Basics of Thermodynamics Some of the material covered here is also covered in the chapter/topic on: Equilibrium MATERIALS SCIENCE & A Learner’s Guide ENGINEERING Part of Instructors/students may download the appropriate files and delete the portion not needed. This will help tailor the contents for any specific syllabus or need. (I.e. you may copy left, right and centre!!). Reading 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 Four Laws that Drive the Universe Peter Atkins* Oxford University Press, Oxford, 2007 Physical Chemistry Ira N Levine Tata McGraw Hill Education Pvt. Ltd., New York (2002). Web resource: https://www.khanacademy.org/science/physics/thermodynamics *It is impossible for me to write better than Atkins- his lucid (& humorous) writing style is truly impressive- paraphrasing may lead to loss of the beauty of his statements- hence, some parts are quoted directly from his works. Thermodynamics versus Kinetics Thermodynamics deals with stability of systems. It tells us ‘what should happen?’. ‘Will it actually happen(?)’ is not the domain of thermodynamics and falls under the realm of kinetics. At –5C at 1 atm pressure, ice is more stable then water. Suppose we cool water to –5C. “Will this water freeze?” (& “how long will it take for it to freeze?”) is (are) not questions addressed by thermodynamics. Systems can remain in metastable state for a ‘long-time’. Window pane glass is metastable– but it may take geological time scales for it to crystallize! At room temperature and atmospheric pressure, graphite is more stable then diamond– but we may not lose the glitter of diamond practically forever! * The term metastable is defined in the chapter on equilibrium. Thermodynamics (TD): perhaps the most basic science One branch of knowledge that all engineers and scientists must have a grasp of (to some extent or the other!) is thermodynamics. In some sense thermodynamics is perhaps the ‘most abstract subject’ and a student can often find it very confusing if not ‘motivated’ strongly enough. Thermodynamics can be considered as a ‘system level’ science- i.e. it deals with descriptions of the whole system and not with interactions (say) at the level of individual particles. I.e. it deals with quantities (like T,P) averaged over a large collection of entities (like molecules, atoms)*. This implies that questions like: “What is the temperature or entropy of an atom?”; do not make sense in the context of thermodynamics (at lease in the usual way!). TD puts before us some fundamental laws which are universal** in nature (and hence applicable to fields across disciplines). TD parameters are measureable macrocopic quantities, which are characterize (/associated with) the system. These include: P, T, V, H (magnetic field). (Note: non-bolded H will be used for enthalpy) * Thermodynamics deals with spatio-temporally averaged quantities. ** They apply to the universe a whole as well! (Though the proof is lacking!). The language of TD To understand the laws of thermodynamics and how they work, first we need to get the terminology right. Some of the terms may look familiar (as they are used in everyday language as well)- but their meanings are more ‘technical’ and ‘precise’, when used in TD and hence we should not use them ‘casually’. System is region where we focus our attention (Au block in figure). A TD system is a macroscopic system. Surrounding is the rest of the universe (the water bath at constant ‘temperature’). Universe = System + Surrounding (the part that is within the dotted line box in the figure below) More practically, we can consider the ‘Surrounding’ as the immediate neighborhood of the system (the part of the universe at large, with which the system ‘effectively’ interacts). In this scheme of things we can visualize: a system, the surrounding and the universe at large. Things that matter for the surrounding: (i) T, (ii) P, (iii) ability to: do work, transfer heat, transfer matter, etc. Parameters for the system: (i) Internal energy, (ii) Enthalpy, (iii) T, (iv) P, (v) mass, etc. The surrounding does not change in any way during any process that the system undergoes (i.e. its T, P, etc. remain the same). I.e. the surrounding is not transmutable. In TD we usually do not worry about the universe at large! Open, closed and isolated systems To a thermodynamic system two ‘things’ may be added/removed: energy (in the form of heat &/or work) matter. An open system is one to which you can add/remove matter (e.g. a open beaker to which we can add water). When you add matter- you also end up adding heat (which is contained in that matter). A system to which you cannot add matter is called closed. Though you cannot add/remove matter to a closed system, you can still add/remove heat (you can cool a closed water bottle in fridge). A system to which neither matter nor heat can be added/removed is called isolated. A closed vacuum ‘thermos’ flask can be considered as isolated. Type of boundary Interactions Open All interactions possible (Mass, Work, Heat) Closed Matter cannot enter or leave Semi-permeable Only certain species can enter or leave Insulated Heat cannot enter or leave Rigid Mechanical work cannot be done* Isolated No interactions are possible** * By or on the system ** Mass, Heat or Work Mass Interactions possible Work Heat Matter is easy to understand and includes atoms, ions, electrons, etc. Energy may be transferred (‘added’) to the system as heat, electromagnetic radiation etc. In TD the two modes of transfer of energy to the system considered are Heat and Work. Heat and work are modes of transfer of energy and not ‘energy’ itself. Once inside the system, the part which came via work and the part which came via heat, cannot be distinguished*. More sooner on this! Before the start of the process and after the process is completed, the terms heat and work are not relevant. From the above it is clear that, bodies contain internal energy and not heat (nor work!). Matter when added to a system brings along with it some energy. The ‘energy density’ (energy per unit mass or energy per unit volume) in the incoming matter may be higher or lower than the matter already present in the system. * The analogy usually given is that of depositing a cheque versus a draft in a bank. Once credited to an account, cheque and draft have no meaning. (Also reiterated later). Variables in a TD system The TD state is specified by a set of values of all the TD parameters required for the description of the system. The state of a system is determined by ‘Potentials’, which is analogous to the potential energy of the block under gravity (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). 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). Macroscopic and Microscopic Variables Macroscopic Variables The macroscopic variables defining a state are the State or 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. * To be discussed later Microscopic Variables In addition, we can have microscopic variables associated with a system. I.e. these are associated with the description of the system, via states of individual particles (like position, velocity, kinetic energy, etc.). These variable change continuously, even for a system in equilibrium (and hence are typically not considered in classical thermodynamics). The macroscopic variables are defined only under equilibrium conditions (or during a quasistatic process*). The state variables like T & P are not defined during ‘transients’ (transient states of the system). However, even during transients the microscopic variables are well defined (but changing in value continuously). Thermodynamic Equilibrium* & Thermodynamic Transformation If the TD state of a system does not change with time, then the system is in TD equilibrium. Often the term state in TD implies a state in equilibrium. A TD Transformation is a change of the state of a TD system. Equation of State** Is a functional relation between the TD parameters of the system in equilibrium. If P, V, T are TD parameters of the system, the equation of state can be written as: f ( P,V , T ) 0 The existence of a such a relation reduces the number of independent variable by one. A state of the system is a point in the P-V-T space. The equation of state gives us a surface in the P-V-T space and any point on the surface is state in equilibrium. * Much more on the chapter on equilibrium. ** Will learn a lot more about this later. Q&A What is meant by microscopic in the context of thermodynamics? In TD we often take recourse to macroscopic and microscopic viewpoints. E.g. if we think of an gas, we can visualize ‘T’ as the parameter, which drives heat transfer between two bodies (heat flows from high-T to low-T). This is the macroscopic picture. In the microscopic picture (which is arises from statistical TD), we visualize energy levels available for the species to populate and the distribution of species across these levels. Similarly, we can think of pressure macroscopically as the causative agent for driving the piston (direction from high-P to low-P). Microscopically, it is the momentum transferred per unit area per unit time by the species of the medium (e.g. gas molecules). When we talk about entropy, again we invoke the microscopic and macroscopic pictures. Macroscopically, we sit at the system boundary and track heat transfer (Qrev) & S = Qrev/T. Microscopically, we ‘worry’ about the species occupying (e.g.) certain configurations (microstates). Hence, in TD micro-scopic does not concern with a lengthscale, but with the details. In the microscopic picture, we look at the species comprising of the system, like molecules and track their configurations, energy states they occupy, vibrations, etc. Thermodynamic Process If a system is in a equilibrium state, then a TD transformation can be brought about only by changes in the external conditions (parameters) of the system. I.e., actions of the surrounding can only bring about the change to a system in TD equilibrium. The change from one TD state to another is considered as a process. Different types of Processes in TD We will deal with some of these in detail later on Here is a brief listing of a few kinds of processes, which we will encounter in TD: Isothermal process → the process takes place at constant temperature (e.g. freezing of water to ice at –10C) Isobaric → constant pressure (e.g. heating of water in open air→ under atmospheric pressure) Isochoric → constant volume (e.g. heating of gas in a sealed metal container) Quasi-static process → the process occurs so gradually, that the system is in internal equilibrium throughout the process. The macroscopic state variables (like P & T) are well defined during the process. (e.g. removal of sand, grain by grain, from a piston loaded by the sand) Transient process → the process occurs so fast, that the internal equilibrium is not maintained during the process. The macroscopic state variables (like P & T) are not well defined during the process. This process is not part of the realm of equilibrium TD. (e.g. expansion of a gas from one part of a system to another, when the partition is removed) Reversible process → the system is close to equilibrium at all times (and infinitesimal alteration of the conditions can restore the universe (system + surrounding) to the original state. Most quasi-static processes are reversible. Cyclic process → the final and initial states are the same. However, q and w need not be zero. Adiabatic process → dq is zero during the process (no heat is added/removed to/from the system during the process). A system undergoing an adiabatic process is thermally isolated by adiabatic walls. A combination of the above are also possible: e.g. ‘reversible adiabatic process’. Funda Check What is the relation between ‘quasi-static’ and ‘reversible’ processes? We have noted before that: actions of the surrounding can only bring about the change to a system in TD equilibrium, the change from one TD state to another occurs via a process & during a quasi-static process the system is in internal equilibrium throughout the process. Also we can think of a reversible process as follows. A process is reversible if the transformation retraces path in time, when the external conditions retraces its path in time. E.g. if the external pressure is increased in steps of P each time, the piston will move in & pressure will equilibrate after each step. I.e. after the first step the internal pressure (Pint = Pext = P0 + P). We can conceive a series of similar steps to increase the internal pressure to Pf. Now, if we decrease the external pressure by P, the piston will move out and an equilibrium pressure will be established (Pint = Pext = Pf P). By following such steps the physical and TD path can be retraced. A reversible transformation is quasi-static, but the converse may not be true. 1 Pt=0 = P0 2 Pext = (P0 +P) A reversible process can be shown as a continuous path in the P-V diagram (as we shall see soon). Also, a process which is not reversible cannot be shown as a continuous path in the P-V diagram. Other Processes In chemistry and physics may processes exist. Some of them are listed below. Phase Transitions. In phase transitions the composition does not change. A super set of phase transitions is Phase Transformations. General: α phase → β phase. Fusion: Solid → Liquid. Vaporization: Liquid → Gas. Sublimation: Solid → Gas. Mixing. Pure A + Pure B → Mixture. Solution/dissolution: Solute + Solvent → Solution. Reaction. Reactants → Products. Combustion: Element/Compound + Oxygen → Oxide. Formation. Elements → Compound. Activation. Reactants → Activated complex. State functions in TD A property which depends only on the state of the system (as defined by T, P, V etc.) is called a state function. This does not depend on the path used to reach a particular state. Analogy: one is climbing a hill- the potential energy of the person is measured by the height of his CG from ‘say’ the ground level. If the person is at a height of ‘h’ (at point P), then his potential energy will be mgh, irrespective of the path used by the person to reach the height (paths C1 & C2 will give the same increase in potential energy of mgh- in figure below). In TD this state function is the internal energy (U or E). (Every state of the system can be ascribed to a unique U). Hence, the work needed to move a system from a state of lower internal energy (=UL) to a state of higher internal energy (UH) is (UH) (UL). W = (UH) (UL) The internal energy of an isolated system (which exchages neither heat nor mass) is constant this is one formulation of the first law of TD. A process for which the final and initial states are same is called a cyclic process. For a cyclic process change in a state function is zero. E.g. U(cyclic process) = 0. Q &A When will a process occur? Under equilibrium conditions ‘nothing’ will take place (at least macroscopically). For a process to occur there has to be a ‘causative agent’ (typically of a critical magnitude). The common driving forces are differences in temperature, pressure and chemical potential. One of the processes of interest, which we will deal with repeatedly in TD, is the reversible process; which occurs close to equilibrium conditions. In mechanics these are also referred to as quasi-static processes. Continued on the next slide A spontaneous process is one which occurs ‘naturally’, ‘down-hill’ in energy*. I.e. the process does not require input of work in any form to occur. Melting of ice at 50C is a spontaneous process. A driven process is one which wherein an external agent takes the system uphill in energy (usually by doing work on the system). Freezing of water at 50C is a driven process (you need a refrigerator, wherein electric current does work on the system). Later on we will note that the entropy of the universe will increase during a spontaneous change. (I.e. entropy can be used as a single Spontaneous process (Click to see) parameter for characterizing spontaneity). Spontaneous and Driven processes * The kind of ‘energy’ we are talking about depends on the conditions. As in the topic on Equilibrium, at constant temperature and pressure the relevant TD energy is Gibbs free energy. System and Process Related Complexities Starting with an ideal gas# in a closed system at constant T, P at equilibrium, we can progressively relax the conditions to obtain more ‘realistic’ and complicated systems. A broad picture of these is shown below. We will consider a part of this picture in the current e-book. Currently we restrict ourselves to P-V work only, noting that other types of work are possible. System level complexity Close System Ideal Gas Single phase* T, P Equilibrium Chemical non-equilibrium (Ir-reversible Process) Close System Ideal Gas Single phase* Near Equilibrium (Reversible Process) Material level complexity Close System Ideal Gas Single phase* T, P, Material Equilibrium Close System Real Gas/Solid/Liquid Single phase* Single component T, P, Material Equilibrium Close System Real Gas/Solid/Liquid Two phases* Single component T, P, Material