Posts Tagged ‘equilibrium’

gibbs

It was the American mathematical physicist Josiah Willard Gibbs who introduced the concepts of phase and chemical potential in his milestone monograph On the Equilibrium of Heterogeneous Substances (1876-1878) with which he almost single-handedly laid the theoretical foundations of chemical thermodynamics.

In a paragraph under the heading “On Coexistent Phases of Matter” Gibbs mentions – in passing – that for a system of coexistent phases in equilibrium at constant temperature and pressure, the chemical potential μ of any component must have the same value in every phase.

This simple statement turns out to have considerable practical value as we shall see. But first, let’s go through the formal proof of Gibbs’ assertion.

An important result

pe01

Consider a system of two phases, each containing the same components, in equilibrium at constant temperature and pressure. Suppose a small quantity dni moles of any component i is transferred from phase A in which its chemical potential is μ’i to phase B in which its chemical potential is μ”i. The Gibbs free energy of phase A changes by –μ’idni while that of phase B changes by +μ”idni. Since the system is in equilibrium at constant temperature and pressure, the net change in Gibbs free energy for this process is zero and we can write

pe02

hence

pe03

This result can be generalized for any number of phases: for a system in equilibrium at constant temperature and pressure, the chemical potential of any given component has the same value in every phase.

pe04

– – – –

Visualizing variance

The equality of pressure P, temperature T and component chemical potentials μn between coexistent phases in equilibrium provides a convenient way to visualize variance, or the number of degrees of freedom a system possesses. For example, the triple point of a single component system can be visualized as the array

pe05

where the solid, liquid and vapor phases are indicated by one, two and three primes respectively.

Each row represents a single variable, so the number of rows equates to the total number of variables. Each column lists the variables in a single phase. All but one of these may be independently varied; the last is determined by the Gibbs-Duhem relation

pr02

There are one of these for each phase, so the number of columns equates to the number of relations (=constraints) to which the system variables are subject. The variance, or number of degrees of freedom (f) of the system is defined

pe06

For arrays of the kind presented above, this transposes into

pe07

For the triple point of a single component system, there are three rows and three columns, so f =0. With zero degrees of freedom, the triple point is not subject to independent variation and is represented by a fixed point in the PT plane.

The above rule implies that a system of coexistent phases in equilibrium cannot have more phases than intensive system variables.

– – – –

Generating useful equations

For a component present in any pair of coexistent phases in equilibrium at constant temperature and pressure, the chemical potential of that component has the same value in both phases

pe03

From this general relation, equations may be deduced for computing various properties of thermodynamic systems such as ideal solutions, for example the elevation of boiling point, the depression of freezing point, and the variation of the solubility of a solute with temperature.

The key point to grasp is that μi is the chemical potential of component i in an arbitrary state, i.e. in a mixture of components. In order to compute this potential we need to know two things: the chemical potential of the pure substance μi0 at a pressure p (such as that of the atmosphere), and the mole fraction (xi) of the component in the mixture. Assuming an ideal solution, use can then be made of the textbook formula

dce12 … (1)

where for a given phase, μi is the arbitrary chemical potential of i in the mixture, μ°i is the chemical potential of the pure substance, and xi is the mole fraction of the component.

As an example, let us take the relation

pe08 …(2)

where the chemical potential of the solid solvent is necessarily the standard potential because the mole fraction x is unity. The above relation will generate an equation for the depression of the solvent freezing point in a solution at a fixed pressure (p).

Substituting (1) for the liquid phase in (2) gives

pe09

pe10

where by convention the subscript 1 refers to the solvent. Differentiating with respect to T at constant pressure

pe11

using the quotient rule for ΔG/T gives

pe12 … (3)

Now since

pe13

equation (3) simplifies to

pe14

Integrating from the pure solvent state, where the mole fraction x1=1 and T0fus is the freezing point of the pure solvent, to the solution state where the mole fraction x1= x1 and Tfus is the freezing point of the solvent in the solution

pe15

yields the equation for the depression of the solvent freezing point in a solution at a fixed pressure (p)

pe16

Since x1<1 in a solution, the logarithm is negative and therefore the freezing point of the solvent in the solution must be lower than the freezing point of the pure solvent.

– – – –

Ok, so maybe that wasn’t the simplest procedure for generating a useful thermodynamic equation. But the point to be made here is that the same procedure applies in the other cases, so you only need to understand the principle once.

For example, the equation for elevation of solvent boiling point in solution with a non-volatile solute at a fixed pressure (p) is

pe17

The similarity to the previous equation is evident.

