Ctrl+K

Fugacity

Fugacity#

Additional Readings for the Enthusiast#

Tester and Modell [4], Ch. 8.5, 9.7, 9.1, 9.2

Goals for Today’s Lecture#

  • Why is it insufficient to calculate the chemical potential with respect to a reference state at \(P=0\)?

  • What is the fugacity, and what does it conceptually represent?

  • How do interactions within a system affect the fugacity?

Single-component phase equilibrium: defining fugacity#

We have now described departures from ideality with respect to the \(PVT\) behavior of systems by defining the compressibility factor, \(Z = PV/RT\), and deriving the general form for a departure function. In principle, we now have the framework necessary to compute the chemical potential and thus determine conditions of phase equilibrium. Recall that for a single-component system, phase equilibrium is reached at constant temperature and pressure if the following condition is met:

\[\mu^{(1)}(T,P) = \mu^{(2)}(T,P)\]

where \((1)\) and \((2)\) refer to the two phases that coexist in the system. We can relate the chemical potential to the temperature and pressure via the Gibbs-Duhem equation in intensive form:

\[\begin{split}\begin{aligned} d\mu &= - SdT + VdP \\ \therefore \mu &= f_1(T,P) \end{aligned}\end{split}\]

Therefore, in order to understand phase equilibrium, we need a procedure for calculating \(\mu\). We’ll start with vapor-liquid equilibrium in a single-component system, as this serves as the most straightfoward example.

To calculate \(\mu\) in either the liquid or vapor phase, we can integrate the Gibbs-Duhem equation from a reference state to the chemical potential of interest.

What would be a good choice for our reference state?

Click for answer

Just as in our discussion of departure functions, we can use the ideal gas state at the same temperature \(T\).

This reference state is convenient because we can obtain the value of the chemical potential for any given temperature at constant pressure for an ideal gas if we know the temperature dependence of \(C_P^0\) (with the \(^0\) indicating the ideal gas state as above); this follows from the relationship between the entropy and the heat capacity derived in the last two lectures. To obtain the chemical potential for any given \(T\) and \(P\) we can thus integrate the Gibbs-Duhem equation at constant temperature \(T\) from \(P^{\textrm{ref}}\) to \(P\), yielding:

\[\mu(T, P) - \mu(T, P^{\textrm{ref}}) = \int_{P^{\textrm{ref}}}^{P} V dP\]

This integral is straightforward if we know the \(PVT\) behavior of the system either through tabulated compressibility factors or equations of state. Again, note that we only have the \(dP\) term because we are assuming a isothermal path between a reference state at the same temperature (and hence \(dT=0\)).

What happens to \(\mu(T, P) - \mu(T, P^{\textrm{ref}})\) as \(P^{\textrm{ref}}\to 0\)?

Click for answer

Given that our reference state is an ideal gas, we can write

\[\begin{split}\begin{aligned} \mu^0(T, P) - \mu^0(T, P^{\textrm{ref}}) &= \int_{P^{\textrm{ref}}}^{P} \frac{RT}{P} dP \\ &= RT \int_{P^{\textrm{ref}}}^{P} \frac{1}{P} dP \\ &= RT \ln\left(\frac{P}{P^{\textrm{ref}}}\right) \end{aligned}\end{split}\]

This shows the system will behave like an ideal gas as \(P^\textrm{ref} \rightarrow 0\), but the chemical potential will approach negative infinity in the same limit.

This issue, which also appeared in our treatment of departure functions, motivates the definition of a new quantity, the fugacity, which is better behaved.

We define the fugacity \(f\) in terms of the isothermal departure function of the molar Gibbs free energy (or, equivalently, chemical potential):

fugacity#
\[\begin{aligned} f = P\exp^{\frac{G(T,P) - G^0(T,P)}{RT}} = P\exp^{\frac{\mu(T,P) - \mu^0(T,P)}{RT}} \end{aligned}\]

such that

\[\begin{aligned} RT \ln \frac{f}{P} \equiv G(T,P) - G^0(T,P) = \mu(T,P) - \mu^0(T,P) \end{aligned}\]

Here, \(f\) is the fugacity and is a function of temperature and pressure. \(G^0\) is the molar Gibbs free energy of the ideal gas reference state. The fugacity has units of pressure, and by inspection of this definition we see that

If the system is an ideal gas…

f is less than P. Try again!
f is equal to P.

Correct! Because \(f=P\) and therefore \(\mu(T,P) = \mu^0(T,P)\) for an ideal gas, we can also think of the fugacity as the pressure of an ideal gas that has the same chemical potential as the real fluid of interest at the same temperature.

f is greater than P. Try again!

