CHEM 331 Physical Chemistry Fall 2017 Legendre Transforms, Thermodynamic Potentials and Equilibrium We now pivot to the development of new Thermodynamic Potentials which will allow us to more naturally apply the 2 nd Law to chemical problems. As a prelude to this discussion, we dive into two examples of the application of the entropy maximum principle to chemical reactions, examples drawn from Atkin's book The 2 nd Law: Energy, Chaos, and Form. The book is written for a general audience, so it lacks many numerical details. But the examples are quite illustrative of the difficulties associated with trying to apply the entropy maximum principle to chemical reactions. And will thus motivate us to develop friendlier ways for applying the 2 nd Law to chemistry First, to a review of the Entropy. When defined according to: ds = the entropy can be used to write the Internal Energy. du = Q + W = T ds - P dv This means the internal energy is a function of S and V; U(S,V). This can be inverted to write the Entropy as a function of U and V; S(U,V). ds = du + dv Hence U(S,V) and S(U,V) are simply inverted forms of the same thermodynamic information. Further, recall, one consequence of the Clausius Inequality, which is the 2 nd Law codified in terms of S, is that spontaneous processes occurring within an isolated system, like the Universe, must increase the entropy of the system. Thus, a chemical reaction can proceed spontaneously toward products only if the entropy of the Universe concomitantly increases. For his first example, Atkins' considers the burning of Iron to form Ferric Oxide: 4 Fe(s) + 3 O 2 (g) 2 Fe 2 O 3 (s) As Atkins points out, ".. a kilogram of iron in fact requires about 300 liters of oxygen for its complete combustion under atmospheric conditions. The product is a little pile of oxide."
Additionally, considerable heat is dumped into the surroundings as the reaction proceeds. Atkins identifies two contributions to the entropy change of the Universe for this reaction: i. S surr = (Heat released into the Surroundings as a result of the reaction.) ii. S rxn (A change due to confinement of the Oxygen in the Oxide and due to the fact that the product bonds are less energetic than those of the reactant bonds.) Hence; S Univ = S surr + S rxn > 0 (~large positive) (~small negative) where Atkins has attached qualitative values to each term. The result is that S Univ > 0 as the reaction proceeds and therefore the reaction should occur spontaneously; which it does. As Atkins puts it, "Hence steel artifacts are intrinsically unstable, and automobiles have an ineluctable tendency to smolder to death." In his second example, Atkins analyzes an endothermic reaction; the decomposition of Dinitrogen Tetroxide: N 2 O 4 (g) 2 NO 2 (g) In this case heat is drawn into the system; meaning S surr < 0. Two other contributions to the entropy of the Universe involve S rxn and an entropy increase because of the increase in the system volume as the reaction proceeds, S vol > 0. Now to the nub of things; these three contributions to the overall entropy are slightly negative. This means that a reaction we know proceeds in a forward direction would not occur if these were only contributions to S Univ. Something is missing. It turns out this reaction does not proceed to completion. Instead it settles in on an equilibrium position; a mixture of N 2 O 4 and NO 2. At the equilibrium point, the gases mix together and we have to consider this contribution to S Univ ; S mix. This last contribution will be reasonably positive. So, we have: S Univ = S surr + S vol + S rxn + S mix > 0 where S Univ > 0 only as long as we do not proceed beyond the equilibrium position for the reaction. N 2 O 4 (g) 2 NO 2 (g) Now to why we are considering these examples. The central point is that we cannot predict the tendency of the reaction to proceed naturally unless we consider each contribution to S Univ. In particular, we do need to know how the reaction interacts with its surroundings; requiring an
