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Equilibrium requirement

If a system is eoupled with its enviromnent tlirough an adiabatie wall free to move without eonstraints (srieh as the stops of the seeond example above), meehanieal equilibrium, as diseussed above, requires equality of the pressure p on opposite sides of the wall. With a diathemiie wall, themial equilibrium requires that the temperature 0 of the system equal that of its surroundings. Moreover, it will be shown later that, if the wall is pemieable and pemiits exehange of matter, material equilibrium (no tendeney for mass flow) requires equality of a ehemieal potential p. [Pg.332]

It follows that, because phase equilibrium requires that the chemical potential p. be the same in the solution as in the gas phase, one may write for the chemical potential in the solution ... [Pg.360]

Ashton solved this problem approximately by recognizing that the differential equation, Equation (5.32), is but one result of the equilibrium requirement of making the total potential energy of the mechanical system stationary relative to the independent variable w [5-9]. An alternative method is to express the total potential energy in terms of the deflections and their derivatives. Specifically, Ashton approximated the deflection by the Fourier expansion in Equation (5.29) and substituted it in the expression for the total potential energy, V ... [Pg.292]

Now to complete the solution of the problem one would need to know the solution laws for iron, and a small amount of nickel, in each of these carbide phases, since equilibrium requires that and a , in the segregated carbide must be 0-74 and 0-08 respectively as well as a., being 0-18. At present nothing is known about these laws except that the metal atoms might well be randomly distributed in the carbide phase, in which case, as an example. [Pg.1110]

Seemingly other problems arise in the treatment of the rate of adsorption, viz. the probability that a gas particle will reach on impact a free adsorption site on a partially covered surface, and the probability that it will remain attached there. Since, however, the rates of adsorption and desorption are connected by the equilibrium requirements, essentially the same problems have to be solved in a theoretical evaluation of both the rates. In practice,... [Pg.352]

The liquid sample must be large enough that all does not evaporate before equilibrium is achieved, The condition of phase equilibrium requires that two (or more) phases be present. [Pg.225]

Conditions of phase equilibrium require that the chemical potential of polymer in each phase and that of solvent in each phase be equal ... [Pg.181]

Another general type of behavior that occurs in polymer manufacture is shown in Figure 3. In many polymer processing operations, it is necessary to remove one or more solvents from the concentrated polymer at moderately low pressures. In such an instance, the phase equilibrium computation can be carried out if the chemical potential of the solvent in the polymer phase can be computed. Conditions of phase equilibrium require that the chemical potential of the solvent in the vapor phase be equal to that of the solvent in the liquid (polymer) phase. Note that the polymer is essentially involatile and is not present in the vapor phase. [Pg.183]

From the equilibrium requirement that the chemical potential involving all ionic species be uniform throughout the phase boundary, the distribution of ions within the electrical double layer can be expressed by the Boltzmann equation ... [Pg.457]

The paradox arises because (1) and equilibrium require a formulation as in (3), where most of Xt that is formed returns to reactants and only a small... [Pg.140]

Hi/9)2, so that material equilibrium requires pi)i = 1)2- The statistical parameters thus possess the essential property of chemical potentials. [Pg.482]

Phase equilibrium requires that A2 = Al and hence that the integral vanish. All conditions are satisfied if the points 1 and 2 are located such that the areas A = B. This geometry defines the Maxwell construction. It shows that stable liquid and vapour states correspond to minima in free energy and that AL = Ay when the external pressure line cuts off equal areas in the loops of the Van der Waals isotherm. At this pressure that corresponds to the saturated vapour pressure, a first-order phase transition occurs. [Pg.510]

Let us apply Equation (6.8) to the two-phase liquid-vapor equilibrium requirement for a pure substance, namely p = p T) only. This applies to the mixed-phase region under the dome in Figure 6.5. In that region along a p-constant line, we must also have T constant. Then for all state changes along this horizontal line, under the p—v dome, dg = 0 from Equation (6.8b). The pure end states must then have equal Gibbs functions ... [Pg.142]

It will be obvious from the description of Lewis s and Donnan and Barker s experiments that equilibrium is assumed to establish itself during the time of contact between the mercury or air surface and the liquid in fact this point was checked by increasing the time and showing that the result was not affected, i.e., that no further quantity of the solute was removed from solution. Experiments to decide this question had, however, been made at an earlier date by Wilhelm Ostwald. The strict definition of an equilibrium requires that it should be independent of the mass of the phases in contact thus, a soluble substance and its concentrated solution are in equilibrium at a given temperature and pressure, and this obviously remains unaffected by altering the quantity of either solid substance or solution. Ostwald placed a quantity of charcoal in a given volume of dilute hydrochloric acid and determined the decrease in concentration after a short time. If, then, a part of either the charcoal or the dilute solution was... [Pg.50]