– – – –

pr01

The Phase Rule formula was first stated by the American mathematical physicist Josiah Willard Gibbs in his monumental masterwork On the Equilibrium of Heterogeneous Substances (1875-1878), in which he almost single-handedly laid the theoretical foundations of chemical thermodynamics.

In a paragraph under the heading “On Coexistent Phases of Matter”, Gibbs gives the derivation of his famous formula in just 77 words. Of all the many Phase Rule proofs in the thermodynamic literature, it is one of the simplest and shortest. And yet textbooks of physical science have consistently overlooked it in favor of more complicated, less lucid derivations.

To redress this long-standing discourtesy to Gibbs, CarnotCycle here presents Gibbs’ original derivation of the Phase Rule in an up-to-date form. His purely prose description has been supplemented with clarifying mathematical content, and the outmoded symbols used in the single equation to which he refers have been replaced with their modern equivalents.

– – – –

Gibbs’ derivation

Gibbs begins by introducing the term phase to refer solely to the thermodynamic state and composition of a body (solid, liquid or vapor) without regard to its quantity. So defined, a phase cannot be described in terms of extensive variables like volume and mass, since these vary with quantity. A phase can only be described in terms of intensive variables like temperature and pressure, which do not vary with quantity.

To derive the Phase Rule, Gibbs chooses as his starting point equation 97 from his treatise, now known as the Gibbs-Duhem equation

pr02

This general thermodynamic equation, which relates to a single phase, connects the intensive variables temperature T, pressure P, and chemical potential μ where μn is the potential of the nth component substance. Any possible variations of these quantities sum to zero, indicating a phase in internal equilibrium.

If there are n independent component substances, the phase has a total of n+2 variables

pr03

These quantities are not all independently variable however, because they are related by the Gibbs-Duhem equation. If all but one of the quantities are varied, the variation of the last is given by the equation. A single-phase system is thus capable of (n+2) – 1 independent variations.

Now suppose we have two phases, each containing the same n components, in coexistent equilibrium with each other. Signifying one phase by a single prime and the other by a double prime we may write

pr04

since this is the definition of equilibrium between phases. So in the two-phase system the total number of variables remains unchanged at n+2, but there are now two Gibbs-Duhem equations, one for each phase. It follows that if all but two of the quantities are varied, the variations of the last two are given by the two equations. A two-phase system is thus capable of (n+2) – 2 independent variations.

It is evident from the foregoing that regardless of the number of coexistent phases in equilibrium, the number of variables will still be n+2 while the number subtracted (called the number of constraints) will be equal to the number of Gibbs-Duhem equations i.e. one for each phase.

A system of r coexistent phases is thus capable of n+2 – r independent variations, which are also called degrees of freedom (F). Therefore

pr05

This is the Phase Rule as derived by Gibbs himself. In contemporary textbooks it is usually written

pr06

where C is the number of independent components and P is the number of phases in coexistent equilibrium.

– – – –

Why don’t we use Gibbs’ original derivation of the Phase Rule?

This is a question for science historians with better information resources at their disposal than mine. But I can offer a couple of indicators.

The first point to make is that although Gibbs was undoubtedly the first to discover that a coexistent r-phase system containing n independent components has n + 2 – r degrees of freedom, he did not draw any attention to it; in fact his derivation is almost ‘hidden away’ in the text of his milestone monograph.

Nor did Gibbs apply the sobriquet ‘phase rule’; this seems to have originated in Europe. The Dutch chemist Hendrik Roozeboom, who in the 1880s began research into verifying Gibbs’ theoretical predictions of phase equilibria at the University of Amsterdam, certainly introduced the term “Phasenlehre”. But the earliest dated literature reference I can find is from 1893 when the German chemist Wilhelm Meyerhoffer, who was also working at the University of Amsterdam, published a paper entitled “Die Phasenregel”.

The second point concerns the two books which established chemical thermodynamics as a modern, practical science, and set the study curriculum at countless colleges around the world. One was American – the famous Thermodynamics by G.N. Lewis and Merle Randall, published in 1923. The other was European – Modern Thermodynamics by Edward Guggenheim, published in 1933.

The extraordinary thing about Lewis and Randall’s 600+ page book, the pivotal work which first made Gibbs’ powerful ideas accessible to students of physical science, is that it devotes barely a page to the Phase Rule and – crucially – does not even state the equation, let alone its derivation.