Equilibrium Open System Ideal Gas Single phase* T, P Equilibrium Chemical non-equilibrium (Ir-reversible Process) Close System Real Gass/Solids/Liquids Single phase* Multi-component T, P, Material Equilibrium Close System Real Gass/Solids/Liquids Two phases* Multi-component T, P, Material Equilibrium Open System Ideal Gas Single phase* T, P Equilibrium No (T, P, Chemical) equilibrium (Ir-reversible Process) Various types of complexities can be combined * Gases always form single phase system. # We will deal with ideal gases in detail soon. In the current set of notes we will follow the path as below. Closed system at equilibrium Closed system Reversible process Single phase P-V work only Open System Reversible process Single phase P-V work only Phase Equilibrium Open System Reversible process Multi-phase P-V work only Reaction Equilibrium Funda Check How to understand the ‘macroscopic’ versus ‘microscopic variables? Let us consider a gas expanding from a high pressure chamber (at pressure P1) into a chamber under vacuum (via a nozzle). Let the system be insulated, so that no heat can enter the system. As the gas expands the pressure falls from P1 to a lower value. Let us track the process starting from t = 0 to infinitesimal times. As the gas expands the number of collisions (with each other and a fictitious wall) decreases. The position and velocity (the microscopic variables) of each molecule can be ‘measured/known’ (say on plane AB, Fig. below). However, under these transient conditions macroscopic variables like pressure are not defined (the pressure on the left of AB is different from that on the right of AB) due to the non-equilibrium conditions. Nozzle A P1 Vacuum B Temperature Though we all have a feel for temperature (‘like when we are feeling hot’); in the context of TD temperature is technical term with ‘deep meaning’. As we know (from a commons sense perspective) that temperature is a measure of the ‘intensity of heat’. ‘Heat flows’ (energy is transferred as heat) from a body at higher temperature to one at lower temperature. (Like pressure is a measure of the intensity of ‘force applied by matter’→ matter (for now a fluid) flows from region of higher pressure to lower pressure). That implies (to reiterate the obvious!) if I connect two bodies (A)-one weighing 100 kg at 10C and the other (B) weighing 1 kg at 500C, then the ‘heat will flow’ from the hotter body to the colder body (i.e. the weight or volume of the body does not matter). But, temperature comes in two important ‘technical’ contexts in TD: 1 it is a measure of the average kinetic energy (or velocity) of the constituent entities (say molecules) 2 it is the parameter which determines the distribution of species (say molecules) across various energy states available. A 10C Heat flow direction B 500C How is constant temperature maintained (isothermal conditions)? The systems is in contact with the surroundings via dia-thermal walls (walls which conduct heat). The surroundings acts like a thermal reservoir (i.e. is so large that input or withdrawal of heat (Q) does not change its temperature). System Thermal reservoir Temperature as a parameter determining the distribution of species across energy levels Let us consider various energy levels available for molecules/species in a system to be promoted to. Let the system be in thermal equilibrium. At low temperatures the lower energy levels are expected to be populated more, as compared to higher energy levels. (Fig.1). In Fig.1 the energy levels are assumed to be equally spaced for simplicity (this will not be true for an real system). As we heat the system, more and more ‘molecules’ will be promoted to higher energy levels. The distribution of molecules across these energy levels is given by: P( E ) P0 e P0 e 1 kT P(E) is the population of species at an energy level E. is the single parameters which controls the distribution across energy levels. Note that is the only parameter which determines the distribution. The numerical value of decreases as ‘environment’ gets colder. Hence, we define ‘T’ which is the inverse of ; such that as the hotter temperatures have a higher numerical value of a parameter. T could have been just the inverse of , but to keep the magnitude of 1C equal to 1 K, we introduce a constant k (= kB), which is the Boltzmann constant. This implies that kB is not a fundamental constant like many others. At 0 K only the ground state is populated, while at infinite temperature all states are populated equally. With increasing T, progressively the population of higher energy increases Fig.1 At 0 K only the ground state is filled Few points about temperature scales and their properties Celsius (Farenheit, etc.) are relative scales of temperature and zero of these scales do not have a fundamental significance. Kelvin scale is a absolute scale. Zero Kelvin and temperatures below that are not obtainable in the classical sense.* Classically, at 0 K a perfect crystalline system has zero entropy (i.e. system attains its minimum entropy state). However, in some cases there could be some residual entropy due to degeneracy of states (this requires a statistical view point of entropy). At 0 K the kinetic energy of the system is not zero in the quantum mechanical picture. There exists some zero point energy due to fluctuations arising from the Heisenberg uncertainty principle. * In systems with population inversion, we have a negative Kelvin temperature (which is hotter than infinity, rather than being colder than zero)! Pressure Pressure* is force per unit area (usually exerted by a fluid on a wall**). It is the momentum transferred (say on a flat wall by molecules of a gas) per unit area, per unit time. (In the case of gas molecules it is the average momentum transferred per unit area per unit time on to the flat wall). P = momentum transferred/area/time. Pressure is related to momentum, while temperature is related to kinetic energy. Can we define pressure inside the container (not on the walls as it is easy to visualize pressure on a wall, but not inside the container)? Pressure is a ‘hydrostatic’, ‘homogeneous’ and ‘isotropic’ quantity$ i.e. it is same in each direction and throughout the inside of the container (including the walls). In the interior it is best visualized by introducing a hypothetical wall and computing the momentum transferred per unit area, per unit time. an ideal gas, about which we will talk soon. Wall of a container $ for ‘Crude schematic’ of particles impinging on a wall. P force / area momentum transferred / area / time p mv A Area P mv [ Kg ] [ L] [ Kg ] 2 At [ L ][ s ] [ s] [ L][ s 2 ] * ‘Normal’ pressure is also referred to as hydrostatic pressure. ** Other agents causing pressure could be radiation, macroscopic objects impinging on a wall, etc. Pressure is a parameter best suited for gases and liquids. For solids stress is the appropriate parameter (though pressure may also be used). Kinds of Equilibrium The topic of equilibrium is dealt with in detail elsewhere. A system in complete equilibrium satisfies 4 types of equilibrium as listed below. Mechanical equilibrium Thermal equilibrium Reaction Equilibrium Material Equilibrium Phase Equilibrium Units of pressure Heat and Work Work (W) in mechanics is displacement (d) against a resisting force (F). W = F d. Work has units of energy (Joule, J). Work can be expansion work (PV), electrical work, magnetic work etc. (many sets of stimuli and their responses). Heat as used in TD is a tricky term (yes, it is a very technical term as used in TD). The transfer of energy as a result of a temperature difference is called heat. “In TD heat is NOT an entity or even a form of energy; heat is a mode of transfer of energy” [1]. “Heat is the transfer of energy by virtue of a temperature difference” [1]. “Heat is the name of a process, not the name of an entity” [1]. “Bodies contain internal energy (U) and not heat” [2]. The ‘flow’ of energy down a temperature gradient can be treated mathematically by considering heat as a mass-less fluid [1] → this does not make heat a fluid! Expansion work P To give an example (inspired by [1]) Assume that you start a rumour that there is ‘lot of’ gold under the class room floor. This rumour ‘may’ spread when persons talk to each other. The ‘spread of rumor’ with time may be treated mathematically by equations, which have a form similar to the diffusion equations (or heat transfer equations). This does not make ‘rumour’ a fluid! [1] Four Laws that Drive the Universe, Peter Atkins, Oxford University Press, Oxford, 2007. [2] Physical Chemistry, Ira N Levine, Tata McGraw Hill Education Pvt. Ltd., New York (2002). Work is coordinated flow of matter. Lowering of a weight can do work Motion of piston can do work Flow of electrons in conductor can do work. Heat involves random motion of matter (or the constituent entities of matter). Like gas molecules in a gas cylinder Water molecules in a cup of water Atoms vibrating in a block of Cu. Energy may enter the system as heat or work. Once inside the system: it does not matter how the energy entered the system* (i.e. work and heat are terms associated with the surrounding and once inside the system there is no ‘memory’ of how the input was received and the energy is stored as potential energy (PE) and kinetic energy (KE). This energy can be withdrawn as work or heat from the system. Q) Why is work done at constant ‘P’ equal to PV. Work done at constant pressure (isobaric) (leading to a volume change V): W = PV. This can be understood easily. Consider a ideal gas in a cylinder with a piston of area A. Let the gas expand, such that the piston moves by x. The work done: W = F . x. W * As Aktins put it: “money may enter a back as cheque or cash but once inside the bank there is no difference”. F A x P V A Q&A Give examples of a few types of work. Work done should have units of energy. Mechanical work could be in 3D, 2D or 1D. Type Sub-type Formula Comments Mechanical 3D P.V P is Pext 2D (e.g. surface tension () work) .dA is the surface tension on a fluid surface 1D (e.g. line tension (f) work) f. dl Electrical Magnetic Q.V Reversible P-V work on a closed system In a closed system (piston in the example figure below), if infinitesimal pressure increase causes the volume to decrease by V, then the work done on the system is: The system is close to equilibrium during the whole process thus making the process reversible. dWreversible PdV As V is negative, while the work done is positive (work done on the system is positive, work done by the system is negative). If the piston moves outward under influence of P (i.e. ‘P’ and V are in opposite directions, then work done is negative. Note that the ‘P’ is the pressure inside the container. For the work to be done reversibly the pressure outside has to be P+P (~P for now). Since the piston is moving in a direction opposite to the action of P, the work done by the surrounding is PV (or the work done by the system is PV, i.e. negative work is done by the system). 1 P (P+P) 2 ‘Ultimately’, all forms of energy will be converted to heat!! One nice example given by Atkins: consider a current through a heating wire of a resistor. There is a net flow of electrons down the wire (in the direction of the potential gradient) i.e. work is being done. Now the electron collisions with various scattering centres leading to heating of the wire i.e. work has been converted into heat. What is the significance of the term ‘reversible’ in the context of work? dWreversible PdV Fig.1 As we shall soon see, maximum work is done in a reversible process. Typically, in irreversible process, we are far from equilibrium. Example of a irreversible process is the expansion of a gas as in Fig.1. Here, there is a partition separating two regions, one with gas at ‘P1’ and another with vacuum and the then the partition vanishes. During the expansion of the gas (states between S1 and S2), macroscopic variables like ‘P’ are not defined and hence it is not prudent to use formulae like PV for work. Funda Check What is a thermal bath or surrounding (in general)? State-1 (S1) T T, P1 V1 V1 NDia-thermal0 Partition Funda Check walls Heat reservoir State-2 (S2) T 2V1 N Q Surrounding remains unchanged in any process that the system undergoes. The surrounding is ‘un-transmutable’. The surrounding can act like a (a) thermal, (b) pressure or (c) chemical reservoir. I.e. (a) if heat is transferred (in or out, Q can be +ve or ve) from the surrounding to the system, the temperature of the surrounding does not change. (b) Similarly, if the surrounding does P-V work (+ve or ve) on the system, its pressure does not change. (c) If the surrounding gives some species (one or more) to the system (or equivalently takes species from the system), the concentration* of the species does not change in the surrounding. * Or chemical potential (a quantity which we will see later and which is responsible for mass transfer) Reversible process ‘Reversible’ is a technical term (like many others) in the context of TD. A reversible process is one where an infinitesimal change in the conditions of the surroundings leads to a ‘reversal’ of the process. (The system is very close to equilibrium and infinitesimal changes can restore the system and surroundings to the original state). If a block of material (at T) is in contact with surrounding at (TT), then ‘heat will flow’ into the surrounding. Now if the temperature of the surrounding is increased to (T+T), then the direction of heat flow will be reversed. If a block of material (at 40C) is contact with surrounding at 80C then the ‘heat transfer’ with takes place is not reversible. Though the above example uses temperature differences to illustrate the point, the situation with other stimuli like pressure (differences) is also identical. Consider a piston with gas in it a pressure ‘P’. If the external pressure is (P+P), then the gas (in the piston) will be compressed (slightly). The reverse process will occur if the external (surrounding pressure is slightly lower). Maximum work will be done if the compression (or expansion) is carried out in a reversible manner. Reversible process Heat flow direction Heat flow direction T TT NOT a Reversible process Heat flow direction T T+T 40C 80C Funda Check Why is the work done maximum in a reversible process? Let us consider two cases (Fig.1) with pressure inside the cylinder is 100 bar: (C1) the outside pressure is lowered infinitesimally to 50 bar and (C2) the outside pressure is constant at 50 bar. The gradual reduction in pressure in C1 can be achieved by removal of sand grains as in Fig.1a. For simplicity we replace the infinitesimals with small finite quantities (). Let us assume an ideal gas in the cylinder. In C1 the initial resisting pressure is (100 ) bar or ~100 bar and the work done is: PV = 100 V. This expansion leads to a drop in the pressure to P2 (< 100 bar) say 99 bar. Further expansion will lead to a drop in pressure to (99 ) bar and the work done in this step will be ~ 99V. We can carry on this process and the net work will be given by the summation (integration in reality). Wnet (C1) = 100 V + 99V + 98V + ... + 50 V. (We could have done 99.5V +...) In C2 the opposing force to the expansion is 50 bar. Hence the work done is: W (C2) = Pext V + Pext V + Pext V + ...= 50 V + 50 V + 50 V + ... Clearly, WC1 > WC2. Pext = (100 ) bar Pext = (50) bar If the piston is in equilibrium at the start 2 and finish then we can calculate: Wirreversible Pext dV (a) Fig.1 C2 C1 Sand grains 100 bar Warning: thermodynamics cannot be used to calculate work during a irreversible process. Above is some kind of ‘crude’ rationalization. Kinetic energy gained by the piston, pressure gradients within the gas and turbulence in the gas will have to be considered. 1 100 bar (b) Note: in PdV or PV, P = Pexternal = Presisting is the resisting pressure (against which work is done) How to visualize a ‘reversible’ equivalent to a ‘irreversible’ processes? Let us keep one example in mind as to how we can (sometimes) construct a ‘reversible’ equivalent to a ‘irreversible’ processes. Let us consider the example of the freezing of ‘undercooled water’* at –5C (at 1 atm pressure). This freezing of undercooled water is irreversible (P1 below). Yes, it is possible to obtain water below its freezing point. This is referred to as ‘under-cooled’ water. In fact freezing always occurs with some undercooling, due to the existence of a ‘nucleation barrier’. We can visualize this process as taking place in three reversible steps hence making the entire process reversible (P2 below). P1 Water at –5C Irreversible Ice at –5C Freezing Ice at 0C Water at –0C P2 Heat Water at –5C Reversible Cool Ice at –5C * ‘Undercooled’ implies that the water is held in the liquid state below the bulk freezing point! How is this possible?