Equation second Gibbs-Duhem

At the addition of not all components or all but in other proportions, the composition of the source solution noticeably changes, but the addend in the equation (1.11) Ndg. 0. Equating equations (1.11) and (1.12), we abtain the second Gibbs-Duhem Equation ... 

It is strictly for convenience that certain conventions have been adopted in the choice of a standard-state fugacity. These conventions, in turn, result from two important considerations (a) the necessity for an unambiguous thermodynamic treatment of noncondensable components in liquid solutions, and (b) the relation between activity coefficients given by the Gibbs-Duhem equation. The first of these considerations leads to a normalization for activity coefficients for nonoondensable components which is different from that used for condensable components, and the second leads to the definition and use of adjusted or pressure-independent activity coefficients. These considerations and their consequences are discussed in the following paragraphs. 

In experimental investigations of thermodynamic properties of solutions, it is common that one obtains the activity of only one of the components. This is in particular the case when one of the components constitutes nearly the complete vapour above a solid or liquid solution. A second example is when the activity of one of the components is measured by an electrochemical method. In these cases we can use the Gibbs-Duhem equation to find the activity of the second component. 

The set of basic equations is completed by the Gibbs-Duhem (the local formulation of the second law of thermodynamics) and the Gibbs relation (which connects the pressure P with the other thermodynamic quantities), which we will use in the following form ... 

Finally, consider a two-phase, two-component system in which the two phases are separated by an adiabatic membrane that is permeable only to the first component. In this case we know that the temperatures of the two phases are not necessarily the same, and that the chemical potential of the second component is not the same in the two phases. The two Gibbs-Duhem equations for this system are... 

Two methods may be used, in general, to obtain the thermodynamic relations that yield the values of the excess chemical potentials or the values of the derivative of one intensive variable. One method, which may be called an integral method, is based on the condition that the chemical potential of a component is the same in any phase in which the component is present. The second method, which may be called a differential method, is based on the solution of the set of Gibbs-Duhem equations applicable to the particular system under study. The results obtained by the integral method must yield... 

In the second case the liquid and vapor are at equilibrium in a closed vessel without restrictions. An inert gas is pumped into the vessel at constant temperature in order to increase the total pressure. For the present we assume that the inert gas is not soluble in the liquid. (The system is actually a two-component system, but it is preferable to consider the problem in this section.) The Gibbs-Duhem equations are now... 

The expressions for the derivatives become somewhat simpler when one of the components is not present in one of the phases. For the purposes of discussion we assume that the second component is not present in the double-primed phase. Then the two Gibbs-Duhem equations become... 

The first term on the right-hand side f(° is the standard chemical potential of component i, the second term is the ideal mixture contribution, and the third term is the nonideality caused by the hard-sphere interactions. The chemical potential of the hard-sphere droplet (component g) is obtained using the Gibbs—Duhem equation... 

In practice AG is known for a given T, p and x2 so that the other quantities based on the ratio a must be calculated from the excess functions for the mixture. Differentiation of eqn (26) with respect to yields, using the Gibbs-Duhem equation, In (fi/f2), and hence a. A second differentiation yields d In a/dxj. If eqn (27) is used to fit the Ge data then these quantities can be calculated from the /-coefficients. The arithmetic is tedious but a computer program can be used to advantage here (Blandamer et al., 1975b). Because the... 

Although the correlations provided by the Margules equations for the two sets of VLE data presented here are satisfactory, they are not perfect. The two possible reasons are, first, that the Margules equations are not precisely suited to the data set second, that the data themselves are systematically in error such that they do not conform to the requirements of the Gibbs/ Duhem equation. 

There are two ways to derive the Gibbs-Duhem equation for a system (1) Subtract the fundamental equation from the total differential of the thermodynamic potential. (2) Use a complete Legendre transform. As an example of the second method, eonsider fundamental equation 3.3-10 for a single reactant ... 

Tills is a special fonn of the Gibbs/Duhem equation. Substitutionof —dx2 for dxi in the second tenn produces ... 

This equation, which is one example of the Gibbs-Duhem equation, shows that changes in the partial molar volume of one component may be related to changes in the same quantity for the other component. Experimentally, it means that one only has to measure one partial molar volume as a function of composition provided one has a value of the second partial molar volume at a reference point. In order to illustrate this point, equation (1.4.8) is written in a form suitable for calculating ua from ug ... 

The first term in the last equation vanishes because % is unchanged. The second and third terms add up to zero according to the Gibbs-Duhem equation. There is left only... 

These data can be studied in two ways. The first is to use the Gibbs-Duhem equation and numerical integration methods to calculate the vapor-phase mole fractions, as considered in Problem 10.2-6. A second method is to choose a liquid-phase activity coefficient model and determine the values of the parameters in the model that give the best fit of the experimental data. We have, from Eq. 10.2-2b, that at the jth experimental point... 

AH and AV are molar enthalpy and molar volume of mixing, respectively. The heat of mixing is negligible in most cases. In an isobaric system the second term vanishes, and the Gibbs-Duhem equation takes the form ... 

The second term on the rhs can be replaced with the Gibbs-Duhem equation (3.4.10), so (3.6.5) becomes... 

But at fixed T and P, the first term on the rhs is zero by the Gibbs-Duhem equation, and the second term can be rewritten using (7.4.15). The result is... 

Figure 3.16 Partial molar Gibbs energy of Au in the alloy Ag u, as function of the mole fraction of Ag. Two sets of values are shown The first set (squares) was calculated with the Gibbs-Duhem equation, as described in Section 3.2.3, and the second set of data (circles) was calculated with the Duhem-Margules equation, temperature 500 °C.

The second quantity obtained from the Gibbs-Duhem equation is the relative surface excess which is the difference of the amount of a substance in the interphase and the relative amount of solvent (s) expressed by the ratio of the mole fractions r, and x. ... 

A second coefficient can be derived from the Gibbs-Duhem equation if, instead of the potential, the chemical potential is kept constant... 

The second equation is Young-Laplace equation for a bubble and P e is the vapor pressure of an embryo in equilibrium with the liquid. Using the above equation along with Gibbs-Duhem equation it can be shown that the radius of an embryo at equilibrium conditions are ... 

Now, derived from the second law of tliermodynamics, the Gibbs-Duhem equation relates the activities of the components in a phase (McQuarrie and Simon 1997 Berry, Rice, and Ross 2000 Levine 2008 Atkins and Paula 2009). At constant temperature and pressure, for a two-component system, it reads... 

A second relation between pi and pg i supplied by the Gibbs-Duhem equation for an athermal binary system ... 

By using Gibbs-Duhem s equation, it follows that the sum of the terms on the left hand side of the second line of eq 14.17 times c is zero. As the second line of eq 14.17 holds for an arbitrary measurable heat and molar fluxes it follows that the resistivities in the matrix are dependent, and satisfy... 

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