# Mixtures and Phases

• Andreas Hofmann
Chapter

## Abstract

In every-day life, we frequently encounter systems that consist of more than one component and thus represent mixtures. An important fundamental question in this context is whether two different components will spontaneously mix.

## 3.1 Why Do or Don’t Things Mix

In every-day life, we frequently encounter systems that consist of more than one component and thus represent mixtures. An important fundamental question in this context is whether two different components will spontaneously mix.

### 3.1.1 Gases

Experience tells us that two gases contained in the same physical container will inter-mix. Since all spontaneous processes (at constant pressure) must be accompanied by a negative change of the Gibbs free energy (Eq. ), we can assess the every-day experience by thermodynamic means.

The Gibbs free energy change for the mixing process can be calculated as per:
$$\Delta {G}_{\mathrm{mix}}={G}_{\mathrm{final}}-{G}_{\mathrm{initial}}$$
(3.1)
The Gibbs free energies for the initial state of the system is given by the sum of the chemical potentials of each of the components that are about to mix, A and B, multiplied by their respective molar amounts nA and nB. The chemical potential of a gas at a pressure p was introduced with Eq. , and μØ indicates represents the chemical potential under standard conditions:
$${G}_{\mathrm{initial}}={n}_{\mathrm{A}}\cdot {\upmu}_{\mathrm{A}}\left({p}_{\mathrm{initial}}\right)+{n}_{\mathrm{B}}\cdot {\upmu}_{\mathrm{B}}\left({p}_{\mathrm{initial}}\right)={n}_{\mathrm{A}}\cdot \left({\upmu}_{\mathrm{A}}^{\mathrm{\O}}+\mathrm{R}\cdot T\cdot \ln \frac{p}{{\mathrm{p}}^{\mathrm{\O}}}\right)+{n}_{\mathrm{B}}\cdot \left({\upmu}_{\mathrm{B}}^{\mathrm{\O}}+\mathrm{R}\cdot T\cdot \ln \frac{p}{{\mathrm{p}}^{\mathrm{\O}}}\right)$$
When the mixing process starts (i.e. in the initial state), we are releasing two gases A and B into a container with an atmosphere of pressure p. In the final state, when the inter-mixing of the gases is complete, we know that each component can be described by its partial pressure, pA and pB (see page 28). The chemical potential for each component can thus be calculated by using the partial pressures, and one thus obtains for the Gibbs free energy in the final state:
$${G}_{\mathrm{final}}={n}_{\mathrm{A}}\cdot {\upmu}_{\mathrm{A}}\left({p}_{\mathrm{A}}\right)+{n}_{\mathrm{B}}\cdot {\upmu}_{\mathrm{B}}\left({p}_{\mathrm{B}}\right)={n}_{\mathrm{A}}\cdot \left({\upmu}_{\mathrm{A}}^{\mathrm{\O}}+\mathrm{R}\cdot T\cdot \ln \frac{p_{\mathrm{A}}}{{\mathrm{p}}^{\mathrm{\O}}}\right)+{n}_{\mathrm{B}}\cdot \left({\upmu}_{\mathrm{B}}^{\mathrm{\O}}+\mathrm{R}\cdot T\cdot \ln \frac{p_{\mathrm{B}}}{{\mathrm{p}}^{\mathrm{\O}}}\right)$$
Both expressions can now be used to substitute in Eq. 3.1:
$$\Delta {G}_{\mathrm{mix}}={n}_{\mathrm{A}}\cdot \left({\upmu}_{\mathrm{A}}^{\mathrm{\O}}+\mathrm{R}\cdot T\cdot \ln \frac{p_{\mathrm{A}}}{{\mathrm{p}}^{\mathrm{\O}}}\right)-{n}_{\mathrm{A}}\cdot \left({\upmu}_{\mathrm{A}}^{\mathrm{\O}}+\mathrm{R}\cdot T\cdot \ln \frac{p}{{\mathrm{p}}^{\mathrm{\O}}}\right)+{n}_{\mathrm{B}}\cdot \left({\upmu}_{\mathrm{B}}^{\mathrm{\O}}+\mathrm{R}\cdot T\cdot \ln \frac{p_{\mathrm{B}}}{{\mathrm{p}}^{\mathrm{\O}}}\right)-{n}_{\mathrm{B}}\cdot \left({\upmu}_{\mathrm{B}}^{\mathrm{\O}}+\mathrm{R}\cdot T\cdot \ln \frac{p}{{\mathrm{p}}^{\mathrm{\O}}}\right)$$
The molar amounts nA and nB, the chemical potentials in the standard state, as well as R and T are all constants and therefore separated from the unknown parameters p, pA and pB:
$$\Delta {G}_{\mathrm{mix}}={n}_{\mathrm{A}}\cdot {\upmu}_{\mathrm{A}}^{\mathrm{\O}}-{n}_{\mathrm{A}}\cdot {\upmu}_{\mathrm{A}}^{\mathrm{\O}}+{n}_{\mathrm{A}}\cdot \mathrm{R}\cdot T\cdot \left(\ln \frac{p_{\mathrm{A}}}{{\mathrm{p}}^{\mathrm{\O}}}-\ln \frac{p}{{\mathrm{p}}^{\mathrm{\O}}}\right)+{n}_{\mathrm{B}}\cdot {\upmu}_{\mathrm{B}}^{\mathrm{\O}}-{n}_{\mathrm{B}}\cdot {\upmu}_{\mathrm{B}}^{\mathrm{\O}}+{n}_{\mathrm{B}}\cdot \mathrm{R}\cdot T\cdot \left(\ln \frac{p_{\mathrm{B}}}{{\mathrm{p}}^{\mathrm{\O}}}-\ln \frac{p}{{\mathrm{p}}^{\mathrm{\O}}}\right)$$
We see that the above expression can be simplified since some terms cancel:
$$\Delta {G}_{\mathrm{mix}}={n}_{\mathrm{A}}\cdot \mathrm{R}\cdot T\cdot \left(\ln \frac{p_{\mathrm{A}}}{{\mathrm{p}}^{\mathrm{\O}}}-\ln \frac{p}{{\mathrm{p}}^{\mathrm{\O}}}\right)+{n}_{\mathrm{B}}\cdot \mathrm{R}\cdot T\cdot \left(\ln \frac{p_{\mathrm{B}}}{{\mathrm{p}}^{\mathrm{\O}}}-\ln \frac{p}{{\mathrm{p}}^{\mathrm{\O}}}\right)$$
With the logarithm rule of $$\log a-\log b=\log \frac{a}{b}$$ (see Appendix A.1.2) one obtains:
$${\displaystyle \begin{array}{l}\Delta {G}_{\mathrm{mix}}={n}_{\mathrm{A}}\cdot \mathrm{R}\cdot T\cdot \ln \frac{p_{\mathrm{A}}\cdot {\mathrm{p}}^{\mathrm{\O}}}{{\mathrm{p}}^{\mathrm{\O}}\cdot p}+{n}_{\mathrm{B}}\cdot \mathrm{R}\cdot T\cdot \ln \frac{p_{\mathrm{B}}\cdot {\mathrm{p}}^{\mathrm{\O}}}{{\mathrm{p}}^{\mathrm{\O}}\cdot p}\\ {}\Delta {G}_{\mathrm{mix}}={n}_{\mathrm{A}}\cdot \mathrm{R}\cdot T\cdot \ln \frac{p_{\mathrm{A}}}{p}+{n}_{\mathrm{B}}\cdot \mathrm{R}\cdot T\cdot \ln \frac{p_{\mathrm{B}}}{p}\end{array}}$$
As we have introduced earlier (Eq. ), the quotient of partial pressure and total pressure yields the mole fraction x, when dealing with gases. The above equation can thus be re-written in terms of mole fractions xA and xB:
$$\Delta {G}_{\mathrm{mix}}={n}_{\mathrm{A}}\cdot \mathrm{R}\cdot T\cdot \ln {x}_{\mathrm{A}}+{n}_{\mathrm{B}}\cdot \mathrm{R}\cdot T\cdot \ln {x}_{\mathrm{B}}$$
The molar amounts of A and B, nA and nB, are related to the mole fractions via the total molar amount of compounds in the system:
$$\frac{n_{\mathrm{A}}}{n}={x}_{\mathrm{A}}\kern0.5em \Rightarrow \kern0.5em {n}_{\mathrm{A}}=n\cdot {x}_{\mathrm{A}}\kern0.5em \mathrm{and}\kern0.5em \frac{n_{\mathrm{B}}}{n}={x}_{\mathrm{B}}\kern0.5em \Rightarrow \kern0.5em {n}_{\mathrm{B}}=n\cdot {x}_{\mathrm{B}}$$
Therefore we arrive at:
$$\Delta {G}_{\mathrm{mix}}=\left(n\cdot \mathrm{R}\cdot T\right)\cdot {x}_{\mathrm{A}}\cdot \ln {x}_{\mathrm{A}}+\left(n\cdot \mathrm{R}\cdot T\right)\cdot {x}_{\mathrm{B}}\cdot \ln {x}_{\mathrm{B}}<0$$
(3.2)
Since xA and xB are positive numbers between 0 and 1, it becomes clear that lnxA and lnxB are negative numbers, since any logarithmic function is negative between 0 and 1 (Fig. 3.1). With (n ⋅ R ⋅ T) yielding a positive value, we can then conclude that ΔGmix from Eq. 3.2 will always deliver a negative value. Since all processes with ΔG < 0 are spontaneous processes, the mixing of two gases is a spontaneously occurring process!
Considering that the Gibbs free energy change at constant pressure and constant temperature is composed of enthalpic and entropic changes as per:
$$\Delta G=\Delta H-T\cdot \Delta S$$
(2.42)
we can conclude that the mixing two ideal gases is a purely entropic effect, since there is no enthalpy change (ΔH = 0) when two ideal gases intermix. This follows from the requirement of the gas being ideal; in the ideal gas there are no intermolecular forces between the molecules.

In contrast, in real gases inter-molecular interactions do occur and there is therefore a change of enthalpy upon mixing (ΔH ≠ 0), due to intermolecular forces between molecules. It also needs to be considered that ΔH and ΔS depend on p and T. However, the Gibbs free energy change for the inter-mixing of real gases is still generally negative.

