Published on *PNG 520: Phase Behavior of Natural Gas and Condensate Fluids* (https://www.e-education.psu.edu/png520)

Intuitively, the concept of equilibrium conveys the message that something “balances out.” Equilibrium describes a state of vanishing driving forces or gradients, where everything remains as it is. If a system is in equilibrium, it retains its current state because there are no driving forces causing anything to change.

If two materials have the same temperature, we say that they are in *thermal equilibrium*. No exchange of heat takes places because there are no thermal gradients. For instance, a liquid and a vapor phase are in *thermal *equilibrium when:

$${T}_{l}=Tv$$

We achieve *mechanical equilibrium* if two substances are found at the same pressure. No bulk movement of fluids takes place because there are no *pressure gradients*. A liquid and a vapor phase are in *mechanical* equilibrium when:

$${P}_{l}=Pv$$

For a thermodynamic system to be in equilibrium, all intensive (temperature, pressure) and extensive thermodynamic properties (U, G, A, H, S, etc) must be constants. Hence, the total change in any of those properties ($d\Im $ ) must be zero at equilibrium.

Now we would like to have a concept of thermodynamic equilibrium for a vapor-liquid equilibrium. Let us consider a closed, heterogeneous vapor-liquid system. Any changes in a total property of the system will be the result of the changes of that property in the liquid phase plus the changes of that property in the vapor phase.

$$d{\Im}^{\left(total\right)}=d{\Im}^{\left(liquid\right)}+d{\Im}^{\left(vapor\right)}$$

In this case, liquid and vapor by themselves are not closed systems; they can exchange matter between themselves but not with the surroundings. To elaborate more upon the concept of equilibrium, let’s look at equation (15.26c). Because it is written in terms of changes in pressure and temperature, two measurable laboratory quantities, it is the “friendliest” of all fundamental equations. We write it for both of the phases:

$$d{\Im}^{\left(liquid\right)}={\left(VdP-SdT\right)}^{l}+{\displaystyle \sum _{i}{\mu}_{i}^{l}d{n}_{i}^{l}}$$

$$d{G}^{\left(vapor\right)}={\left(VdP-SdT\right)}^{v}+{\displaystyle \sum _{i}{\mu}_{i}^{v}d{n}_{i}^{v}}$$

In (16.3), for $\Im =G$ , we get:

$$d{G}^{\left(total\right)}=d{G}^{\left(liquid\right)}+d{G}^{\left(vapor\right)}$$

Hence,

$$d{G}^{\left(total\right)}={\left(VdP-SdT\right)}^{l}+{\left(VdP-SdT\right)}^{v}+{\displaystyle \sum _{i}{\mu}_{i}^{l}}d{n}_{i}^{l}+{\displaystyle \sum _{i}{\mu}_{i}^{v}}d{n}_{i}^{v}$$

Since at equilibrium all extensive properties, such as G, must remain constant, dG^{(total)} must be zero. For this to hold true, and by inspection of equation (16.6), the conditions for thermodynamic equilibrium are:

$$dP=0\text{}\left[\text{Mechanicalequilibrium}\right]$$

E$$dT=0\text{}\left[\text{Thermalequilibrium}\right]$$

$$\sum _{i}{\mu}_{i}^{l}}d{n}_{i}^{l}+{\displaystyle \sum _{i}{\mu}_{i}^{v}}d{n}_{i}^{v}=0\text{}\left[{\mu}_{i}\text{criteriaforequilibrium}\right]$$

It can be also proven that, at equilibrium, the total free energy of the system (G^{(total)}) must take a minimum value; this reinforces the fact that dG^{(total)}=0 at equilibrium. The minimum Gibbs energy criterion for equilibrium is a restatement of the second law of thermodynamics, from which we know that the entropy of a system in equilibrium must be at its maximum, considering all of the possible states for equilibrium.

It is somehow reasonable that for a true equilibrium condition there should be neither pressure nor temperature gradients (equations 16.7 and 16.8). This is because equilibrium is, at the very least, a state of lack of gradients. But what is equation (16.9) trying to tell us? To demystify equation (16.9), we recall that we are dealing with a closed system, hence, the total amount of moles per species:

$${n}_{i}^{(total)}={n}_{i}^{(l)}+{n}_{i}^{(v)}$$

must be constant (we do not allow for chemical reactions within the system). Thus we write:

$$d{n}_{i}^{(total)}=d{n}_{i}^{(l)}+d{n}_{i}^{(v)}=0$$

Therefore,

$$d{n}_{i}^{(v)}=-d{n}_{i}^{(l)}$$

(16.12) into (16.11) yields:

$$\sum _{i}\left({\mu}_{i}^{(l)}-{\mu}_{i}^{(v)}\right)}d{n}_{i}^{(v)}=0$$

For equation (16.13) to hold true,

$${\mu}_{i}^{(l)}={\mu}_{i}^{(v)}{\text{foralli=1,2,...n}}_{c}$$

We have then arrived at the criteria for vapor-liquid equilibria for a system at constant pressure and temperature: *the chemical potential of every species must be the same in both phases.* We may generalize this finding to any number of phases, for which the chemical potential of every species must be the same in all phases. The chemical potential being the driving force which moves a species from one phase to the other, equation (16.14) is physically reasonable. If the chemical potential of a species in one phase is the same as that in the other, there is zero driving force and thus a zero net transfer of species at equilibrium.