That job was left to Edward Guggenheim in Europe. In Chapter 1 of Modern Thermodynamics – Introduction and Fundamental Laws – he states the Phase Rule and gives the derivation.

But it is not Gibbs’ derivation based on a single intensive factor relation. Guggenheim’s method involves counting component concentrations, which are related by an equation of condition within each phase, and are also subject to individual constraints between phases since the chemical potential of any component is the same in all phases at equilibrium.

Two separate sets of constraints are thus imported into Guggenheim’s calculation of degrees of freedom

pr11

which to this writer’s eyes, lacks the simplicity of Gibbs’ approach.

Nevertheless, Guggenheim’s method is the one which has been taught to generations of students (including me), and will in all likelihood be taught for generations to come…

… unless CarnotCycle succeeds with this post in awakening interest in Gibbs’ own derivation, which is surely the original and arguably the best!

– – – –

cse01

In terms of simplicity, purely mechanical systems have an advantage over thermodynamic systems in that stability and instability can be defined solely in terms of potential energy. For example the center of mass of the tower at Pisa, in its present state, must be higher than in some infinitely near positions, so we can conclude that the structure is not in stable equilibrium. This will only be the case if the tower attains the condition of metastability by returning to a vertical position or absolute stability by exceeding the tipping point and falling over.

cse02

Thermodynamic systems lack this simplicity, but in common with purely mechanical systems, thermodynamic equilibria are always metastable or stable, and never unstable. This is equivalent to saying that every spontaneous (observable) process proceeds towards an equilibrium state, never away from it.

If we restrict our attention to a thermodynamic system of unchanging composition and apply the further constraint that the system is closed, i.e. no mass transfer takes place between system and surroundings, the fundamental relation of thermodynamics can be written as an exact differential expression exhibiting two independent variables: dU = TdS–PdV. There are therefore 22–1 possible Legendre transformations, each representing a new function. We thus arrive at a set of four state functions

cse03

where the independent variables, shown in parentheses, are the natural (canonical) variables for that function [S = entropy, V = volume, T = temperature, P = pressure].

If the above pairs of natural variables are held constant, the energy function is predictably a minimum at equilibrium. But what if the energy function and one of its natural variables are instead held constant? Which extremum will the other natural variable reach at equilibrium – minimum or maximum? There are eight permutations, giving a grand total of twelve sets of equilibrium conditions.

Of these twelve, only four have straightforward proofs based on the fundamental criteria [derived in my previous blogpost Reversible and Irreversible Change] that

cse04

The proofs of the other eight involve a lengthier train of logical argument which is now rarely seen either in textbooks or on the internet. To redress this issue, CarnotCycle herewith presents all twelve proofs in their entirety.

– – – –

The Twelve Conditions of Equilibrium and Stability

Here are the four straightforward proofs

1. For given U and V that S is a maximum

For a closed system, the First Law may be written

cse05 (3)

Consider an adiabatic change: dq = 0 so the above criteria for equilibrium (1) and observable change (2) become dS = 0 and dS > 0 respectively. If the volume is also kept constant during the change, it follows from the First Law that dU = 0. The criteria for these conditions may be expressed as follows

rev07

Since observable change always proceeds towards equilibrium, S will increase towards a maximum at equilibrium when U and V are held constant.

2. For given H and P that S is a maximum

For a closed system, the First Law may be written

cse05

Consider an adiabatic change: dq = 0 so the above criteria for equilibrium (1) and observable change (2) become dS = 0 and dS > 0 respectively. If the pressure P of the system is kept constant during the change, and is made to differ only infinitesimally from that of the surroundings, equation (3) becomes

cse06

By the definition of enthalpy, we find

cse07

In this case therefore, H and P are constant, and the criteria for these conditions may be expressed as follows

cse08

Since observable change always proceeds towards equilibrium, S will increase towards a maximum at equilibrium when H and P are held constant.

3. For given T and V that A is a minimum

For a closed system undergoing a change at constant temperature, we may write the differential expression

cse09

But by the first law

cse10

so that

cse11 (4)

Suppose the system is capable only of PV work. If the volume is kept constant, no work can be done and therefore

cse12

Applying criteria (1) and (2) for equilibrium (dq = TdS) and observable change (dq < TdS) respectively, we obtain the criteria in terms of A:

cse13

Since observable change always proceeds towards equilibrium, A will decrease towards a minimum at equilibrium when T and V are held constant.