→ read chapter on phase transformations Ideal Gas Kinetic theory of gases: Pressure, Temperature, Molecular speeds. We will take up this concept here briefly and return to it later. The kinetic theory gives some ‘nice results’, which are to be treated as approximate in many circumstances. When we want to understand any new subject/concept, it is best to start with some idealizations. In TD some of these idealizations include: ‘reversible’, ‘quasi-static’, etc. In an ideal gas: (i) the particles are point particles (no size) and (ii) there are NO interactions between the particles. The ideal gas obeys the ideal gas law (which is a simplified equation of state*). Many real gases (like the noble gases, oxygen, hydrogen, nitrogen, etc.) under certain conditions of P & T behave close to an ideal gas. All gases tend to behave more like an ideal gas at high T and low P. In an ideal gas ALL the Internal energy is due to the translational motion (velocity) of the particles (molecules). The velocity is a function of the temperature (T). Higher the temperature, higher the kinetic energy and higher the internal energy. Three kinds of ideal gases are differentiated in the literature: (i) the classical or Maxwell– Boltzmann ideal gas (which follows the Maxwell-Boltzmann distribution), (ii) the ideal quantum Bose gas (which is composed of Bosons, Bose-Einstein statistics), (iii) the ideal quantum Fermi gas (composed of Fermions, Fermi-Dirac statistics). * For gases, equation of state relates P, V & T. The equation may have one or more parameters. (More about this later). The equation of state obeyed by an ideal gas is given by the Boyle’s law. N is no. of molecules. PV Constant The value of the constant depends on the experimental scale of temperature used N The equation of state of an ideal gas can be used to define a temperature scale the ideal gas temperature ‘T’. PV kT N k kB (Boltzmann Constant) =1.380 1023 J / K / mole 8.617 105 eV / K / mole The value of the Boltzmann constant is determined by the choice of temperature intervals, usually chosen as 1C. The universality of the scale arises from the universal character of the ideal gas. The procedure to define an ideal gas temperature scale is as follows. Determine the value of PV/Nk at the freezing (P1) and boiling (P2) points of water. These points are plotted with PV.Nk and T as the axes. A straight line is drawn through the points, which intersects the T axis at T = 0. The interval between P1 & P2 is divided into 100 divisions (as we are using the degree Celsius scale). The resulting scale is the Kelvin scale. To measure the T of an unknown system, it is brought into thermal contact with an ideal gas and PV/Nk is determined for the ideal gas. Then, the T is read off from the plot. The ideal gas law is the combination of the Charles law, the Boyle’s law and the Avogadro's law and equivalently we can write the equation of state for an ideal gas as PV = nRT. n R N PV Nk T NA k T n RT NA Ideal gas law PV nRT P1 V1 P2 V2 nR This leads to T1 T2 NA is the Avogadro's no. (=6.023 1023 atoms/mole), n is the number of moles, R is the gas constant (= 8.315 J/C) As the molecules of a ideal gas do not interact with each other, the internal energy of the system is expected to be ‘NOT dependent’ on the volume of the system. U Internal energy (a state function) is normally a function of T & V: U = U(T, V). For an ideal gas: U = U(T) only. 1 Mole of N2 Typically, we plot P versus V at constant temperature (isotherms). (Fig.1). The curves are asymptotic to the x and y axis. Fig.1a V1 0 V T At a constant volume (say V1) if heat the gas (say from 75C to 100C) then the pressure increases (vertical dashed line). Similarly, at constant pressure if we heat the gas, the volume increases (a horizontal line). (Fig.2 & 3). For an ideal gas the plot of PV versus P should be a horizontal line at constant temperature. However, real gases (like N2) show marked deviation at high P and low T. At room temperature (RT), for gases like H2 and He the variation of PV/nRT with P is nearly linear. The behaviour for other gases (like CO and CH4) is more complicated. We will consider the details of the behaviour later, when we discuss real gases and the compressibility factor (Z). Fig.3 Fig.2 1 Mole of N2 Increasing deviation from Ideality: P T Ideal gas In an idea gas the following assumptions are made: (i) there is no (negligible) attractive force between the molecules, (ii) the molecules are point particles (volume occupied by the molecule is << the volume of the container), (iii) the collisions are perfectly elastic, (iv) the duration of the collisions is negligible as compared to the time between the collisions. Pressure is force/area = change in momentum/area/time. Calculation of pressure • Let the velocity of a ‘typical’ molecule be ‘c’ and its components along x, y, z be u, v, w, respectively. This implies: c2 = u2 + v2 + w2. Let the box be a cube of dimensions: L L L. • For this typical molecule, if we consider the velocity along y-direction and elastic collision with the wall, the change in momentum is: mv (mv) = 2 mv. • The time (t) taken for the molecule to return to this wall is: 2L/v = t. The number of impacts per second (the rate) on this wall due to this molecule is: v/2L. • Momentum change due to one molecule per second: (v/2L).(2mv) = mv2/L = Force on +X face. • Pressure on +X face due to one molecule = F/area = (mv2/L)/L2 = mv2/L3 = P. • If there are ‘N’ molecules in the chamber, each with a velocity vi (i = 1 to N), the total pressure due to these ‘N’ molecules is: mvi2 Nm n vi2 Nm 2 P 3 3 3 v (1) L i 1 N L i 1 L n Continued… • Since none of the axis is special it is reasonable to assume: Hence: u v w 2 c 2 u 2 v 2 w2 & c2 3v 2 Hence from (1): c2 is the mean square speed 2 P 2 Nm 2 v L3 Nm c 2 1 Nm c 2 P 3 (2) L 3 3 L3 • So far we assumed that the molecules do no suffer any collisions and are free to move from end to end in the container. In reality, the molecules will collide and redistribution of speeds occur. However, it is reasonable to assume that the average speed of the molecules does not change with time. Root mean square (RMS) speed 1 Nm 2 1 2 P The pressure can be written in terms of density: 3 c 3 c (3) 3 L The square root of the mean square speed can be written in terms of macroscopically measurable quantities P and : Incredible speeds! 3P 2 Of the same order as that of speed of sound. (4) c For H2 at STP if we substitute the values of P and we get: RMS speed c 2 1840 m / s • James Prescott Joule (1818–1889) had first done this calculation in 1848. This incredible speed is if the same order of magnitude as the speed of sound in in H2 at 0C (~1.3 km/s). • The equation implies that for as the density of the gas increases RMS speed decreases. O2 (with a molecular mass of 32 amu has a RMS speed given by: 2 c2 1840 460 m / s 32 Oxygen Continued… 1 Nm P 3 c 2 13 c 2 (3) 3 L Temperature From (3): PL3 1 Nm c 2 3 PV 1 Nm c 2 3 According to the ideal gas equation PV = nRT: nRT 1 Nm c 2 3 3 2 n R T 12 m c 2 (5) N • This is an important result, that for an ideal gas the T in Kelvin is proportional to the mean square velocity of the molecules. I.e. the average K.E. associated with the translation of a molecule of the gas is proportional to the absolute temperature (T). If we consider an Avogadro no. of gas molecules (N = NA & n = 1), then R/NA = k is the gas constant per molecule and is the Boltzmann constant (k = kB). 1 2 m c2 3 2 R T 23 kT NA K .E. per molecule 12 m c 2 23 kT (6) We had noted before that: NA = 6.023 1023 R (Molar gas constant) = 8.31 J/mol/K k (Boltzmann constant) = 1.38 1023 J/K c2 3v2 ( 3u2 3w2 ) Hence, the kinetic energy associated with one ‘degree of freedom’ (DOF) is: K.E. per molecule per degree of freedom 12 m v 2 12 kT Continued… Degrees of freedom associated with a molecule (& contributions to the kinetic energy) • In the discussions so far, the molecule was mono-atomic (like a hard ‘point-like’ sphere). Hence, the only contribution to the kinetic energy is the translational motion. This true for molecules like Ar & Xe. Such molecules do not have rotational DOF. • Molecules like H2 and H2O have additional degrees of freedom at the molecular level, which can contribute to the kinetic energy. • These molecules can vibrate and rotate. Hence, contributions to the K.E. and hence the internal energy (U) arise from these DOF. I.e. the internal energy of a polyatomic molecule is shared between translation, vibration and rotation. We will ignore vibration for now. Translation + Vibration + Rotation Contributions to the K.E. (& hence Internal Energy (U) • Let us consider H2 (a liner molecule) and H2O (an angular molecule). • Let us further assume (due to Maxwell) that the K.E. per degree of freedom per molecule is ½kT. This is referred to as the principle of equi-partition of energy. This principle is approximately valid at ‘high’ temperatures. • DOF of H2 (lying along the x-axis): rotation along y-axis & z-axis. • DOF of H2O rotations along all 3 axis. K.E. monoatomic molecule (3 DOF) 3 12 kT 23 kT K.E. diatomic molecule (3 tranlation+2 rotation DOF) Translation 3 2 K.E. polyatomic molecule (3 tranlation+3 rotation DOF) Rotation kT 22 kT 25 kT Translation 3 2 Rotation 3 2 kT kT 62 kT Continued… Internal Energy (U) Note again that these are under the assumptions' of the kinetic theory of gases • The kinetic theory of gases, assuming an ideal gas, gave us the kinetic energy (K.E.) of monoatomic, diatomic and polyatomic molecules. • The internal energy of an ideal gas is the K.E. of a mole of gas molecules (as there are no other contributions to U*). Internal Energy (U ) Monoatomic N A ( K .E. / molecule) N A 23 kT 23 RT U Monoatomic 23 RT Internal Energy (U ) Diatomic N A ( K .E. / molecule) N A 25 kT 25 RT U Diatomic 25 RT Internal Energy (U ) Polyatomic N A ( K .E. / molecule) N A 62 kT 3RT U Polyatomic 3RT Molar heat capacities (CV & CP) • The molar heat capacity at constant volume (CV) is the heat required to increase the internal energy of 1 mole of a gas through 1 K. CV Monoatomic U Monoatomic 3 2 R CP CV R T CP Monoatomic 23 R R 25 R • The ratio of molar heat capacities (CP/CV ) is given the symbol (known as the adiabatic index or the isentropic expansion factor). 5 CP 5 2 R Monoatomic 3 CV 2 R 3 * Internal Energy (U) in other systems can have contributions from the following. Molecular translational energy Molecular rotational energy Molecular vibrational energy Energy due to electronic states Interaction energy between molecules Relativistic rest- mass energy (mc2) arising from electrons and nucleons. Q&A Why is CP greater than CV ? We have noted that: CP CV R (which implies that CP > CV). This can be understood as follows. The amount of heat supplied at constant pressure is utilized for: (i) increasing the internal energy (and hence temperature since internal energy is a function of temperature) and (ii) for doing work. In contrast at constant volume, no work can be done. This implies that the heat supplied at constant volume is utilized only for increasing the internal energy. Hence, more heat has to be supplied at constant pressure and CP > CV. Mean Free Path (MFP, ) • The ‘average’ distance travelled by a gas molecule before suffering a collision is called the MFP (). • Let: (i) there be ‘nV’ molecules of gas per unit volume, (ii) the radius of the molecule be ‘r’ (diameter ‘d’); then the MFP is given by the formula below (we do not derive it here). • If ‘n’ is the number of moles of the gas, then: nV n N A V 4 1 2 nV r 2 2 nV d 2 For an ideal gas 1 2 d2 RT NA P • This implies that MPF decreases with increasing ‘n’ or the pressure. • Using data for H2, r = 3 1010 m*, at STP (1 bar, 0C) 6.023 1023 molecules occupy 22.4 L, we get: 22.4 103 8 9 10 m 90 nm An extremely short distance! 23 20 2 6.023 10 9 10 • We have seen earlier that the mean speed is 1840 m/s (~2000 m/s) and the number of collisions per second is: Collisions per second = mean speed * Noting that hydrogen molecule is not spherical! 2000 10 10 A really large number! 9 108 Distribution of molecular speeds • The distribution of the no. of molecules with speed ‘c’ (N(c)) as a function of ‘c’ is given by the Maxwell-Boltzmann distribution (1). • The no. of molecules N with speeds in the range c & (c + c) is the area of the shaded portion of the curve. (Fig.1). N = N(c) c. • The following quantities are marked in the figure: (i) the most probable speed (c0), (ii) the mean speed (cm) & (iii) RMS speed (cr). • For a Maxwell-Boltzmann distribution: c0 : cm : cr = 1.00 : 1.13 :1.23. • The RMS speed is related to macroscopic gas properties like pressure and specific heat capacity. The mean speed is related to macroscopic gas properties like diffusion through porous partitions. 3 2 m 2 N ( c) 4 c e 2 RT c0 2RT m cm 8RT m cr 3RT m mc 2 2 RT (1) Continued… • We had noted earlier (eq.(4)) that RMS speed is an inverse function of the density (and hence the atomic mass). Hence, molecules of lighter gases have (crudely speaking) a higher speed on the average (i.e. the Maxwell-Boltzmann distribution is shifted to higher speeds). • However, the pressure exerted (P1) by n1 moles of these gases at a given T1 is same in a constant volume (V1) container. P1 n1 R T1 V1 T = RT c2 Funda Check 3P (4) What is the relevance of the fact that there is a distribution of speeds? There are many important implications of the fact that there are molecules “hotter”* than the average molecule (and similarly “colder”*). If all molecules had the same KE, then we have to heat (say water) above the boiling point to take water to the vapour state. However, given the distribution, some of the molecules have a higher speed to escape the water surface and hence evaporation can occur at lower temperatures (than the boiling point). * We already know that hot and cold cannot be defined for ‘some/few molecules’ here hotter implies “faster” molecules. PV diagrams Many important concepts and processes in TD can be understood using P-V (PV) diagrams. Case-A. Isobaric process. Constant Pressure Process. Let an ideal gas at P1 & T1 be enclosed in a chamber with a frictionless movable piston (of initial volume V1). Let the piston be loaded with sand giving rise to an external pressure Pext = P1. This is initial state of the sytem in equilibrium (S1). Let us heat the system to a temperature T2 (very gradually), such that the gas expands inside the chamber to a new volume V2. In this new state (S2) the Pext = Pint = P1 (due to equilibrium). Due to the expansion the system does mechanical work on the surrounding, which is the area under the P-V curve (= P(V2 V1) = PV). As the temperature has changed the internal energy of the system has changed (by U). Pext = P1 Pext = P1 Closed Ideal System gas Closed System N, T1, P1, V1 N, T2, P1, V2 Dia-thermal walls Heat reservoir Ideal gas Q As work is done by the system U Q W Q PV State-2 (S2) State-1 (S1) P1 V1 P1 V2 V1 T2 V2 T1 T2 T1 Q is given throughout the process and not in state-2 N, T1, P1, V1 N, T2, P1, V2 In the current case (as we go from S1 to S2) the work is done In the current case (as we go from S1 to S2) the work is done by the system on the surrounding. Hence, as far as the system goes, it is negative. If the gas is compressed (i.e. we go from S2 to S1), then work is done on the system and PV term is positive. Case-B. For an arbitrary process in a closed system from S1 to S2, the work done by the system during the process can be computed by dividing the area under the curve into small parts, with the assumption that during the ‘small’ change in volume by V, the P remains constant. Hence, the work for each segment is given by: Wi = Pi V. n Pi V The net work is the sum of all the rectangular areas: W i 1 N, T1, P1, V1 V2 For infinitesimal areas the summation is W P dV replaced by integration: V 1 N, T2, P2, V2 Funda Check Work done by the system versus work done on the system. We had written the first law as: U Q W We used a positive (+) sign for W. In this sign convention, ‘anything’ (Q or W) which goes on to increase the internal energy of the system is given a positive sign. The opposite sign convention for work is also found in literature. From S1 to S2 (expansion of the system), work is being done by the system (the sign of W is negative, i.e. W is a negative quantity), while if we go from S2 to S1 (compression of the system), work is being done on the system and hence W has a +ve sign. Let the pressure inside the cylinder be: P = PInternal PExternal = PResisting Let the decrease in the volume be V (i.e. V is a negative quantity). .Work done on the system is: WReversible = P.V This is a positive quantity as V is negative. c0 2RT m The colour coding is w.r.t to the effect on the internal energy Case-C1a. Isothermal process. Let us consider a system at constant temperature (T) with N molecules (particles) of an ideal gas. Let the container (the system) have two chambers: C1 and C2 (each of V1 (Fig.1)). All the ideal gas is in C1 and C2 is under vacuum in state-1 (S1). Now let the partition ‘vanish’, such that the gas expands into C2 also (with total volume 2V1). In the new state (S2) the volume of the system is 2V1 (for simplicity we have assumed 2V1 in general can be any V2), the temperature is T (which is maintained via contact with a heat reservoir heat is transferred from the reservoir