EQUILIBRIUM

1. EQUILIBRIUM MATERIALS SCIENCE & ENGINEERING Part of A Learner’s Guide AN INTRODUCTORY E-BOOK Anandh Subramaniam & Kantesh Balani Materials Science and Engineering (MSE) Indian Institute of Technology, Kanpur- 208016 Email:anandh@iitk.ac.in, URL:home.iitk.ac.in/~anandh http://home.iitk.ac.in/~anandh/E-book.htm Introduction to Thermodynamics of Materials David R Gaskell Taylor & Francis, New York (2003). Thermodynamics and an Introduction to Thermostatics Herbert B Callen John Wiley and Sons, New York (2006). Recommended website http://hyperphysics.phy-astr.gsu.edu/hbase/heacon.html#heacon

2. What will you learn in this chapter? The fields of Thermodynamics and Kinetics are vast oceans and Chapter 2 will introduce the bare essentials required to understand the remaining chapters. • Stable, Metastable, Unstable & Neutral equilibrium states • Thermodynamic variables and potentials In this text only some aspects will be dealt with, readers may consult standard texts on thermodynamics/thermostatics for a other aspects/detailed account.

3. Motivation These slides are intended to set the stage for understanding the ‘purpose’ and ‘power’ of thermodynamics and its ‘quantities’. Gibbs Free Energy (G) and Entropy (S) will be in special focus. 1 • Let us start by performing the following (thought) experiment:Heat a rod of Al from room temperature to 500C  As expected the rod will expand (A → B in figure below). • The expansion occurs because of two reasons: 1 Vibration of atoms (leading to an increase in average spacing between atoms→ the usual reason)(A → M in figure below). • 2 Increase in the concentration of vacancies* (a vacancy is created when a Al atom goes to the surface and for every 4 vacancies created the volume equal to 1 unit cell is added). (M → B in figure below). • The 2nd reason is of subtler origin and must be surprising to many readers. Additionally, it is a smaller effect in terms of its contribution to the overall increase in length of the specimen (see solved example link below- it is about 1 in 10000 effect). Click here for solved example Metal expands on heating due to 2 different physical reasons! * It costs energy for the system to put vacancies (broken bonds, distortion to the lattice) then why does the system tolerate vacancies?

4. Now let us perform another (thought) experiment to put in perspective the previous experiment:Heat a elastomer (cut rubber band) which has been stretched by a small weight by about 20C (room temperature + 20C)  the stretched rubber band will contract! • The 2nd reason for the expansion of the Al rod is closely related to the contraction of the stretched rubber band!  occurs because of thermodynamic reasons (quantities like Gibbs Free Energy (G) and Entropy (S)), which we shall learn in this chapter. • In the case of the heating of the Al rod- “how the vacancies form” is an issue of kinetics. Kinetics will be dealt with in the topic of kinetics and chapter on Diffusion. A ‘stretched’ elastomer contracts on heating!

5. Motivation 2 • Let us next consider the melting of a pure metal at its melting point (MP) (at constant T and P) → by supplying heat to the sample of metal (so that the metal sample is only partly molten). At the MP the liquid metal is in equilibrium with the solid metal. • The liquid has higher potential energy as well as higher kinetic energy than the solid. • Then why does the liquid co-exist with the solid? • The answer to this question lies in the fact that internal energy is not the measure of stability of the system (under the circumstances). • We will learn in this chapter that it is the Gibbs Free Energy (G). The molten metal has higher energy (internal energy and enthalpy), but also higher Entropy. So the melting is driven by an increase in Entropy of the system. The molten metal and the crystalline solid metal have the same G → hence they co-exist in equilibrium.

6. Stability and Equilibrium • Equilibriumrefers to a state  wherein there is a balance of ‘forces’*(as we shall see equilibrium points have zero slope in a energy-parameter plot) • Stability relates to perturbations (usually small perturbations** about an equilibrium state) (as we shall see stable relates to the curvature at the equilibrium points). * Force has been used here in a generalized sense (as an agent which can cause changes) ** Perturbation is usually a small ‘force/displacement’ imposed in a short span of time.