estimate of S surr. This makes the use of the entropy maximum principle somewhat cumbersome. We must remember that the Clausius Inequality tells us the entropy of an isolated system, constant U and V, tends toward a maximum. But our chemical reactions are not isolated systems. In fact, a reaction occurring in a beaker sitting on a stir plate is distinctly not isolated from its surroundings. Instead occurs under conditions of constant temperature and pressure, where the surroundings provide a temperature and pressure reservoir for the reacting species. Therefore, we need to find a new system property or properties that allow us to make predictions based on the 2 nd Law without all the overhead of needing to specifically consider the interaction of the system and the surroundings in the manner above. Chemical reactions are rarely studied under conditions of constant entropy or constant energy. Usually the chemist places his systems in thermostats and investigates them under conditions of approximately constant temperature and pressure. Sometimes changes at constant volume and temperature are followed, as in bomb calorimeters. It is most desirable, therefore, to obtain criteria for thermodynamic equilibrium that are applicable under these practical conditions. A system under these conditions is called a closed system, since no matter can be transferred across the boundary of the system, although transfer of energy is allowed. Physical Chemistry, 4 th Ed. Walter J. Moore It turns out we need to carry out a Legendre Transform of the entropy or the internal energy into new thermodynamic potentials, which will allow for the more convenient application of the 2 nd Law to systems of interest. This type of transform carries all of the thermodynamic information of the original function into the transformed function. The Legendre Transform takes a function X(u 1, u 2, ) and transforms it into a new function Y(v 1, v 2, ). If Y is subsequently transformed appropriately, X is returned. Hence the transform Y preserves all the information of X. So, the transform is defined by: X(u 1, u 2, ) Y(v 1, v 2, ) where: v i = Y = - X In what follows we will carry-out a general Legendre transform and illustrate it's use for a case we have already considered, the transform of U(S,V) into H(S,P); although this is not exactly how we viewed things when we originally introduced the Enthalpy.
General Case Specific Case X(u 1, u 2, ) Y(v 1, v 2, ) U(S,V) H(S,P) v i = - P = Y = - X H = - PV - U (U + PV for consistency) dy = dh = ds + dp = - dx = VdP + PdV + du dx = du = ds + dv = = VdP + (P + )dv + ds = 0 = 0 = = TdS + VdP where u i = where V = So, we can invert S(U,V) into U(S,V) and then transform this into H(S,P); where the variables of interest are now S and P. We will now consider the additional cases of transforms to A(T,V) and G(T,P). H(S,P) = U + PV Enthalpy dh = TdS + VdP S(U,V) U(S,V) A(T,V) = U - TS Helmholtz Free Energy da = - SdT - PdV G(T,P) = U + PV - TS Gibbs Free Energy dg = - SdT + VdP
For each of these new thermodynamic potentials: dh = du + PdV + VdP = TdS + VdP = ds + dp = T & = V da = du - TdS - SdT = - SdT - PdV = dt + dv = - S & = - P dg = du + PdV + VdP - TdS - SdT = - SdT + VdP = dt + dp = - S & = V Each of these new thermodynamic potentials has a purpose. However, it is the Gibbs Function which will allow us to consider chemical systems operating under constraints of constant temperature and pressure. A summary of the differentials of all of our state functions is provided in the appendix below. Additionally, for chemical systems, we typically tabulate measurements of S, H and G as their measured values when the substance is in its standard state. How to define these standard states is sometimes tricky; especially when dealing with gases. So, how to define the standard states for chemical systems is provided in an appendix below. With this housekeeping out of the way, we now consider how to determine the state of a system at equilibrium. Determining a system's equilibrium configuration is in fact the fundamental problem of thermodynamics. This will be of particular interest when we begin to consider chemical problems. Let us suppose that two simple systems are contained within a closed cylinder, separated from each other by an internal piston. Assume that the cylinder walls and the piston are rigid, impermeable to matter, and adiabatic and that the position of the piston is firmly fixed. Each of the systems is closed. If we now free the piston, it will, in general, seek some new position. Similarly, if the adiabatic coating is stripped from the piston so that heat can flow between the two systems, there will be a redistribution of energy between the two systems. Again, if holes are punched in the piston, there will be a redistribution of matter (and also of energy) between the two systems.