into the system). As the number of molecules (N) & T of the system is constant, the kinetic energy (velocity) of the molecules is constant. However, since the volume has doubled, the molecules will hit the walls of the container less frequently and hence a pressure would be lower. Note that we cannot draw a path from S1 to S2, as the system is not under equilibrium between S1 and S2 (Fig.2). Macroscopic thermodynamic variables like P & T are not defined (and so too V), during the expansion. Between S1 and S2 (after the partition vanishes) the gas is expanding, the P is falling and the T is tending to fall; but in parallel, heat is flowing into the system in an attempt to maintain the temperature. The system is in transient condition from S1 to S2. State-1 (S1) T 2V1 N Partition Heat reservoir C P V This is like y = 1/x (asymptotic to both x & y axis) State-2 (S2) T T V1 V1 NDia-thermal0 walls PV = nR T = Constant N, T1, P1, V1 Fig.2 Fig.1 Q N, T1, P2, 2V1 P1 V1 P2 V2 P2 2V1 T T T P1 P2 2 Continued… Case C1: isothermal process During the isothermal process We can generalize the process for expansion from V1 to V2 T 0 (PV) = nR Tisothermal = Constant ( PV ) 0 U 0 Case-C1b. The system previously considered can be visualized as that with a piston. S1: N, T1, P1, V1. S2: N, T1, P2, V2. We go from S1 to S2 via removal of sand grains gradually, such that the system is always under equilibrium. The temperature of the system will tend to fall due to the expansion, but heat transfer from the heat reservoir (via the dia-thermal walls) will maintain the temperature. In this process there are no transients and hence macroscopic TD variables like P, T & V are defined throughout the process. (Fig.2). As T is constant, U is constant and U = 0. The work done can be calculated using the first law as below, which is the area under the PV curve. U 0 Q W P1 V1 P2 V2 T1 T1 PV nRT P nRT V Qinto the system = W work done by the system W P dV V1 V2 nRT dV V V1 W W nRT ln(V )V2 V 1 N, T1, P1, V1 Pext = P1 Pext = P2 Fig.1 Closed Ideal System gas Gradually remove sand grains N, T1, P1, V1 Dia-thermal walls Heat reservoir Closed System N, T1, P2, V2 Ideal gas Q V WIsothermal nRT ln 2 reversible V1 process Fig.2 State-2 (S2) State-1 (S1) Q W V2 Hence we can use the ideal gas equation at each point on the curve N, T1, P2, V2 This is the work done by the system (a negative quantity as far as the system goes) Case-C2. Isothermal process. Let us consider a system at constant temperature (T). If volume of a container (the system) reduces (say ‘magically’ to half its original volume, i.e. from V1 to V2 (= V1/2)), the pressure will increase. As the number of molecules (N) & T of the system is constant, the kinetic energy (velocity) of the molecules is constant. However, since the volume has reduced, the molecules will hit the walls of the container more frequently and hence a higher pressure. This is similar to the previous case, but going from S2 to S1. Again in this case, between S1 and S2 the system is in transients and hence macroscopic thermodynamic variables like P & T are not defined (and so too V). State-1 State-2 T, V1, N T V1/2 N Case-C3. Isothermal process. Let us consider two isothermal processes at two different temperatures. Case-D. Isochoric process. Let us consider a system at a temperature T1 with N molecules (particles) of an ideal gas. Let the container (the system) have a volume V1 which is constant throughout the process (Fig.1)). Let heat Q be transferred gradually to the chamber, such that the system is under equilibrium throughout. This will lead to an increase in the temperature (gradually to T2). The molecules of the gas will impinge with higher velocity on the walls, which will increase the internal pressure. The system will try to expand this will have to be countered (to maintain constant volume) via gradual addition of sand grains (to increase the external pressure and maintain equilibrium). The states are as follows. S1: N, T1, P1, V1. S2: N, T2, P2, V1. The area under the curve (Fig.2) is zero, hence NO PV work is done during the process. All the heat Q goes into an increase in the internal energy (as seen by an increase in the temperature of the system). Another way to visualize the process is to introduce Q gradually into a rigid container with dia-thermal walls. P1 V1 P2 V1 T1 T2 N, T2, P2, V1 P1 P2 T1 T2 Pext = P2 Pext = P1 Fig.1 Closed Ideal System gas N, T1, P1, V1 Closed Ideal System gas Gradually add sand grains Dia-thermal walls Heat reservoir (at T2) N, T2, P2, V1 Q State-2 (S2) State-1 (S1) U Q W Q PV U Q N, T1, P1, V1 Fig.2 Case-E. Adabatic process. Let us consider an insulated & closed system with an ideal gas at S1: N, T1, P1, V1. (Fig.1)). The walls are insulated. Let us increase the volume of the system by moving the piston (by removing grains of sand) to expand the gas to a volume V2. Since the walls are insulated, no heat can enter the system to equilibrate the temperature (Q = 0). This implies that molecules will impinge on the walls less frequently and the pressure will drop. The system has done work in the expansion, which will lead to a decrease in the internal energy (and the T). Hence, U is negative. The work done by the system is the area under the PV curve. S2: N, T2, P2, V2. If this is compared with an isothermal expansion to V2, it will be seen that P2 (adiabatic) > P2 (isothermal). (Fig.2) The adiabatic process can be visualized as jumping from one isotherm to another. And as the T is changing during the process, we cannot determine the work using the relation PV = C. (Fig.3). U Q W U W Pext = P1 Pext = P2 Fig.2 Fig.3 Fig.1 Closed Ideal System gas Gradually remove sand grains N, T1, P1, V1 Closed System N, T2, P2, V2 Ideal gas Insulated walls State-2 (S2) State-1 (S1) N, T1, P1, V1 N, T2, P2, V2 Continued… Work done in an adiabatic process Work done is the area under the PV curve. During the expansion this is the work done by the system, leading to the reduction in the internal energy. For an adiabatic process the equation of state is: PV Constant = C P1 V1 P2 V2 Eq.(1) Here is ratio of the specific heats γ = CP /CV. It is a factor which determines the speed of sound in a gas. (γ = 1.66 for an ideal monoatomic gas and γ = 1.4 for air, which is predominantly a diatomic gas). V2 W P dV V1 2 V2(1 ) V1(1 ) C V (1 ) CV2(1 ) CV1(1 ) W dV W C W W C (1 ) V (1 ) (1 ) V1 V1 V2 V P2 V2 V2(1 ) P1 V1 V1(1 ) Using Eq.(1), with two different combinations of PV: W (1 ) WAdibatic P2 V2 P1 V1 (1 ) U W A comparison of the 4 processes (isobaric, isothermal, isochoric &adiabatic) on the PV diagram (S1) N, T1, P1, V1 N, T2, P1, V2 Conditions P = 0 WIsobaric PV U Q PV Work done First Law dU TdS PdV V = 0 WIsochoric 0 U Q N, T2, P2, V1 (PV) = 0 T = 0 U = 0 Q W (PV) = Constant N, T1, P2, V2 V Q WIsothermal nRT ln 2 S T V1 P V P1 V1 P Q = 0 U W WAdibatic 2 2 CV (1 ) N, T2, P2, V2 C PV = Constant The adiabatic curve (adiabat) lies below the isotherm; as, the adiabat follows PV (=C), while the isotherm follows PV (=C). [ > 1, hence the adiabat falls more steeply as compared to the isotherm]. Suppose that the volume doubles in an (i) isothermal and (ii) adiabatic process. Let = 5/3 and let the inital pressure be 100 bar. Then: (i ) Isothermal : P2 V2 P1 V1 Isothermal 2 P 1 100 50 bar 2 (ii ) Adiabatic : P2 V2 P1 V1 Adiabatic 2 P 1 100 2 5/3 32 bar T-S diagrams 18 Carnot Cycle Details are in the upcoming slides The Carnot cycle is a conceptual ideal thermodynamic cycle (due to Nicolas Léonard Sadi Carnot). The cycle effectively does one of the following: (M1) use transfer of heat from a hot source to a cold source to produce work (act like a work engine) or (M2) use work as an input to transfer heat from a cold source to a hot source (act like a heat engine of refrigerator) and gives the upper limit on the efficiency that any classical thermodynamic engine can achieve. The cycle can be represented in a PV diagram and consists of four steps (Fig.1); such that start and finish states are the same. In the ‘work engine mode’ a complete cycle consists of: (P1) Isothermal Expansion, (P2) Adiabatic Expansion, (P3) Isothermal Compression, (P4) Isothermal Compression. At the end of the cycle the internal energy (U) remains unchanged (as it is a state variable and we are back to the same P1 & V1). Fig.1 In M1 (clock-wise operation), the combined effect of the 4 processes (P1 to P4) is to transfer heat from a hot source (Q1, +ve magnitude) to a cold sink (Q2, ve magnitude) and to produce work. The amount of work produced is given by the area enclosed by the curve. In P1 and P2 work is done by the system on the surrounding (W1 & W2), while in P3 and P4 work is done on the system (W3 & W4) by the surrounding. In the clockwise cycle, working as a work engine. As |[(W1) + (W2)]| > | (W3 + W4)|. The net effect is that the system does (produces) work during the clockwise operation. The net effect of the cycle is to covert (Q1 Q2) amount of heat into [W1 W2 + W3 + W4] amount of work. P1 & P2 are expansions, while P3 & P4 are compressions (P1) Isothermal expansion at T1 (= Th). The system takes in Q1 amount of heat and does work on the surrounding of W1. Being an isothermal process, the internal energy remains unchanged (U = 0). The heat input (Q1) can be conceived as coming from a hot bath at T1. (P2) Adiabatic expansion. No heat is exchanged with the surrounding. The system does work (W2) on the surrounding and this comes at an expense of the internal energy (decreases, U < 0). (P3) Isothermal compression at T2 (= Tc). The system rejects Q2 amount of heat and the surrounding does work on the system of W3. Being an isothermal process, the internal energy remains unchanged (U = 0). The rejection of heat (Q2) can thought of going to a cold bath at T2. (P4) Adiabatic compression. No heat is exchanged with the surrounding. The surrounding does work on the system (W4) and this leads to an increase in the internal energy (U > 0). The net effect of the cycle is the following. Take heat Q1 from a hot source and reject Q2 to a cold sink. Produce [W1 W2 + W3 + W4] amount of work. Lose internal energy in P2 (adiabatic expansion) and gain the same in P4 (adiabatic compression), such that the net change in internal energy is zero. Gain entropy during the isothermal expansion and lose an equal amount of entropy during the isothermal compression. Q1 Work done at each step is the area under each one of these curves • Since work is involved in all the steps, this implies that we must have some kind of piston connected to the system (not shown). W1 P1 - U P1 Hot source W1 Cold Sink Q1 > Q2 S1 W2 P2 Q1 - U U P2 P3 - U Hot source W3 - U S4 U P3 U W3 W2 W1 P1 S2 W4 P2 P2 P3 W2 Cold Sink Q2 S3 Q2 P4 W4 U • The source and sink are both reservoirs, whose temperature does not change by the addition or removal of Q. • The internal energy lost in P2 is the same as that gained in P4. U P 2 U P 4 W3 P4 W4 Carnot Cycle: T-S diagrams The Carnot cycle can be drawn with T & S as axes. Clearly, the isothermal sections are horizontal. S1 S2 S4 S3 The interesting point to note is that the adiabatic processes are isentropic and hence vertical. During the adiabatic process, the T is changing. If we consider an infinitesimal section where T is constant (say Ti), then Si = Qrev/Ti. Qrev = 0, for the entire adiabatic process and for each section. Hence, Si = 0. Hence, the adiabatic processes are vertical in the T-S diagram. During the isothermal expansion (S1S2): Which is the area under the curve U = 0 Q W S S 1S 2 Q W T S From S3S4 the work done is the area as shown Negative quantity Negative quantity S4 S3 S S 3 S 4 Q2 T2 Q1 T1 S1 S2 W12 T1 ( S2 S1 ) Note: symbol S is being used for both state and entropy (pl do not get confused) SS 1S 2 Q Q1 SS 3S 4 2 T1 T2 The Carnot cycle (operating as a work engine) can be compact drawn as below (Fig.1). The cycle operating as a heat engine can be drawn as Fig.2. The heat engine is also called a heat pump. Terms body and reservoir are used interchangeably (essential the properties, including T of the reservoir does not change). ‘Work engine’ ‘Heat engine’ Like a steam engine Like a refrigerator Heat reservoir Source Hot body Heat Q1 Heat Q1 Cyclic engine Work (W) Heat Q2 Cold Reservoir Sink Cyclic engine Work (W) Heat Q2 Fig.1 Fig.2 Cold Body The efficiency of a heat engine The efficiency of a heat engine is the amount of work output divided by the amount of heat input. The efficiency can be calculated using the equation of the first law applied to the whole cycle. By definition Carnot engine heat engine woutput qinput Carnot engine Output Wnet Input Q1 U cycle 0 Qnet Wnet Qnet Q2 Q1 Wnet Q Wnet Q1 Q2 1 2 Q1 Q1 Q1 V2 W nRT ln As seen earlier, the work done during an isothermal process is: Isothermal V1 Carnot engine V3 RT2 ln V4 Now*: V3 V2 1 V4 V1 V2 RT1 ln V1 T2 T1 Hence: Carnot engine 1 Tsink Tsource max heat engine 1 Nicolas Léonard Sadi Carnot in 1824. “Reflections on the Motive Power of Fire”, Chapman & Hall ltd, London, 1897. * Not shown here. Continued… The efficiency of a heat engine T2 T1 Carnot engine 1 Tsink T source max heat engine 1 This is surprising as: there is no mention of the medium of the system (or its properties), the formula has only temperatures and the temperature of the sink seems to play a major role (as the presence of the sink is usually not intentional or obvious→ in a steam engine sink is the air around the engine and source is the hot steam). Important message Sink (characterized by its temperature) is as important as the source. To increase the maximum possible efficiency of a heat engine, either the temperature of the source has to be increased on the temperature of the sink has to be decreased. The Carnot engine is the most efficient engine Let us assume there exists an Engine E2 which is of higher efficiency (Fig.2) than the Carnot Work Engine (CWE) (Fig.1). This implies that this engine can produce more work (say W(1+x)) than the CWE. Let us couple this higher efficiency E2 with a CHE* (reverse of CWE), which needs W(1+x) amount of work as input (Fig.3). In this engine all the variables are scaled up by (1+x). The net effect of such an coupled system (Fig.4) is to transfer heat from a cold source to a hot source (by an amount Q1 x) without any other influence. This violates the second law. Hot Hot Q1 Q1 CWE Q2 = Q1 W Cold E2 W Hot Q1 (1+x) W(1+x) Q2 = Q1 (W(1+x)) Cold Fig.1 Hot Hot Fig.2 CHE Cold Q1 E2 W(1+x) Q2 = (Q1 W)(1+x)) Fig.3 Q1 (1+x) W(1+x) Q2 = Q1 (W(1+x)) Cold Hence, an engine with an efficiency higher than the Carnot engine is not possible. Q1 x E2+CHE Q1 x Cold Hot Fig.4 * Carnot Heat Engine (CHE) CHE Q2 = (Q1 W)(1+x)) Cold Fig.4 Funda Check What is the work done for an arbitrary closed cycle? Let us consider an arbitrary path from S1 to S2 and back to S1, forming a closed cycle. The work done for the path from S1 to S2 is the area under the curve (Eq.1). The work done during the cyclic process is the area enclosed by the curve. V2 WS 1S 2 P(V ) dV (1) V1 What is the entropy change for an arbitrary closed cycle? Entropy is a state function and hence for a close cycle (like the Carnot cycle) the entropy change is ZERO. Sclosed cycle dQrev 0 T closed cycle Heat Capacity Heat capacity is the amount of heat (measured in Joules or Calories) needed to raise an unit amount of substance (measured in grams or moles) by an unit in temperature (measured in C or K). As mentioned before bodies (systems) contain internal energy and not heat. This ‘heating’ (addition of energy) can be carried out at constant volume or constant pressure. At constant pressure, some of the heat supplied goes into doing work of expansion and less is available with the system (to raise it temperature). Heat capacity at constant Volume (CV): E C V It is the slope of the plot of internal energy with temperature. T V Heat capacity at constant Pressure (CP): H It is the slope of the plot of enthalpy with temperature. CP T P Units: Joules/Kelvin/mole, J/K/mole, J/C/mole, J/C/g. Heat capacity is an extensive property (depends on ‘amount of matter’) If a substance has higher heat capacity, then more heat has to be added to raise its temperature. Water with a high heat capacity (of CP = 4186 J/K/mole =1 Cal/C/Kg) heats up slowly as compared to air (with a heat capacity, CP = 29.07J/K/mole) this implies that oceans will heat up slowly as compared to the atomosphere. As T0K, the heat capacity tends to zero. I.e near zero Kelvin very little heat is required to raise the temperature of a sample. (This automatically implies that very little heat has to added to raise the temperature of a material close to 0K. This is of course bad news for cooling to very low temperatures small leakages of heat will lead to drastic increase in temperature). The Laws of Thermodynamics There are 4 laws of thermodynamics. Like any of the laws, there are generally found to be true (i.e. no violation of the laws have been found so far); but, there is no proof for any of the laws. The laws are numbered from the 0th to the 3rd (for historical reasons). The second law is considered as one of the most (if not the most) profound laws of nature. it introduces the concept of Entropy. The second law further gives us the direction for the spontaneity of a process. Applied to the universe at large, it gives us the arrow of time and Cosmological concepts like the ‘heat-death theory’ (of the Universe). The third law sets the scale for entropy. The First Law The internal energy of an isolated system is constant. A closed system may exchange energy as heat or work. Let us consider a close system at rest without external fields. There exists a state function U such that for any process in a closed system: U = Q + W [1] (For an infinitesimal change: dU = (U2 U1) = dQ + dW) Q → heat flow into the system. W → work done on the system (work done by the system is negative of above- this is just ‘one’ sign convention). U is the internal energy. Being a state function for a process U depends only of the final and initial state of the system. U = Ufinal – Uinitial. Hence, for an infinitesimal process it can be written as dU. In contrast to U, Q & W are NOT state functions (i.e. depend on the path followed). q and w have to be evaluated based on a path dependent integral. For an infinitesimal process eq. [1] can be written as: dU = dQ + dW. The change in U of the surrounding will be opposite in sign, such that: Usystem + Usurrounding = 0 Actually, it should be E above and not U {however, in many cases K and V are zero (e.g. a system at rest considered above) and the above is valid- as discussed elsewhere}. It is to be noted that in ‘w’ work done by one part of the system on another part is not included. The experimental foundation of the first law is the Joule’s demonstration of the equivalence between heat and mechanical energy. * Depending on the sign convention used there are other ways of writing the first law: dU = dq dW, dU = dq + dW A note on exact and in-exact differentials Previously, we had dealt with the concept of state functions and path functions. Internal energy (U) (& H, A, G, S) are state functions and the change between two states is not dependent on the path of the process. E.g. if we go from state ‘1’ to state ‘2’, the change in the internal energy (U) is given by: 2 U dU And we refer to dU as an exact differential* 1 In contrast, the Q and W for a process are path dependent. 