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Open Access 2024 | OriginalPaper | Chapter

# 4. The Gibbs Free Energy

Author : Jack J. Middelburg

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## Abstract

This chapter introduces the Gibbs free energy and its relevance for the direction of phase changes and reactions. The dependence of Gibbs free energy on temperature and pressure is presented and used to derive the (Clausius-)Clapeyron relation. One-component phase diagrams and Gibb’s phase rule are presented.
The second law provides a robust framework for predicting whether processes will be spontaneous or not. However, the assessment is made at the level of the universe rather than the system alone. To resolve this, the Gibbs free energy, another state function, must be introduced.
Starting with Eq. 3.​11:
$$\Delta {S}_{Tot}=\Delta {S}_{Sys}+ \Delta {S}_{Surr}> 0$$
we can replace $$\Delta {S}_{Surr}$$ using the definition of entropy (Eq. 3.​5):
$$\Delta {S}_{Surr}=\frac{{q}_{surr}}{T}= \frac{\Delta {H}_{Surr}}{T}$$
Noting that the latter equality only applies under constant pressure conditions (see Sect. 2.​3). The enthalpy change of the surroundings is the reverse of that of the system (Eq. 3.​10): $$\Delta {H}_{Surr}= - \Delta {H}_{Sys}$$. Hence, we can obtain an equation for the system only:
$$\Delta {S}_{Sys} -\frac{{\Delta H}_{Sys}}{T}>0$$
$${\Delta H}_{Sys}-T\Delta {S}_{Sys}<0$$
and the subscripts can be dropped to eventually obtain an equation for spontaneous reaction based on systems properties only
$$\Delta H - T\Delta S < 0.$$
(4.1)
The Gibbs (free) energy (G) is formally defined as
$$G\equiv H-TS$$
(4.2)
and is a state function because the constituent terms (H, T, S) are state variables. The change in Gibbs free energy ($$\Delta G)$$ can thus be evaluated based on the difference between final and initial states.
Accordingly at constant temperature and pressure,
$$\Delta {\varvec{G}} = \Delta {\varvec{H}} - {\varvec{T}}\Delta {\varvec{S}}.$$
(4.3)
This useful equation provides the basis for predicting spontaneity, hence the direction of processes, a criterion for equilibrium and quantifies the energy available for work or metabolism of organisms. Comparing Eq. 4.3 with Eq. 3.​12, it is clear that at constant P and T,
$$\Delta G < 0\,\, \text{for a spontaneous process}$$
(4.4a)
$$\Delta G > 0{\text{ for a non{-}spontaneous process}}$$
(4.4b)
$$\Delta G=0\text{ for equilibrium},\text{ no tendency of the system to change}$$
(4.4c)
The Gibbs free energy function also provides first-order guidance for the temperature dependence of chemical and phase equilibria. The change in Gibbs free energy $$\Delta G=\Delta H-T\Delta S$$ contains two terms: a change in enthalpy or heat of reaction ($$\Delta H)$$ and the product of temperature and the change in entropy ($$T\Delta S).$$ The temperature dependence of the direction of phase transformations and chemical reactions relate to this second term. Processes can be divided into four categories based on the temperature response (Table 4.1).
Table 4.1
Temperature and spontaneity of processes
$$\Delta H$$
$$\Delta S$$
$$-T\Delta S$$
$$\Delta G$$
High T
Low T
+
Spontaneous
Spontaneous
+
±
Nonspontaneous
Spontaneous
+
+
±
Spontaneous
Nonspontaneous
+
+
+
Nonspontaneous
Nonspontaneous
The temperature at which processes switch from being spontaneous to becoming non-spontaneous can be obtained from $$\Delta G=\Delta H-T\Delta S=0$$ which after re-arrangement leads to
$$T=\frac{\Delta H}{\Delta S}$$
(4.5)
The Gibbs function can also be interpreted in the context of efficiency analogue to potential energy in mechanics and the ratio of work done to heat supplied in energy analysis. The Gibbs free energy gives the maximal energy available to perform work, while the enthalpy gives the total amount of energy released due to phase transformations or chemical reactions and the T $$\Delta S$$ represents the loss term or non-useable energy needed for expansion work. The efficiency is then defined as $$\frac{\Delta G}{\Delta H}$$.
