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Process hypothetical path

The material in Chapter 1 forms the conceptual foundation on which we will construct our understanding of thermodynamics. We will formulate thermodynamics by identifying the state that a system is in and by looking at processes by which a system goes from one state to another. We are interested in both closed systems, which can attain thermodynamic equilibrium, and open systems. The state postulate and the phase rule allow us to identify which independent, intensive thermodynamic properties we can choose to constrain the state of the system. If we also know the amount of matter present, we can determine the extensive properties in the system. Thermodynamic properties are also called state functions. Since they do not depend on path, we may devise a convenient hypothetical path to calculate the change in their values between two states. Conversely, other quantities, such as heat or work, are path functions. [Pg.30]

Figure 2.3 Plot of a process that takes a system from state 1 to state 2 in Tv space. Three alternative paths are shown the real path as well as two convenient hypothetical paths. Figure 2.3 Plot of a process that takes a system from state 1 to state 2 in Tv space. Three alternative paths are shown the real path as well as two convenient hypothetical paths.
The properties on the left-hand side of Equation (2.12a) depend only on the initial and final states. They can be calculated using the real path or any hypothetical path we create. The terms on the right-hand side are process dependent and the real path of the system must be used. [Pg.56]

We have also developed these equations in intensive forms, on a mass and a molar basis, and for differential increments. Given a physical problem, we must determine which form to use and which terms in these equations are important and which terms are negligible or zero. We must also identify whether the ideal gas model or property tables are needed to solve the problem. For some processes, it is convenient to define a hypothetical path so that we can use available data to solve the problem. [Pg.109]

This section illustrates how to calculate the change in entropy of an ideal gas between two states if P and T for each state are known. We will define the initial state as state 1, at Pi and Ti, and the final state as state 2, at P2 and T. Since entropy is a state function, we can construct any path that is convenient between state 1 and state 2 to calculate As. Figure 3.6 illustrates such a hypothetical path. We choose a reversible process for our hypothetical path so that we can apply the definition of entropy. The first step consists of isothermal expansion, while the second step is isobaric heating. To find As, we will calculate the entropy change for each step and add them together. Details of the analysis for each step follow. [Pg.151]

Figure 3.6 Plot of a process in which a system goes from state 1 to state 2 in TP space. The change in entropy is calculated along a reversible hypothetical path. Figure 3.6 Plot of a process in which a system goes from state 1 to state 2 in TP space. The change in entropy is calculated along a reversible hypothetical path.
Consider a gas that undergoes a process from state 1 to state 2. You know the ideal gas heat capacity and an equation of state. Which of the following hypothetical paths would be most appropriate to chose to calculate Am Explain. [Pg.306]

To calculate the enthalpy of mixing, it is usefid to consider another hypothetical path. In this case, we need to recognize the thermochemical data are in the form of the difference between the pure species (unmixed state) and the mixture in the form of the given equation for A/jot. However, both the initial state and the final state contain mixtures of more than one species. Therefore, we pick the hypothetical path shown in Figure E6.9C where we first unmix the initial state into its pure species components and then mix all three components to arrive at the final state. As a procedural note, in this case it is useful to formulate the problem in terms of extensive properties (i.e., AH, ) rather than intensive properties (i.e., Ah ) because the unmixing process and the mixing process contain different numbers of moles (8 vs. 10 mol). [Pg.351]

Table 1 gives the components present in the crude DDSO and their properties critical pressure (Pc), critical temperature (Tc), critical volume (Vc) and acentric factor (co). These properties were obtained from hypothetical components (a tool of the commercial simulator HYSYS) that are created through the UNIFAC group contribution. The developed DISMOL simulator requires these properties (mean free path enthalpy of vaporization mass diffusivity vapor pressure liquid density heat capacity thermal conductivity viscosity and equipment, process, and system characteristics that are simulation inputs) in calculating other properties of the system, such as evaporation rate, temperature and concentration profiles, residence time, stream compositions, and flow rates (output from the simulation). Furthermore, film thickness and liquid velocity profile on the evaporator are also calculated. [Pg.692]

Analogous to the oxidation process discussed above, the conversion of Of back to Ra may follow six different reduction paths, depending on the overpotential. Consequently, a redox cycle may involve any of 36 possible mechanisms for the hypothetical case of a single start state and a single end state. [Pg.497]