2 Q 1, path dQ Here, dQ as an in-exact differential and hence the LHS is Q and not Q Q Q2 Q1 2 dQ Q 1, path * An exact differential is an infinitesimal quantity which upon integration gives a result, that is independent of the path taken. The Second Law The second law comes in many equivalent forms It is impossible to build a cyclic machine* that converts heat into work with 100% efficiency Kelvin’s statement of the second law. Another way of viewing the same: it is impossible to construct a cyclic machine** that completely (with 100% efficiency) converts heat, which is energy of random molecular motion, to mechanical work, which is ordered motion. The unavailable work is due to the role of Entropy in the process. Heat reservoir Heat q Cyclic engine 100% Work (w) Not possible Heat reservoir Called the source Heat q H Cyclic engine G Heat q’ T S Cold Reservoir Called the sink Work (w) * For now we are ‘building’ ‘conceptual machines’ ! ** These ‘engines’ which use heat and try to produce work are called heat engines. Kelvin’s statement of the second law Possible Another statement of the second law → the Clausius statement Heat does not ‘flow*’ from a colder body to a hotter body, without an concomitant change outside of the two bodies Clausius’s statement of the second law.(a) This automatically implies that the spontaneous direction of the ‘flow of heat*’ is from a hotter body to a colder body.(b) The Kelvin’s and Clausius’s statements of the second law are equivalent. I.e. if we violate Kelvin’s statement, then we will automatically violate the Clausius’s statement of the second law (and vice-versa). Heat cannot spontaneously flow from a cold (low temperature) body to a hot body. To make heat flow from a cold body to a hot body, there must be accompanying change elsewhere (work has to be done to achieve this). Hot body H Hot body Not possible Spontaneous flow not possible Cold body G T S Cold body * Used here in the ‘common usage sense’. (b) is obvious, but not (a) → though they represent the same fact. Work A combined (Kelvin + Clausius) statement of the II Law The entropy* of a closed system will increase during any spontaneous change/process. If we consider the Universe to be a closed system (without proof!!)**, then, the entropy of the universe will increase during any spontaneous change (process). The universe can be thought of as an isolated system. The entropy of an isolated system will increase during an spontaneous process. You may want to jump to chapter on equilibrium to know about Entropy first Entropy sets the direction for the arrow of time ! * Soon we will get down to this mysterious quantity. ** For all we know the Universe could be ‘leaky’ with wormholes to other parallel Universes! Q&A What is the difference between ‘heat engine’ and ‘work engine’? Actually both the engines we are going to describe here are usually known as heat engines. We are differentiating two types of engines to see which one produces work and which one actually transfers heat. In the heat engine as the temperature of the cold body tends to zero Kelvin, more and more work has to be done to transfer the heat from the cold body to the hot body. ‘Work engine’ ‘Heat engine’ Like a steam engine Like a refrigerator Heat reservoir Source Heat q Heat q Cyclic engine Work (w) Cyclic engine Heat q’ Heat q’ Cold Reservoir Hot body Sink The main objective here is to produce work Cold Body The main objective here is to transfer heat from a cold body to a hot body Work (w) The Third Law For substances in internal equilibrium, undergoing an isothermal process, the entropy change goes to zero as T (in K) goes to zero. lim S 0 T 0 The law is valid for pure substances and mixtures. Close to Zero Kelvin, the molecular motions have to be treated using quantum mechanics → still it is found that quantum ideal gases obey the third law. Phenomenological description of the third law. There does not exist any finite sequence of cyclical process, which can cool a body to zero Kelvin (absolute zero). Other statements of the third law. For a closed system in thermodynamic equilibrium, the entropy of the system approaches a constant value as the temperature goes to absolute zero. If there is a unique ground state with minimum energy at zero Kelvin, then the entropy at zero Kelvin is ZERO. However, if there is a degeneracy with respect to the number of microstates at absolute zero, then there will be some Residual Entropy. Humorous look at the three laws The first law says: “you cannot win”. The second law says: “you can at best break even- that too at zero Kelvin”. Third law says: “zero Kelvin is unattainable”. Q&A What is the difference in the ‘status’ of quantities like T, U, S, H, A & G? T, U & S are fundamental quantities of thermodynamics. H, A & G do not give us new fundamental concepts, but are for better ‘accounting’ in thermodynamics. Chemical Thermodynamics Equilibrium and Non-equilibrium Conditions Some ‘thingamajigs’ we will encounter: Material Equilibrium Phase equilibrium Reaction equilibrium o Gibbs Equations Non-equilibrium systems Chemical Potential * Etymology. [Ergon (Greek) = Work] + [Hodos (Greek) = Way] = [Ergoden (German)] + [ic (English)] = [Ergodic] a dynamical systems term Equilibrium and Non-equilibrium Systems Nothing will happen if a system is in equilibrium. This implies that non-equilibrium conditions are required for a process to occur. The first step is to consider processes in closed systems, wherein the system is close to equilibrium conditions these are the reversible processes. Also, we restrict ourselves to P-V work only. Later on we can consider open systems and non-equilibrium processes which are not reversible (i.e. the process occurs irreversibly and hence is not close to equilibrium). Reversible processes passing through equilibrium states in a closed system First, using the first law and the definition of entropy we derive the equation for dU. The using definitions for H, A, G, CV and CP we get set of 6 basic equations. Reversible composition changes are included in the processes (e.g. when we heat a two phase mixture and the proportions of the two phases changes). The closed system condition precludes the introduction of additional species, leading to a change in the composition. Reaction of species, leading to a composition change is also not included in the above. This is because chemical reactions occurs spontaneously, which implies that the system is not under equilibrium. Next, akin to that for U (i.e. dU), we will derive expressions for dH, dA and dG. Collectively these are known as the Gibbs equations. Thermodynamic Relations for a System in Equilibrium A system undergoing a reversible process passes through equilibrium states (only). The first law for a closed system is: dU dQ dW For only P-V work done reversibly dW dWrev PV Again, for a reversible process (second law) From (1-3) (combining the first and second laws) Definition of A H U PV A U TS Definition of G G H TS Definition of H Definition of CV Definition of CP U cV T V H cP T P (1) (2) dQ dQrev TdS dU TdS PV We are assuming here that work is done by the system & hence the ve sign (3) (e1) (e2) (e3) (e4) (e1) to (e6) are basic thermodynamic equations of a closed system in equilibrium (e5) Close system in equilibrium, P-V work only (e6) Close system in equilibrium, P-V work only We have considered reversible processes, i.e. system is passing through equilibrium states. Reversible composition changes are included in the processes (e.g. when we heat a two phase mixture and the proportions of the two phases changes). The closed system condition precludes the introduction of additional species, leading to a change in the composition. Reaction of species, leading to a composition change is also not included in the above. This is because chemical reactions occurs spontaneously, which implies that the system is not under equilibrium. The Gibbs equations We had derived the (e1) in differential form for U. Similar equations can be derived for dH, dA & dG. Collectively, e1, e7, e8, e9 are known as the Gibbs equations. TdS PdV dH d (U PV ) dU PdV VdP TdS VdP TdS PdV dA d (U TS ) dU TdS SdT SdT PdV TdS VdP dG d ( H TS ) dH TdS SdT SdT VdP dU TdS PV dH TdS VdP (e7) dA SdT PdV (e8) dG SdT VdP (e9) (e1) Non-equilibrium Systems The primary question is: “to what extent can we use concepts of equilibrium TD to nonequilibrium systems?” or equivalently “to what extent can we assign definite values of TD properties to non-equilibrium systems?”. In general systems, the non-equilibrium can arise from variations in (i) time or (ii) space (location within the system) or both time and space. The variations could be in one or more of the following. (a) T (thermal non-equilibrium), (b) P (mechanical non-equilibrium), (c) Phase fractions (system not in phase equilibrium) or (d) Chemical species (system not in chemical equilibrium). We can achieve the understanding of non-equilibrium systems (at least partial) in the following steps. C1 Assume that there is no spatial variation in quantities within a phase (i.e. any variations are only in time). C2 System is in T and P equilibrium, but not in phase equilibrium. C3 System is in T and P equilibrium, but not chemical equilibrium. C4 Consider system with spatial variations in quantities as well. (E.g. T, P, composition is varying spatially within the system). Q&A How can the composition of a system change? The Gibbs equations are not valid for open systems, wherein the composition can change due to entry of matter. Composition can change due to any of the following. P1 Entry of matter (open system). P2 Inter-phase transport of matter (within the system). P3 Chemical reaction. P1 above is easy to understand. E.g. if we introduce alcohol into a beaker with water. P2. Let us consider a two phase mixture in the micrograph below (Pb-Sn eutectic system*). The phase in light contrast is a Pb rich phase (labelled as the -phase, with Sn dissolved in it). The dark phase is Sn rich phase (labelled as the -phase, with Pb dissolved in it). Now if Pb diffuses from dark phase to the light phase, then the transport of matter is within the system; i.e. inter-phase transport of matter. P3. In a chamber with H2 and O2, if we introduce a catalyst, then the reaction will begin to give rise to H2O, which is composition change due to chemical reaction. If P1, P2, P3 occur, then equations we considered before have to be modified with additional term(s) * We will learn about this in the chapter on phase diagrams. dU TdS PdV (terms) Let us start with conditions C1 & C2. Consider a system under thermal and mechanical equilibrium (i.e. chemical reactions do not occur at fast rate (explosively)). Assume that the composition within each phase is uniform (i.e. the diffusion rate within each phase is rapid compared to the rate of transport of components from one phase to another). C1 & C2 System Not in Phase Equilibrium We consider first a system which is not in phase equilibrium (Fig.1). In this system a remove able partition separates NaCl (crystal) from water (or unsaturated solution of NaCl in water). The internal energy and entropy of the system, which are extensive quantities, can be obtained by adding the values for the two phases (salt and salt solution). U System U NaCl U Solution S System S NaCl S Solution Let the partition vanish ‘magically’ (or equivalently is removed reversibly and adiabatically, such that Q and W are zero for the process, along with U and S). Fig.1 Once the partition is removed the system is not in phase equilibrium anymore and salt starts to dissolve in the solution. (Assume that saturation of the solution does not occur during our discussions!). C1 Assume that there is no spatial variation in quantities within a phase (i.e. any variations are only in time). C2 System is in T and P equilibrium, but not in phase equilibrium. C3 System is in T and P equilibrium, but not chemical equilibrium. C4 Consider system with spatial variations in quantities as well. (E.g. T, P, composition is varying spatially within the system). In spite of the fact that the system is not in equilibrium, at any stage of dissolution we can define USystem and SSystem. This can be understood as follows. At any point in time* (or any stage in the dissolution) we can freeze a frame (or equivalently re-insert the thin partition in the ‘ideal’ way we removed it) and measure the amount of solid and solution and hence compute U and S. (Noting that the solution has changed, while the solid is the same). This implies that, in spite of the non-equilibrium situation, we can use the concepts and parameters of equilibrium TD (proviso we are able to visualize a sequence of equilibrated steps). C1 & C3 System Not in Chemical Equilibrium We consider first a system which is not in reaction equilibrium (Fig.2). In this system there is mix of H2, O2 and H2O gases. The relative fractions of these gases will not change with time at RT as H2 will not react with O2. We can introduce a catalyst to start the reaction, which will lead to an increase in the amount of H2O at the expense of H2 and O2. (I.e. the composition of the mixture will change with time). The quantities U and S can be calculated at the initial state. At any stage of the reaction if the catalyst vanishes (magically) /is removed then the reaction can be stopped and U and S for the system can be calculated for the composition at that ‘time’. Fig.2 * Unfortunately, there is no time in thermodynamics (it is the domain of kinetics)! The two examples we considered (systems in phase and reaction non-equilibrium conditions), serve to illustrate an important point. For systems in thermal and mechanical equilibrium (with well defined P, V & T) and such that the composition of each phase is uniform, we can assign/determine U and S values. This is in spite of the fact that the system is not in material equilibrium. C4 What about systems with varying T and composition? For such systems, first we have to take an instant of time (freeze a frame of the video). Then, we will have to locate a small region (maybe a thin slice) where we can assume the T to be constant. For the case of reactions, we have to assume that composition is constant in that region of the system. We can then assign U and S values to these regions and then sum up the quantities to obtain S and U for the whole system. USystem =U Region-1 +U Region-2 U Region-3... GSystem =GRegion-1 +GRegion-2 G Region-3... Gibbs equations for non-equilibrium systems Let us consider a single phase system in thermal and mechanical equilibrium (but not material equilibrium). If the system is made of ‘k’ components with ni moles of each component (i goes from 1 to k), then the TD state of the system can be defined by a set of values