Similar to the introduction of standard enthalpies of formation (see Sect. 4.2), we can also define standard Gibbs free energies of formation ($$\Delta {G}_{f}^{o}$$) which is the change in Gibbs free energy for the formation of one mol of that compound from its elements under standard conditions (STAP, 25 °C and 1 bar). These $$\Delta {G}_{f}^{o}$$ values are tabulated, reported in kJ mole−1 and provide information on the stability of compounds. For instance, the $$\Delta {G}_{f}^{o}$$ values of water in liquid and gas forms are − 237.1 and − 228.6 kJ mol−1 respectively, indicating that liquid water is the stable phase at 25 °C and 1 bar pressure.
These $$\Delta {G}_{f}^{o}$$ values can also be used to calculate the Gibbs free energy of a reaction ($${\Delta }_{r}{G}^{o}$$) at 298 K and 1 bar:
$$\Delta_{r} G^{o} = \sum v_{i} \Delta G_{f}^{o} \left( {products} \right) - \sum v_{i} \Delta G_{f}^{o} \left( {reactants} \right),$$
(4.6)
where $${v}_{i}$$ is the stoichiometric coefficient of the reaction.
Returning to reaction (Eq. 3.​9) discussed before for which we have provided the thermodynamic data (Table 3.​1):
$$2{{\text{H}}}_{2}\left(g\right)+{{\text{O}}}_{2}\left(g\right)\to 2{{\text{H}}}_{2}{\text{O}}(l)$$
Like the entropy ($${\Delta }_{r}{S}^{o}$$) and enthalpy ($${\Delta }_{r}{H}^{o}$$) of reactions, the Gibbs free energy ($${\Delta }_{r}{G}^{o})$$ can be calculated from the product-minus-reactants (Eq. 4.6).
$${\Delta }_{r}{G}^{o}=\left[2\times \Delta {G}_{f}^{o}\left({{\text{H}}}_{2}{\text{O}}(l)\right)\right]-\left[2\times \Delta {G}_{f}^{o} \left({{\text{H}}}_{2}\left(g\right)\right)+ 1\times \Delta {G}_{f}^{o} \left({{\text{O}}}_{2}\left(g\right)\right)\right]=-474.26 {\text{kJ}},$$
indicating that water is the stable phase at earth surface conditions.
This Gibbs free energy change $${(\Delta }_{r}{G}^{o})$$ can also be calculated based on the changes in entropy ($${\Delta }_{r}{S}^{o}$$) and enthalpy ($${\Delta }_{r}{H}^{o}$$) of reactions (Eq. 4.3):
$${\Delta }_{r}{G}^{o}={\Delta }_{r}{H}^{o} -T{\Delta }_{r}{S}^{o}= -571.66\times 1000-298.15\times -326.68=-474.26\text{ kJ}$$
and the results are identical.
The first approach via the products-minus-reactants approach is simpler but is restricted to SATP conditions because $$\Delta {G}_{f}^{o}$$ values are tabulated for SATP conditions, while the second approach via enthalpies and entropies can be used at other temperatures as well. However, calculation of Gibbs free energy of reactions at non-standard temperatures via the second approach only holds if enthalpies and entropies do not vary significantly with temperature. In this introduction course, we make this assumption, being aware that both enthalpy and entropy vary with temperature.
Changes in Gibbs free energy during chemical reactions can be used to perform work, to heat our homes or to support the metabolism of organisms. As an example of the latter, we calculate the $${\Delta }_{r}{G}^{o}$$ for microbially mediated reactions. For aerobic (involving oxygen) and anaerobic (in the absence of oxygen, involving sulphate) degradation of model organic matter CH2O ($$\Delta {G}_{f}^{o}=-128.7$$ kJ mol−1) we have the following reactions:
$${{\text{CH}}}_{2}{\text{O}} \left(s\right)+ {{\text{O}}}_{2}\left(g\right)\to {{\text{CO}}}_{2}\left(aq\right)+ {{\text{H}}}_{2}{\text{O}}(l)$$
$${2{\text{CH}}}_{2}{\text{O}} (s)+ {{\text{SO}}}_{4}^{2-}(aq)\to {{\text{H}}}_{2}{\text{S}}(g)+2{{\text{HCO}}}_{3}^{-} (aq)$$
The $${\Delta G}_{r}^{o}$$ values for these two reactions are − 494 and − 205 kJ, respectively, or normalized to one mole of CH2O: − 494 and − 102 kJ. First, both aerobic respiration and sulphate reduction provide enough energy needed for growth and maintenance of the microbes involved (about − 10 kJ per mol). Second, about five times more energy is liberated during aerobic respiration than during sulphate reduction. Where is the missing energy? It is transferred to the reaction products: the hydrogen sulphide produced can react with oxygen:
$${{\text{H}}}_{2}{\text{S}} \left(g\right)+ 2{{\text{O}}}_{2}\left(g\right)\to {{\text{H}}}_{2}{{\text{SO}}}_{4}(aq)$$
and the $${\Delta }_{r}{G}^{o}$$ value for this reaction is − 329 kJ. Microbial communities are highly efficient in using energy and the product of one organism is often the reactant for another organism, i.e., these reactions are coupled, with the result that most of the energy potentially available is eventually used.