If the system has repulsive interactions that increase the chemical potential…

f is less than P. Try again!
f is equal to P. Try again!
f is greater than P. Correct!

If the system has attractive interactions that decrease the chemical potential…

f is less than P. Correct!
f is equal to P. Try again!
f is greater than P. Try again!

Thus, the fugacity effectively captures departures of the pressure from ideality. We can also write the fugacity in differential form as:

\[\begin{aligned} d \ln(f) = \frac{d\mu}{RT} \end{aligned}\]

Finally, we can also define a related, unitless quantity, the

fugacity coefficient#
\[\begin{aligned} \phi \equiv \frac{f}{P} \end{aligned}\]

Returning to the condition of phase equilibrium, if two phases (e.g., the liquid and vapor phases of a pure component) are in equilibrium, then the chemical potentials are equal, leading to:

\[\begin{split}\begin{aligned} \mu^{\textrm{liq}}(T,P) &= \mu^{\textrm{vap}}(T,P) \\ \mu^{\textrm{liq}} - \mu^0(T,P) &= \mu^{\textrm{vap}}(T,P)- \mu^0(T,P) \\ RT \ln \frac{f^{\textrm{liq}}}{P} &= RT \ln \frac{f^{\textrm{vap}}}{P}\\ %\frac{\mu^{\textrm{liq}} - \mu^{\textrm{vap}}}{RT} &= \ln \left ( \frac{f^{\textrm{liq}}}{f^{\textrm{vap}}}\right ) = 0 \\ \therefore f^{\textrm{liq}} &= f^{\textrm{vap}} \\ \therefore \phi^\textrm{liq} &= \phi^\textrm{vap} \end{aligned}\end{split}\]

Since the ideal gas reference state is the same for both phases, the reference state chemical potentials ,\(\mu^0(T,P\)), can be subtracted from each phase’s chemical potential and we find that fugacities are equal between two phases at equilibrium. We also establish the equivalence of fugacity coefficients since pressures are equal at equilibrium.

Concept check:

The phase with the _____ fugacity will be present at equilibrium.

lower Correct!
higher Try again!

At phase equilibrium, the ______ of the two phases must be equal.

fugacities Try again!
fugacity coefficients Try again!
fugacities and fugacity coefficients Correct!

Calculating the fugacity#

Next, let’s consider how to calculate the fugacity as a function of the pressure in analogy to what was discussed above for the chemical potential. Our eventual goal is to try to identify a single pressure, the saturation pressure, for which the fugacities of the liquid and vapor phases of a material are equal, which would then satisfy the condition of phase equilibrium. First, we can write from the definition:

\[\begin{aligned} \left ( \frac{\partial \ln f}{\partial P} \right )_{T} &= \frac{1}{RT}\left ( \frac{\partial \mu}{\partial P} \right )_{T} \end{aligned}\]

The intensive form of the Gibbs-Duhem relation is:

\[\begin{split}\begin{aligned} d\mu &= -S dT + V dP \\ \therefore \left ( \frac{\partial \mu}{\partial P} \right )_{T} &= V \\ \left ( \frac{\partial \ln f}{\partial P} \right )_{T, N} &= \frac{V}{RT} \\ \end{aligned}\end{split}\]

and thus

\[\ln \left [ \frac{f(P)}{f(P^{\textrm{ref}})} \right ] = \frac{1}{RT} \int_{P^{\textrm{ref}}}^P V dP\]

Note that this integral is performed at constant \(T\) and \(N\). We thus obtain a similar expression as was originally developed for the chemical potential in terms of an ideal gas. If our reference state is defined such that the molar volume \(V\) is not a strong function of pressure, then we can directly integrate this expression. For example, if we consider a change in the fugacity of a condensed state (i.e., a liquid), then the molar volume is largely invariant with respect to pressure since most liquids/solids are incompressible. In this case, a reasonable reference pressure is the saturation pressure, \(P^{\textrm{sat}}\), where the vapor fugacity is equal to the liquid fugacity, and we integrate to the pressure of interest at constant volume and temperature:

\[\begin{aligned} f^{\textrm{liq}}(P) = f^{\textrm{sat}}(P^{\textrm{sat}}) \exp\left [ \frac{V^{\textrm{liq}}(P-P^{\textrm{sat}})}{RT} \right ] \end{aligned}\]

The exponential term is called the Poynting correction and accounts for the change in the fugacity associated with pressures that are higher than the saturation pressure. Recall that compressibility factors for liquids are often quite small (\(Z\approx 0.01 - 0.3\)) and the term in the exponential is equivalent to the difference in the compressibility factor of the liquid at a pressure \(P\) relative to the saturation pressure. Therefore, the term in the exponential is also typically \(\approx 0\) and the Poynting correction is typically close to 1 unless the pressure is significantly larger than the saturation pressure, allowing the correct to be safely neglected.