### 3.1.2 Liquids

A similar approach can be taken, when considering the mixing of two liquids. As in the previous section, we establish the Gibbs free energy of the system before mixing occurs. This can be expressed as the sum of the chemical potentials of the pure liquids (indicated by the asterisk ‘*’), multiplied with the respective molar amounts of each liquid in the system (since the chemical potential is the Gibbs free energy per 1 mol of substance):
$${G}_{\mathrm{initial}}={n}_{\mathrm{A}}\cdot {\upmu}_{\mathrm{A}\left(\mathrm{l}\right)}^{\ast }+{n}_{\mathrm{B}}\cdot {\upmu}_{\mathrm{B}\left(\mathrm{l}\right)}^{\ast }$$
When applying Eq. to liquid systems, we appreciate that the chemical potential of a liquid at any concentration can be expressed with reference to the chemical potential in its pure state (‘*’) by adjusting for the concentration of the liquid with the term R ⋅ T ⋅  ln x, where x is the mole fraction of the liquid:
$$\upmu ={\upmu}^{\ast }+\mathrm{R}\cdot T\cdot \ln x$$
(3.3)
The final state of the liquid mixture is attained when both liquids are present in their respective concentrations, expressed as mole fractions xA and xB. With Eq. 3.3 we thus obtain:
$${G}_{\mathrm{final}}={n}_{\mathrm{A}}\cdot \left({\upmu}_{\mathrm{A}\left(\mathrm{l}\right)}^{\ast }+\mathrm{R}\cdot T\cdot \ln {x}_{\mathrm{A}}\right)+{n}_{\mathrm{B}}\cdot \left({\upmu}_{\mathrm{B}\left(\mathrm{l}\right)}^{\ast }+\mathrm{R}\cdot T\cdot \ln {x}_{\mathrm{B}}\right)$$
and can now proceed to calculate the change in the Gibbs free energy during the mixing process:
$${\displaystyle \begin{array}{l}\Delta {G}_{\mathrm{mix}}={G}_{\mathrm{final}}-{G}_{\mathrm{initial}}\Delta {G}_{\mathrm{mix}}={n}_{\mathrm{A}}\cdot {\upmu}_{\mathrm{A}\left(\mathrm{l}\right)}^{\ast }+{n}_{\mathrm{A}}\cdot \mathrm{R}\cdot T\cdot \ln {x}_{\mathrm{A}}\\ {}+{n}_{\mathrm{B}}\cdot {\upmu}_{\mathrm{B}\left(\mathrm{l}\right)}^{\ast }+{n}_{\mathrm{B}}\cdot \mathrm{R}\cdot T\cdot \ln {x}_{\mathrm{B}}-{n}_{\mathrm{A}}\cdot {\upmu}_{\mathrm{A}\left(\mathrm{l}\right)}^{\ast }-{n}_{\mathrm{B}}\cdot {\upmu}_{\mathrm{B}\left(\mathrm{l}\right)}^{\ast}\end{array}}$$
Since several terms cancel, this simplifies to:
$$\Delta {G}_{\mathrm{mix}}={n}_{\mathrm{A}}\cdot \mathrm{R}\cdot T\cdot \ln {x}_{\mathrm{A}}+{n}_{\mathrm{B}}\cdot \mathrm{R}\cdot T\cdot \ln {x}_{\mathrm{B}}$$
We then express the individual molar amounts nA and nB as mole fractions xA and xB (and thus with respect to the total molar amount n in the system):
$$\frac{n_{\mathrm{A}}}{n}={x}_{\mathrm{A}}\kern0.5em \Rightarrow \kern0.5em {n}_{\mathrm{A}}=n\cdot {x}_{\mathrm{A}}\kern0.5em \mathrm{and}\kern0.5em \frac{n_{\mathrm{B}}}{n}={x}_{\mathrm{B}}\kern0.5em \Rightarrow \kern0.5em {n}_{\mathrm{B}}=n\cdot {x}_{\mathrm{B}}$$
which yields for the Gibbs free energy change of the mixing process:
$${\displaystyle \begin{array}{l}\Delta {G}_{\mathrm{mix}}=n\cdot \mathrm{R}\cdot T\cdot {x}_{\mathrm{A}}\cdot \ln {x}_{\mathrm{A}}+n\cdot \mathrm{R}\cdot T\cdot {x}_{\mathrm{B}}\cdot \ln {x}_{\mathrm{B}}\\ {}\Delta {G}_{\mathrm{mix}}=n\cdot \mathrm{R}\cdot T\cdot \left({x}_{\mathrm{A}}\cdot \ln {x}_{\mathrm{A}}+{x}_{\mathrm{B}}\cdot \ln {x}_{\mathrm{B}}\right)\end{array}}$$
(3.4)
As we have discussed in the previous section, due to the mole fractions xA and xB taking values between 0 and 1, Eq. 3.4 we always deliver a negative value for the Gibbs free energy change, and the mixing of two liquids should thus always occur spontaneously. However, as in the case of gases, this is only true for ideal liquids, i.e. such that do not possess any intermolecular forces between molecules. With
$$\Delta G=\Delta H-T\cdot \Delta S$$
(2.42)
we can then conclude that in ideal solutions, at constant temperature and constant pressure, there is no enthalpy change (ΔH = 0) and the mixing of ideally behaving liquids is thus a purely entropic effect: ΔGmix =  − T ⋅ ΔS.

In contrast to gases, non-ideal effects are a lot more common in liquids, and therefore we frequently need to consider changes in ΔH due to intermolecular forces, which differ when comparing interactions between A–A, B–B and A–B. A positive change of the enthalpy during the mixing (ΔH > 0) of two liquids is therefore possible and may eventually exceed the entropic effect. The Gibbs free energy of mixing of non-ideal liquids may thus be positive or negative; this will determine whether or not the liquids mix.

## 3.2 Liquids

### 3.2.1 Chemical Potential of Liquid Solutions

A closer look at liquids and liquid solutions shows that there is not only the liquid phase that needs to be considered, but also a vapour phase above the liquid. If we envisage a container in which there is a liquid in equilibrium with its vapour, we can describe the entire contents of the container as one system. Since we assume equilibrium conditions, the chemical potential must be uniform throughout the system, i.e. the chemical potential in the liquid phase is the same as in the vapour phase:
$${\upmu}_{\left(\mathrm{g}\right)}={\upmu}_{\left(\mathrm{l}\right)}$$
For an ideal gas A, we know that chemical potential at any pressure ($${p}_{\mathrm{A}}^{\ast }$$) can be described as per Eq. :
$${\displaystyle \begin{array}{l}{\upmu}_{\mathrm{A}\left(\mathrm{g}\right)}^{\ast }={\upmu}_{\mathrm{A}}^{\mathrm{\O}}+\mathrm{R}\cdot T\cdot \ln \frac{p_{\mathrm{A}}^{\ast }}{{\mathrm{p}}^{\mathrm{\O}}}\\ {}{\upmu}_{\mathrm{A}\left(\mathrm{g}\right)}^{\ast }={\upmu}_{\mathrm{A}}^{\mathrm{\O}}+\mathrm{R}\cdot T\cdot \ln \frac{p_{\mathrm{A}}^{\ast }}{{\mathrm{p}}^{\mathrm{\O}}}\kern0.5em =\kern0.5em {\upmu}_{\mathrm{A}\left(\mathrm{l}\right)}^{\ast}\end{array}}$$
(3.5)
where the asterisks ‘*’ indicate the pure component (i.e. here pure component A in either liquid or vapour form).
When two components A and B are mixed, they no longer exist in their pure forms, and we indicate this by omitting the asterisk ‘*’. However, for each component A and B in the liquid mixture, the chemical potential in the different phases can still be calculated as in Eq. 3.5; we thus obtain:
$${\upmu}_{\mathrm{A}\left(\mathrm{l}\right)}={\upmu}_{\mathrm{A}\left(\mathrm{g}\right)}={\upmu}_{\mathrm{A}}^{\mathrm{\O}}+\mathrm{R}\cdot T\cdot \ln \frac{p_{\mathrm{A}}}{{\mathrm{p}}^{\mathrm{\O}}}$$
(3.6)
$${\upmu}_{\mathrm{B}\left(\mathrm{l}\right)}={\upmu}_{\mathrm{B}\left(\mathrm{g}\right)}={\upmu}_{\mathrm{B}}^{\mathrm{\O}}+\mathrm{R}\cdot T\cdot \ln \frac{p_{\mathrm{B}}}{{\mathrm{p}}^{\mathrm{\O}}}$$
(3.7)

The pressures pA and pB now indicate the partial pressure of component A and B in the vapour above the liquid.

Inspection of Eqs. 3.5 and 3.6 shows that the chemical potential of component A is different when we compare the pure state with a mixture where A is not the exclusive component of the system. The difference in the chemical potential between both states is available by subtracting Eq. 3.5 from 3.6:
$${\displaystyle \begin{array}{l}{\upmu}_{\mathrm{A}\left(\mathrm{l}\right)}-{\upmu}_{\mathrm{A}\left(\mathrm{l}\right)}^{\ast }={\upmu}_{\mathrm{A}\left(\mathrm{g}\right)}-{\upmu}_{\mathrm{A}\left(\mathrm{g}\right)}^{\ast}\\ {}{\upmu}_{\mathrm{A}\left(\mathrm{l}\right)}-{\upmu}_{\mathrm{A}\left(\mathrm{l}\right)}^{\ast }={\upmu}_{\mathrm{A}}^{\mathrm{\O}}+\mathrm{R}\cdot T\cdot \ln \frac{p_{\mathrm{A}}}{{\mathrm{p}}^{\mathrm{\O}}}-\left({\upmu}_{\mathrm{A}}^{\mathrm{\O}}+\mathrm{R}\cdot T\cdot \ln \frac{p_{\mathrm{A}}^{\ast }}{{\mathrm{p}}^{\mathrm{\O}}}\right)\\ {}{\upmu}_{\mathrm{A}\left(\mathrm{l}\right)}-{\upmu}_{\mathrm{A}\left(\mathrm{l}\right)}^{\ast }=\mathrm{R}\cdot T\cdot \ln \frac{p_{\mathrm{A}}}{{\mathrm{p}}^{\mathrm{\O}}}-\mathrm{R}\cdot T\cdot \ln \frac{p_{\mathrm{A}}^{\ast }}{{\mathrm{p}}^{\mathrm{\O}}}\\ {}{\upmu}_{\mathrm{A}\left(\mathrm{l}\right)}-{\upmu}_{\mathrm{A}\left(\mathrm{l}\right)}^{\ast }=\mathrm{R}\cdot T\cdot \left(\ln \frac{p_{\mathrm{A}}}{{\mathrm{p}}^{\mathrm{\O}}}-\ln \frac{p_{\mathrm{A}}^{\ast }}{{\mathrm{p}}^{\mathrm{\O}}}\right)\end{array}}$$
Using the log rule of log b (uv) = log b u + log b v and $${\log}_b\left(\frac{u}{v}\right)={\log}_bu-{\log}_bv$$ (Appendix A.1.2), this yields:
$${\upmu}_{\mathrm{A}\left(\mathrm{l}\right)}-{\upmu}_{\mathrm{A}\left(\mathrm{l}\right)}^{\ast }=\mathrm{R}\cdot T\cdot \ln \frac{p_{\mathrm{A}}\cdot {\mathrm{p}}^{\mathrm{\O}}}{{\mathrm{p}}^{\mathrm{\O}}\cdot {p}_{\mathrm{A}}^{\ast }}$$
which simplifies to
$${\upmu}_{\mathrm{A}\left(\mathrm{l}\right)}={\upmu}_{\mathrm{A}\left(\mathrm{l}\right)}^{\ast }+\mathrm{R}\cdot T\cdot \ln \frac{p_{\mathrm{A}}}{p_{\mathrm{A}}^{\ast }}$$
(3.8)

This equation delivers a relationship between the chemical potential of a particular component in a liquid mixture (μA) and its partial vapour pressure (pA). It is found that pA always takes smaller values than $${p}_{\mathrm{A}}^{\ast }$$ (see next section), which results in $$\frac{p_{\mathrm{A}}}{p_{\mathrm{A}}^{\ast }}<1$$ and thus $$\ln \frac{p_{\mathrm{A}}}{p_{\mathrm{A}}^{\ast }}<0$$; the chemical potential μ of a solvent is therefore lowered as a result of the presence of the solute.