Thus, the removal of a constraint in each case results in the onset of some spontaneous process, and when the systems finally settle into new equilibrium states they do so with the new values of the parameters U (1), V (1), N 1 (1) and U (2), V (2), N 1 (2). The basic problem of thermodynamics is the calculation of the equilibrium values of these parameters. [More generally], given two or more simple systems, they may be considered as constituting a single composite system. The composite system is termed closed if it is surrounded by a wall that is restrictive with respect to the total energy, the total volume, and the total mole numbers of each component of the composite system. The individual simple systems within a closed composite system need not themselves be closed. Thus, in the particular example referred to, the composite system is closed even if the internal piston is free to move or has holes in it. Constraints that prevent the flow of energy, volume, or matter among the simple systems constituting the composite system are known as internal constraints. If a closed composite system is in equilibrium with respect to certain internal constraints and if some of these constraints are then removed, the system eventually comes into a new equilibrium state. That is, certain processes which were previously disallowed be allowed or, in the terminology of mechanics, become virtual processes. The basic problem of thermodynamics is the determination of the equilibrium state that eventually results after the removal of internal constraints in a closed composite system. H.B. Callen Thermodynamics Castellan approaches the problem of equilibrium in a slightly different manner. Our aim now is to find out what characteristics distinguish irreversible (real) transformations from reversible (ideal) transformations. We begin by asking what relation exists between the entropy change in a transformation and the irreversible heat flow that accompanies it. At every stage of a reversible transformation, the system departs from equilibrium only infinitesimally. The system is transformed, yet remains effectively at equilibrium throughout a reversible change in state. The condition for reversibility is therefore a condition of equilibrium; from the defining equation for ds, the condition of reversibility is that T ds = Q rev The condition placed on an irreversible change in state is the Clausius inequality, which we write in the form T ds > Q Irreversible changes are real changes or natural changes or spontaneous changes. We shall refer to changes in the natural direction as spontaneous changes, and the inequality [above] as the condition of spontaneity. The two relations [above] can be combined into T ds Q
where it is understood that the equality sign implies a reversible value of Q. Gilbert W. Castellan Physical Chemistry, 3 rd Ed. Now to cases that are of interest to us; an isolated system, a system in contact with a temperature reservoir and a system in contact with temperature and pressure reservoirs. Isolated System 2 nd Law Statement Consider a system surrounded by an adiabatic, rigid and impermeable wall. This wall is restrictive with respect to energy, volume and matter. The system is completely isolated. For any system process du = 0 and dv = 0 along the entire process path. In this case, we can write: Q = du - W = du + P op dv = 0 So, the Clausius Inequality becomes: Or, ds = 0 S 0 This means that the entropy tends toward a maximum for the spontaneous processes of an isolated system. Equilibrium Now allow the system be made up of two subsystems; and. Each is separated by an adiabatic, rigid and impermeable wall and is at a specified temperature. The internal constraint of adiabaticity is removed and heat is allowed to flow from one subsystem to another.
If heat flows, it is specified that it will flow in the direction indicated above. Now, d S 0 Since, ds = ds + ds, we have: ds + ds 0 This gives, + 0 Since, as specified above, Q = - Q and Q = + Q, we have: Q 0 If heat flows, then T > T. If the system is at equilibrium, then T = T. Constant Temperature Processes 2 nd Law Statement Now our system is contained within a rigid diathermal wall and placed in a temperature reservoir. Thus, all internal processes will be isothermal. Again we start with the Clausius Inequality, ds Writing Q = du - W, we have: or, T ds du - W - du + T ds - W Now, since our processes are isothermal: So, - du + T ds = - du + d(ts) = - d(u - TS) = - da
- da - W Or, finally: da W This provides us with an interpretation for the Helmholtz Energy; A represents the maximum work available as a result of an isothermal process. Thus, A is sometimes referred to as the "Work Function". If W = - P op dv, then: or, da - P op dv da P op dv Thus, for isothermal, constant volume process: Or, da 0 A 0 This means that the Helmholtz Energy tends toward a minimum for spontaneous isothermal, isochoric processes. Equilibrium Now allow the system be made up of two subsystems; and. Each is separated by an diathermal, rigid wall and is at a specified pressure. The internal constraint of rigidity is removed and the volume of each subsystem is allowed to change according to the diagram below, if a change does in fact occur.