7. Equilibrium in a Mechanical System • Let us start with a simple mechanical system → a rectangular block (Figure in next slide)(under an uniform gravitational potential). • The potential energy (PE) of the system depends on the height of the centre of gravity (CG). • The system has higher PE when it rests on face-A, than when it rests on face-B. • The PE of the system increases when one tilts it from C1 → C2 configuration. • In configurations such as C1,C2 & C3 the system will be in equilibrium (i.e. will not change its configuration if there are no perturbations). • In configuration C2 the system has the highest energy (point B) and any small perturbations to the system will take it downhill in energy → Unstable state. • Configuration C3 has the lowest energy (point C) and the system will return to this state if there are small perturbations → the Stable state. • Configuration C1 also lies in an ‘energy well’ (like point C) and small perturbations will tend to bring back the system to state C1. However this state is not the ‘global energy minimum and hence is called a Metastable state. • Additionally, one can visualize a state of neutral equilibrium, like a ball on a plane (wherein the system is in a constant energy state with respect to configurations).

8. Mechanical Equilibrium of a Rectangular Block We start by considering the mechanical equilibrium of a block- this is to get a first feel- additional concepts will be required when dealing with condensed matter systems. A Ball on a plane Neutral Equilibrium B Centre Of Gravity C2 C3 C1 B Unstable Potential Energy = f(height of CG) Stable A Lowest CG of all possible states Metastable state C Configuration

9. Points to be noted: A system can exist in many states (as seen even for a simple mechanical system: block on a plane) These states could be stable, metastable or unstable Using the relevant (thermodynamic) potential the stability of the system can be characterized(In the case of the block it is the potential energy, measured by the height of the CG for the case of the block on the plane) System will ‘evolve’ towards the stable state provided ‘sufficient activation’ is provided(in the current example the system will go from C1 to C3 by ‘sufficient jolting/shaking’ of the plane)

10. Kinds of Stability (Equilibrium) Three kinds of equilibrium (with respect to energy) • Global minimum → STABLE STATE • Local minimum → METASTABLE STATE • Maximum → UNSTABLE STATE • Constant energy → Neutral State/Equilibrium Also next slide • Kind of equilibrium can be understood by making perturbations to the system • For the mechanical system (block) this corresponds to tilting the block • If the system changes its state after small perturbations then the system→ is in an unstable state • If the system returns to its original state after a small perturbation (tilt) then the system → is in a stable or metastable state (lies in an energy minimum) • If the system returns to its original position after small perturbations but does not do so for large perturbations then the system → is in a metastable state (not in the global energy minimum) • If there is no change in energy for any kind of perturbation then the system → is in a state of neutral equilibrium (e.g. the case of the ball on a plane) • In a 2D system where perturbations are possible in more than one direction (i.e. the energy landscape is a surface), perturbations in one direction may be stable and in another direction it may be unstable (like on a surface with negative Gaussian Curvature)

11. If the system tends to return to the original state after the perturbation Stable/Metastable state In Metastable state the system goes to a new state if the amplitude of the perturbation is large. • If the system goes to a different state on perturbation (the perturbation will tend to take the system to a new state ‘very different’ from the original state) Unstable state • If the system goes to the ‘new perturbed state’ without a change in energy (the perturbation will tend to take the system to a new state ‘close to’ the original state, as imposed by the perturbation)  Neutral state

12. Condensed Matter systems • In Materials Science we are mainly interested with condensed matter systems (solids and liquids) (also sometimes with gases) • The state of such a system is determined by ‘Potentials’ analogous to the potential energy of the block (which is determined by the centre of gravity (CG) of the block).These potentials are the Thermodynamic Potentials (A thermodynamic potential is a Scalar Potential to represent the thermodynamic state of the system). • The relevant potential depends on the ‘parameters’ which are being held constant and the parameters which are allowed to change. More technically these are the State/Thermodynamic Variables (A state variable is a precisely measurable physical property which characterizes the state of the system- It does not matter as to how the system reached that state). Pressure (P), Volume (V), Temperature (T), Entropy (S) are examples of state variables. • There are 4 important potentials (in some sense of equal stature). These are: Internal Energy (U or E), Enthalpy (H), Gibbs Free Energy (G), Helmholtz Free Energy (A or F). Of these internal energy can be conceived as a fundamental quantity, while enthalpy a quantity keep track of the available heat. G and A arise due to the fact that the entropy of the universe increases.****