### 3.2.2 Raoult’s Law

Eq. 3.8 can be transformed with some basic algebra to yield:
$$\frac{p_{\mathrm{A}}}{p_{\mathrm{A}}^{\ast }}={\mathrm{e}}^{\frac{\upmu_{\mathrm{A}\left(\mathrm{l}\right)}-{\upmu}_{\mathrm{A}\left(\mathrm{l}\right)}^{\ast }}{\mathrm{R}\cdot T}}$$
(3.9)

The French chemist Francois-Marie Raoult (1830–1901) discovered experimentally that there is a relationship between the partial vapour pressure of component and its mole fraction in the liquid mixture, and summarised his results in 1886 in a rule known as Raoult’s law:

The partial vapour pressure of a component in a liquid mixture is equal to the vapour pressure of the pure component multiplied by its mole fraction in the liquid phase
$${p}_{\mathrm{A}}={x}_{\mathrm{A}}\cdot {p}_{\mathrm{A}}^{\ast }$$
(3.10)
Equation 3.9 can thus be extended to read:
$$\frac{p_{\mathrm{A}}}{p_{\mathrm{A}}^{\ast }}={\mathrm{e}}^{\frac{\upmu_{\mathrm{A}\left(\mathrm{l}\right)}-{\upmu}_{\mathrm{A}\left(\mathrm{l}\right)}^{\ast }}{\mathrm{R}\cdot T}}={x}_{\mathrm{A}}$$
(3.9)

Since the mole fraction xA takes values less or up to 1 (0 ≤ xA ≤ 1), an important effect of Raoult’s law is that the vapour pressure of a solution is always lower than that of the pure solvent at any particular temperature. This effect is also known as vapour pressure depression.

Considering that the boiling point of a solution is the temperature at which the vapour pressure of the liquid is equal to the surrounding environmental pressure, another consequence of the above effects is the boiling point elevation observed with liquid solutions. A solution therefore always has a higher boiling point than the pure solvent. More generally, several properties of liquids (Tm, Tb, p, etc) depend on the presence of solute molecules, in particular the number of particles in solution. Such properties are called colligative properties and will be further discussed in Sect. .

### 3.2.3 Activity: Mole Fraction for a Non-ideal Solution

In Sect. , we have defined the fugacity f for a non-ideal gas to ensure that the chemical potential of the gas at any pressure can be calculated according to the equation
$$\upmu (p)={\upmu}^{\mathrm{\O}}+\mathrm{R}\cdot T\cdot \ln \frac{f}{{\mathrm{p}}^{\mathrm{\O}}}.$$
(2.72)
In Sect. 3.2.1 of this chapter, we established a relationship between the chemical potential of a liquid component (μA(l)) in a mixture and its partial vapour pressure (pA) above the liquid mixture:
$${\upmu}_{\mathrm{A}\left(\mathrm{l}\right)}={\upmu}_{\mathrm{A}\left(\mathrm{l}\right)}^{\ast }+\mathrm{R}\cdot T\cdot \ln \frac{p_{\mathrm{A}}}{p_{\mathrm{A}}^{\ast }}$$
(3.8)
For an ideal ideal solution, it then follows with Raoult’s law that $$\ln \frac{p_{\mathrm{A}}}{p_{\mathrm{A}}^{\ast }}=\ln {x}_{\mathrm{A}}$$, and therefore:
$${\upmu}_{\mathrm{A}\left(\mathrm{l}\right)}={\upmu}_{\mathrm{A}\left(\mathrm{l}\right)}^{\ast }+\mathrm{R}\cdot T\cdot \ln {x}_{\mathrm{A}}$$
(3.11)
Analogous to the fugacity in the case of gases, we can now define a property that will account for non-ideal behaviour in the liquid mixtures. This property is called activity (a) and replaces the mole fraction xA in the above case, such that the chemical potential of compound A at any concentration in a non-ideal solution varies as per the relationship:
$${\upmu}_{\mathrm{A}\left(\mathrm{l}\right)}={\upmu}_{\mathrm{A}\left(\mathrm{l}\right)}^{\ast }+\mathrm{R}\cdot T\cdot \ln {a}_{\mathrm{A}}$$
(3.12)
The activity coefficient γ indicates the deviation from the ideal behaviour:
$${a}_{\mathrm{A}}={\upgamma}_{\mathrm{A}}\cdot {x}_{\mathrm{A}}.$$
(3.13)

### 3.2.4 Gibbs Free Energy Change for a Reaction (Part 2)

In Sect. , we discussed the Gibbs free energy change of the following reaction
$${\upnu}_{\mathrm{A}}\;{\mathrm{A}}_{\left(\mathrm{l}\right)}+{\upnu}_{\mathrm{B}}\;{\mathrm{B}}_{\left(\mathrm{g}\right)}\to {\upnu}_{\mathrm{C}}\;{\mathrm{C}}_{\left(\mathrm{l}\right)}+{\upnu}_{\mathrm{D}}\;{\mathrm{D}}_{\left(\mathrm{g}\right)}$$
by means of the reaction quotient Q, before introducing the chemical potential.

After discussing mixtures of components by means of their chemical potentials in the preceding sections, we can now apply this knowledge and revise our discussion of the Gibbs free energy change for a reaction.

As before, the change in free energy is given by
$$\Delta G={G}_{\mathrm{products}}-{G}_{\mathrm{reactants}}=\left({G}_{\mathrm{C}}+{G}_{\mathrm{D}}\right)-\left({G}_{\mathrm{A}}+{G}_{\mathrm{B}}\right)$$
This free energy change is an extensive property, as it depends on the actual amounts of compounds. We can introduced the corresponding intensive state functions which are the molar Gibbs free energies (Gm):
$$\Delta G=\left({\upnu}_{\mathrm{C}}\cdot {G}_{\mathrm{mC}}+{\upnu}_{\mathrm{D}}\cdot {G}_{\mathrm{mD}}\right)-\left({\upnu}_{\mathrm{A}}\cdot {G}_{\mathrm{mA}}+{\upnu}_{\mathrm{B}}\cdot {G}_{\mathrm{mB}}\right)$$
The molar Gibbs free energies G m are indeed the chemical potentials, so we can substitute and obtain:
$${\displaystyle \begin{array}{ll}\Delta G=& \left({\upnu}_{\mathrm{C}}\cdot {\upmu}_{\mathrm{C}}^{\ast }+{\upnu}_{\mathrm{C}}\cdot \mathrm{R}\cdot T\cdot \ln {x}_{\mathrm{C}}+{\upnu}_{\mathrm{D}}\cdot {\upmu}_{\mathrm{D}}^{\mathrm{\O}}+{\upnu}_{\mathrm{D}}\cdot \mathrm{R}\cdot T\cdot \ln {p}_{\mathrm{D}}\right)\\ {}& -\left({\upnu}_{\mathrm{A}}\cdot {\upmu}_{\mathrm{A}}^{\ast }+{\upnu}_{\mathrm{A}}\cdot \mathrm{R}\cdot T\cdot \ln {x}_{\mathrm{A}}+{\upnu}_{\mathrm{B}}\cdot {\upmu}_{\mathrm{B}}^{\mathrm{\O}}+{\upnu}_{\mathrm{B}}\cdot \mathrm{R}\cdot T\cdot \ln {p}_{\mathrm{B}}\right)\end{array}}$$
By combining the constant terms ($${\upnu}_i\cdot {\upmu}_i^{\ast, \mathrm{\O}}$$ represent the chemical potentials of the components in their pure or standard states), we obtain:
$$\Delta G={\upnu}_{\mathrm{C}}\cdot {\upmu}_{\mathrm{C}}^{\ast }+{\upnu}_{\mathrm{D}}\cdot {\upmu}_{\mathrm{D}}^{\mathrm{\O}}-{\upnu}_{\mathrm{A}}\cdot {\upmu}_{\mathrm{A}}^{\ast }-{\upnu}_{\mathrm{B}}\cdot {\upmu}_{\mathrm{B}}^{\mathrm{\O}}+\mathrm{R}\cdot T\cdot \left({\upnu}_{\mathrm{C}}\cdot \ln {x}_{\mathrm{C}}+{\upnu}_{\mathrm{D}}\cdot \ln {p}_{\mathrm{D}}-{\upnu}_{\mathrm{A}}\cdot \ln {x}_{\mathrm{A}}-{\nu}_{\mathrm{B}}\cdot \ln {p}_{\mathrm{B}}\right)$$
The constant terms $${\upnu}_i\cdot {\upmu}_i^{\ast, \mathrm{\O}}$$ represent a standard Gibbs free energy change for this reaction and are thus substituted by ΔGØ. Using the logarithm rules of a ⋅  log x =  log x a , log a +  log b =  log (a ⋅ b) and $$\log a-\log b=\log \frac{a}{b}$$ (see A.1.2) one obtains:
$$\Delta G=\Delta {G}^{\mathrm{\O}}+\mathrm{R}\cdot T\cdot \ln \frac{x_{\mathrm{C}}^{\upnu_{\mathrm{C}}}\cdot {p}_{\mathrm{D}}^{\upnu_{\mathrm{D}}}}{x_{\mathrm{A}}^{\upnu_{\mathrm{A}}}\cdot {p}_{\mathrm{B}}^{\upnu_{\mathrm{B}}}}$$
which is a relationship between the Gibbs free energy change for a reaction at varying concentrations of components, here expressed as mole fractions for the liquids A and C, and as partial pressures for the gases B and D. The argument of the logarithm is the reaction coefficient Q introduced earlier. Based on the above rigorous assessment of the chemical potentials, we can thus confirm the relationship introduced in Sect. :
$$\Delta G=\Delta {G}^{\mathrm{\O}}+\mathrm{R}\cdot T\cdot \ln Q$$
(2.58)
We recall that when the reaction has reached equilibrium, there is no change the Gibbs free energy observed, therefore:
$$\Delta G=0$$
Also, at equilibrium, the reaction coefficient Q becomes the equilibrium constant
$${Q}_{\mathrm{eq}}=K$$
One thus obtains from Eq. 2.58 for equilibrium conditions:
$${\displaystyle \begin{array}{l}\Delta G=\Delta {G}^{\mathrm{\O}}+\mathrm{R}\cdot T\cdot \ln K=0\\ {}\ln K=-\frac{\Delta {G}^{\mathrm{\O}}}{\mathrm{R}\cdot T}\kern0.5em \mathrm{and}\kern0.5em K={\mathrm{e}}^{-\frac{\Delta {G}^{\mathrm{\O}}}{\mathrm{R}\cdot T}}\end{array}}$$
(2.59)

This means that if we can determine the change in the Gibbs free energy of a process, we can calculate the equilibrium constant.

## 3.3 Phase Equilibria

### 3.3.1 Phase Diagrams and Physical Properties of Matter

Phase diagrams tell us the state of a system under various conditions. The states of a system at various pressures and temperatures can be depicted in a p-T diagram (see for example Fig. 3.2). In phase diagrams, lines separate regions where various phases are thermodynamically stable. These lines are called phase boundaries and show values of p and T where two phases coexist in equilibrium. When changing one of the two parameters (p or T), in order to maintain the state of equilibrium, the other parameter needs to follow suit. Phase boundaries are therefore univariant (one parameter can be freely chosen).
From the ideal gas equation, we know that
$$p\cdot V=n\cdot \mathrm{R}\cdot T$$
(2.6)

If we assume a system with constant molar amount n, then the above equation relates the three variables p, V and T. For a given state, the third variable is determined by choosing the two other—a characteristic of equations of state that we have introduced earlier. As a consequence, the lines in a p-T phase diagram need to describe conditions of constant volumes; these lines are thus called isochores.

There are several points of interest in a phase diagram, illustrated in the p-T diagram in Fig. 3.2.