Now, da + da 0 Since, da = - P dv for an isothermal process, we have: - P dv - P dv 0 If the volume changes of the subsystems occur according to the diagram above, then dv = + dv and dv = - dv. So: (P - P ) dv 0 If the volumes change, then P > P. If the system is at equilibrium, then P = P. Constant Temperature and Pressure Processes 2 nd Law Statement Finally, we consider a system that is contained within a piston surrounded by a diathermal wall and placed in a temperature reservoir. The piston works against a pressure reservoir. Thus, all internal processes will be isothermal and isobaric. Again, we start with the Clausius Inequality, ds We now write Q = du - W or Q = du + P dv - W a, where have split the work into terms representing PV-work and other "available" forms of work, W a. These other forms of work may be electrical, chemical, gravitational, etc. The Clausius Inequality is now: or, T ds du + P dv - W a - du + T ds + P dv - W a As before, we can write TdS as d(ts) because all processes are isothermal. Similarly, we can write PdV as d(pv) because of the constant pressure constraint. Thus, our terms on the left in the above equation can be written as: - d(u + TS - PV) - W a
Invoking the definition of G, we have: or, - dg - W a dg W a This gives us a physical interpretation for the Gibbs Free Energy. G represents the maximum non-pv work available during an isothermal, isobaric process. If no additional work is available, then W a = 0 and we have: or, - dg 0 G 0 This means that the Gibbs Free Energy tends toward a minimum for spontaneous isothermal, isobaric processes. Changing our viewpoint slightly, we can write this minimization principle as involving both enthalpic and entropic considerations: So, G = U +PV - TS = H - TS G = H - (TS) If we have a constant temperature and pressure process, this gives us: G = H - T S which is how the Gibbs Free Energy is introduced to freshmen chemistry students. This mean that the requirement for G < 0 for a spontaneous, isothermal and isobaric process can be achieved by having H < 0 and/or S > 0. Many of our chemical reactions occur open to the atmosphere in a vessel such that the reaction system is in thermal equilibrium with its surroundings. Thus, G rxn = H rxn - T S rxn
In order that the reaction occur spontaneously: G rxn < 0. If G rxn = 0, then the reaction is at equilibrium. Atkins, in his book on the 2 nd Law explains: Consider a reaction that liberates energy as heat (such reactions are called exothermic). Suppose the reaction also reduces the entropy of the system itself. For instance, this is true for the oxidation of metallic iron; we saw that the reaction liberates heat, but reduces the entropy of the substances overall (largely because the large volume of gaseous oxygen collapses into the tiny heap of oxide). Suppose, furthermore, that we want to harness the energy that the reaction produces, not merely to heat the world, but to work in it. For instance, we might be burning iron in a furnace and using the energy to drive some kind of vehicle (burning coal would be a more familiar example). Since transporting the energy released by the reaction to the outside world as (quasistatic) work does not change the entropy of the surroundings, we are now confronted with an overall decrease of the Universe's entropy, because the reaction substances undergo a reduction of entropy, but there is no change in the surroundings. It follows that the conversion of all the energy released by this tye of reaction into work is not a natural process. Note carefully the following distinction. All the energy released by a reaction may emerge into the surroundings as heat, for that increases their entropy; not all the energy released may emerge into the surroundings as work, for if it did the overall change of entropy would be negative, and the Universe would have shifted spontaneously to a less