The terms on the right-hand sides of Eqs. (6.62) and (6.63) are readily associated with steps in a calculation path leading from an initial to a final state of a system. Thus, in Fig. 6.14, the actual path from state 1 to state 2 (dashed line) is replaced by a three-step calculational path. Step 1 - 1 represents a hypothetical process that transforms a real gas into an ideal gas at T, and Pi. The enthalpy and entropy changes for this process are... [Pg.109]

Figure 10. Hypothetical mechanisms for the oxidative chemistry ofl and the reaction of the one-electron oxidation product, l+m. Left from l+% The process corresponding to the slow loss of the EPR signal and two paths for the formation of disulfides upon a second one-electron oxidation. Path A Retaining a dimeric structure, the two-electron reduction of the disulfide regenerates 1. Path B The second one-electron oxidation cleaves the dimer leading to the formation of a mononuclear disulfide complex, in analogy with that shown in Figure 5 and a mononuclear Ni(II) complex that rapidly dimerizes to form 2, in analogy with chemistry known for similar complexes (85). Reduction of the disulfide leads to production of the same mononuclear Ni(II) complex. Figure 10. Hypothetical mechanisms for the oxidative chemistry ofl and the reaction of the one-electron oxidation product, l+m. Left from l+% The process corresponding to the slow loss of the EPR signal and two paths for the formation of disulfides upon a second one-electron oxidation. Path A Retaining a dimeric structure, the two-electron reduction of the disulfide regenerates 1. Path B The second one-electron oxidation cleaves the dimer leading to the formation of a mononuclear disulfide complex, in analogy with that shown in Figure 5 and a mononuclear Ni(II) complex that rapidly dimerizes to form 2, in analogy with chemistry known for similar complexes (85). Reduction of the disulfide leads to production of the same mononuclear Ni(II) complex.
Once we know how to calculate and for these five types of processes, we can calculate these quantities for any process by taking advantage of the fact that 0 and H are state properties. Tlie procedure is to construct a hypothetical process path from the initial state to the final state consisting of a series of steps of the given five types. Having done this, we calculate A/ for each of the steps, and then add the A/ s for the steps to calculate A// for the total process. Since H is a state property, tsH calculated for the hypothetical process path—which we constructed for convenience—is the same as for the path actually followed by the process. The same procedure can be followed to calculate Af for any process. [Pg.360]

However, we do not have such a table. Our task is then to construct a hypothetical process path from the solid at 25°C and 1 atm to the vapor at 300 C and 3 atm. To do so, we will look ahead a bit and note that Table B.l gives enthalpy changes for the melting of phenol at 1 atm and 42.5°C (the normal melting point of phenol) and for the vaporization of phenol at 1 atm and 181.4 C (the normal boiling point of phenol). We therefore choose the following hypothetical process path ... [Pg.360]

To calculate the four unknown specific enthalpies in the table, we construct hypothetical process paths from the reference states to the states of the species in the process and evaluate AH for each path. This is the part of the calculation you have not yet learned to do. We will show you the calculation of Hi to illustrate the method, give the results of the other calculations, and go into detail about the required procedures in Sections 8.2-8.5. [Pg.363]

Phase changes often occur at temperatures other than the temperature for which the latent heal is tabulated. When faced with this situation, you must select a hypothetical process path that permits the available data to be used. [Pg.379]

This reaction cannot be performed simply in the laboratory. If 1 mol graphite is heated with j mol oxygen, almost half the carbon burns to COxig) and the remainder is left as unreacted carbon. Nevertheless, thermodynamics allows us to predict the heat that would evolve if we could perform the reaction as written. This is possible because H is a state function, and thus AH for the reaction is independent of the path followed from reactants to products. We are free to select any path for which we have all the data needed for the calculation. In Figure 12.14, we illustrate the path in which 1 mol C is burned with O2 to CO2 (with AH = —393.5 kJ), and to this is added the calculated enthalpy change for the (hypothetical) process in which CO2 is converted to CO and O2 AH = +283.0 kJ). The total AH is the algebraic sum of the two known enthalpy changes, —393.5 kJ + 283.0 kJ = — 110.5 kJ. To see this more clearly, the reactions are written out as follows ... [Pg.506]


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Hypothetical paths

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