### Boiling Point

The boiling point is the temperature at which the vapour pressure of the liquid is equal to the external pressure. Note that the frequently used term ‘boiling point’ is strictly speaking not correct, as it is not a single point in the phase diagram. It is a temperature extrapolated from the isochore branch that separates liquid and vapour phases. If the external pressure is the normal pressure
$${\mathrm{p}}_{\mathrm{normal}}=1\ \mathrm{atm}=101.3\ \mathrm{kPa}\kern1em \mathrm{then}\ \mathrm{this}\ \mathrm{is}\ \mathrm{called}\ \mathrm{the}\ \mathrm{normal}\ \mathrm{boiling}\ \mathrm{point}.$$
If the external pressure is the standard pressure
$${\mathrm{p}}^{\mathrm{\O}}=1\ \mathrm{bar}=100.0\ \mathrm{kPa}\kern1em \mathrm{then}\ \mathrm{this}\ \mathrm{is}\ \mathrm{called}\ \mathrm{the}\ \mathrm{standard}\ \mathrm{boiling}\ \mathrm{point}.$$

### Melting Point

The melting point is the temperature at which the liquid and solid phases coexist; it equals the freezing temperature. As above, please note that the frequently terms ‘boiling/freezing points’ are strictly speaking not correct, as it is they are not single points in the phase diagram. It is a temperature extrapolated from the isochore branch that separates solid and liquid phases. If the external pressure is the normal pressure
$${\mathrm{p}}_{\mathrm{normal}}=1\ \mathrm{atm}=101.3\ \mathrm{kPa}\kern1em \mathrm{then}\ \mathrm{this}\ \mathrm{is}\ \mathrm{called}\ \mathrm{the}\ \mathrm{normal}\ \mathrm{freezing}\ \mathrm{or}\ \mathrm{melting}\ \mathrm{point}.$$
If the external pressure is the standard pressure
$${\mathrm{p}}^{\mathrm{\O}}=1\ \mathrm{bar}=100.0\ \mathrm{kPa}\kern1em \mathrm{then}\ \mathrm{this}\ \mathrm{is}\ \mathrm{called}\ \mathrm{the}\ \mathrm{standard}\ \mathrm{freezing}\ \mathrm{or}\ \mathrm{melting}\ \mathrm{point}.$$

### Critical Point

At the critical point, there is no physical interface between the liquid and the vapour; both phases coalesce and there is no liquid phase. At pressures and/or temperatures beyond the critical point, the system is said to be in the supercritical state, which is neither vapour nor liquid. Note that the isochore branch stops at the critical point, i.e. this isochore branch does not continue beyond the critical point. Since the critical point is defined by two discrete values on the p- and T-axes, it is indeed a point in the p-T diagram. The critical point of water occurs at:
$${T}_{\mathrm{crit}}=647.10\ \mathrm{K},{p}_{\mathrm{crit}}=22.1\ \mathrm{MPa}$$

### Triple Point

At a single definite pressure and temperature, three phases can exist in equilibrium; this is called a triple point. There may be more than one triple points in a phase diagram. Since triple points are defined by two discrete values on the p- and T-axes, they are indeed points in the p-T diagram. One important triple point of water is observed at
$${T}_{\mathrm{triple}}=273.16\ \mathrm{K},{p}_{\mathrm{triple}}=611\ \mathrm{Pa}$$

Here, solid, liquid and gaseous water exist at the same time; this triple point has a general importance as it is used to define the thermodynamic temperature scale.

### 3.3.2 Phase Diagrams with Isotherms

Phase diagrams may also be constructed by plotting the pressure versus the volume of a system; a useful method especially when analysing gases. As discussed in the previous section, the third parameter, now the temperature, needs to be constant when characterising a system under varying pressures and volumes. The lines in the p-V diagram are thus called isotherms; the temperature is the same all the way along (see Fig. 3.3).
In Fig. 3.3 (left), the behaviour of a real gas is compared to that of an ideal gas. The real gas undergoes condensation. At a certain pressure, the volume collapses (B–C–D) and the gas turns into a liquid, but the system maintains the same pressure (Fig. 3.4). The pressure at B–C–D is called the vapour pressure of the liquid. For condensation to occur, the molecules must be close enough and slow enough to aggregate.

High temperatures imply high molecular velocities. Therefore, at sufficiently high temperatures (above Tcrit), the individual gas molecules possess too high velocity in order to engage in intermolecular interactions and thus no condensation will occur, no matter how small the volume is made.

At the critical point, when the boundaries between liquid and gas phase vanish, the isotherm in the p-V diagram has zero slope (Fig. 3.3, right; red isotherm). Since the critical point is indeed a point in the phase behaviour of compounds, it possesses three discrete values for pressure, temperature and volume: pcrit, Tcrit and Vcrit. The critical point is a characteristic of a particular substance (Table 3.1).
Table 3.1

Critical constants for selected gases

Gas

pcrit (kPa)

Vcrit (cm3 mol−1)

θcrit (°C)

Ar

4862

75.3

−123

CO2

7385

94.0

31.0

He

228.9

57.8

−268

O2

5075

78.0

−118

### 3.3.3 Phase Transitions

A form of matter that is uniform throughout in chemical composition and physical state of matter is called a phase. The fundamentally different phases are solid, liquid and gas; with water, for example, ice, liquid water and water steam. However, there may also be various solid phases, conducting and superconducting phases, superfluid phases. A phase transition is the conversion of one phase into another. The possible phase transitions between the three states of matter—solid, liquid and gas—are illustrated in Fig. 3.5. Some compounds possess more than one liquid or solid phase; transitions are then also possible between those.

When two or more phases are in equilibrium, the chemical potential of a substance is the same in each of the two phases and at all points in each of the phases. This is a consequence of the 2nd law of thermodynamics. For a given pressure, the temperature at which two phases are in equilibrium (and thus matter spontaneously transitions from one to the other) is called transition temperature.

### 3.3.4 The Gibbs Phase Rule

In Sect. 3.3.1, we introduced phase boundaries is univariant regions in the phase diagram: it is possible to freely choose one parameter in order to maintain this particular state of the system. The Gibbs phase rule allows calculation of the number of intensive parameters (i.e. independent of the amount of substance) that can be varied independently (F) while the number of phases (P) remains constant, given a system with a particular number of components (C) of a system:
$$F=C-P+2$$
(3.14)
For example, if two phases (P = 2) in a system consisting of one component (C = 1) are in equilibrium, then
$$F=1-2+2=1$$
parameter (e.g. T) can be changed, but the other parameters (e.g. p) will follow suit. Systems with F = 1 are called univariant.

### Gibbs phase rule examples

Water steam

Water steam describes one phase (P = 1) in the one-component system water (C = 1). Therefore:
$$F=1-1+2=2$$

With two degrees of freedom, water steam constitutes a bivariant system. It corresponds to an area in an xy plot (for example, in a p-T diagram). We can vary the temperature of the steam without having to change the pressure at the same time, but still maintain the gas phase.

Liquid water in equilibrium with its vapour

Here, we again deal with the one-component system water (C = 1), but now have to consider two phases, liquid and vapour, i.e. P = 2. Therefore:
$$F=1-2+2=1$$

With one degree of freedom, this constitutes a univariant system. It corresponds to a line in an xy plot. The temperature can be varied, but the pressure needs to be varied accordingly to in order to preserve the equilibrium between liquid and vapour phase.

### 3.3.5 One Component Systems: Carbon Dioxide

The phase diagram of carbon dioxide is shown in Fig. 3.6, annotated with the three states of matter. as well as the triple and critical points. The degrees of freedom for the different areas in the phase diagram are also shown.

A closer look at the numerical values of the p- and T-axes shows that solid CO2 sublimes (i.e. transitions to the vapour phase). Since the solid and liquid phases of CO2 do not co-exist at normal pressure, solid CO2 is also called dry ice. The liquid phase does not exist below a pressure of 518 kPa. In other words, if liquid CO2 is required, a pressure of at least 518 kPa must be applied.

At a pressure of 101.3 kPa, which constitutes the normal atmospheric pressure, the transition temperature between solid and gas phase is 195 K (−78 °C), so solid CO2 evaporates under normal conditions.

Inspection of the phase boundary between solid and liquid CO2 shows that an increase in pressure results in an increase in the melting temperature; the slope of the solid–liquid isochore is positive. This is due to solid CO2 being denser than the liquid. An increase in pressure promotes the denser phase, so higher pressures stabilise the solid state. This behaviour is typical of most substances.

The phase changes when approaching the critical point and formation of supercritical propane is illustrated in Fig. 3.7. In the supercritical fluid, one homogenous phase is formed which shows properties of both liquids and gases.

### 3.3.6 One Component Systems: Water

The p-T phase diagram for water is shown in Fig. 3.8. The thick lines indicate the main phase boundaries between solid, liquid and vapour states. Especially at high pressures, water can form several different solid phases, the boundaries of which are shown by the thin lines. Therefore, the phase diagram contains several triple points, where three phases are in co-existence. The main triple point, where solid, liquid and vapour water are in equilibrium, is observed at T = 273.16 K and p = 611 Pa. This triple point has fundamental importance, as it is used to define the zero point on the Celsius temperature scale.

Of particular importance is the rather exceptional property of water arising from the steep slope of the solid–liquid phase boundary (the melting temperature curve): it not only has a very steep, but a negative slope. The negative slope indicates that the liquid phase has a higher density than the solid phase. Therefore, solid ice floats on liquid water (due to the lower density of the former; see also Sect. 3.4.6), and the possibility of ice skating also arises form this unusual behaviour. A skater on ice of about 70 kg exerts a pressure of about 7 MPa (assuming a contact area of 1 cm2). At that high pressure, the melting temperature of ice is no longer 0 °C but −1 °C. The high pressure leads to disruption of hydrogen bonds that hold the water molecules in the solid structure. The generated film of liquid water enables the smooth skating process.

### Water: not one, but two liquids?

The list of anomalies of water is constantly increasing, and currently includes some 72 properties that distinguish water from conventional liquids (Chaplin 2014). Among the best known anomalies are the density maximum of water in its liquid state at 3.98 °C, as well as the so-called Mpemba effect which describes the phenomenon that hot water freezes faster than cold water (Mpemba and Osborne 1969).

The reason for such extra-ordinary behaviour lies in discontinuities in the heat capacity C p (see also Sect. 3.4.9). Such a discontinuity gives rise to the critical point, where fluctuations happen at all length scales and thus light cannot penetrate the system—it becomes opaque. The critical point indicates the end of the boiling curve, and liquid and gas state are no longer distinguishable (supercritical fluid). In supercritical water, salts can no longer be dissolved, but mixture with apolar solvents is observed. However, the C p discontinuity at the critical point cannot explain anomalies that occur at lower pressures and temperatures.

Based on simulations (Poole et al. 1992), it has thus been proposed that a so far non-identified equilibrium curve exists in water that separates the liquid phase into two liquids, a high-density and a low-density liquid. Similar to the boiling curve, this postulated equilibrium curve may contain a critical point at approx. −50 °C and atmospheric pressure.

This hypothesis is a matter of ongoing debate, but may have notable implications. If this second critical point exists, then water under standard conditions would constitute a supercritical fluid that fluctuates between a low- and a high-density liquid state. At lower temperatures, the mixture of the two phases should spontaneously separate into a low-density liquid that flows on top of the high-density liquid.

### 3.3.7 One Component Systems: Helium

Compared to the two substances we have looked at so far, the phase diagram for helium shows further unusual behaviour. The two isotopes of helium, 3He and 4He, have different phase diagrams; the p-T diagram of 4He is illustrated in Fig. 3.9. In contrast to its lighter isotope, 4He has two liquid phases, called He-I and He-II, with a transition between them. This transition line is called the λ-line. Whereas the liquid phase He-I forms a normal liquid, He-II has properties of a superfluid: it flows with zero viscosity (Figure 3.10). At the low temperature triple point of helium, the phases He-II(l), He-I(l) and He(g) coexist. Another unusual observation is that solid and gas are never in equilibrium, due to the light He atoms having large amplitude vibrations. Therefore, helium needs very high pressures to form a solid.