probable state. Although not all the energy released by a reaction is available for doing work, perhaps if we allow some of the energy to escape as heat, enough entropy may be generated in the surroundings for the process to be spontaneous, even though we withdraw the remainder for the change of energy as work. We can then ask the following question; what is the minimum amount of energy that must leak into the surroundings as heat in order to generate enough entropy there to allow the reaction to proceed spontaneously? Suppose the reaction reduces the entropy of the system by an amount Entropy change. In order for the reaction to proceed spontaneously, at least this amount of entropy must be generated in the surroundings But we have seen that the entropy generated there is always given by the expression (Heat supplied)/temperature. Therefore the minimum amount of energy that must be supplied as heat to the surroundings by the exothermic reaction is obained by equating these two expressions and solving for Heat supplied. Clearly the minimum heat that must be released to the surroundings is the product of temeprature and the reduction of entropy: Minimum heating = Temperature x (Entropy change). It follows that the energy not available for doing work when the reaction occurs is equal to the expression on the right. This is normally written symbolically as T S. On the other hand, the energy that is available for doing work is the difference between the total energy released and the amount we have just calculated. In other words, the free energy, the energy available for doing work, is given by: Free energy = (Total energy) - Temperature (Entropy change).
The free energy* ofa reaction is its single most important thermodynamic property, and it will now stand at the center of our stage, just as the man who introduced it, Josiah Gibbs, is the single most important contributor to chemical thermodynamics. *The quantity normally considered by the chemist is the Gibbs free energy: it relates to changes taking place when the pressure is constant. A slightly different property, the Helmholtz free energy, arises when that changes are taking place at constant volume. We are troubling to make the distinction here, just as we are not distinguishing between internal energy and enthalpy. P.W. Atkins The 2 nd Law: Energy, Chaos, and Form Most chemical reactions are Exothermic ( H rxn < 0) and are driven forward by their exothermicity; S rxn being relatively unimportant. Historically this led to the mistaken assumption that chemical reactions were driven by heat evolution. Thermodynamics was not quickly applied to chemistry even though there had long been an interest in the heat liberated during chemical reactions. Lavoisier and Laplace had studied heat output, both in combustion and respiration. Germain Henri Hess (1802-1850) had enunciated a limited form of the law of conservation of energy with his law of heat summation, in which he concluded that the heat liberated in a chemical process is independent of the path by which the process is carried out. Beginning in 1852 more extensive measurements of heats of reaction were undertaken by Julius Thomsen (1826-1909) in Copenhagen and Marcelin Berthelot in Paris, who considerably refined their equipment and the techniques of thermochemical measurements during the next decade. The Berthelot bomb for measuring heats of combustion, developed in 1881, is essentially the one used today. For a time these studies were based on the assumption that chemical forces were proportional to the heat evolved during a chemical reaction. Aaron J. Ihde The Development of Modern Chemistry Bertholet and Thomsen codified their initial observations in the Bertholet- Thomsen Principle: All chemical changes are accompanied by the production of heat and those processes which occur will be ones in which the most heat is produced.
