to get,

(8)             .

Take the antilog of both sides,

(9)             .

In Equation 9 we can let p1 be a constant and let p2 range over the pressures of interest. There is no reason why we have to keep the subscript "2" on p2 so change p2 to just p. This gives us V as a function of p,

(10)             .

On the far right of Equation 10 the constant parts have been separated from the variable part to make it easy to integrate. When this expression for V is plugged into Equation 4 only the need stay inside the integral.


Ideal Gases

Equation 4 is also valid for gases, only here we put in the value of V for an ideal gas.

(11)              .

With this substitution Equation 4 becomes,

(12)              ,

which is an integral we have done many times. After integration Equation 12 becomes.

(13)              .

It is customary (and useful) to make several changes in Equation 13. We let p2 range over the pressures of interest to us and call it just p, we let p1 be some standard state pressure and call it po, and finally we divide through by the number of moles of gas, n. With these changes equation 13 is written,

(14)              .

One more change: the quantity G/n turns out to be so important that it is given a special symbol and its own name. Strictly speaking G/n is just the Gibbs free energy per mole of substance, but the simplicity belies its importance. This quantity is called the chemical potential and it is given the symbol, μ . Our final version of what used to be Equation 13 is now,

(15)              .

We have replaced G(po )/n , the molar Gibbs free energy at the standard state pressure, with its chemical potential symbol, μ o. In most cases we will set the standard state pressure equal to one atmosphere. It is not unusual to see Equation 15 written,

(16)              ,

but when it is written like this we have to remember that there is an implied po = 1 atm dividing the p in the ln p, otherwise the argument of the log function would not be unitless.


Nonideal Gases

Equation 15 was derived assuming the gas is ideal. It does not apply to real gases or approximations to a real gas, like the van der Waals equation of state. If we know the equation of state we can go back to Equation 4 and make the appropriate modifications in our notation. That is, we divide Equation 4 by the number of moles, n, let p1 equal the standard state pressure, po and note that V/n is the molar volume to get,

(17)             .

(We have also let p2 range over the pressures of interest and called it just plain p, which means that we have to change the dummy variable of integration from p to p'.) Equation 17 would provide the correct answer in numerical calculations, but it would wreak havoc in some of the later developments of thermodynamics, namely the equilibrium constant expression, as we will see later. G. N Lewis (the same Lewis of the Lewis dot structures and Lewis acid/base theory) proposed to preserve the form of Equation 15 by writing the chemical potential as,

(18)            .

This equation defines a quantity f (p) called the fugacity. The fugacity has units of pressure and it is a function of pressure. It contains all the information on the nonideality of the gas. For an ideal gas the fugacity is the same as the pressure. Since all real gases become ideal in the limit as pressure goes to zero we must have.

(19)             .

We would like to have a way of calculating the fugacity from the equation of state for a gas. To do this go all the way back to Equation 2 and divide it by the number of moles, n,

(20)              ,

or, in our new notation,

(21)              .

We can get another expression for by taking the differential of Equation 18 (Remember that R, T, po , and μ o are constants.)

(22)              .

The in Equations 21 and 22 must be the same, so we can set them equal to each other

(23)             .

Rearrange this to get,


We could integrate this equation directly, but that would sort of take us back to where we started from. Instead, we use a mathematical trick before we integrate it. Add and subtract to the right hand side of Equation 24,

(25)             .

You can see that we didn't really change anything. Regroup the terms in Equation 25,

(26)             .

Now integrate from po to p . (We will have to call our dummy variable of integration p' so as not to conflict with the limits of integration.) We get,

(27)             ,

where fo is the fugacity at po. Move the to the right hand side,

(28)              .

Now we can take the limit where po goes to zero. We know that fo goes to po as po → 0 so the last two terms in parentheses on the right cancel each other in this limit. Equation 28 becomes,

(29)              .

Equation 29 will suffice to calculate the fugacity, but it is customary to take the antilog of both sides to get,

(30)             .

In the last segment of Equation 29, , is the so-called compressibility factor. In either version of Equation of 29 it is easy to see that if the gas is ideal f = p. It requires an equation of state or experimental data to calculate a fugacity from either Equation 30 or Equaton 29. From the right-hand side of Equation 30 we can see that the second form of the virial expansion,

(31)              ,

would be the best choice for calculating fugacity.


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Last updated 5 Nov 04