for T, P and ni. As discussed before, in spite of the system not being in material equilibrium, we can assign values to U and S. Further, since T, P, S and U are defined, so are other TD state functions (H, A, G). The state functions U, H, A & G can be expressed as functions of T, P, ni. At any instant during the progress of a chemical process (a ‘frame’ of the movie), we can write: G G(T , P, ni ) Let T, P, ni (‘k’ values) change by infinitesimal amounts due to an irreversible process (chemical reaction or transport of matter into the system). The changes will be: dT, dP, dn1, dn2, ..., dnk. The aim is to compute dG for this irreversible process. However, keeping in view that G is state function, we substitute the irreversible process with a reversible one and calculate dG for this reversible process. Substitute dGIrreversible GState 2 GState 1 dGReversible State-2 is only slightly different from State-1 Assume that the process occurring is a chemical reaction. To perform the process reversibly, first we stop the chemical reaction (say by removing the catalyst). Then, we add dn1 moles of species ‘1’, dn2 moles of species ‘2’ and so forth. Further, we change the T by dT and P by dP. The change in G can be written as: G G G G dGRev dT dP dn ... dnk (1) 1 T P n n P ,ni T ,ni 1 T ,P ,n j 1 k T ,P ,n j k P & Composition is held constant T. P & amount of all species except ‘1’ is held constant T & Composition is held constant We already know that for a reversible process where no change in composition occurs: dG SdT VdP G S T P ,ni G dG SdT VdP dni n i 1 i T , P ,n j i G V P T ,ni Single phase Open system Reversible process P-V work only k (2) (3) The term in the brackets has a special meaning it is referred to as the Chemical Potential. Chemical Potential (i) G i (4) n i T ,P ,n j i G ni T ,P ,n i Single Phase System j i k Using i Eq. (3) can be written as: dG SdT VdP i dni (5) i 1 The chemical potential is an important quantity in TD. It is the change in the Gibbs free energy of a system, when a small amount of a given species (say ‘1’ or ‘A’) is added into the system. I.e. it is the slope of the G vs composition curve at a given composition. (When we say added, we also can considered removal of a given species from the system). If i is positive (G increases on addition of ‘A’ to the system), it implies that the system does “not like” to accommodate ‘A’. k dG SdT VdP i dni (5) i 1 Single phase Open system Reversible process P-V work only The equation we saw before (eq. (5)) is an important one which is applicable to open systems (single phase). The equation is applicable to systems in thermal and mechanical equilibrium but not in material equilibrium. The equation is valid during irreversible chemical reaction taking place in the system and during transport of matter into (or out of) the system. Similarly, we can write down equations for the other thermodynamic potentials (U, H, A). These are the Gibbs equations for open systems. dU TdS PV i dni i dH TdS VdP i dni i dA SdT PdV i dni i dG SdT VdP i dni (e1a) (e7a) (e8a) Single phase system Mechanical & Thermal equilibrium P-V work only Not in Chemical equilibrium (dni arises due to irreversible chemical reaction or k i i1 irreversible transport of matter into the system) (e9a) i In the above discussions, the system was open, but was a single phase system. We now need to generalize the equation(s) to multi-phase systems. Gibbs free energy is s state function which is extensive and hence we can add the G’ for individual phases to obtain that for the whole system. G Multi-phase =G Phase-1 G Phase-2 G Phase-3 .... (6) Generalization of the Gibbs equation for a single phase open system to a multi-phase system Let the phases be labelled , , , etc. Then the total ‘G’ for the system can be written as below. ( has been used as the label/name for a single phase and also as an index for the summation over all the phases). GTotal G G G ... G (6a) Given that the differential of a sum is the sum of the differentials (i.e. d(u + v)= du + dv), we can write: dGTotal d G dG dG dG dG ... (7) Eq. (5), the formula for dG can be written for a given phase () as follows (the equation is basically the same, but written for the -phase). S is the entropy of the -phase dG S dT V dP i dni (8) V is the volume of the -phase i 1 ni is the no. of moles of the ‘ith’ species in the -phase i is the chemical potential of the ‘ith’ species in the -phase The term i is: G k i ni T ,P , nj i (9) From (7) & (8) dGTotal k S dT V dP i dni (10) i 1 We have assumed T and P to be the same across the phases. ‘S’ and ‘V’ are extensive quantities and hence we can write: S S Total (11a) VTotal V (11b) Hence, we can k i dni write from (10): dGTotal STotal dT VTotal dP i 1 (12) Multi-phase system Mechanical & Thermal equilibrium P-V work only Not in Chemical equilibrium (dni arises due to irreversible chemical reaction or irreversible transport of matter into the system) Example of a two component, two phase system Let us consider a system with water and alcohol. Further, let the vapour phase containing both water and alcohol (in certain fractions) be in equilibrium with the liquid phase (again containing both alcohol and water). Keeping in view eq. (11, 12) in view we can write. STotal S Liquid S Gas VTotal V Liquid V Gas Now the system can be taken out of equilibrium by one (or more) of the following methods. (i) Change the T, (ii) change the P or (iii) introduce water or alcohol into the system (or both). [We can also do a combination of the aforementioned]. Changes in one of the quantities will lead to changes in other parameters of the system (e.g. if heat the system, then the P will change and the phase fractions and composition of the liquid and gas phases will change). The change in ‘G’ due to the process as above can be written as below. dGTotal k STotal dT VTotal dP i dni i 1 The term in the brackets can be expanded as (say is the liquid phase): k i 1 Liquid i Liquid Liquid Liquid Liquid dniLiquid Water dnWater Alcohol dn Alcohol The double sum can be written as: k Liquid Liquid Liquid Liquid Gas Gas Gas Gas i dni Water dnWater Alcohol dn Alcohol Water dnWater Alcohol dn Alcohol i 1 Vapour/Gas phase Water + Alcohol T, P T, P Liquid Phase Water + Alcohol Multi-phase System in Material Equilibrium Material equilibrium includes phase and reaction equilibrium. If we derive a condition for material equilibrium, we can use it for subsets of the same (i.e. only reaction or only phase equilibrium). We had written down the equation for the (infinitesimal) change in G (dG) for a multi-phase system (eq. (12)). Also as below dGTotal k STotal dT VTotal dP i dni (12) i 1 Multi-phase system, Mechanical & Thermal equilibrium P-V work only, Not in Chemical equilibrium (dni arises due to irreversible chemical reaction or irreversible transport of matter into the system) Now, let the system proceed towards equilibrium. If the process is spontaneous, then G will decrease and reach a minimum at equilibrium. At equilibrium, dG =0 at constant T & P and eq. (12) becomes: dGTotal k 0 STotal dT VTotal dP i dni i 1 k i dni 0 (13) i 1 Multi-phase closed system, T, P constant P-V work only, Material equilibrium Note that eq.(13) does not contain P, T or V. Hence, eq. (13) is applicable at constant T and V (where, dA = 0 under material equilibrium conditions). Further notes on the Chemical Potential We had noted earlier that the chemical potential of a single phase system is given by eq. (4) as below. Since, G = G (T, P, n1, n2, ...), the chemical potential, which is a partial derivative of G w.r.t to ni is also a function of these variables (i.e. T, P, n1, n2, ...). G ni T ,P ,n i (4) Single phase G G(T , P, n1, n2 ,...) Hence i i (T , P, n1, n2 ,...) j i Since the chemical potential (i) is the ratio of infinitesimal changes in two extensive properties, it is a intensive property. Hence, we can replace the functionality with no. of moles with mole fractions. i i (T , P, X 1, X 2 ,...) Xi ni n1 n2 ... For the case of a multi-phase system, we can write down the chemical potential of each component in each phase as below (e.g. shown for the -phase). i i (T , P , X 1 , X 2 ,...) It is to be noted that even if a given component ‘i’ is absent from a phase, the chemical potential of that component can be defined with respect to that phase as we can add dni of the phase and monitor the change in G (i.e. dG). [i = dG/dni]. An example of this situation is to add Ni to pure Cu (or a Cu-Au alloy). Phase Equilibrium For simplicity let us consider a two-phase system (or keep our focus on two phases in a multiphase system), wherein the -phase is in equilibrium with the -phase. Let dnj moles of species ‘j’ flow from to . We had derived the condition for multi-phase material equilibrium before (eq. (13)). Fig.1. k i dni 0 i 1 (13) j This gives us: dnj j dn j 0 j Whatever loses, gains. Hence: dnj dn j , dn j dn j j j dn j j dn j 0 Phase equilibrium ClosedSystem P-V work only This implies that if two phases are in equilibrium that the chemical potential of a chemical species will be the same in both the species. To generalize, if multiple phases are in equilibrium, then the chemical potential of any species will the same in all the phases. (Note: this is an important result). Example-1. If water (Liquid) is in equilibrium with its vapour (water (gas)) then the chemical potential of water in liquid and gas (vapour) phases will be equal. Gas HLiquid O H O 2 2 P ,T Water Vapour Water Water + Alcohol Gas T, P Water + Alcohol Liquid Gas HLiquid O H O 2 Liquid Alcohol 2 Gas Alcohol P ,T P ,T Fig.1 T, P Example-2. If a mixture of alcohol + water (Liquid) is in equilibrium with vapour of alcohol + water (gas) then: (i) the chemical potential of water in liquid and gas phases will be equal and (ii) the chemical potential of alcohol in liquid and gas phases will be equal. -phase Components: i, j T, P dnj -phase Components: i, jp What if the chemical potential of a species is not equal in two phases? A species will ‘flow’ (‘tend to move’*) spontaneously from a phase with higher chemical potential to a phase with lower chemical potential. For simplicity let us consider a two-phase system of Liquid () and Gas (, vapour). Let both the phases contain two components (species): water (‘i’) and alcohol (‘j’). If the system is not under phase equilibrium and let the chemical potential of alcohol be greater in the liquid phase: Liquid Gas Alcohol Alcohol P ,T Then, to establish phase equilibrium, alcohol will (tend to*) move spontaneously from the liquid phase to the gas phase. The process will continue till: Liquid Gas Alcohol Alcohol P ,T In a closed system in thermodynamic equilibrium, the chemical potential of a species will be same in every phase, in which the substance is present. Funda Check Stimulus and response pairs. Difference in Drives T Heat transfer Fig.1 Gas () phase Components: Water + Alcohol dnAlcohol P Work Mass transfer * We already know well that TD can talk only about ‘tendencies’, the actual ‘happenings’ is the realm of kinetics. dnj Liquid () phase Components: Water + Alcohol T, P Reaction Equilibrium Let us consider a reaction between two species, giving rise to two products (this can be generalized to multiple reactants and mulitple products). aA bB cC dD This can also be written as: (Progressively looking more obscrure!)* aA1 bA2 cA3 dA4 0 aA1 bA2 cA3 dA4 A C2 H 6 B O2 E. g. : 2C2 H 6 15O2 12CO2 6H 2O C CO D H O2 2 0 v1 A1 v2 A2 v3 A3 v4 A4 0 vi Ai i A N a 1 v1 1 2 E. g. : N 2 3H 2 2 NH 3 B H 2 b 3 v2 3 C NH 3 c 2 v3 2 A C2 H 6 B O2 C CO2 D H 2O a2 b 15 c 12 d 6 a 2 v1 2 vC2 H6 2 b 15 v2 15 vO2 15 c 12 v3 12 vCO2 12 d 6 v4 6 vH 2O 6 The a, b, c, d define the stoichiometry of the reaction. These coefficients with some ‘sign adjustments’ as above are refered to as the stoichiometric coefficients (vi). The stoichiometric coefficients are positive for the products and negative for the reactants. * And you thought that tensorial notations were bad! A N a 1 v1 1 2 E. g. : N 2 3H 2 2 NH 3 B H 2 b 3 v2 3 C NH 3 c 2 v3 2 Stoichoimetric coefficient (vi) Change in no. of For the general case moles during reaction with vi and (n) (ni) Label Species Moles reacting A N2 20 1 20 (= 1 20) 1 = v1 B H2 60 3 60 (= 3 20) 3 = v2 C NH3 40 2 +40 (= +2 20) 2 = v3 In the reaction above, if we start with 20 moles of A (N2) and 60 moles of B (3H2), then we will get 40 moles of C (NH3). During a reaction, the change in the no. of moles of each species is proportional to the stoichiometric coefficients (vi). The proportionality constant is refered to as the extent of reaction (). For a small extent in the reaction we can write: d. The extent of reaction for the above example is 20. For the general reaction 0 vi Ai i ni vi Condition for reaction equilibrium (in terms of the chemical potentials and the stoichiometric coefficients) Earlier we had seen the condition for material equilibrium (eq. (13)): k i dni 0 (13) i 1 Multi-phase closed system, T, P constant P-V work only, Material equilibrium Now, let us obtain the condition for reaction equilibrium assuming that material equilibrium exists and is maintained throughout. Under material equilibrium the chemical potential of each species/component is the same in all the phases; hence, it does not matter in which phase each component exists (both for the reactants and the products). Further, this implies I can drop out the summation of phases from eq. (13) and write as below. k i 1 i dni 0 (14) Multi-phase closed system, T, P constant P-V work only, Phase equilibrium For a finite extent of reaction we had noted before: ni vi For an infinitesimal extent of reaction (d) we can write: dni vi d (15) k dni k Substituting eq. (15) in eq. (14) we get: i vi d i vi d 0 i 1 i 1 k Since this equation is valid for any arbitrary infinitesimal value of d, we can write: i vi 0 i 1 v 0 k Hence, the condition for reaction equilibrium is: i i 1 i (16) Multi-phase closed system, T, P constant P-V work only, Phase equilibrium Reaction equilibrium Eq. (16) states that: at reaction equilibrium the chemical potentials of the products balance that of the reactants. k k i 1 i 1 dni We had noted earlier that at constant T, P: dG i dni i i d T ,P k Hence: dG i i d i 1 At equilibrium G is minimized and dG/d = 0 (i.e we obtain Eq. (16)) T ,P The quantity on the LHS of eq. (16) is written also as: GReaction GR rG i vi k i 1 The variation of G versus can be plotted as in figure below. G eq. What happens when we mix two elements (say Ag and Au→ two crystals*)? When we mix two (or for that matter more) elements (A & B), the stable phase will be that with the lowest G. There are three options here (as we have seen in Chapter 4a): 1 Phase separation → A & B do not want to talk to each other 2 Formation of solid solution → A & B do not care about their environment 3 Compound formation → A & B prefer each other’s environment as compared to their own environment In a compound the each one of the components are fixed to their sub-lattices and hence the configurational entropy of the compound is zero. This is true in the case of a complete phase separation as well (i.e. the configurational entropy is zero). The solid solution is also called a disordered solid solution, in which case each component is randomly occupies a lattice point without any preference. In practice, there might be some tendency for ‘ordering’ (i.e. compound formation) or ‘clustering’ (phase separation) and in that case the ‘random configuration’ assumption will be violated. 