Marcelin Berthelot Julius Thomsen http://en.wikipedia.org/wiki/marcellin_berthelot http://en.wikipedia.org/wiki/hans_ Peter_J%C3%B8rgen_Julius_Thomsen Of course, this Principle could not account for the fact that Endothermic reactions, H rxn > 0, do occur. So, it was rather short lived and is now mostly of historical interest. Endothermic reactions do occur because they can be driven by a favorable entropy change. We must keep in mind that it is the Gibbs function that determines the spontaneity of chemical reactions. Two examples of entropically driven chemical reactions are the reaction of Barium Hydroxide and Ammonium Nitrate and the combustion of Peroxyacetone. Ba(OH) 2 8H 2 O(s) + 2 NH 4 NO 3 (s) Ba(NO 3 ) 2 (aq) + 2 NH 3 (aq) + 10 H 2 O 2 C 9 H 18 O 6 (s) + 21 O 2 (g) 18 CO 2 (g) + 18 H 2 O(g) The first of these is very endothermic ( = +62.3 kj/mol at 298.15K) and can produce temperatures as low as -25 o C to -30 o C for even small reaction mixtures. This reaction is entropically driven ( = 406 J/K mol at 298.15K) by the large number of aqueous products; these will be much more entropically favored than the relatively ordered solid reactants. This reaction proceeds with (298.15) = -60.2 kj/mol. A nice demonstration of the endothermicity of this reaction can be found at http://www.youtube.com/watch?v=9krs8ty7ajy. The second reaction, the combustion of Peroxyacetone, is also entropically driven; S rxn > 0. This reaction is an example of a heatless explosion; H rxn ~ 0. In this case it is the very large number of gaseous products, with their large entropy, that cause the reaction to be spontaneous. A demonstration of this reaction can be found at http://www.youtube.com/watch?v=gglbzfusirm.
We are now in a position to look back and examine why the state functions H(S,P), A(T,V) and G(T,P) were introduced. In both the energy [U] and entropy [S] representations the extensive parameters play the roles of mathematically independent variables, whereas the intensive parameters [T and P] arise as derived concepts. This situation is in direct contrast to the practical situation dictated by convenience in the laboratory. The experimenter frequently finds that the intensive parameters are the more easily measured and controlled and therefore is likely to think of the intensive parameters as operationally independent variables and of the extensive parameters as operationally derived quantities. The extreme instance of this situation is provided by the conjugate variables entropy and temperature; [U(S,V) vs. G(T,P)]. No practical instruments exist for the measurement and control of entropy, whereas thermometers and thermostats, for the measurement and control of the temperature, are common laboratory equipment. It is, perhaps, superfluous at this point to stress again that thermodynamics is logically complete and self-contained within either the entropy [S(U,V)] or the energy [U(S,V)] representations and that the introduction of the transformed representations is purely a matter of convenience. This is, admittedly, a convenience without which thermodynamics would be almost unusably awkward, but in principle it is still only a luxury rather than a logical necessity. H.B. Callen Thermodynamics This last is most important. We have identified a new property of systems, which is useful for constant T and P processes. And, the 2 nd Law can now be codified in the statement G 0. This does for us what we want. From now on, we will not need to referred to heat engines or consider only isolated systems. We can now examine the chemical reaction occurring in a beaker, open to the atmosphere and regulated by a thermostat.
Appendix - Thermodynamic State Functions Internal Energy U(S,V) * du = ds + dv = T ds - P dv U(T,V) du = dt + dv = C v dt + T dv Enthalpy H(S,P) * dh = ds + dp = T ds + V dp H(T,P) dh = dt + dp = C p dt - C p dp Entropy S(U,V) * ds = du + dv = du + dv S(T,V) ds = dt + dv = dt + dv S(T,P) ds = dt + dp = dt - V dp