### 3.3.8 Other Phases

In addition to the three fundamental phases (solid, liquid, gas), there a few unusual phases:
• Plasma is a phase formed by an ionised gas when electrons are stripped from atoms at high temperatures. Plasmas are an important phase in high-temperature processes such as nuclear fusion and stellar atmospheres. They have a practical importance in spectroscopic instruments where they are used as ionisation devices (e.g. inductively-coupled plasma atomic emission spectroscopy, ICP- AES).

• Supercooled liquids constitute a phase that is established by cooling a liquid below the freezing temperature but without crystallisation. The most prominent example for a supercooled liquid is glass. Albeit these materials appear solid, the particles form an amorphous (formless) structure.

• Liquid crystals flow like ordinary liquids, but the molecules form swarms with some low-order structure. As such, the liquid crystalline phase is an intermediate between the crystalline and liquid phase. The liquid crystalline phase is of eminent importance for the physics of membranes. It also has important practical applications since the low-order structure of liquid crystalline materials can be changed by orienting the molecules by applying a small electric potential; the change of orientation results in different optical properties of the liquid crystal. These phenomena are used in liquid crystalline displays ( LCDs), which makes this phase a highly important phenomenon for microelectronic devices (displays, computer and TV screens).

## 3.4 Thermodynamic Aspects of Phase Transitions

The locations of phase boundaries (lines) in phase diagrams are determined by the relative thermodynamic stability of the individual phases, i.e. by the chemical potential μ (molar Gibbs free energy Gm). The chemical potential can thus be thought of as the potential of a substance to bring about a physical change. Under different conditions, different forms of the same system become more stable. A system that is in equilibrium has the same chemical potential throughout in all actually existing phases at that time.

In this section, we will discuss how the stability of phases are affected by particular environmental conditions. Knowledge about these relationships will allow us to predict phase transitions based on thermodynamic parameters.

### 3.4.1 Temperature Dependence of Phase Stability

We have previously () learned that the Gibbs free energy of a system is given by:
$${G}_{\mathrm{sys}}={H}_{\mathrm{sys}}-T\cdot {S}_{\mathrm{sys}}$$
(2.44)
and thus its derivative is calculated as per
$$\mathrm{d}{G}_{\mathrm{sys}}=\mathrm{d}{H}_{\mathrm{sys}}-\mathrm{d}\left(T\cdot {S}_{\mathrm{sys}}\right)$$
(2.45)
which can be resolved by considering the product rule (A.2.3):
$$\mathrm{d}{G}_{\mathrm{sys}}=\mathrm{d}{H}_{\mathrm{sys}}-{S}_{\mathrm{sys}}\cdot \mathrm{d}T-T\cdot \mathrm{d}{S}_{\mathrm{sys}}$$
After dividing the equation by dT, one obtains:
$$\frac{\mathrm{d}{G}_{\mathrm{sys}}}{\mathrm{d}T}=\frac{\mathrm{d}{H}_{\mathrm{sys}}}{\mathrm{d}T}-\frac{S_{\mathrm{sys}}\cdot \mathrm{d}T}{\mathrm{d}T}-\frac{T\cdot \mathrm{d}{S}_{\mathrm{sys}}}{\mathrm{d}T}$$
which simplifies to:
$$\frac{\mathrm{d}{G}_{\mathrm{sys}}}{\mathrm{d}T}=\frac{\mathrm{d}{H}_{\mathrm{sys}}}{\mathrm{d}T}-{S}_{\mathrm{sys}}-\frac{T\cdot \mathrm{d}{S}_{\mathrm{sys}}}{\mathrm{d}T}$$
(3.15)
In Sect. , it was established that for processes at equilibrium, there is no net change in the entropy of the universe:
$$\mathrm{d}{S}_{\mathrm{universe}}=0$$
(2.51)
The entropy of the universe was given by Eq. :
$$\mathrm{d}{S}_{\mathrm{universe}}=\mathrm{d}{S}_{\mathrm{sys}}-\frac{\mathrm{d}{H}_{\mathrm{sys}}}{T}$$
which combines to:
$$\mathrm{d}{S}_{\mathrm{universe}}=\mathrm{d}{S}_{\mathrm{sys}}-\frac{\mathrm{d}{H}_{\mathrm{sys}}}{T}=0$$
From this equation, it follows that
$$\mathrm{d}{S}_{\mathrm{sys}}=\frac{\mathrm{d}{H}_{\mathrm{sys}}}{T}\kern0.5em \mathrm{and}\kern0.5em \mathrm{thus}$$
$$\mathrm{d}{H}_{\mathrm{sys}}=T\cdot \mathrm{d}{S}_{\mathrm{sys}}$$
We can now use this expression to substitute dHsys in Eq. 3.15 and obtain:
$$\frac{\mathrm{d}{G}_{\mathrm{sys}}}{\mathrm{d}T}=\frac{T\cdot \mathrm{d}{S}_{\mathrm{sys}}}{\mathrm{d}T}-{S}_{\mathrm{sys}}-\frac{T\cdot \mathrm{d}{S}_{\mathrm{sys}}}{\mathrm{d}T}$$
which simplifies to
$$\frac{\mathrm{d}{G}_{\mathrm{sys}}}{\mathrm{d}T}=-{S}_{\mathrm{sys}}$$
(3.16)
When considering a system consisting of 1 mol of substance (so we can use Gm = μ), and remembering that the Gibbs free energy is not only dependent on the temperature, but also on the pressure, we obtain the following expression from Eq. 3.16 and also introduce the molar entropy Sm:
$${\left(\frac{{\updelta G}_{\mathrm{m}}}{\updelta T}\right)}_p={\left(\frac{\updelta \upmu}{\updelta T}\right)}_p=-{S}_{\mathrm{m}}$$
(3.17)

Since G (as well as Gm, μ) is also dependent on the pressure, we need to request constant pressure when calculating the temperature differential; this is done by calculating a partial differential (indicated by ‘δ’), and denoting the pressure as a subscript.

The molar entropy Sm is a characteristic parameter of a substance and always positive. Therefore, the change of the chemical potential μ with increasing temperature at constant pressure is always negative.

The phase with the lowest chemical potential μ at a particular temperature is the most stable one for that temperature. The transition temperatures (Tm/Tf and Tb) are the temperatures, at which the chemical potentials of the two interfacing phases are equal and the phases are thus at equilibrium.

### 3.4.2 Entropies of Substances

In the previous section, we have introduced the molar entropy S m in the context of different phases of a one-component system. We also noted that molar entropies are characteristic parameters of a substance.

In gases, molecules can freely diffuse through the volume they are contained in and the individual molecules (as well as their energies) are dispersed across the entire volume. Therefore, the degree of disorder is rather large, compared to that of liquids or solids. The molar entropy Sm of gases is thus larger than that of liquids or solids (see Table 3.2).
Table 3.2

Molar entropies Sm in J K−1 mol−1 of selected solids, liquids and gases at 25 °C

Solids

Liquids

Gases

C (diamond)

2.4

Hg

76.0

H2

130.6

C (graphite)

5.7

H2O

69.9

N2

192.1

Fe

27.3

H3C–COOH

159.8

O2

205.0

Cu

33.1

C2H5OH

160.7

CO2

213.6

AgCl

96.2

C6H6

173.3

NO2

239.9

Fe2O3

87.4

NH3

192.3

CuSO4·5 H2O

300.4

CH4

186.2

sucrose

360.2

N2O4

304.0

In contrast, the molecules in a solid are confined to a small volume and their degrees of freedom are restricted to vibrational motion. The molar entropies Sm of solids are thus fairly low. Solids comprising of large molecules (e.g. sucrose) or complexes (CuSO4·5 H2O) have a large number of atoms and may thus possess comparatively high molar entropies, since the energy may be shared among the many atoms.

We remember that we have defined the explicit entropy change dS in Sect. as:
$$\mathrm{d}S=\frac{\mathrm{d}Q}{T}$$
(2.30)
where dQ is the reversibly transferred heat at a particular temperature T. From this equation, it becomes clear that there is a smaller change of entropy when a given quantity of energy (dQ) is transferred to an object at high temperature than at low temperature. More disorder is induced when the object is cool rather than hot.
The ability of a substance to distribute energy over its molecules is related to the heat capacity. As we have discussed in Sect. , this link can be used to establish a relationship between the entropy change and the heat capacity C p :
$$\Delta S=\underset{T_{\mathrm{start}}}{\overset{T_{\mathrm{end}}}{\int }}\frac{C_p}{T}\mathrm{d}T$$
(2.39)
which forms the basis for entropy determination of substances by heat capacity measurements (i.e. calorimetrically). Alternatively, electrochemical cells can be used to determine entropies of substances.

### 3.4.3 Pressure Dependence of Melting

In Sect. above, we looked at the change of the Gibbs free energy with temperature (at constant pressure) and derived the definition of the molar entropy Sm. From the Maxwell equations describing the relations between different state variables (Sect. ), the variation of the Gibbs free energy with pressure (at constant temperature) can be derived, leading to the definition of the molar volume Vm:
$${\left(\frac{{\updelta G}_{\mathrm{m}}}{\updelta p}\right)}_T={\left(\frac{\updelta \upmu}{\updelta p}\right)}_T={V}_{\mathrm{m}}$$
(2.65)

Like the molar entropy, the molar volume Vm is a characteristic parameter for a substance and always positive. Therefore, the change of the chemical potential μ (= Gm) with increasing pressure at constant temperature is always positive. Generally, the molar volume Vm is larger for liquids than for solids (exception: water).

We remember that the transition temperatures (Tm/Tf and Tb) are the temperatures, at which the chemical potentials of the two interfacing phases are equal and the phases are thus at equilibrium. The transition temperature between the solid and liquid phases (Tm/Tf) is generally larger at higher pressures (exception: water).

Conceptually, this is illustrated in Fig. 3.11. If the molar volume Vm of the solid is smaller than that of the liquid (Fig. 3.11 left)—an observation made for most substances—the chemical potential of the solid phase μ(s) increases less than the chemical potential of the liquid phase, μ(l), when the pressure is increased. This situation leads to a shift of the intersect between the two branches to higher temperatures; therefore, the melting temperature increases when the pressure is increased.

In contrast, water, for example, shows the behaviour illustrated in the right panel of Fig. 3.11: the molar volume Vm of the solid is larger than that of the liquid. Therefore, when the pressure is increased, the liquid phase experiences a lesser change of the chemical potential than the solid phase. The intersect between the two branches thus migrates to lower temperatures. This means that at higher pressure, substances like water freeze at lower temperatures.