## 4.1 How Gibbs Free Energy Depends on Conditions

The Gibbs free energy change during a process provides information whether a reaction or phase transformation will be spontaneous at constant temperature and pressure ($$\Delta G<0)$$, but chemists and earth system scientists are often interested how the natural variables temperature or pressure impact the direction of processes or the stability of certain substances. We will therefore derive an equation, the fundamental Gibbs equation, expressing the dependence of Gibbs energy on temperature and pressure. The term fundamental is used because P and T are the basic natural variables. We will use basic calculus for the derivations.
Starting from the definition of the Gibbs free energy (Eq. 4.2),
$$G\equiv H-TS$$
and substituting the formal definition of H (Eq. 2.​11):
$$H\equiv U+PV,$$
We arrive at,
$$G= H-TS= U+PV-TS.$$
Next, we take the total differential (i.e., the differential of terms of each variable in turn)
$$dG= dU+PdV+VdP-TdS-SdT$$
(4.7)
Making use of the first law of thermodynamics:
$$dU=q-PdV$$
and the definition of entropy (Eq. 3.​5):
$$dS=\frac{q}{T} ,$$
rewritten as $$TdS=q,$$
we arrive at
$$dU=TdS-PdV$$
Substituting this equation into Eq. 4.7 leads to
$$dG= TdS-PdV+PdV+VdP-TdS-SdT$$
Simplifying by cancelling terms, we eventually arrive at the fundamental Gibbs equation which expresses the dependence of Gibbs free energy on changes in pressure and temperature:
$$dG= VdP-SdT$$
(4.8)
Another way to quantify this dependency is to use the total differential (see Eq. 1.​9 in Box 1):
$$dG= {\left(\frac{\partial G}{\partial P}\right)}_{T}dP+{\left(\frac{\partial G}{\partial T}\right)}_{P}dT$$
(4.9)
Comparing these equations, the partial derivative with respect to temperature is minus entropy
$${\left(\frac{\partial G}{\partial T}\right)}_{P}= -S$$
(4.10)
Hence, the entropy determines the temperature sensitivity of the Gibbs free energy. The larger the entropy, the greater the temperature dependence. Moreover, the Gibbs free energy of a compound declines (becomes more negative) with increasing temperature (because S values of compounds are always positive).
Similarly, the partial derivative with respect to pressure is volume
$${\left(\frac{\partial G}{\partial P}\right)}_{T}=\text{ V}$$
(4.11)
Consequently, the volume of a substance determines how its Gibbs free energy changes with pressure. Solids and liquids have small molar volumes (compared to gases) and the pressure dependence of the Gibbs free energy for solids and liquids is rather small. This is evidently not the case for gases. For an isothermal process, we can rewrite Eq. 4.8 as follows:
$$dG=V dP= \left(\frac{nRT}{P}\right)dP$$
Integrating this equation to find $$\Delta G$$
$$\Delta G={\int }_{{P}_{i}}^{{P}_{f}}dG= {\int }_{{P}_{i}}^{{P}_{f}} \left(\frac{nRT}{P}\right)dP=nRT {\int }_{{P}_{i}}^{{P}_{f}} \frac{1}{P}dP=nRTln\left(\frac{{P}_{f}}{{P}_{i}}\right)$$
(4.12a)
where we have used the ideal gas law to eliminate V and $$\int \frac{dx}{x}={\text{ln}}x$$ for the integration.
This equation expresses the isothermal change in Gibbs free energy of a gas when pressures changes from Pi to Pf.