13. Intensive and Extensive Properties • Intensive properties are those which are independent of the size of the system P, T • Extensive Properties are dependent on the quantity of material V, E, H, S, G

14. Thermodynamic potentials and the relation between them • There are 4 important potentials (in some sense of equal stature). These are: Internal Energy, Enthalpy, Gibbs Free Energy, Helmholtz Free Energy • The relation between these potentials and the state variables is as below.  TS +PV

15. The relevant thermodynamic potential determining the stability of the system • In terms of the stimulus and response*: P → V (Pressure difference drives volume changes) T → S (Temperature difference drives entropy changes)  → Ni (Chemical potential difference drives mass transfer) (subscript i stands for the ith species) • The relevant thermodynamic potential which characterizes the stability of the system is dependent on the state variables which are held constant (as in the table below). • The most conditions are one of Constant P and Constant T→ Hence G (Gibbs Free Energy) is the most relevant thermodynamic potential. • However, we should remember that depending on the state variables being held constant any one of the four potentials (U, H, F, G) could be the relevant potential. (i.e. all of the potentials have the same ‘stature’). * The main response.

16. 1 Internal Energy (U) • Internal Energy (U or E) = Kinetic Energy (KE) + Potential Energy (PE). • The origin of Kinetic Energy → Translations, Rotations, Vibrations. • The origin of Potential Energy → Bonding between atoms (interactions in the solid). • Internal energy is the total of the energies of molecules (or atoms or ions..) and their interactions. • The increase in internal energy on heating from 0 to T Kelvin is given by the equation below; where CV is the specific heat at constant volume and E0 is the internal energy of the system at 0K.

17. 2 Enthalpy (H) H = U + PV • Enthalpy (H) = Internal Energy + PV (work done by the system). • Measure of the heat content of the system. • At constant pressure the heat absorbed or evolved is given by H. • Transformation / reaction will lead to change of enthalpy of system. • Enthalpy is also known as the ‘heat content’ of the system. • Why do we need a quantity like enthalpy?Suppose we burn a fuel (say petrol) in open air all the energy of the combustion is not available to us, as some energy is consumed in ‘making space’ for the products of the combustion. I.e. work is done by the system against the surrounding (atmosphere at 1atm pressure), to accommodate the products of combustion (and this work is not available to us for ‘useful’ purposes). If the combustion were to be carried out in a rigid (closed) container, then no work of expansion would have to be accounted for. • So, if internal energy (U) is the quantity of energy in a system (and say U is being made available during an experiment), then H is a measure of the part available to us (some gets ‘wasted’ in accommodating the products of the reaction→ i.e. out of the U only H is available to us). This is true only if the ‘entropy’ is constant (else we will have to get the amount available from the free energies (G or A)).

18. Gaseous state is considered as the reference state with no interactions. • For condensed phases PV << E  H ~ E. • The increase in enthalpy on heating from 0 to T Kelvin is given by the equation below; where CP is the specific heat at constant pressure and H0 is the internal energy of the system at 0K (H0 represents energy released when atoms are brought together from the gaseous state to form a solid at zero Kelvin). • Enthalpy is usually measured by setting H = 0 for a pure element in its stable state at 298 K (RT).