4. For given T and P that G is a minimum

For a closed system undergoing a change at constant temperature, we may write the differential expression (cf. equation 4 above)

cse14

Suppose the system is capable only of PV work. If the pressure P of the surroundings is kept constant during the change, and is made to differ only infinitesimally from that of the system, then

cse15

Applying criteria (1) and (2) for equilibrium (dq = TdS) and observable change (dq < TdS) respectively, we obtain the criteria in terms of G:

cse16

Since observable change always proceeds towards equilibrium, G will decrease towards a minimum at equilibrium when T and P are held constant.

– – – –

A word of introduction to the remaining eight not-so-straightforward proofs.

On page 2 of his masterwork On the Equilibrium of Heterogeneous Substances, under the heading Criteria of Equilibrium and Stability, J. Willard Gibbs constructs the foundations of chemical thermodynamics on the equivalence of two propositions relating to an isolated system – that for given U and V that S is a maximum, and for given S and V that U is a minimum.

Gibbs shows by neat argument that the truth (according to Clausius) of the first proposition necessitates the truth of the second proposition. In parallel fashion, the proof already given for the first proposition (condition 1 above) will be used to prove that for given S and V that U is a minimum. This sequitur argument repeats throughout, with condition 2 used to prove condition 6, condition 3 used to prove condition 7 etc.

The train of logical argument for these remaining proofs is somewhat labyrinthine, but the method is same in each case. It gets easier to follow as you go from one to the next.

– – – –

5. For given S and V that U is a minimum

Imagine a closed system in some state of equilibrium for which the values of the entropy, volume and internal energy are S1, V1 and U1. Then in any non-equilibrium state with the same values V1 and U1, the entropy S2 must be less than S1 because observable change always proceeds towards equilibrium, and the condition (1) of observable change at constant U and V requires entropy to increase.

Now imagine a second equilibrium state for which the values of the entropy, volume and internal energy are S2, V1 and U2. Compared with the first equilibrium state the volume is unaltered but S2 < S1. Now U and S always change in the same direction since dU = TdS – PdV implies (∂U/∂S)V = T which is always positive. Hence in the second equilibrium state U2 < U1. Comparing this second equilibrium state with the above-mentioned non-equilibrium state, the entropy and volume are the same but U2 < U1. Since observable change always proceeds towards equilibrium, U will decrease towards a minimum at equilibrium when S and V are held constant.

The foregoing arguments can be summarized in tabulated form as follows:

cse18

6. For given S and P that H is a minimum

Imagine a closed system in some state of equilibrium for which the values of the entropy, pressure and enthalpy are S1, P1 and H1. Then in any non-equilibrium state with the same values P1 and H1, the entropy S2 must be less than S1 because observable change always proceeds towards equilibrium, and the condition (2) of observable change at constant H and P requires entropy to increase.

Now imagine a second equilibrium state for which the values of the entropy, pressure and enthalpy are S2, P1 and H2. Compared with the first equilibrium state the pressure is unaltered but S2 < S1. Now H and S always change in the same direction since dH = TdS + VdP implies (∂H/∂S)P = T which is always positive. Hence in the second equilibrium state H2 < H1. Comparing this second equilibrium state with the above-mentioned non-equilibrium state, the entropy and pressure are the same but H2 < H1. Since observable change always proceeds towards equilibrium, H will decrease towards a minimum at equilibrium when S and P are held constant.

The foregoing arguments can be summarized in tabulated form as follows:

cse19

7. For given A and V that T is a minimum

Imagine a closed system in some state of equilibrium for which the values of the Helmholtz free energy, volume and temperature are A1, V1 and T1. Then in any non-equilibrium state with the same values V1 and T1, the Helmholtz free energy A2 must be greater than A1 because observable change always proceeds towards equilibrium, and the condition (3) of observable change at constant T and V requires the Helmholtz free energy to decrease.

Now imagine a second equilibrium state for which the values of the Helmholtz free energy, volume and temperature are A2, V1 and T2. Compared with the first equilibrium state the volume is unaltered but A2 > A1. Now A and T always change in opposite directions since dA = –SdT – PdV implies (∂A/∂T)V = –S which is always negative. Hence in the second equilibrium state T2 < T1. Comparing this second equilibrium state with the above-mentioned non-equilibrium state, the Helmholtz free energy and volume are the same but T2 < T1. Since observable change always proceeds towards equilibrium, T will decrease towards a minimum at equilibrium when A and V are held constant.