1 Phase separation + = 2 Formation of solid solution 3 Compound formation * For the case of Au and Ag, solid solution will form (irrespective of the amount of Au or Ag). The Gibbs free energy change on mixing (for now we visualize mixing– soon we will see if they actually mix!) is: Gmix = (Gmixed state – Gunmixed state) = Hmix – T Smix. Hmix = (Hmixed state – Hunmixed) Hence, if we know two numbers (Hmix, Smix) our job is done! The game-plan is to find these numbers (especially, Hmix). Various models are used for this purpose and that can be quite confusing! Each one of these models come with their own baggage of assumptions (& hence approximations). The simplest model of mixing is the formation of the ideal solution. In an ideal solution AB bonds are energetically no different from the A-A or B-B bonds. This implies that (Hmix)Ideal solution = 0. If (Hmix)Ideal solution 0, which is usually found in practice (i.e. usually the mixing process is endothermic or exothermic), then we need a more ‘realistic’ computation of Hmix. One of the popular models is the regular solution model (which is based on the quasi-chemical approach). In real alloys the following factors come into the picture, which can lead to substantial deviation from the some of the models considered: (i) ordering (if Hmix is very negative), (ii) clustering (leading to deviation from the random configuration model, (iii) strain in the lattice due to size difference between the atoms (the quasi-chemical model will underestimate the change in internal energy on mixing), (iv) substantial size difference leading to the formation of a interstitial solid solution. Ideal solution In an ideal solution: (Hmix)ideal solution = 0 Hence Gmix Ideal solution H mix T .Smix Smix R( X A ln X A X B ln X B ) Click here to see derivation Gmix Ideal solution RT ( X A ln X A X B ln X B ) Gmix → 0 A Increasing T XB → B Regular solution model (quasi-chemical approach) No change in volume before and after mixing The regular solution model makes the following assumptions. A + B = AB (i) The enthalpy of mixing is only due to nearest neighbour atoms, (ii) The volume of A and B (per mole) is the same and there is no volume change due to the mixing process (interatomic distances and bond energies are independent of the composition), (iii) Point (ii) above also implies that there is no strain in the lattice. If no. of A-A bonds (per mole) is NAA, the no. of B-B bonds is NBB and the no. of A-B bonds is NAB and the corresponding bond energies (per bond) are given by EAA, EBB & EAB the internal energy of the solution is given by (Esolution): Esolution N AA EAA N BB EBB N AB EAB The change in internal energy on mixing (noting that since there is no change in volume): H P U PV H mix N AB E N AB E Where E EAB 12 ( EAA EBB ) Three scenarios are possible regarding the sign of E E < 0 → Hmix negative AB bonds are preferred over AA or BB bonds Sign of E E = 0 → Hmix is zero Ideal solution (no difference between AA, AB or BB bonds E > 0 → Hmix is positive AB bonds less preferred over AA or BB bonds Let us consider the scenarios a little further If E = 0 → Hmix is zero This implies an ideal solution (no difference between AA, AB or BB bonds For an ideal solution it can be shown that: N AB z N0 X A X B N0 → Avogadro’s No. z → No. of bonds per atom If E is not too negative or E is not too positive The equation in the case of the ideal solution for NAB can still be used as an approximation N AB z N0 X A X B This implies: H mix z N0 X A X B E X A X B H mix X A X B z N0 E E EAB 12 ( EAA EBB ) Gmix regular solution X A X B RT ( X A ln X A X B ln X B ) H mix T Smix This implies that for regular solutions is the key parameter determining the behaviour of the regular solution ( in turn depends on E, which is ‘some’ measure of the relative values of AA and AB bond energies). We can further consider the liquid and solid phases to be two regular solutions with it own ‘ parameter’: S & L. The effect of (& T) High T Could be –T or ‘nothing’ Low T Could be H, S or G <0 >0 We will understand these figures in the coming slides Understanding the G- composition curves: General aspects The parameter determines the sign of the Hmix is obviously zero for pure (unmixed) components As mixing leads to an increase in entropy and T is always positive (in K) –TSmix term is always negative The Gmix is determined just by the addition of the Hmix & –TSmix for each composition The Smix term only depends on the composition for a random solid solution <0 Hsolution < Hpure components High T As Hmix & –TSmix are both negative Gmix is always negative. Gmix gets more negative with increasing T due to the –TSmix. Low T The phase diagram of such a system will show complete solubility at high T and phase separation for a range of compositions (in the middle) at low T. Hsolution > Hpure components High T Hmix and –TSmix oppose each other at high ‘enough’ T, –TSmix wins at all compositions and Gmix is always negative At ‘low’ T, Hmix wins over –TSmix for some compositions (in the ‘middle’) and Gmix turns positive for this range of compositions. Low T Except at absolute zero (T), Gmix always decreases on the addition of a small amount of solute (even if Hmix gets ‘very’ positive). >0 Funda Check For the case where > 0 (Hmix > 0), why does mixing occur? In the regular solution model if > 0, then (Hmix > 0). This implies that A atoms prefers A neighbours and B atoms prefer B neighbours and this should lead to phase separation purele based on enthalpy considerations. However, at constant T & P it is G which determines the ‘energetics’ (stability) of the system and the term TS is negative for the mixing process. At ‘high enough’ temperatures, this TS offsets the Hmix term and Gmix is negative (i.e. the mixed state is energetically preferred). Funda Check How to understand the case where > 0 (Hmix > 0), at ‘low T’? In this case for a range of compositions Gmix is positive. If we plot the G-XB curves, we observe two points of inflection. All compositions between that labelled XM & XN in the G-XB plot will decompose to phases with compositions XM and XN. X1 and X2 are determined by the common tangent construction. As we shall see in the chapter on phase transformations, that the transformation (splitting into two phases) can occur by a spinodal mechanism or a nucleation and growth mechanism. The locus of the points like XM & XN at various temperatures will give us a region of phase separation (miscibility gap) in the phase diagram (the binodal curve). The locus of the points like XP & XQ will give us the spinodal curve. We will learn about many of the concepts in this page in later chapters Common tangent to the G-XB curve The Molar Free Energy So far what we have seen is the change in the quantities (H, S, G) on mixing. To obtain the molar free energy (G per mole) of an ‘alloy’ of a certain composition, we have to add the linear composition dependent term to the Gmix. The molar free energy of the system before mixing (Gunmix) is a linear combination of that of the pure elements (GA and GB). Gunmix X AGA X BGB Gmix Gunmix Gmix V PVm PVm nRT 1 n RT An ideal gas obeys in the equation: PV = nRT. The following assumptions are violated in the case of a ‘real’ gas: (i) the particles are point particles (no size) and (ii) there are no interactions between the particles. I.e. in a real gas, (i) the gas ‘atoms/molecules’ have a finite size & (ii) there are repulsive and attractive interactions between the atoms/molecules of the gas (intermolecular forces). The deviation from ideality can be measured using the Compressibility Factor (Z). The compressibility factor is a function of P & T (Z = Z(P,T)) (eq. (1)). Z can also be defined in terms of ‘V’ and ‘P’ (Eq. (2) , (3)). For an ideal gas Z = 1 for all T & P. In the limit that: (ci) P 0 or (cii) T , ideal behaviour is retrieved (Z 1) for all gases. In these cases the volume goes to infinity and the density goes to zero. The molecules are so far apart that their size does not matter & there are no forces between molecules to reckon with. Real Gases Ideal gas : PV nRT Z ( P, T ) PVm RT (1) P V ( Molar Volume) Vm n Z ( P, T ) Vm Vmideal (2) Z ( P, T ) P P ideal (3) • Here, Vmideal molar volume of an ideal gas at the same T and P as the real gas and Pideal is the pressure of an ideal gas at the same T and Vm as the real gas. Eq. (3) implies that, when Z < 1, the gas exerts a lower pressure than an ideal gas would. Fig.1 shows the plot of Z (= PVm/RT) verses P for some gases at 0C. Fig.2 shows Z vs P for CH4 at various temperatures. The following can be noted from the plots. At 0C some gases like H2 have only positive deviations from ideality, while some have negative deviations only and some have both positive and negative deviations. For a single gas like CH4 the deviations from ideality increase with a decrease in temperature (and the plot which looks akin to that for H2 in Fig.1 at 727C, starts to look like that for CH4 shown in Fig.1 at lower T (73C). Note the marked deviation from ideality for CH4 at 200K in Fig.2 (which is close to the critical temperature). We will see soon that when we reduced variables, the plots for most gases will coincide (known as the law of corresponding states, which was proposed by Van der Waals). Note the marked deviation from ideality for CH4 at 200K in Fig.2 (which is close to the critical temperature). 0C Fig.1 Increasing deviation from ideality: P T Fig.2 Equations of State for a Real Gas Many equations of state have been proposed. The most common one is that van der Waals equation. The constants ‘a’ and ‘b’ are different for different gases. The source of these terms is as marked in the equation. Note the term (Vm b), which arises due to the non-zero volume of the molecules and due to this the volume available for molecules to move around is less and is accounted for by the subtraction of ‘b’. The constant ‘b’ is approximately same as the molar volume of the solid or liquid. Z < 1 implies that attractive forces are dominant; while, Z > 1 implies that repulsive forces are dominant. nRT n a P a (1b) P V 2 Vm b RT (1a) V nb V m 2 RT a P 2 Vm b Vm (1c) Correction term for volume of the molecules Van der Waals coefficients Gas a (atm.L2/mol2) b (L/mol) Ar 1.337 3.2 102 CO2 3.610 4.29 102 He 0.0341 2.38 102 Xe 4.137 5.16 102 H2 0.2453 2.651 102 CH4 2.3 4.301 102 Correction term for intermolecular attractive forces Eq. (1c) can be written as (by multiplying by Vm/RT): V a PVm Z m RT Vm b RTVm Intermolecular repulsions make Z >1 RT a P V 2 V b m m (1c) 1 a b RTV 1 m Vm Intermolecular attraction makes Z < 1 <1 The denominator in the first term is less than one and hence the presence of ‘b’ and intermolecular repulsions tend to make Z >1. The second term (with ‘a’ & a negative sign) tends to make Z <1. The parameter ‘b’ is the liquid molar volume and hence b < Vm and b/Vm <1. Hence, we can use the series expansion using the formula below. 1 1 x x 2 x 3 ... 1 x PVm a 1 b2 b3 Z 1 b 2 3 ... RT RT Vm Vm Vm At low P, Vm is much larger than ‘b’ and we can write (ignoring the higher order terms) a 1 Z 1 b RT Vm At low T (& P), a/RT > b and [b(a/RT)] is negative and Z <1. Thus, P < Pideal and hence at low T intermolecular attractions are more important than intermolecular repulsions in determining the value of P. At high T and low P, [b(a/RT)] > 0 and Z >1. Thus, P > Pideal and this arises from the fact that at high T molecules (which have a higher velocity) crash into each other harder and this leads to a dominance of repulsions on value of P. Condensation of real gases An ideal gas will never condense. However, real gases with intermolecular forces, will condense if the T is less than a critical value (TC) and the pressure is ‘high enough’. The P-Vm diagrams for H2O can be used to understand the deviations from ideality and condensation. The solid curves in Fig.1 are isotherms. AT high-T (> 400C) the curve will be asymptotic to x, y-axis and we will have ideal behaviour. At 400C, curve M-N shows deviations from ideality. Fig.1 At the special T of 374C, the P-Vm curve shows marked deviations from ideality. The curve passes through the critical point ‘C’. The temperature corresponding to ‘C’ is called the Critical Temperature (TC). The corresponding molar volume is referred to as the Critical Volume (Vm,C) and the pressure as the Critical Pressure (PC). The critical values (TC, PC, Vm,C) are important parameters for a particular gas (and are tabulated later for a few gases). At T < TC, the gas begins to condense and we have a two phase field in the P-Vm diagram. Continued… We shall see later that the properties of the fluid change abruptly at around TC. In the part of the curve C-B, the fluid displays a liquid like density, but its viscosity is much lower than a typical liquid. This state of fluid is referred to as the supercritical fluid. Any fluid existing in a state with T > TC and P > PC is a supercritical fluid. The diffusion coefficients of species in such a fluid are much higher than that in typical liquids. Let us track the P-VM curve at 300C (which is < TC) along G-H-I-J-K-L. From G to H we have the gas phase and from K to L we have the liquid phase of water. This can be visualized using a piston with a ‘real’ gas like water vapour as below. From K to L we have compression of the liquid and hence the curve is steeper as compared to the curve GH, which corresponds to the compression of the gas. Between H and K we have a two phase mixture of vapour and liquid water. Any point between H and K (say I or J) ‘does not exist’. If the pressure is infinitesimally greater than the pressure corresponding to the horizontal line (PL+G (300C)) then we have only liquid. And if the pressure is infinitesimally less than (PL+G (300C)) then we have only gas. At 300C we can imagine liquid at K (with PL+G & Vm, Liquid) to be in equilibrium with gas at H (with PL+G & Vm, Gas). The physical meaning of a point like I arises as follows. If we follow the path Z to I, then the relative fractions of gas and liquid is given by the tie-line K-H and the lever rule (Fig.2b), where point I acts like the fulcrum.* Still, the point I does not exist it is used only for the computation of the phase fractions. * Refer to chapter on phase diagrams for explanation of the lever rule and the tie-line. Fig.2 Steep (a) (b) P slightly less than PL+G (300C)) P slightly greater than PL+G (300C)) Jump KH The process of conversion of gas to liquid at 300C along G-H-I-J-K-L involves abrupt start of liquefaction at H (Fig.2). This involves a sudden density change from the gas state to the liquid state, which is what we are accustomed to all this while. Interestingly, we can visualize a path X-M-N-Y (or an equivalent path), wherein X-M & N-Y are isochores and M-N is a isotherm. Fig.3. Here, we can go from the gaseous state (at X) to the liquid state (at Y), without causing a sudden change in the density. Fig.3 Fig.2 The Van der Waals (VdW) equation in the two phase region It is interesting to note that the VdW, which is meant for a gas, is able to able to capture the P-V behaviour (isotherm) in the liquid state (albeit approximately). The dotted curve O-P-Q-R-S in the two phase field corresponds to the plot of the Van der Waals (VdW) equation of state (Fig.1). Clearly the VdW equation is not applicable to the two-phase region (even more so that the equilibrium states in the two phase region do not exist!). However, there is a certain utility to the VdW equation in the two-phase region in the context of metastable equilibrium. The part R-Q-P is unphysical as volume is increasing with increase in pressure. The part of the curve S-R in Fig.1. can be conceived as a metastable state of super-heated liquid. At these pressures, the state should have been a gas, but is a metastable liquid. The part P-Q as super-cooled gas. At these pressures, the phase must have been a liquid, but persists as a metastable gas. Fig.1 The Critical State And regions close to it A fluid at the critical point is at its critical state. Critical constants (corresponding to the critical state) for some gases can be found in Table-1. The slope of the P-Vm curve is zero at the critical point and is negative on either side of it. Hence, we can write the equations as below. Note that the isothermal compressibility () diverges at the critical point. Vm At the critical point P 0 V P T Critical Point m T Critical point 1 Vm Vm P T Critical Point P Experimentally it is observed that T is finite & positive at the Critical Point V m P Also T V m Since, = infinity at the critical point, / can be finite if = at the critical point. This implies that CP ,m 2 P CV ,m TVm CV ,m TVm Critical Point T V m Gas PC (atm) Vm, C Cm3/mol TC K ZC TB Ar 48 75.3 150.7 0.292 411.5 CO2 72.9 94 304.2 0.274 714.8 He 2.26 57.8 5.2 0.305 22.64 O2 50.14 78 154.8 0.308 405.9 H2O 217.8 56 647.1 H2 12.83 33.16 Continued… The Critical State Fig.1 shows the variation in CP for water (saturate liquid water and saturated water vapour) versus T. Note that the y-axis is in log scale and that CP,m for both the phases diverge as the critical point (TC = 647 K = 374C) is approached. This implies that the value of CP,m very large close to the critical point. Fig.1 Continued… When we plot specific volume versus P isotherms at 380C, which is just above TC (= 374C) we observe a steep increase in V at nearly constant pressure (part AB in the 380C curve in Fig.2). Note that Fig.2a and 2b are the same, but with axes orientation changed. At temperatures much above TC (like 400C), steep variation which was observed at 380C, is not observed (i.e. the P-V variation is smoother). At T < TC we observe condensation, at low P we have the gaseous phase and at high P we have the liquid phase. Fig.3 shows the region close to the critical point ‘C’ in the P-T diagram. CC’ is an isochore, which is an extension of the X-C liquid-gas co-existence line. AB is an isotherm and corresponds to the same points as that marked in Fig.2. From A to B the fluid shows a rapid change in density and compressibility (i.e. a change from a ‘gas-like’ to ‘liquid-like’ behaviour). Similar steep change is also observed in entropy (S) and internal energy (U). (a) Fig.3 Fig.2 (b) Critical constants and reduced variables Using the critical constants we can define dimensionless quantities as: Tr V P T Vr m Pr Vm ,C PC TC If these reduce variables are used to express the states of gases, then to a good approximation, all gases shows the same P-Vm-T behaviour. I.e. if two different gases, say H2O & CO2, have the same Pr and Tr, then they will have the same Vr values. This is refered to as the law of Corresponding States. (Proposed by Van der Waals). This can be written as: Vr f ( Pr , Tr ) The function ‘f’ is the same for all gases to a good approximation Further, it can be shown that if the law of corresponding states is true, then the compressibility (Z) is a universal function of Pr and Tr. The approximation is true within a few percent for all gases, except those with large dipole moments. Z g ( Pr , Tr ) The function ‘g’ is the universal The origin of the law of corresponding states is related to the fact that the interatomic interactions can be expressed to a good approximation for a spherical molecules as a two parameter interatomic potential (e.g. the Lennard-Jones potential). This principle can be used to create, Z vs Pr plots for various values of Tr. I.e. separate Z vs P curves for each gas (at a given T) collapses into a single curve, as shown in Fig.1. Fig.1 Other Equations of state Many equations of state have been proposed. The most accurate two parameter equation is considered to be the Redlich-Kwong equation (Eq.