### 3.4.4 Pressure Dependence of Vapour Pressure

In Sect. 3.2.2, we arrived at an expression of Raoult’s law that relates the vapour pressure of a solution with that of the pure solvent as per:
$${p}_{\mathrm{A}}={p}_{\mathrm{A}}^{\ast}\cdot {\mathrm{e}}^{\frac{\upmu_{\mathrm{A}\left(\mathrm{l}\right)}-{\upmu}_{\mathrm{A}\left(\mathrm{l}\right)}^{\ast }}{\mathrm{R}\cdot T}}$$
(3.11)
μA(l) and $${\upmu}_{\mathrm{A}\left(\mathrm{l}\right)}^{\ast }$$ describe the chemical potentials of the solution and the pure solvent, respectively. In the previous section, we have seen that the differential of the chemical potential with respect to pressure changes is
$${\left(\frac{\updelta \upmu}{\updelta p}\right)}_T={V}_{\mathrm{m}}.$$
(2.65)
We then realise that the chemical potential difference in Eq. 3.11 can be expressed in terms of the molar volume and the pressure change:
$${\upmu}_{\mathrm{A}\left(\mathrm{l}\right)}-{\upmu}_{\mathrm{A}\left(\mathrm{l}\right)}^{\ast }=\Delta \upmu ={V}_{\mathrm{m}}\cdot \Delta p$$
which we can substitute in Eq. 3.11 and obtain:
$$p={p}^{\ast}\cdot {\mathrm{e}}^{\frac{V_{\mathrm{m}}\cdot \Delta p}{\mathrm{R}\cdot T}}$$
(3.18)
The pressure difference Δp in the exponential term describes the pressure of two different states. Let the initial state be at pinitial = pØ, and the final state at a much high pressure. In that case, Δp > 0 and thus the exponent is positive and the entire exponential term a factor greater than 1 (see Fig. 3.12). This means that the vapour pressure of a pressurised liquid is higher than that of the system under standard pressure.

Similarly, if we consider the case where pfinal describes the system at lower pressure (e.g. the evacuated system), then we appreciate that Δp < 0, the exponent becomes negative and thus the exponential factor has a value less than 1. Therefore, the vapour pressure will be less in an evacuated system than in one that contains ambient atmosphere.

### 3.4.5 Phase Boundaries

As we have established earlier, the lines in a phase diagram separating two neighbouring phases are called phase boundaries. Along those boundaries, two phases (we will here call them ‘a’ and ‘b’) co-exist. Being lines in a two-dimensional plot, the phase boundaries are best discussed in terms of their slopes. So for a p-T diagram, phase boundaries are characterised by the differential
$$\left(\frac{\mathrm{d}p}{\mathrm{d}T}\right).$$
Since the two phases co-exist, there must be equilibrium and the changes in the chemical potentials must be equal:
$$\mathrm{d}{\upmu}_{\mathrm{a}}\left(p,T\right)=\mathrm{d}{\upmu}_{\mathrm{b}}\left(p,T\right)$$
From earlier discussions we know that
$$\mathrm{d}\upmu =\mathrm{d}{G}_{\mathrm{m}}={V}_{\mathrm{m}}\cdot \mathrm{d}p-{S}_{\mathrm{m}}\cdot \mathrm{d}T$$
(2.45)
and can therefore substitute this expression on both sides of the equilibrium equation above:
$${V}_{\mathrm{m},\mathrm{a}}\cdot \mathrm{d}p-{S}_{\mathrm{m},\mathrm{a}}\cdot \mathrm{d}T={V}_{\mathrm{m},\mathrm{b}}\cdot \mathrm{d}p-{S}_{\mathrm{m},\mathrm{b}}\cdot \mathrm{d}T$$
We group together the volumes and entropies on opposing sides:
$$\left({S}_{\mathrm{m},\mathrm{b}}-{S}_{\mathrm{m},\mathrm{a}}\right)\cdot \mathrm{d}T=\left({V}_{\mathrm{m},\mathrm{b}}-{V}_{\mathrm{m},\mathrm{a}}\right)\cdot \mathrm{d}p$$
and consider that the differences of the volumes and entropies of phases ‘a’ and ‘b’ describe the transition from one phase to the other:
$$\Delta {S}_{\mathrm{m},\mathrm{trans}}\cdot \mathrm{d}T=\Delta {V}_{\mathrm{m},\mathrm{trans}}\cdot \mathrm{d}p$$
This equation can be re-arranged to yield the slope of the phase boundary through the differential of pressure and temperature:
$$\left(\frac{\mathrm{d}p}{\mathrm{d}T}\right)=\frac{\Delta {S}_{\mathrm{m},\mathrm{trans}}}{\Delta {V}_{\mathrm{m},\mathrm{trans}}}$$
(3.19)

This equation is of fundamental importance as it describes the phase transitions (phase boundaries) in p-T phase diagrams; it is also known as the Clapeyron equation, developed by the French physicist an engineer Benoît Clapeyron in 1834 (Clapeyron 1834). Equipped with this equation, we can now discuss the three phase boundaries solid–liquid, liquid–vapour and solid–vapour.

### 3.4.6 Phase Boundaries: Solid–Liquid

We can apply the general form of the Clapeyron equation above to particular phase transitions, such as the melting or fusion process where solid and liquid phases are in equilibrium. The slope of the solid–liquid phase boundary in a p-T diagram is described as the differential of pressure (y-axis) with respect to temperature (x-axis), $$\frac{\mathrm{d}p}{\mathrm{d}T}$$. The differences in the molar entropies and volumes between solid and liquid states are macroscopically measurable; we thus use ‘Δ’ instead of ‘d’, and obtain the Clapeyron equation for the melting (fusion) process:
$$\frac{\mathrm{d}p}{\mathrm{d}T}=\frac{\Delta {S}_{m,\mathrm{melt}}}{\Delta {V}_{m,\mathrm{melt}}}$$
(3.19)
We remember from earlier discussions that
$$\Delta S=\frac{\Delta H}{T}$$
(2.35)
and therefore can express the phase transition in terms of the molar enthalpy change:
$$\frac{\mathrm{d}p}{\mathrm{d}T}=\frac{\Delta {S}_{m,\mathrm{melt}}}{\Delta {V}_{m,\mathrm{melt}}}=\frac{\Delta {H}_{m,\mathrm{melt}}}{T_{\mathrm{melt}}\cdot \Delta {V}_{m,\mathrm{melt}}}$$
(3.20)
For the melting process, we can evaluate the Clapeyron equation in an approximative fashion. The change in molar enthalpies
$$\Delta {H}_{m,\mathrm{melt}}={H}_{m,\mathrm{liquid}}\hbox{--} {H}_{m,\mathrm{solid}}$$
is generally positive, and the change in molar volumes
$$\Delta {V}_{m,\mathrm{melt}}={V}_{m,\mathrm{liquid}}\hbox{--} {V}_{m,\mathrm{solid}}$$
is generally positive and rather small (the liquid and solid states of most substances have a similar volume, with the liquid phase typically having a slightly larger volume; exception: water). Also, ΔVm,melt can be considered independent of the temperature. Therefore, the differential
$$\frac{\mathrm{d}p}{\mathrm{d}T}$$
is generally positive and large. This means we will obtain steep boundaries at the solid–liquid interfaces in a p-T diagram (Fig. 3.13), with this phase boundary having a positive slope.

### 3.4.7 Phase Boundaries: Liquid–Vapour

In analogy to the discussion in the previous section, we can formulate, based on the Clapeyron equation, an expression for the vapourisation process, i.e. the transition from the liquid to the gas phase:
$$\frac{\mathrm{d}p}{\mathrm{d}T}=\frac{\Delta {S}_{m,\mathrm{vap}}}{\Delta {V}_{m,\mathrm{vap}}}=\frac{\Delta {H}_{m,\mathrm{vap}}}{T\cdot \Delta {V}_{m,\mathrm{vap}}}$$
(3.19)
For the vapourisation process, we can not assume that ΔVm,vap is independent of the temperature. We remember that for an ideal gas, there is a relationship between the gas volume and the temperature and pressure, given by the ideal gas equation:
$$V=\frac{n\cdot \mathrm{R}\cdot T}{p}$$
(2.6)
When we consider the molar volume V m , we normalise the volume with respect to the molar amount n ($${V}_m=\frac{V}{n}$$), so we obtain:
$${V}_m=\frac{\mathrm{R}\cdot T}{p}$$
Since the gas phase of a substance occupies a much larger volume than the liquid state, we can assume that the volume of the liquid is negligible compared to the volume of the gas:
$${\displaystyle \begin{array}{l}{V}_{\mathrm{m},\mathrm{gas}}\gg {V}_{\mathrm{m},\mathrm{liquid}}\kern0.5em \Rightarrow \kern0.5em \Delta {V}_{\mathrm{m},\mathrm{vap}}={V}_{\mathrm{m},\mathrm{gas}}-{V}_{\mathrm{m},\mathrm{liquid}}\approx {V}_{\mathrm{m},\mathrm{gas}}\\ {}\Delta {V}_{\mathrm{m},\mathrm{vap}}\approx {V}_{\mathrm{m},\mathrm{gas}}=\frac{\mathrm{R}\cdot T}{p}\end{array}}$$
Note that we now assume conditions of an ideal gas and negligible volume of the solid compared with the gas. This expression for ΔVm,vap can be used to substitute in Eq. 3.19 which then yields:
$$\frac{\mathrm{d}p}{\mathrm{d}T}=\frac{\Delta {H}_{\mathrm{m},\mathrm{vap}}}{T\cdot \Delta {V}_{\mathrm{m},\mathrm{vap}}}=\frac{\Delta {H}_{\mathrm{m},\mathrm{vap}}}{T\cdot \left(\frac{\mathrm{R}\cdot T}{p}\right)}$$
This simplifies to
$$\frac{\mathrm{d}p}{\mathrm{d}T}=\frac{p\cdot \Delta {H}_{\mathrm{m},\mathrm{vap}}}{\mathrm{R}\cdot {T}^2}.$$
We isolate the two independent variables, p and T, on opposite sides of the equation, and use the formality of $$\int \frac{1}{x}\mathrm{d}x=\ln x\kern0.5em \iff \kern0.5em \frac{1}{x}\mathrm{d}x=\mathrm{d}\left(\ln x\right)$$ to achieve a more convenient notation (see Appendix A.3.1). This results in a relationship known as the Clausius-Clapeyron equation:
$$\frac{\mathrm{d}p}{p\cdot \mathrm{d}T}=\frac{\mathrm{d}\left(\ln p\right)}{\mathrm{d}T}=\frac{\Delta {H}_{\mathrm{m},\mathrm{vap}}}{\mathrm{R}\cdot {T}^2}.$$
(3.21)

This equation was first derived by the German physicist and mathematician Rudolf Clausius in 1850 (Clausius 1850).

### 3.4.8 Phase Boundaries: Solid–Vapour

The phase boundary between the solid and vapour phases describes the sublimation (or deposition) process. For this process, the Clapeyron equation yields:
$$\frac{dp}{dT}=\frac{\Delta {S}_{\mathrm{m},\mathrm{subl}}}{\Delta {V}_{\mathrm{m},\mathrm{subl}}}=\frac{\Delta {H}_{\mathrm{m},\mathrm{subl}}}{T\cdot \Delta {V}_{\mathrm{m},\mathrm{subl}}}$$
(3.19)
As in the vapourisation process (previous section), we can not assume that ΔV m is independent of the temperature, because a gas phase is involved. Therefore, we again replace V m with the expression from the ideal gas equation. We thus assume conditions of an ideal gas and negligible volume of the solid compared with the gas. The substitution yields:
$$\frac{\mathrm{d}p}{\mathrm{d}T}=\frac{\Delta {H}_{\mathrm{m},\mathrm{subl}}}{T\cdot \Delta {V}_{\mathrm{m},\mathrm{subl}}}=\frac{\Delta {H}_{\mathrm{m},\mathrm{subl}}}{T\cdot \frac{\mathrm{R}\cdot T}{p}}$$
Which simplifies to the following equation, also called the Clausius-Clapeyron equation for the sublimation process:
$$\frac{\mathrm{d}p}{p\cdot \mathrm{d}T}=\frac{\mathrm{d}\left(\ln p\right)}{\mathrm{d}T}=\frac{\Delta {H}_{\mathrm{m},\mathrm{subl}}}{\mathrm{R}\cdot {T}^2}.$$
(3.22)

### 3.4.9 The Ehrenfest Classifications

In Fig. 3.11, we visualised the change of the chemical potential μ with temperature in the region where the solid–liquid phase transition occurs. At the transition, the chemical potential function shows a kink (the function is continuous, but the first derivative of the function is not), so the change in the chemical potential μ is not smooth with a change in the temperature T. This is observed at all major phase transitions
• solid↔liquid (melting/freezing)

• liquid↔gas (vapourisation/condensation)

• solid↔gas (sublimation/deposition)

which are therefore called first order phase transitions. At a molecular level, first-order transitions involve the relocation of atoms, molecules or ions, accompanied by a change of the interaction energies.