The change in Gibbs free energy ($$\Delta G)$$ at pressure P can then be expressed relative the standard Gibbs free energy change at 1 bar ($$\Delta {G}^{o})$$
$$\Delta G =\Delta {G}^{o}+ nRTln\left(\frac{P}{1}\right)= {\Delta G}^{o}+ nRTln\left(P\right)$$
(4.12b)
This latter equation is only valid if pressure is expressed in bar units, but the ratio $$\left(\frac{{P}_{f}}{{P}_{i}}\right)$$ is dimensionless and Eq. 4.12a can be used when other units or other reference levels are adopted.
Equation 4.12 has been derived for gases expressed in pressures, but an equivalent expression can be derived when considering ideal liquid or solid solutions. In this case, the concentration [c] in mol L−1 ($$\left[c\right]= \frac{n}{V}$$) is the analogue of pressure and the change in Gibbs free energy is:
$$\Delta G= \Delta {G}^{o}+ nRTln\left[\frac{c}{{c}^{o}}\right]$$
(13a)
where $$\left[{c}^{o}\right]$$ = 1 mol L−1 for an ideal solution. For non-ideal solutions, we use thermodynamic activity a rather than concentration [c]:
$$\Delta G = {\Delta G}^{o}+ nRTln a$$
(13b)
Thermodynamic activities for pure liquids and solids are 1 and the second term then becomes equal to zero and disappears as it should be because $$\Delta G = {\Delta G}^{o}$$ for pure phases. Later in the course (Chap. 5) we return to Eqs. 4.12 and 4.13 when we discuss equilibria of solutions.

## 4.2 Phase Equilibria

In the previous section we have derived the Gibbs free energy equation for a single phase and compound as a function of temperature and pressure. Now we are going to study the relationships between phases and how these depend on pressure and temperature conditions. A phase is a part of a system which is homogenous and separated from other phases by a definite boundary. As discussed earlier we deal with gas, liquid, and solid phases. Gases are always present as single phase because they mix well. A liquid containing a single component will only be one phase, but a mixture of liquid may be one phase when well mixed (ethanol in water) or multiphase when they do not mix (water–oil). Solids are often present in multiple phases, even a single component may be present in multiple phases (e.g., carbon in the form of graphite, diamond or fullerene, CaCO3 as calcite or aragonite, Al2SiO5 as the minerals kyanite, sillimanite or andalusite).
For phases A and B co-existing in equilibrium we can write:
$$\Delta G={dG}_{A}-{dG}_{B}=0$$
(4.14)
where $${dG}_{A}$$ and $${dG}_{B}$$ are the Gibbs free energies of phases A and B. For both phases we can write the Gibbs free energy using the fundamental equation (Eq. 4.8)
$$d{G}_{A}= {V}_{A}dP-{S}_{A}dT\text{ and }d{G}_{B}= {V}_{B}dP-{S}_{B}dT$$
Since phases A and B are in equilibrium, $${dG}_{A}={dG}_{B}$$, hence
$${V}_{A}dP-{S}_{A}dT= {V}_{B}dP-{S}_{B}dT$$
(4.15)
$${V}_{A}dP-{V}_{B}dP= {S}_{A}dT-{S}_{B}dT$$
$$\left({V}_{A}-{V}_{B}\right)dP= {(S}_{A}-{S}_{B}) dT$$
Introducing $$\Delta V=\left({V}_{A}-{V}_{B}\right)$$ and $$\Delta S{=(S}_{A}-{S}_{B})$$, we can rewrite this as
$$\Delta VdP=\Delta SdT$$
or
$$\frac{dP}{dT}=\frac{\Delta S}{\Delta V}$$
(4.16a)
Since we are at equilibrium $$\Delta H=T\Delta S$$ we arrive at
$$\frac{dP}{dT}=\frac{\Delta H}{T\Delta V}$$
(4.16b)
This is the Clapeyron equation, and it describes the slope of the line between two phases in equilibrium (also called coexistence curve) in the pressure-temperature domain as a function of changes in molar volume ($$\Delta V)$$ and molar enthalpy ($$\Delta H)$$ or entropy ($$\Delta S)$$. These slopes are usually positive (Fig. 4.1) because entropy and volume changes are positive going from a solid to a liquid or liquid to a gas/vapour.