(1) with ‘c’ and ‘d’ as parameters). RT c P (1) Vm d Vm Vm d T 1 2 The virial equation of state (2) is also commonly used. The virial coefficients (B(T), C(T), D(T), ... are function of T only and are determined from experimental P-V-T data for gases. B(T ) C (T ) D(T ) PVm RT 1 2 .... (2) 3 V V V m m m Phase Diagrams for Gases (P-V, P-T, T-V diagrams) (c) Schematic Diagram Fig.1 (b) We have already seen P-V diagrams showing gas to liquid transition and the supercritical state. Fig.1(a) is more realistic, while (b) & (c) are schematic. Let us consider the P-T phase diagram for CO2 (Fig.1(a). If the pressure is further increased (along the blue curve isotherm (T2) in Fig.1(b) we can observe the transition to the solid state. At lower temperature, along the green curve isotherm (T1), the gas directly condenses into a solid. (a) Schematic Diagram Continued… Pressurization above the critical temperature* (TC) isothermally (Fig.1b) leads to the following sequence of transformations: Gas Supercritical fluid Solid. Between the single phase regions there exists a two phase regions (e.g. the gas + liquid region or the solid + gas region). In (b), the dashed maroon line corresponding to the triple point in (a) is the the three phase coexistence line). The critical point in (a) can be thought of as a co-existence point between the gas, the liquid and the supercritical fluid. The usually expected sequence of transformations (i.e, Solid Liquid Gas on heating at a constant pressure), occurs only between the critical point pressure (PC) and triple point pressure (PTP). (b) * Temperature corresponding to the critical point. (a) Properties of Ideal versus Real Gases We have noted earlier that the compressibility factor (Z) gives us the deviation from ideality in terms of volume and pressure. We are also interested in this deviation between an ideal gas and a real gas with respect to enthalpy, entropy and other TD properties. The deviation from ideality for molar enthalpy (Hm) can be expressed as: H Deviation H D H mIdeal H mReal V H Deviation T m V dP ' 0 T P P T ,P T ,P Similarly, the deviation from ideality for molar entropy (Sm) can be expressed as: SDeviation SD Sm Sm Ideal Real T ,P SDeviation R V m dP ' T P P ' P’ is the dummy variable for integration T ,P One of the forms of the virial equation of state, is convenient to evaluate (Vm/T)P and compute the integrals as above. Currently, we will graphically see the difference between real and ideal gas behaviour. Gas-Solid Equilibrium (Pd-H system) Let us consider chamber with a piston containing H2 at pressure ‘P’ and pure solid Pd. Let the temperature be in the range 0C to 400C. (Fig.1) At any pressure, H2 molecule will first adsorb on the surface of Pd and then dissociate into atomic ‘H’ at the surface. Further, it will enter the solid and form an interstitial solid solution (usually ‘H’ is present in the octahedral void). Keeping the temperature constant, if the pressure of the gas is increased, then the concentration of ‘H’ in Pd (H/Pd ratio) will increase as shown in Fig.2. At low ‘P’ the -phase will exist, followed by the + mixture. The phases & are solid solutions. The existence of a pressure plateau (corresponding to the two phase mixture), implies that more hydrogen (as ‘H’) can be dissolved in the solid at relatively low pressures. If the pressurization is carried out above the critical temperature, then the + phase region is absent. Tc = 295C, PC = 19.8 atm. P Fig.1 H2 gas T Fig.2 Pd(Solid) Statistical Thermodynamics Clearly we are not going to do much justice to the subject here just a spattering! Some ‘thingamajigs’ we will encounter: Statistical Physics Macrostate Microstate Quantum mechanical picture of a microstate Statistical physics view of temperature The equilibrium state The steady state The Postulates of statistical thermodynamics The equal probability postulate The ergodic* hypothesis Ensemble, average over ensembles * Etymology. [Ergon (Greek) = Work] + [Hodos (Greek) = Way] = [Ergoden (German)] + [ic (English)] = [Ergodic] a dynamical systems term Statistical Thermodynamics Equilibrium statistical mechanics is also referred to as statistical thermodynamics[1]. This branch of science pertains to systems in thermodynamic equilibrium. The thermodynamic state also known as the macrostate of a system can be described/fixed by macroscopic parameters like P, T, V & ni (sufficient number of these). In contrast, the quantum state of the system (or the microstate) requires a large a large number of variables to describe. Given that system consists of a large number of particles*, it requires a large number of variables to describe the microstate [if ‘j’ is the microstate then the wave function (j) will be a function of a large number of spatial and spin coordinates]. In special cases, wherein the particles do not interact with each other, like the molecules in a ideal monoatomic gas; it would suffice to specify the quantum states available to each particle. Classically a microstate can be described by giving the positions and velocities of all molecules. The macrostate is experimentally accessible (observable), while the microstates are usually not (at least not easily!). [1] Founding fathers of the field include: J.C. Maxwell, L.E. Boltzmamm, J.W. Gibbs, A. Einstein. * Defined soon. At the heart of statistical thermodynamics are the concept of an ensemble and couple of postulates. In the macroscopic description of a system we carry out spatio-temporal averaging of quantities. Pressure for instance is obtained by averaging the momentum transferred per unit area per unit time by the molecules (1023 impacts occur per cm2 per second at 25C and 1 atm pressure). Instead of time averaging, we typically use the concept of a ensemble and the postulate: the measured time average of a macroscopic property in a system is equal to the average value of the property in the ensemble. An ensemble is a collection of a ‘infinite’ number of isolated systems, which are in the same macrostate as the system under consideration. Another way of looking at this is to consider an infinite system. What happens at different times in a portion of this system, already exists in elsewhere in the infinite system. So instead of time averaging, we can average over these spatially separated identical systems (in terms of their macrostate). Statistical Thermodynamics A brief overview... especially with respect to Entropy A system typically consists of a large number (~mole) of entities (atoms, molecules, ions,..) sometimes refereed to as ‘particles’. This implies that the system is associated with a large number of degrees of freedom (DoF) (microscopically). These DoF correspond to the translations, rotations, vibrations, electronic, etc. of the particles. This ‘implies’ that there are a large number of microstates corresponding to a single macrostate (i.e. there is a degeneracy with respect to the microstates). The macrostate, unlike the microstates, can be described with a few variables (P, T, ni, ...). The Ergodic theorem (a.k.a Gibbs postulate or second postulate of statistical thermodynamics) says that the macroscopic properties of a system can be found as probability weighed average of the values for microstates. In this scenario the internal energy (U) is given as: U pi i i pi → probability of microstate ‘i’. i → energy of the ith microstate. i → summation is over the ‘i’ microstates. This simplifies the matters as now we need to know only the probability of finding the ‘ith’ microstate, along with it energy (instead of all the degrees of freedom of the particles). The concept of the ensemble and types of ensembles Depending on the conditions we consider various kinds of ensembles. Canonical ensemble NVT constant A closed & rigid system cannot exchange particles with its surroundings, but it can exchange energy (in form of heat). The following parameters are constant: temperature T, volume V and the number of particles N. Micro-Canonical ensemble NVE constant An isolated system can neither exchange particles nor energy with its surroundings. The energy E, the volume V and the number of particles N are constant in these systems. Grand-canonical ensemble VT An open system exchanges particles and heat with its surroundings. The following parameters are constant temperature T, volume V and chemical potential µ. Isothermal-isobaric ensemble NPT In a closed system which exchanges energy with its surrounding via heat and work the following parameters are constant: temperature T, Pressure P and the number of particles N. NPH No special name WORK HEAT MASS Exchange work (not heat or mass) Surroundings Exchange mass and heat (not work) Surroundings Closed Work System Heat Open Mass System Material & Thermal reservoir E, P, N T, V, Grand-canonical ensemble Mathematical concept (Adiabatic piston paradox) Ensemble Exchange work & Heat (not mass) Surroundings You cannot have a system which exchanges mass and work, but no heat Closed Work Heat System T, P, N Isothermal-isobaric ensemble Thermal reservoir Exchange heat (not mass) No exchange of heat, mass or work Surroundings Surroundings Isolated System Heat Closed System Thermal reservoir Canonical ensemble T, V, N Micro-Canonical ensemble E, V, N The probability of a microstate as a function of its energy can be computed using the Boltzmann distribution. pi e i kT e i kT T → in Kelvin k → Boltzmann constant (kB) i This can be combined with the Gibbs Entropy equation to compute the Entropy: S k pi ln pi i Hence, if we know the probability distributions of the microstates, we can calculate the entropy of the system. Now, let us consider as system with constant volume and number of particles. Further, if a given macrostate corresponds to microstates of equal energy, then we can invoke the Laplace principle to assume that the all the microstates are equally probable. The Laplace principle is also knows as the principle of equal a priori probabilities or the first postulate of statistical thermodynamics. If is the total number of microstates, then the probability of occurrence of a given microstate (the ith state) is: (1/). pi 1/ Substituting into the Gibbs Entropy Equation and summing over all the number of states: 1 1 1 1 S k pi ln pi k ln k ln k ln i 1 i 1 S k ln This is the Boltzmann equation or the Boltzmann-Planck equation: S k ln S k ln S k ln w Often written with ‘small Omega’ or ‘w’ instead of the ‘capital Omega’. (or ) is the number of microstates available to the system. Currently, we are not asking questions like “will all these microstates will actually be explored by the system?” or “how long will it take to explore all these microstates?”. The system may have multiple macrostates and the entropy of the Mth macrostate is computed using the number of microstates corresponding to that macrostate. S M k ln M The number of microstates includes both ‘disordered’ and ‘ordered’ microstates. Since the number of ‘disordered’ microstates >> number of ‘ordered’ microstates. Hence, logarithm of the number of microstates can be approximated to the logarithm of the number of ‘disordered’ microstates. S k ln ordered disordered k ln disordered Residual Entropy The ‘classically’ defined entropy (macroscopic interpretation) is zero at zero Kelvin. In the statistical picture of entropy, there could exist a multiplicity of microstates even at zero Kelvin. These degeneracy/multiplicity of states gives rise to configurational entropy even at zero Kelvin. Let us consider an example. Argon crystal at 0K versus CO crystal at 0K. Unlike the Ar crystal, the linear CO molecule has two orientations (CO or CO). In a CO crystal (at 0K) if the up and down states are randomly positions on the lattice, this will give rise to configurational entropy. This will be the residual entropy of CO at RT. This is assuming that the ordered and disordered states have the same energy. The microstates for entropy at +ve Kelvin temperatures are given by the Maxwell-Boltzmann statistics, while residual entropy is described by Normal distribution. End Pd Polytropic process A polytropic process is a thermodynamic process that obeys the relation: PV^n =C where P is the pressure, V is volume, n is the polytropic index, and C is a constant. The polytropic process equation can describe multiple expansion and compression processes which include heat transfer. The term "polytropic" was originally created to describe any reversible process on any open or closed system of gas or vapor which involves both heat and work transfer, such that a specified combination of properties were maintained constant throughout the process. In such a process, the expression relating the properties of the system throughout the process is called the polytropic path. There are an infinite number of reversible polytropic paths between two given states; the most commonly used polytropic path is TdS/dT = C, which is a constant and is equal to zero for an adiabatic process. T is Temperature, S is Entropy. For ideal gases PV = constant. Cp/Cv is a measure of how much internal energy is getting converted to work in any compression/expansion process. Every change in internal energy impacts entropy changes and brings irreversibility in the process. Thus, it can also be said that the heat capacity ratio is the ratio between the enthalpy to the internal energy: y = Cp/Cv = H/U [ Enthalpy divided by internal energy] The value of n can vary from zero to infinity. n=0 for an isobaric process, in which the pressure of the system stays constant: ΔP = 0. n= infinity for an isochoric process, when volume stays constant i.e., ΔV = 0. The value of n is different in different thermodynamic processes. Some examples are given below. The polytropic index is a measure of the work done by the system. If you have a value for n, then you can determine the heat of compression by Eq. below. It is common to look at the measured suction and discharge temperatures determine the polytropic index. T discharge = T suction ^ n-1/n Typical example How high Cp of gas helps more work? To understand this relation, consider the following thought experiment. A closed pneumatic cylinder contains air. The piston is locked. The pressure inside is equal cylinder is heated to a certain target temperature. Since the piston cannot move, the volume is constant. The temperature and pressure will rise. When the target te heating is stopped. The amount of energy added equals CV ΔT, with ΔT representing the change in temperature. The piston is now freed and moves outwards, stopping as the pressure inside the chamber reaches atmospheric pressure. We assume the expansion occurs witho (adiabatic expansion). Doing this work, the air inside the cylinder will cool to below the target temperature. To return to the target temperature (still with a free pisto no longer under constant volume, since the piston is free to move as the gas is reheated. This extra heat amounts to about 40% more than the previous amount ad of heat added with a locked piston is proportional to CV at constant volume, whereas the total amount of heat added in a free piston is proportional to CP. The variation for Z with P & V for a few gases is given in Fig.1a,b. Fig.2 gives the variation of Z with P for methane (CH4) at various temperatures. As expected at high T and low P, behaviour tending to that of an ideal gas is observed. The deviation from ideality is more pronounced for methane as compared to that for N2 (which is a relatively inert gas), at 0C. Z < 1 implies that attractive forces are dominant; while, Z > 1 implies that repulsive forces are dominant. Note the marked deviation from ideality for CH4 at 200K in Fig.2 (which is close to the critical temperature). Fig.1 Fig.2 (b) (a) 1 Mole of N2 Nearly ideal T has profound effect < 100K and P has profound effect above 100 bar. 100K: Z increases from ~(1.2 to 1.4) with P = 100 bar 300 bar: Z increases from ~(1.2 to 1.4) with T = 200 K https://chem.libretexts.org/Bookshelves/Gen eral_Chemistry/Map%3A_General_Chemist ry_(Petrucci_et_al.)/12%3A_Intermolecular_ Forces%3A_Liquids_And_Solids/12.4%3A_ Phase_Diagrams