19. Entropy (S) What is entropy a measure of? (We will take this up later). • Entropy is perhaps one of the most profound and subtle concepts of nature. • Entropy can be understood looking at a Macroscopic picture (interpretation) or a Microscopic picture (interpretation) (next slide). • Though these are different approaches to understand entropy– the result is the ‘same’ entropy. • In the Macroscopic view we ‘work’ at the system level and worry about observable average quantities. In the Microscopic view we go into all the ‘details’ about the system. • The entropy of an isolated system will increase in a spontaneous process (cannot spontaneously decrease)→ II law. • Since, the above can be applied to the entire universe (a large isolated system!**)→ the entropy of the Universe will always increase (here we are not using the word ‘spontaneous’ as the Universe cannot be ‘coaxed’ into doing a non-spontaneous processes!). • The microscopic interpretation (view) is the Statistical Physics/Mechanics picture, which is valid for large systems (i.e. systems with a large collections of atoms, molecules etc.). • “Entropy is time’s arrow” → time increases in the direction of increasing entropy*. • For a system the Internal energy (U) measures the quantity of the energy, while Entropy (S) can be thought of a measure of the quality of that energy (lower value of entropy will imply a better quality→ as the entropy of the universe always* increases). *The universe is in a expanding phase now. If it were to stop expanding and start contracting → entropy of the universe is expected to decrease in the contracting phase. ** We are ignoring ‘wormholes’, which connect our universe to other parallel universes for now (or take those universes part of ours)!!

20. Warning! Busy slide → take time to understand • Due to Ludwig Boltzmann • Microscopic view • Statistical Physics Definition • Theoretical understanding • Due to Rudolf Clausius • Macroscopic view • ‘Classical Thermodynamic’ definition • Usually what is experimentally measured Entropy Two views and not two types* Thermal Configurational + other** The entropy change of a system at temperature T absorbing* an infinitesimal amount of heat Q in a reversible way, is: Zero or +ve  if heat input is at lower T (as compared to higher T) then for the same heat input (Q), there will be a higher increase in entropy. Boltzmann constant (1.38  1023 J/K) No. of different configurations the system ** For a system at Constant Energy, many microscopic states can give a macroscopic state of identical energy. These microscopic states could originate from various sources like Configurational, Electronic, Vibrational and Rotational (etc.) states.  In many cases the configurational term may be the predominant one considered. • These two descriptions are equivalent. • Sometimes 0K there are some degenerate states → in which case the configurational entropy is non zero (which can be confirmed by experiments, wherein one measures the ‘macroscopic’ entropy). This is called residual entropy. * If heat is desorbed then entropy change will be negative.

21. A note on the equation • We had noted that the Boltzmann constant arises because we try to make a ‘unit’ Kelvin equal to a unit C. • Suppose we measured temperature in terms of ‘’, then entropy would be a pure number.

22. Entropy (S) A simple understanding of entropy • One way of simply stating the ‘concept behind entropy’ is→ A system will, more often than not, be found in ‘states; with higher probability. (This is nothing but the statement of the obvious!) • However, the implications of the above are ‘profound’. This can be best understood by considering the mixing of two ideal gases (or in the ‘toy model’ below as the mixing of 6 circles- 3 red and 3 blue, on 6 fixed lattice sites). • Assuming that red and blue circles can move about randomly on the fixed sites and also assuming that the probability of the occurrence of each state is identical (i.e. no state is preferred over any other state); there are 20 possible configurations as shown in the next slide. • As seen (from the figure in the next slide) the majority of the states (18/20) are ‘mixed states’ and only two are the ‘totally’ unmixed ones. • Hence, purely from a probabilistic point of view, mixed states occur more often than the unmixed ones. • This implies, if we start with a unmixed configuration as in the figure below and the system can access all possible states with equal probability → the system will go from a unmixed state (of low entropy) to a mixed state (of higher entropy). Two boxes separated by a barrier initially A B Unmixed state

23. Unmixed state Mixed states with ‘various’ degrees of mixing • In the case of two gases initially separated by a barrier, which are allowed to mix on the removal of the barrier: the number of mixed states are very large compared to the unmixed states. Hence, if all configurational states are accessible, the system will more likely found in the mixed state. I.e. the system will go from a unmixed state to a mixed state (worded differently the system will go from ‘order’ to ‘disorder). • On the other hand it is unlikely (improbable) that the system will go from mixed state to a unmixed state. (Though this is not impossible → i.e. a mixed system can spontaneously get ‘unmix’ itself!!) Note: the profoundness of the concept of entropy comes from its connection to free-energy (via T) 18 mixed states 2 unmixed states * We assume that all states have equal probability of occurring and are all accessible Unmixed state