FIGURE 7 Pericyclic reactions in the biosynthesis of giffordene and 7-methyl-cyclooctatriene. (a) The [1.7]-hydrogen shift of the thermolabile undeca-(1,3Z,5Z,8Z)-tetraene generates undeca-(2Z,4Z,6E,8Z)-tetraene (giffordene), the major product of the brown alga G, mitchellae. (b) The thermolabile nona-(l,3Z,5Z,8E)-tetraene cyclizes at ambient temperature rapidly to 7-methylcy-cloocta-l,3,5-triene. At ambient temperature, the bicyclic isomer does not contribute to the equilibrium (requires s=80°C). [Pg.107]

As a function of the new variables, water dissociation equilibrium requires... [Pg.329]

Because the initial condition of equilibrium requires that the first term on the right-hand side of the first equality vanishes, this equation simply becomes... [Pg.361]

We now come to internal metal contacts in ISEs without an internal solution. As discussed above, systems without internal electrolytes are used very often, with both solid and liquid membranes. Obviously, the condition of thermodynamic equilibrium requires that common electrically-charged particles (ions or electrons) be present in electrically-charged phases that are in contact (see chapter 2). ISEs with a silver halide membrane to which a silver contact is attached are relatively simple. In the system... [Pg.70]

If equilibrium in a system is perturbed for some reason, attainment of the new equilibrium requires a certain amount of time t, depending on the rates of the single reactions, and a certain amount of energy spent in activating the reaction process. The relationship between reaction rate k and activation energy has the exponential form... [Pg.598]

Thermodynamic equilibrium requires that we cannot go fiom one side of the equilihrirrm composition to the other in a single process. [Pg.37]

Figure 2-7 Plots of Ca versus 1 for an irreversible reaction for = 0,1, and 1. The kinetics for all reactions must approach first order as the reactant concentration approaches ro to be consistent with equilibrium requirements. Figure 2-7 Plots of Ca versus 1 for an irreversible reaction for = 0,1, and 1. The kinetics for all reactions must approach first order as the reactant concentration approaches ro to be consistent with equilibrium requirements.
The thermodynamic condition of equilibrium requires that the electrochemical potentials for the right- and left-hand sides of the scheme in Equation 15.13 satisfy the condition [3] ... [Pg.423]

The second story differs from the rules of the road in another way too. Each traffic equilibrium requires everybody to behave in the same way - everybody should drive on the right or everybody should drive on the left. Each equilibrium in the second story requires that some people behave differently than others -not because their desires or opponuniiies differ, but because there is a built-in imbalance in their interaction. This is also a feature of frequency-dependent equilibria/ which I referred to in chapter I. Consider the problem of whether it pays to behave honestly in order to build a reputation for honesty, in a population that consisted almost entirely of honest people, it would... [Pg.112]

Both of these steps have the net stoichiometry A- B and therefore have the same free energy change AG° = AG (B) — AG (A). This, in turn, means that both reactions must have the same value for their equilibrium constant Ke. This latter point can also be seen from the equilibrium requirement that the forward and reverse reaction rate should be equal ... [Pg.52]

This linearization, however, is an additional approximation, not inherent in the linearity of the master equation. It may be added that approach to equilibrium requires a i(y ) to be negative, which entails the positivity of the various transport coefficients such as the Ohmic resistance of a circuit. [Pg.125]

The second term on the right hand side of Eqn. (13.5) describes the rate of recombination. In the case of diffusion controlled recombination, fc and k may be calculated in terms of defect diffusivities and steady state concentrations. Without radiation, cd = 0, and the Frenkel equilibrium, requires that cv -cA = K/k. If a steady state is attained under irradiation, the rate of radiation produced defects (cp) add to the thermal production rate, and the sum is equal to the recombination rate. Therefore,... [Pg.318]

Here, the interfacial area has been approximated by that of a thin disc. Because a is held constant, the thermoelastic equilibrium requires that dAQ/dc = 0, and this leads directly to the condition... [Pg.583]

Problem Formulation. The conditions of equilibrium require the equivalence in each phase of temperature, pressure, and chemical potential for each component that is transferable between the phases and are subject to constraints of stoichiometry. A statement of the equivalence of chemical potential is identical to equations 8 and 9. An example is the AJB C D quaternary system. This system contains four binary compounds, AC, BC, AD, and BD, and the conditions of equilibrium allow three equations (of the type given by equation 8) to be written. The fourth possible equation is redundant as a result of the stoichiometric constraint (i.e., equal number of atoms on each sublattice). [Pg.145]


See other pages where Equilibrium requirement is mentioned: [Pg.618]    [Pg.311]    [Pg.271]    [Pg.157]    [Pg.20]    [Pg.100]    [Pg.365]    [Pg.356]    [Pg.650]    [Pg.90]    [Pg.80]    [Pg.83]    [Pg.158]    [Pg.110]    [Pg.170]    [Pg.37]    [Pg.156]    [Pg.241]    [Pg.639]    [Pg.369]    [Pg.4]    [Pg.76]    [Pg.58]   
See also in sourсe #XX -- [ Pg.26 ]




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