What is the implication of this for material properties? As mentioned above, a mathematical function with a kink is characterised by an abrupt change of direction of the plotted function. Whereas the function is continuous and possesses discrete values all the way along, the first derivative is discontinuous. At the kink, the first derivative is infinite.

We remember that the Clapeyron equation (3.19, 3.20) is a function of enthalpy and temperature, in fact $$f=\frac{H}{T}$$, and the first derivative of this function, $${f}^{\hbox{'}}=\frac{\mathrm{d}H}{\mathrm{d}T}$$, leads us to the heat capacity for constant pressure, C p :
$${C}_p={\left(\frac{\updelta H}{\updelta T}\right)}_p$$
(2.29)

At a first-order transition, H changes by a finite amount whereas T changes by an infinitesimal small amount. The heat capacity C p (the first derivative) thus becomes infinite.

Following this classification concept, a transition for which the first derivative of the chemical potential μ with respect to the temperature T is continuous, but the second derivative is discontinuous, is called second order phase transition. This implies that volume, entropy and enthalpy do not change at the transition. The heat capacity C p is discontinuous but not infinite. At a molecular level, second-order transitions are often associated with changes of symmetry in a crystal structure. Rather than the molecular interaction energies, it is the long-range order that varies. Examples for second order phase transitions include the conducting-superconducting transition in metals at low temperatures.

Phase transitions that are not first order, but where nevertheless the heat capacity C p becomes infinite, are called λ-transitions. In such instances, the heat capacity typically increases well before the actual transition occurs. At a molecular level, λ-transitions are associated with order-disorder transitions. Examples include:
• order-disorder transitions in alloys

• onset of ferromagnetism

• fluid-superfluid transition in liquid helium (hence the name λ-line in the He phase diagram, Fig. 3.9)

## 3.5 Mixtures of Volatile Liquids

### 3.5.1 Phase Diagrams of Mixtures of Volatile Liquids

We assume a mixture of two volatile liquids, A and B, where the liquid and vapour phases are in equilibrium. Even though, the compositions in the two phases are not necessarily the same; the vapour phase will contain more of the more volatile component.

Raoult’s law enables calculation of the vapour pressure of a particular liquid (A) in a mixture, for different concentrations of that liquid in the mixture (expressed as molar fraction x):
$${p}_{\mathrm{A}}={x}_{\mathrm{A}}\cdot {p}_{\mathrm{A}}^{\ast }$$
(3.11)

$${p}_{\mathrm{A}}^{\ast }$$ is the vapour pressure of the pure liquid A (i.e. at xA = 1). Since we will have to consider mole fractions for several different phases in the following discussion, in this current chapter, we will denote the mole fraction of substances in the liquid phase as ‘z’ instead of ‘x’.

The total vapour pressure in an ideal mixture of A and B is then given by Dalton’s law (Eq. ), which poses that the total vapour pressure of a vapour mixture is the sum of the partial vapour pressures of all components (here, the partial vapour pressures are given by Raoult’s law for A and B):
$$p={z}_{\mathrm{A}}\cdot {p}_{\mathrm{A}}^{\ast }+{\mathrm{z}}_{\mathrm{B}}\cdot {p}_{\mathrm{B}}^{\ast }={p}_{\mathrm{B}}^{\ast }+\left({p}_{\mathrm{A}}^{\ast}\hbox{--} {p}_{\mathrm{B}}^{\ast}\right)\cdot {z}_{\mathrm{A}}$$
(3.23)
where zA and zB are the mole fractions of A and B in the liquid, and $${p}_{\mathrm{A}}^{\ast }$$ and $${p}_{\mathrm{B}}^{\ast }$$ the vapour pressures of the pure liquids, respectively. This relationship may be plotted in a diagram (see Fig. 3.4, upper left) where the total pressure p provides the ordinate (y-axis) and the mole fraction zA the abscissa (x-axis). It is obvious, that at zA = 0, there is pure liquid B and hence the total vapour pressure is given by the vapour pressure of pure liquid B, $${p}_{\mathrm{B}}^{\ast }$$.
With Dalton’s law informing us that the total pressure of a gas mixture equals the sum of the partial pressures of the individual components, we have access to the mole fraction of the individual components in the vapour phase, which, for clarity, we call ‘y’ instead ‘x’ in this section:
$$\frac{p_{\mathrm{A}}}{p}=\frac{\frac{n_{\mathrm{A}}\cdot \mathrm{R}\cdot T}{V}}{\frac{\left({n}_{\mathrm{A}}+{n}_{\mathrm{B}}\right)\cdot \mathrm{R}\cdot T}{V}}=\frac{n_{\mathrm{A}}}{n_{\mathrm{A}}+{n}_{\mathrm{B}}}={y}_{\mathrm{A}}\kern0.5em \mathrm{and}\kern0.5em \frac{p_{\mathrm{B}}}{p}=\frac{\frac{n_{\mathrm{B}}\cdot \mathrm{R}\cdot T}{V}}{\frac{\left({n}_{\mathrm{A}}+{n}_{\mathrm{B}}\right)\cdot \mathrm{R}\cdot T}{V}}=\frac{n_{\mathrm{B}}}{n_{\mathrm{A}}+{n}_{\mathrm{B}}}={y}_{\mathrm{B}}$$
(3.24)
yA and yB are the mole fractions of A and B in the vapour. This relationship may also be plotted in a phase diagram (Fig. 3.14, upper right) with pressure as the ordinate (y-axis), but this time the mole fraction yA as abscissa (x-axis). It is no surprise that the resulting phase boundary is now a different line than before; after all we are no longer plotting against the mole fraction of A in the liquid, but in the vapour. From Eq. 3.24 it is obvious that at yA = 1, the vapour contains pure substance A, hence the total vapour pressure p is the vapour pressure of A, $${p}_{\mathrm{A}}^{\ast }$$.

Equipped with these relationships, we can now calculate the mole fraction of each component in the vapour phase (y i ) of the liquid mixture, knowing their mole fractions in the liquid phase (z i ).

From Raoult’s law we know:
$${p}_{\mathrm{A}}={\mathrm{z}}_{\mathrm{A}}\cdot {p}_{\mathrm{A}}^{\ast }$$
and from Dalton’s law:
$${y}_{\mathrm{A}}=\frac{p_{\mathrm{A}}}{p}$$
Therefore:
$${y}_{\mathrm{A}}={\mathrm{z}}_{\mathrm{A}}\cdot \frac{p_{\mathrm{A}}^{\ast }}{p}$$
For a mixture consisting of two components only, the mole fraction of the second component is then available as
$${y}_{\mathrm{B}}=1-{y}_{\mathrm{A}}.$$

From a practical perspective, it will be very inconvenient to plot two phase diagrams, one each for liquid and vapour mole fractions. Hence the two diagrams are combined into one (Fig. 3.14, bottom).

The combined phase diagram is plotted with the pressure as ordinate and the mole fraction of the total composition as abscissa. The mole fraction of the total composition (x i ) can easily be obtained from the mole fractions in the liquid (z i ) and vapour phases (y i ); an example is shown in Table 3.3.
Table 3.3

Example calculation to obtain the total composition of a binary system, when the compositions of liquid and vapour phases are known

n(A)

n(B)

Mole fraction of A

Mole fraction of B

Liquid

2 mol

3 mol

2/5 = zA

3/5 = zB

Vapour

1 mol

2 mol

1/3 = yA

2/3 = yB

Total

3 mol

5 mol

(2+1)/(5+3) = 3/8 = xA

(3+2)/(5+3) = 5/8 = xB

From binary phase diagrams such as the one shown in Fig. 3.15, composition data of the system at various conditions can be easily obtained.
For example, a mixture of two liquids A and B, each present with a mole fraction of 0.5 and a phase diagram as shown in Fig. 3.15 is held at a pressure p = p0. In the diagram this situation is depicted as a point at (xA; p) = (0.5; p0). This point lies in the region between the two lines marking the liquid–vapour phase boundary; the upper line which delivers the molar composition of the liquid phase and the lower line which represents the composition of the vapour phase. This area is called the two-phase area, where the liquid and vapour phases co-exist. Therefore, at the identified point, a line parallel to the x-axis is used to extrapolate to the phase boundary lines (such a line is called a conode or tie line). Where the tie line intersects, drop lines to the x-axis are used to determine the molar fraction of A in the vapour and liquid phases. The molar fractions for B are available as per:
$${x}_{\mathrm{B}\left(\mathrm{liquid}\right)}=1-{x}_{\mathrm{A}\left(\mathrm{liquid}\right)}\kern0.5em \mathrm{and}\kern0.5em {x}_{\mathrm{B}\left(\mathrm{vapour}\right)}=1-{x}_{\mathrm{A}\left(\mathrm{vapour}\right)}$$

As a result of the mixing with substance B, the vapour pressure of A in the mixture is lowered as compared to the vapour pressure of pure liquid A. The mole fraction of A in the vapour phase of a 1:1 mixture is therefore larger than 50%, despite it possessing the higher vapour pressure when comparing the pure liquids.

Any point located in the two-phase indicates a mixture composition where separation into the two co-existing phases occurs. The tie line further yields information about the relative molar amounts of substance A (since xA is plotted as abscissa) in the liquid and vapour phases. The ratio of molar amounts of substance A in the liquid and vapour phases is given by the lever rule, graphically illustrated in Fig. 3.16:
$$\frac{n_{\mathrm{A},\mathrm{liquid}}}{n_{\mathrm{A},\mathrm{vapour}}}=\frac{x_{\mathrm{A}}-{x}_{\mathrm{A},\mathrm{vapour}}}{x_{\mathrm{A},\mathrm{liquid}}-{x}_{\mathrm{A}}}=\frac{d_{\mathrm{liq}}}{d_{\mathrm{vap}}}.$$
(3.25)

The examples and illustrations above were all concerned with pressure-composition (p-x) phase diagrams, where phase boundaries are characterised in dependence of pressure at a particular constant temperature.

Of practical importance are also temperature-composition (T-x) phase diagrams, which show phases at a single pressure. A typical T-x diagram found with many real mixtures is shown in Fig. 3.17. The way of determining the mole fractions of a substance in the liquid and vapour phases in T-x diagrams is the same as discussed above for p-x diagrams. The lever rule can also be applied in an analogous fashion.

Temperature-composition phase diagrams are particularly useful when analysing distillation processes.

### 3.5.2 Simple Distillation

Distillation procedures are based on vapour and liquid having different compositions. In a simple distillation apparatus, mixtures comprising two components of low (component A) and high (component B) volatility can be separated to some degree. Since the distillation experiments are carried out at constant pressure (ambient pressure or under vacuum), temperature-composition phase diagrams such as the one shown in Fig. 3.18, can be used to track the process of simple distillation.
We are starting at ambient temperature with a liquid mixture that has relatively low concentration of the high volatility component B (xA ≈ 0.75, so xB ≈ 0.25). The mixture is heated (arrow 1), and at the intersection of arrow 1 with the boiling curve a vapour phase appears. The composition of the vapour can be obtained where the tie line (arrow 2) intersects with the condensation curve. In a basic distillation apparatus, vapour with this composition is condensed on the fractionation side of the apparatus; the condensate consists of a liquid with enriched component B (Fig. 3.19).