The foregoing arguments can be summarized in tabulated form as follows:

cse20

8. For given G and P that T is a minimum

Imagine a closed system in some state of equilibrium for which the values of the Gibbs free energy, pressure and temperature are G1, P1 and T1. Then in any non-equilibrium state with the same values P1 and T1, the Gibbs free energy G2 must be greater than G1 because observable change always proceeds towards equilibrium, and the condition (4) of observable change at constant T and P requires the Gibbs free energy to decrease.

Now imagine a second equilibrium state for which the values of the Gibbs free energy, pressure and temperature are G2, P1 and T2. Compared with the first equilibrium state the pressure is unaltered but G2 > G1. Now G and T always change in opposite directions since dG = VdP – SdT implies (∂G/∂T)P = –S which is always negative. Hence in the second equilibrium state T2 < T1. Comparing this second equilibrium state with the above-mentioned non-equilibrium state, the Gibbs free energy and pressure are the same but T2 < T1. Since observable change always proceeds towards equilibrium, T will decrease towards a minimum at equilibrium when G and P are held constant.

The foregoing arguments can be summarized in tabulated form as follows:

cse21

9. For given U and S that V is a minimum

Imagine a closed system in some state of equilibrium for which the values of the internal energy, entropy and volume are U1, S1 and V1. Then in any non-equilibrium state with the same values S1 and V1, the internal energy U2 must be greater than U1 because observable change always proceeds towards equilibrium, and the condition (5) of observable change at constant S and V requires the internal energy to decrease.

Now imagine a second equilibrium state for which the values of the internal energy, entropy and volume are U2, S1 and V2. Compared with the first equilibrium state the entropy is unaltered but U2>U1. Now U and V always change in opposite directions since dU = TdS – PdV implies (∂U/∂V)P = –P which is always negative. Hence in the second equilibrium state V2 < V1. Comparing this second equilibrium state with the above-mentioned non-equilibrium state, the internal energy and entropy are the same but V2 < V1. Since observable change always proceeds towards equilibrium, V will decrease towards a minimum at equilibrium when U and S are held constant.

The foregoing arguments can be summarized in tabulated form as follows:

cse22

10. For given H and S that P is a maximum

Imagine a closed system in some state of equilibrium for which the values of the enthalpy, entropy and pressure are H1, S1 and P1. Then in any non-equilibrium state with the same values S1 and P1, the enthalpy H2 must be greater than H1 because observable change always proceeds towards equilibrium, and the condition (6) of observable change at constant S and P requires enthalpy to decrease.

Now imagine a second equilibrium state for which the values of the enthalpy, entropy and pressure are H2, S1 and P2. Compared with the first equilibrium state the entropy is unaltered but H2 > H1. Now H and P always change in the same direction since dH = TdS + VdP implies (∂H/∂P)S = V which is always positive. Hence in the second equilibrium state P2 > P1. Comparing this second equilibrium state with the above-mentioned non-equilibrium state, the enthalpy and entropy are the same but P2 > P1. Since observable change always proceeds towards equilibrium, P will increase towards a maximum at equilibrium when H and S are held constant.

The foregoing arguments can be summarized in tabulated form as follows:

cse23

11. For given A and T that V is a minimum

Imagine a closed system in some state of equilibrium for which the values of the Helmholtz free energy, temperature and volume are A1, T1 and V1. Then in any non-equilibrium state with the same values T1 and V1, the Helmholtz free energy A2 must be greater than A1 because observable change always proceeds towards equilibrium, and the condition (3) of observable change at constant T and V requires the Helmholtz free energy to decrease.

Now imagine a second equilibrium state for which the values of the Helmholtz free energy, temperature and volume are A2, T1 and V2. Compared with the first equilibrium state the temperature is unaltered but A2 > A1. Now A and V always change in opposite directions since dA = –SdT – PdV implies (∂A/∂V)T = –P which is always negative. Hence in the second equilibrium state V2 < V1. Comparing this second equilibrium state with the above-mentioned non-equilibrium state, the Helmholtz free energy and temperature are the same but V2 < V1. Since observable change always proceeds towards equilibrium, V will decrease towards a minimum at equilibrium when A and T are held constant.

The foregoing arguments can be summarized in tabulated form as follows:

cse24

12. For given G and T that P is a maximum

Imagine a closed system in some state of equilibrium for which the values of the Gibbs free energy, temperature and pressure are G1, T1 and P1. Then in any non-equilibrium state with the same values T1 and P1, the Gibbs free energy G2 must be greater than G1 because observable change always proceeds towards equilibrium, and the condition (4) of observable change at constant T and P requires the Gibbs free energy to decrease.