The Clapeyron equation can be used to determine if a phase transition is likely to take place under specific conditions. For instance, considering solids with a typical entropy of fusion ($$\Delta {S}_{fusion})$$ of ≈ 22 J mol−1 K−1 and molar volume change during fusion ($$\Delta {V}_{fusion})$$) of ≈ 4 $$\times$$ 10–6 m3 mol−1 (i.e., 4 ml per mol), would yield
$$\frac{dP}{dT}=\frac{\Delta S}{\Delta V}=\frac{22 }{4 \times {10}^{-6}}=5.5\times {10}^{+6} Pa {K}^{-1}=55 bar {K}^{-1}$$
Accordingly, a pressure increase of ≈55 bar is required to change the melting temperature of solids by one degree. The Clapeyron equation is frequently used in solid-earth sciences because it relates the (geo)thermal gradient ($$\frac{{\text{dT}}}{{\text{dP}}}=\frac{{\text{dT}}}{{\text{dz}}})$$ to changes in volume and enthalpy and provides then an estimate for the temperature T at which minerals are transformed or melting (solidus).
This simple calculation can also be applied to water–ice equilibria although the volume change $$\Delta V$$ going from ice to liquid is negative (− 1.63 mL or − 1.63 $$\times {10}^{-6}$$ m3 for one mol of H2O), indicating that the P to T slope is negative (Fig. 4.2). Another application is the calculation of freezing point increases with altitude (see Box 3).
While volume changes for solid–liquid transition are rather limited (in the order of mL per mol), this is not the case for gases (about 24 L per mol) and slopes of liquid–gas co-existence are flatter (Fig. 4.1). The change in volume of the system can thus be approximated with the volume of the gas, i.e. $$\Delta V={V}_{gas}$$. Assuming an ideal gas, we can rewrite Eq. 4.16 to
$$\frac{dP}{dT}=\frac{\Delta H}{T\Delta V}=\frac{\Delta H\;P}{T\;RT}=\frac{\Delta H\;P}{R{T}^{2}}$$
Next, we isolate pressure P on the left-hand side:
$$\frac{dP}{P}=\frac{\Delta H\; dT}{R{T}^{2}}=\frac{\Delta H}{R} \frac{dT}{{T}^{2}}$$
(4.17)
Making use of $$\int \frac{dP}{P}={\text{ln}}P$$ and integrating from P1, T1 to P2, T2 we arrive at the Clausius-Clapeyron equation:
$$ln\frac{{P}_{1}}{{P}_{2}}=-\frac{\Delta H}{R}\left(\frac{1}{{T}_{1}}-\frac{1}{{T}_{2}}\right)$$
(4.18)
The Clausius-Clapeyron equation is very useful for liquid–vapour/gas equilibria such as vapour pressure or boiling points calculation. As an example of the latter, we return to the Himalayas (Box 3).
Box 3 Freezing and Boiling Points in the Himalayas
Consider a mountain high in Himalaya with an atmospheric pressure of 35 kPa (altitude about 7.6 km) and calculate the freezing point increase. The molar enthalpy of melting ($$\Delta {H}_{fusion})$$ is 6.01 kJ mol−1, the volume change from ice to liquid is − 1.63 $$\times {10}^{-6}$$ m3 for one mol of H2O and the pressure difference ($$\Delta P)$$ is 35 $$\times {10}^{3}-1.013 \times {10}^{5}$$= − 6.63 $$\times {10}^{4}$$ Pa.
Using the Clapeyron equation (Eq. 4.16b),
$$\frac{{{\text{dP}}}}{{{\text{dT}}}} = \frac{\Delta H}{{T\Delta V}} = \frac{{6.01 \times { }10^{3} }}{{273.15* - 1.63{ } \times { }10^{ - 6} }} = - 1.35 \times 10^{7} \,{\text{Pa}}\,{\text{K}}^{ - 1}$$
Hence: $$\Delta T=\frac{\Delta P}{-1.35\times {10}^{7}}\approx$$ 0.005 K.
The freezing point is thus 0.005 °C higher.
A similar calculation can be made for the lowering of the boiling point using the Clausius-Clapeyron equation (Eq. 4.18). The boiling point at 1 atm is 100 °C but it will be lower high in the mountains because of lower atmospheric pressure. Using Eq. 4.18, with P1 = 1 atm =$$1.013 \times {10}^{5}$$ Pa and $${T}_{1}=$$ 100 °C = 273.15 K and the standard enthalpy of vaporization of water ($$\Delta {H}_{vap}$$) of 40.7 kJ mol−1:
$${\text{ln}}\frac{1.013 \times {10}^{5}}{35 \times {10}^{3}}=-\frac{\text{40,700}}{8.314}\left(\frac{1}{373.15}-\frac{1}{{T}_{2}}\right)$$
we obtain a boiling point of 345 K (= 72 °C) at 7.6 km altitude. It takes much more time to boil an egg high in the mountains because the heat transfer from the water to your food is less.