Helmholtz Free Energy A(T,V) * da = dt + dv = - S dt - P dv Gibb s Free Energy G(T,P) * dg = dt + dp = - S dt + V dp * Fundamental thermodynamic relationships. According to Bromberg: In mechanics a conservative field is one for which a force is derivable from a potential. An analogous situation exists in thermodynamics in which an intensive property is obtained from the various thermodynamic functions, thus P = - ( U/ V) S. Work terms arise from the product of an intensive property with its associated extensive property, for example, PdV. When we compare these work terms with the definition of work, Fdx, the intensive properties such as P take the form of generalized forces. For this reason, the functions U, H, A, and G are often referred to as potentials. Heat is also measured by the product of an intensive and an extensive property; in the expression TdS, the term T is the intensive and S the extensive property.... the four potentials [U, H, A, G] are written in terms of their natural variables [ * relationships]. The energy U is a function of the extensive properties of the system, S and V. What we have accomplished in constructing H, A, and G from U is to substitute an intensive property for its associated extensive property. Enthalpy is generated from energy by replacing the extensive property V by its associated intensive property P. The Helmholtz energy A is generated by replacing S by T; and G is generated by simultaneously replacing S and V by T and P. These can be regarded as analogous to coordinate transformations such as the transformation from cartesian to polar coordinates. Here the transformation involves replacing an extensive property by its associated intensive property. When viewed in this light, the functions H, A, and G are simply the energy transformed into a different set of variables. In mathematics, such a transformation is known as a Legendre transformation. Physical Chemistry, 2 nd Ed. J. Phillip Bromberg
Appendix - Standard States for U, S, H and G Liquids and Solids The standard state values of our thermodynamic functions for a pure liquid or solid are given by: U o (T) = U(T, P o ) S o (T) = S(T, P o ) H o (T) = H(T, P o ) G o (T) = G(T, P o ) Here P o = 1 bar, as set by the International Committee on Weights and Measures in 1986. Prior to this P o = 1 atm. Gases The standard state values of our thermodynamic functions for gases are defined so as to depend only on the properties of real gases in the limit of zero pressure. However, the standard entropy cannot be simply defined as the zero pressure value because this is infinite. To get around this problem, for any of our thermodynamic functions F(T,P), generally F o (T) is defined as: F o (T) = F ideal (T, P o ) (Eq. 1) where F ideal (T, P) is equivalent to F(T, P) for the real gas in the limit of zero pressure: This definition is not practical in the sense that we are not able to determine F o (T) from measurements on real gases; it requires a measurement of F(T,P) for an Ideal Gas. However, this definition can be made practical. First note that: (Eq. 2) F ideal (T, P) = + (Eq. 3) This can be substituted into Equation 3 to obtain: (Eq. 4)
Since value: is not pressure dependent, we can pull it out of the "limit" and solve for this = (Eq. 5) This allows us to write F o (T) of Equation 1 in terms of measureable quantities: F o (T) = (Eq. 6) Now to our specific thermodynamic functions: Enthalpy As a consequence of Joule's Experiment, we know: = 0 (Eq. 7) for an Ideal Gas. Hence, we can write: H o (T) = = (Eq. 8) And, this is as we defined the standard enthalpy previously. Entropy For an Ideal Gas, we have: = - V = = (Eq. 9) So, (Eq. 10) This allows us to write the standard entropy for a gas as: S o (T) = (Eq. 11)
Entropy (J/K mol) So, although S(T,P) becomes infinite in limit of zero pressure, so does. However, the difference between these values remains finite and tends toward a well defined extrapolated value. Thus, this way of expressing S o (T) makes it clear that the standard value depends only on the properties of the real gas in the low pressure limit. This is usually specified by saying "the standard state is the ideal gas state at pressure P o ". This state of affairs can be seen for CO 2 gas. A plot of S(298.15K, P) vs. P gives us: 234 232 230 Entropy of CO 2 at 298.15K 228 226 224 222 220 218 216 214 0 0.2 0.4 0.6 0.8 1 1.2 Pressure (bar) We see that a plot of vs. P tends to minus infinity:
- Ideal Value of S (J/K mol) R ln (P/P o ) 0-2 -4-6 -8-10 -12-14 -16-18 -20 0 0.2 0.4 0.6 0.8 1 1.2 Pressure (bar) The difference can be extrapolated to zero pressure and will give a result of 213.74 J/K mol. It should be noted that: S ideal (T,P) = (Eq. 12) This gives us the entropy of CO 2, if it were behaving Ideally. All of these curves, plotted on a single graph, schematically give us:
Gibbs' Free Energy For completeness, the standard state value for the Gibbs' Free Energy of a gas is given by: G o (T) = (Eq. 13)