Now imagine a second equilibrium state for which the values of the Gibbs free energy, temperature and pressure are G2, T1 and P2. Compared with the first equilibrium state the temperature is unaltered but G2 > G1. Now G and P always change in the same direction since dG = VdP – SdT implies (∂G/∂P)T = V which is always positive. Hence in the second equilibrium state P2 > P1. Comparing this second equilibrium state with the above-mentioned non-equilibrium state, the Gibbs free energy and temperature are the same but P2 > P1. Since observable change always proceeds towards equilibrium, P will increase towards a maximum at equilibrium when G and T are held constant.

The foregoing arguments can be summarized in tabulated form as follows:

cse25

– – – –

rev01

Reversible change is a key concept in classical thermodynamics. It is important to understand what is meant by the term as it is closely allied to other important concepts such as equilibrium and entropy. But reversible change is not an easy idea to grasp – it helps to be able to visualize it.

Reversibility and mechanical systems

The simple mechanical system pictured above provides a useful starting point. The aim of the experiment is to see how much weight can be lifted by the fixed weight M1. Experience tells us that if a small weight M2 is attached – as shown on the left – then M1 will fall fast while M2 is pulled upwards at the same speed.

Experience also tells us that as the weight of M2 is increased, the lifting speed will decrease until a limit is reached when the weight difference between M2 and M1 becomes vanishingly small and the pulley moves infinitely slowly, as shown on the right.

We now ask the question – Under what circumstances does M1 do the maximum lifting work? Clearly the answer is visualized on the right, where the lifted weight M2 is as close as we can imagine to the weight of M1. In this situation the pulley moves infinitely slowly (like a nanometer in a zillion years!) and is indistinguishable from being at rest.

This state of being as close to equilibrium as we can possibly imagine is the condition of reversible change, where the infinitely slow lifting motion could be reversed by an infinitely small nudge in the opposite direction.

From this simple mechanical experiment we can draw an important conclusion: the work done under reversible conditions is the maximum work that the system can do.

Any other conditions i.e. when the pulley moves with finite, observable speed, are irreversible and the work done is less than the maximum work.

The irreversibility is explained by the fact that observable change inevitably involves some dissipation of energy, making it impossible to reverse the change and exactly restore the initial state of the system and surroundings.

– – – –

Reversibility and thermodynamic systems

The work-producing system so far considered has been purely mechanical – a pulley and weights. Thermodynamic systems produce work through different means such as temperature and pressure differences, but however the work is produced, the work done under reversible conditions is always the maximum work that a system can do.

In thermodynamic systems, heat q and work w are connected by the first law relationship

rev02

What this equation tells us is that for a given change in internal energy (ΔU), both the heat absorbed and the work done in a reversible change are the maximum possible. The corresponding irreversible process absorbs less heat and does less work.

It helps to think of this in simple numbers. U is a state function and therefore ΔU is a fixed amount regardless of the way the change is carried out. Say ΔU = 2 units and the reversible work w = 4 units. The heat q absorbed in this reversible change is therefore 6 units. These must be the maximum values of w and q, because ΔU is fixed at 2; for any other change than reversible change, w is less than 4 and so q is less than 6.

For an infinitesimal change, the inequality in relation to q can be written

rev03

and so for a change at temperature T

rev04

The term on the left defines the change in the state function entropy

rev05

Since reversible conditions equate to equilibrium and irreversible conditions equate to observable change, it follows that

rev06

These criteria are fundamental. They are true for all thermodynamic processes, subject only to the restriction that the system is a closed one i.e. there is no mass transfer between system and surroundings. It is from these expressions that the conclusion can be drawn – as famously stated by Clausius – that entropy increases towards a maximum in isolated systems.

Rudolf Clausius (1822-1888)

Rudolf Clausius (1822-1888)

– – – –

Die Entropie der Welt strebt einem Maximum zu

Consider an adiabatic change in a closed system: dq = 0 so the above criteria for equilibrium and observable change become dS = 0 and dS > 0 respectively. If the volume is also kept constant during the change, it follows from the first law that dU = 0. In other words the volume and internal energy of the system are constant and so the system is isolated, with no energy or mass transfer between system and surroundings.

Under these circumstances the direction of observable change is such that entropy increases towards a maximum; when there is equilibrium, the entropy is constant. The criteria for these conditions may be expressed as follows

rev07

Note:
The assertion that entropy increases towards a maximum is true only under the restricted conditions of constant U and V. Such statements as “the entropy of the universe tends to a maximum” therefore depend on assumptions, such as a non-expanding universe, that are not known to be fulfilled.

– – – –