24. What is entropy a measure of? • In literature entropy is considered to be a measure of: the degree of ‘randomness’ in a system the ‘uncertainty’ about the system the ‘degeneracy’ of the system. • The terms randomness, uncertainty, etc. have other connotations in other contexts, which can lead to confusion to the reader. To avoid this we simply state that entropy is a measure based on the multiplicity of the system (of microstates of a system, corresponding to a macrostate). • If a system can exist in multiple states (corresponding to a particular macrostate), then this leads to an increased entropy. For a gas (with non-interacting atoms) some of these states could be ordered as below (State-A); but more likely these will be some ‘random disordered state’ (State-B). Note that the number of ‘disordered’ states is overwhelmingly large. State-A State-B Gas atoms in a box arriving at a regular configuration by chance. Gas atoms in a box usual (disordered) state we expect.

25. Calculating Configurational Entropy • In the microscopic (statistical mechanics) interpretation of Entropy, we take into account the multiplicity of microstates which give rise to a macrostate • For example for a constant energy system (with a constant number of particles/species and volume), the total energy of the system may be obtained due to a multiplicity of microstates originating from various sources like Configurational, Electronic, Vibrational and Rotational states. • If we consider only configurational entropy for now- the multiplicity of states can arise from various configurations of the atomic species. E.g. If we are talking about a pure crystal of A with just one B atom, this B atom could be in any one of the lattice positions (all of identical energy) giving rise to a multiplicity in the microstates. (for now we ignore the surface states) • Note again that for statistical mechanics to be valid we have to deal with large systems (for illustration purposes we often draw small systems!) Then the configurational entropy is given by Of course there are many more such configurations- 3 are drawn for illustration 3 configurations of equal energy Zero or +ve Boltzmann constant (1.38  1023 J/K) No. of different configurations of equal energy for a constant energy system

26. Entropy change due to mixing of two pure elements • Let us consider the entropy change due to mixing of two pure crystalline elements A & B(a simple case for illustration of the concept of entropy). • The unmixed state is two pure elements held separately. The mixed state (for now assuming that the enthalpy of mixing is negative- i.e. the elements want to mix) represents an atomic level mixing of the two elements. • Let the total number of lattice sites (all equivalent) be N. • The Entropy of the unmixed state is zero (as in pure crystalline elements atoms are indistinguishable and hence represent one state). Spure A = Spure B = k ln(1) =0 • In the mixed state the entropy of the system increases (Smixed state) • The number of permutations possible in the mixed system is  Zero An useful formula for evaluating ln(factorials) is the Stirling’s approximation: ~ asymptotically equal, e = 2.718…

27. Entropy change during melting • At the melting point of a material when heat is supplied (Q) to the material it does not lead to an increase in the temperature. Instead, the absorbed heat leads to melting i.e. the energy goes into breaking of bonds in the solid and consequently a transformation in the state of the material (solid → liquid). The entire process of melting takes place at a constant temperature (Tm). The heat absorbed is called the Latent Heat of Fusion (Hfusion). • Suppose we take a mole of Al atoms melt then the change in entropy can be calculated as below. • In the solid state the atoms are fixed on a lattice (of course with vibrations!) and this represents a ‘low entropy’ state. On melting the entropy of the system increases as the atoms are free to move around and many configurations are possible. From this point of view often Entropy is considered as a measure of disorder (however, it must be clear that the phrase ‘measure of disorder’ is used with the understanding of the context) Data: Enthalpy of fusion (Hf) = 10.67 kJ/mole, Melting Point (Tm) = 933.4 K (660.25C)

28. What is the entropy of mixing 0.5 mole of A with 0.5 mole of B on a mole of lattice sites (N0 sites)? Solved Example Data: Using:

29. ‘Free’ Energies • We will take up a brief discussion about ‘free’ energies before considering G and A. • We all know that there is ‘no free lunch’ so what is meant by ‘free’ energy? It is the energy which is available to do work.From the second law we know that all the available energy cannot be converted into work  some ‘has to be lost’ as heat (to the surrounding/sink). You cannot keep all your earnings some has to be paid as tax (and we are not even including ‘hafta’ here!! • We have noted that all the internal energy change may not be available as heat for us (some may be used up to do work to accommodate the products of say a combustion process) and we introduced the concept of enthalpy for this. • Needless to say, only a spontaneous process can be used for extracting work out of a system. (For a driven process we have to supply work!). • In a spontaneous process the entropy of the universe has to increase. So: (Ssystem+Ssurrounding) > 0. (E.g. If heat eneters the system from the surrounding then Ssystem will be positive, while Ssurrounding will be negative). • To decide the direction of ‘spontaneity of a process’ statement  is enough → but then we have to do two separate computations (one for the system and one for the surrounding). • If we could come up with a quantity, which helps us do calculations only on the system this will help. Further if the quantity is a state function, then I do not have to ‘worry’ about the reversibility of the heat transfer (remember: S = Qreverisble/T).

30. ‘Free’ Energies cotd… • If entropy is constant in a process, then the direction (condition) for spontaneity is obvious → it is the direction in which the internal energy (U) or enthalpy (H) decreases. It is the ‘entropy tax’ which is making things difficult. • Under two conditions this is possible. If the process is carried out under constant T, V. If the process is carried out under constant T, P. •  At constant T,V (let Q enter the system):Condition of spontaneity: At constant V there can be no expansion work and if there is no other form of work (magnetic, electrical etc.) then this energy will lead to an increase in the internal energy of the system (U). • Note that we do not have to write Ureversible as U is a state function (depends only on the initial and final states). • Suppose we introduce a symbol A for UTS. • For a spontaneous process at constant T, V: • A is called the Helmholtz Free Energy → it is a thermodynamic potential (and a state function as U and S are state functions). Hence, if we know the initial and final values of A → we can conclude if the process is spontaneous (at constant T,V). Rigid wallsV constant Heat flow direction T Q T or equivalently Zero at constant T

31.  At constant T,P (let Q enter the system):Condition of spontaneity: At constant P, if there is only expansion work (no other form of work: magnetic, electrical etc.) then this energy will lead to an increase in the internal energy of the system (H). • Note that we do not have to write Hreversible as H is a state function (depends only on the initial and final states). • Suppose we introduce a symbol G for HTS. • For a spontaneous process at constant T, H: • A is called the Gibbs Free Energy → it is a thermodynamic potential (and a state function as H and S are state functions). Hence, if we know the initial and final values of G → we can conclude if the process is spontaneous (at constant T,P). or equivalently Zero at constant T

32. Helmholtz Free Energy (A) 3 • Helmholtz Free Energy (A or F) = E  T.S • S is the entropy of the system • At constant V & T, for a process/reaction* to take place spontaneously the system has to reduce its Helmholtz Free Energy. For a system to go from ‘state*’ 1 → 2 the change in F would be:F2 F1 = F = (E2  E1)  T (S2  S1) = E  TSThis change of ‘state’ would take place spontaneously if F is Negative • This implies that reactions which lead to an increase in the internal energy (E) are allowed (at a ‘sufficiently high’ temperature) if there is a Entropic benefit for the process to occur(the concept of entropy will be dealt with in the context of Gibbs Free Energy) A = E  T S * used in the general sense

33. Gibbs Free Energy (G) 4 And the concept of Entropy • Gibbs Free Energy (G) = H  T.S S is the entropy of the system • For a process/reaction* to take place spontaneously the system has to reduce its Gibbs Free Energy (at constant P & T). For a system to go from ‘state*’ 1 → 2 the change in G would be:G2 G1 = G = (H2  H1)  T (S2  S1) = H  TSThis change of ‘state’ would take place spontaneously if G is Negative • This implies that even Endothermic reactions are allowed (at a ‘sufficiently high’ temperature) if there is a Entropic benefit (weighed in with T) for the process to occur • An example of the above is the presence of (‘equilibrium concentration’ of) vacancies in a crystal (more about this later) • Many a times we are concerned with the relative stability of two phases at a given T and P. We asks questions such as at 1 atm. pressure & 50C, which of the phases is stable: ice, water or steam? (Answer can be found later in the slides). G = H T.S G = H  T S * used in the general sense

34. The Slope of the G vs T curve at 0K is zero. • The curvature is always negative. Melting transition

36. Q & A At 1atm. pressure & 50C, which of the phases is stable: ice, water or steam?