### 3.5.3 Fractional Distillation

In the fractional distillation, vapour is continually removed from the boiling equilibrium system, thus allowing enrichment of the more volatile component to very high purity. This is illustrated in Fig. 3.20.
• Step 1: A mixture of less volatile component (A; higher boiling temperature) and more volatile component (B; lower boiling temperature) is heated. The mole fraction of component B in the initial mixture is xB ≈ 0.15.

• Step 2: The boiling point of the mixture at this molar composition is reached, and a vapour phase with a much higher mole fraction xB is obtained.

• Step 3: The vapour from (2) condenses as it cools at a fractionation plate and reaches the boiling temperature of the liquid mixture at the new molar composition.

• Step 4: A new vapour phase is formed, that is further enriched in component B.

• Steps 5 onwards repeat this process, until an endpoint in the phase diagram is reached.

The efficiency of a fractionating column is expressed in terms of the number of theoretical plates. A theoretical plate is a hypothetical zone in which two phases establish an equilibrium with each other. This is the number of effective condensation and vapourisation steps that are required to achieve a condensate with desired composition from a given distillate. In the example in Fig. 3.20, five theoretical plates are required to obtain pure component B from the initial liquid mixture with xB ≈ 0.15.

In distillation experiments, this process can be achieved by using Vigreux fractionation columns (Fig. 3.21) that contain spikes which form the contact devices (physical plates) between liquid and vapour and thus provide for a number of separation steps. In industrial applications, so-called bubble-cap or valve-cap trays are used. The trays are perforated, thus allowing efficient flow of vapour upwards through the column.
The efficiency of physical plates is non-ideal and therefore the number of physical plates needed for a desired separation step is more than the calculated number of theoretical plates:
$${N}_{\mathrm{a}}=\frac{N_{\mathrm{t}}}{E}$$

Na is the number of actual plates, Nt the number of theoretical plates, and E is the plate efficiency. Obviously, in order to be able to calculate the number of theoretical plates for a distillation process, substantial liquid–vapour equilibrium data (i.e. phase diagrams) need to be available.

### 3.5.4 Mixtures of Volatile Liquids with Azeotropes

The phase diagrams of liquid mixtures we have encountered so far featured monotonous boiling and condensation curves. Some mixtures, though, show additional features in their phase diagrams. The phase diagram of the mixture illustrated in Fig. 3.22 possesses a point where boiling and condensation curves touch in one point. At this point, the liquid and vapour phases have the same composition; the point is called an azeotrope. For example, an ethanol–water mixture with 4% water forms an azeotrope that boils at 78 °C at ambient pressure.

When boiling a liquid mixture that forms an azeotrope, evaporation will proceed without a changing composition of liquid and vapour phases. The mixture behaves as if it were a pure substance. If mixture with azeotropes are subjected to fractionated distillation, the distillation process stops being useful when the azeotrope is reached.

Two types of azeotropes can be distinguished. In low boiling azeotropes, the interactions between the two mixture components are unfavourable compared to ideal mixing. The azeotrope thus boils at the lowest temperature of all possible mixtures of the two components. An example for mixtures with low-boiling azeotropes is the ethanol–water system. In contrast, high boiling azeotropes are mixtures where the interactions between both components are more favourable when compared to the ideal case. Such azeotropes boil at the highest temperature of all possible mixtures of the two components. An example of such behaviour is a system comprising of chloroform and acetone.

### 3.5.5 Mixtures of Immiscible Liquids

Not all liquids can be mixed. If liquids are immiscible, then we can treat the solutions separately, but the vapour pressure of a system that comprises both components is the sum of the two vapour pressures ($${p}_{\mathrm{A}}^{\ast }$$, $${p}_{\mathrm{B}}^{\ast }$$). Boiling occurs when the vapour pressure of the liquid phase equals the atmospheric pressure:
$$p={p}_{\mathrm{A}}^{\ast }+{p}_{\mathrm{B}}^{\ast }={\mathrm{p}}_{\mathrm{normal}}=1\;\mathrm{atm}$$

This results in an interesting consequence: When two immiscible liquids are put together, the pair of them possesses a lower boiling point than either pure liquid alone.

This behaviour is useful when heat-sensitive compounds need to be distilled. When put together with an immiscible liquid, the distillation can proceed at lower temperature than the boiling point of the pure compound. This process is typically carried out as steam distillation.

### 3.5.6 Phase Diagrams of Two-component Liquid/Liquid Systems

In the previous sections, we have discussed liquid mixtures that were either fully or not miscible. The mutual solubility or miscibility of two liquids is a function of temperature and composition. Of course, there are systems where the two liquid components mix under some but not all conditions. When two liquids are partially soluble in each other, two liquid phases can be observed. These are partially miscible liquids (e.g. methanol–cyclohexane, nicotine–water, phenol–water, triethylamine–water, and others). A typical phase diagram for the most common types of partially miscible liquids is shown in Fig. 3.23. The phase diagram indicates that the two liquids are fully miscible and form a one-phase liquid) at high temperatures (above Tuc), but separate into two liquid phases at lower temperatures (below Tuc). Tuc is called the upper critical temperature. The tie line is used to determine the composition of the two phases.
There are two other cases of liquid–liquid mixtures, which are less common. The left panel in Fig. 3.24 shows the phase diagram of a mixture that possesses a lower critical solution temperature Tlc. Below this temperature, the mixture forms one liquid phase, i.e. the two components are fully miscible. Above Tlc, two liquid phases exist. This type of behaviour is observed when there are weak interactions between both components (below the critical solution temperature), such as for example in water-triethylamine.

There also exist some systems that possess both a lower and an upper critical solution temperature, the most prominent example being water and nicotine (Fig. 3.24 right panel). These mixtures are characterised by weak interactions between both components that ensure full miscibility below the lower critical solution temperature. Above Tlc, these interactions are disrupted and there is only partial miscibility, and accordingly, two phases. Above the upper critical solution temperature, the mixture is homogenised and exists as a single liquid phase.

### 3.5.7 Phase Diagrams of Two Component Liquid–Vapour Systems

If we now consider a mixture of two partially miscible liquids and also form a low-boiling azeotrope, we arrive at a fairly common behaviour of real substances. Both properties, partial miscibility and azeotrope formation, emphasise the fact that the molecules of the two components tend to avoid each other.

This behaviour is possible with two different options:
• the liquids may become fully miscible before they boil, i.e. the azeotrope is well separated from the upper critical solution temperature (Fig. 3.25 left), or
• boiling occurs before the two liquids are fully mixed, i.e. the azeotrope and the upper critical solution temperature merge (Fig. 3.25 right).

### 3.5.8 Phase Diagrams of Two-component Solid–Liquid Systems

In the previous sections, we considered systems consisting of two components and discussed their liquid and vapour phase behaviour. In the same fashion, solid and liquid phases can be characterised. Conceptually, there is no difference in the way such phase diagrams are interpreted. As an example, Fig. 3.26 illustrates the phase diagram of two partially miscible solids whose melting point occurs before the two solids are fully mixed.

It is immediately obvious that this phase diagram is highly similar to the one we have seen before (Fig. 3.25 right) in the case of a two-component liquid–vapour system where the two liquids were partially miscible and boiling occurred before full mixing of the liquids. For solids, the mixture with the lowest melting point is called the eutectic point (as opposed to the azeotrope in the liquid–vapour systems).

### 3.5.9 Phase Diagrams of Three-component Systems

Of course, mixtures can be made of more than just two components, but visualisation of phase diagrams for higher order systems becomes challenging. For three-component systems, it is still possible to plot two-dimensional phase diagrams. Instead of a Cartesian xy diagram, a triangular coordinate system can be used, that allows to plot the composition of each of the three components (A, B, C) in the system. However, such phase diagrams are restricted to a particular temperature and pressure.

Like for binary systems before, we can state that the sum of the individual mole fractions is 1:
$${x}_{\mathrm{A}}+{x}_{\mathrm{B}}+{x}_{\mathrm{C}}=1$$
Figure 3.27 illustrates such a triangular phase diagram, showing the mole fractions for a ternary system, at a certain pressure and temperature. The phase separation of the water-butanol-acetic acid system is shown with the grid of the coloured coordinate system in the background.

### 3.5.10 Cooling Curves

Cooling curves show how the temperature in a system changes during the time course of the cooling process. Pure liquid, solid or gas phases have smooth, monotonous changes in temperature, until the process approaches a phase transition.

The phase transitions of pure substances proceed along the pathway of gas → liquid → solid, occur at single temperatures and are exothermic (they are endothermic along the opposite pathway). During the phase transition, the temperature does not change until the phase of the substance has been fully converted. This is illustrated in Fig. 3.28 for a phase diagram with three (left panel) and two phases (middle panel).

Phase transitions during cooling of mixtures also proceed along the pathway of gas → liquid → solid, but occur over a temperature range (see Fig. 3.28 right panel). As with pure substances, the phase transitions during cooling of mixtures are exothermic processes. Therefore, the temperature during these transitions is not constant, but abrupt changes in the overall cooling curve are still visible when the cooling process begins and ends (i.e. the phase transition still leads to occurrence of kinks in the graph). Notably, azeotropes and eutectics behave like a pure substance and show a constant temperature during the phase transition.

## 3.6 Exercises

1. 1.

Is it possible for a one-component system to exhibit a quadruple point?

1. 2.

Henry’s law is valid for dilute solutions. Using the Henry’s law constant for oxygen (solute) and water (solvent) of K(O2) = 781·105 Pa M−1, calculate the molar concentration of oxygen in water at sea level with an atmospheric pressure of patm = pØ.

1. 3.

Below is the T-x phase diagram of the benzene/toluene system acquired at a constant pressure of two bar. A mixture that contains 40% benzene is heated steadily to 122 °C. How many phases are present at this point and what are their compositions? If the total amount of 1 mol of substances was in the initial mixture with 40% benzene, how many moles of substances are in the phase(s) at 122 °C?

1. 4.

A mixture of benzene and toluene with x(benzene) = 0.4 is subjected to fractional distillation at 2 bar (see Exercise 3.3 above). What is the boiling temperature of this mixture? How many theoretical plates are required as a minimum to obtain pure benzene in the distillate?

## References

1. Chaplin M (2014) Water structure and science. http://www1.lsbu.ac.uk/water/. Accessed 7 Nov 2016
2. Clapeyron PBC (1834) Mémoire sur la puissance motrice de la chaleur. Journal de l’École polytechnique 23:153–190Google Scholar
3. Clausius R (1850) Über die bewegende Kraft der Wärme und die Gesetze, welche sich daraus für die Wärmelehre selbst ableiten lassen. Ann Phys 155:500–524
4. Mpemba EB, Osborne DG (1969) Cool? Phys Educ 4:172–175
5. Poole PH, Sciortino F, Essmann U, Stanley HE (1992) Phase behaviour of metastable water. Nature 360:324–328.

© Springer International Publishing AG, part of Springer Nature 2018

## Authors and Affiliations

• Andreas Hofmann
• 1
• 2
1. 1.Eskitis InstituteGriffith UniversityNathanAustralia
2. 2.Faculty of Veterinary and Agricultural SciencesThe University of MelbourneParkvilleAustralia