## 4.3 Phase Diagrams

Materials exist in the solid phase at high pressure and/or low temperature conditions, but eventually transform into gas/ vapor at low pressures or high temperatures. Phase diagrams are a common way to depict which phase is present at a given P and T (Fig. 4.1). The phases (solid, liquid and gas) are separated by boundary lines, i.e., the lines represent conditions at which two phases co-exist. The Clapeyron relation (Eq. 4.16): $$\frac{dP}{dT}=\frac{\Delta H}{T\Delta V}$$ gives the slope of these lines.
There are three types of lines: (1) the line between the solid and liquid phase shows how the melting points (temperature of solid to liquid transformation) change with pressure, (2) the line between liquid and vapor phase shows the dependence of boiling points (temperature of liquid to vapour transformation) on pressure, and (3) the line between solid and vapor delineates the transition from a solid to gas (sublimation and condensation). The triple point represents the P–T combinations at which the three phases co-exist in equilibrium and the critical point represents a combination of temperature and pressure beyond which a gas cannot be liquified regardless of the pressures and temperature. The state of matter beyond the critical point is neither a gas or liquid; it is a supercritical fluid that possesses some typical liquid properties such as high density and ability to act as solvent, but the viscosity and diffusion coefficients are more alike in a gas.
The ‘normal’ melting point is where a horizontal line at 1 bar or 1 atm crosses the solid–liquid equilibrium lines, and the ‘normal’ boiling point is crossed at the liquid–gas (vapor) equilibrium lines (Fig. 4.2). These are called normal melting/boiling points because they correspond to Earth-surface conditions, e.g., 0 and 100 °C for the normal melting and boiling points for water. In Box 3/Sect. 4.2 we have calculated how boiling points drop when pressure drops or increases when pressure increases.
The phase boundary lines constrain the freedom of selecting pressure and temperature conditions for a system to be in thermodynamic equilibrium. Within a single phase, one can freely vary P and T independently. At the boundary between two phases, i.e. at the equilibrium lines, either P or T can be chosen, the other one is then fixed. Gibbs has formalized the number of independent variables (= degrees of freedom) to describe a multicomponent system. Gibbs’ phase rule states that
$$F=C-P+2$$
(4.19)
where F is degrees of freedom, C is the number of components and P is the number of phases in equilibrium. The + 2 term relates to temperature and pressure. Consider a phase diagram for one single component, e.g. H2O (C=1) (Fig. 4.2). Within fluid water, or within the domain of ice or water vapor, pressure and temperature can be combined in multiple ways because there are two degrees of freedom (F = 1–1 + 2 = 2). At the interface of water–ice, there is only one degree of freedom (F = 1–2 + 2 = 1), i.e., if temperature is chosen, pressure is fixed or the other way around. The boundary lines between mineral phases are therefore called univariant curves in petrology and mineralogy. At the triple point, ice, water and water vapour co-exist (water is freezing and boiling at the same time) and Gibb’s phase rule indicates that there are no degrees of freedom (F = 1–3 + 2 = 0). In other words, the triple point has a fixed pressure and temperature for a component, e.g., for water it is 0.0098 °C and 6 $$\times$$ 10–3 atm.
Phase diagrams are instructive to show that melting (solid to liquid) can occur either through increasing the temperature or through lowering the pressure (except for water because of its negative slope). Similarly, boiling of a liquid can not only occur by increasing the temperature, but also induced by lowering the pressure, i.e., it is possible to boil cold water under low pressure conditions. From the quantitative treatment in Sect. 4.2, it is clear that solid–liquid lines are normally steep because the volume changes involved are small, while liquid–gas equilibrium lines are rather flat because volume changes are large (about three orders of magnitude for ideal gases). Moreover, the liquid–gas equilibrium lines usually show substantial curvature because of the temperature dependence of enthalpies of vaporization. Finally, this course is limited to single component phase diagram, but the same principles, including Gibbs’ phase rule, apply to multicomponent systems. However, it can get rather complex.