The prior framework is useful for computing liquid-phase fugacities, but is not particularly helpful in determining conditions of phase equilibrium because it presupposes that we know the saturation pressure, which is exactly the quantity that we seek to determine.

Why can’t we use the Poynting approach for vapor phases?

Click for answer In vapor phases, vapor-phase volume will be a strong function of pressure.

Instead, we can return to our definition of the fugacity coefficient as equal to the isothermal Gibbs departure function and use this to calculate fugacities directly from tabulated or analytical compressibility factors (or equations of state). First, recall the expression for the isothermal Gibbs departure function:

\[\begin{split}\begin{aligned} G(T,P) - G^0(T,P) &= RT \ln \phi \\ &= RT \int_0^P (Z-1)\frac{dP}{P} \\ \therefore \ln \phi &= \int_0^P (Z-1)\frac{dP}{P} \end{aligned}\end{split}\]

In principle we can integrate this expression directly. A problem, however, is that \(Z(P)\) is generally a poorly behaved function for real gases at phase coexistence, since many gas equations of state are cubic with respect to volume and thus would be multivalued for a given value of \(P\) corresponding to volumes of different phases.

For example, the van der Waals equation of state is:

\[P = \frac{RT}{V-b} - \frac{a}{V^2}\]

where

  • \(b\) represents excluded molecular volume

  • \(a\) measures the attraction between molecules

Why would this formulation cause Z(P) to be poorly-defined?

Click for answer

This expression is cubic in \(V\) and when plotted would have pressures that correspond to multiple different volumes. This equation of state is the simplest that properly captures two different molecular volumes corresponding to a vapor and liquid phase if plotted at constant temperature; the smaller volume corresponds to the liquid phase (\(V^L\)), while the larger corresponds to the vapor phase (\(V^V\)) in a two-phase region. Thus, \(Z(P)\) is not a well-defined function because it is multivalued for a given value of \(P\).

Thus we can instead use \(Z(V)\) to calculate the fugacity coefficient. From the definition of \(Z\), we can write a total differential for \(Z\) at constant \(T\) as:

\[\begin{split}\begin{aligned} Z &= PV/RT \\ Z RT &= PV \\ RT dZ &= V dP + P dV \\ \frac{dP}{P} &= \left (\frac{RT}{PV}\right ) dZ - \frac{dV}{V} \\ &= \frac{dZ}{Z} - \frac{dV}{V} \end{aligned}\end{split}\]

Substituting this expression into the fugacity coefficient expression gives:

\[\begin{split}\begin{aligned} \ln \phi &= \int_0^P \left ( Z - 1\right ) \left ( \frac{dP}{P} \right ) \\ &= \int_1^Z \left ( Z - 1\right ) \left ( \frac{dZ}{Z} \right ) - \int_\infty^V \left ( Z - 1\right ) \left ( \frac{dV}{V} \right ) \\ &= \int_1^Z dZ - \int_1^Z \frac{dZ}{Z} - \int_\infty^V \left ( Z - 1\right ) \left ( \frac{dV}{V} \right ) \\ &= Z -1 - \ln Z + \int^\infty_V \left ( \frac{ Z - 1}{V}\right )dV \end{aligned}\end{split}\]

Note that we have chosen appropriate limits of integration for each integral to reflect an ideal gas reference state - at \(P = 0\), \(Z = 1\) since that is the ideal gas state, and also at \(P = 0\), \(V = \infty\) since volume and pressure are inversely related. Also, recall that all of these integrals are along isothermal paths (i.e., at constant \(T\) and \(N\)). We now have an expression in which the integral is related to volume, which would be appropriate for a compressibility factor given as \(Z(V)\). From knowledge of the compressibility factor, we thus have a means of computing fugacities and determining the conditions of equilibrium - we could choose a pressure, determine the corresponding molar volumes of the liquid and vapor phases associated with the isotherm for that pressure (i.e., roots of the cubic equation of state), calculate the fugacity for each molar volume, and iterate until a pressure is found such that the fugacity corresponding to each volume is identical. This pressure would be the saturation pressure. A similar procedure could be performed using tabulated compressibility factor data by interpolating between known data points as needed.

Contents