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Entropy of expansion

This expression is a natural extension of the entropy of expansion of a one-component van der Waals fluid. If the molal free volume v — b) of each component is assumed to be proportional to its molal volume, one can show that Eq. (1.3) reduces to the expression which one can get from Eq. (1.1). However, recent experimental data on iodine solutions show... [Pg.235]

The theoretical values for enthalpy may be read from a Mollier chart, where hj is the enthalpy of the steam at the turbine inlet, and hg is the enthalpy of the steam at the exhaust pressure and at the inlet entropy. The expansion of steam through the turbine is theoretically at constant entropy. These theoretical rates must be corrected for performance inefficiencies of the particular turbine. The calculations presented here are good for the average design, but exact values for a particular make and model turbine must be quoted by... [Pg.674]

Lead, excess entropy of solution of noble metals in, 133 Lead-thalium, solid solution, 126 Lead-tin, system, energy of solution, 143 solution, enthalpy of formation, 143 Lead-zinc, alloy (Pb8Zn2), calculation of thermodynamic quantities, 136 Legendre expansion in total ground state wave function of helium, 294 Lennard-Jones 6-12 potential, in analy-... [Pg.408]

Coefficient of Expansion The change in entropy with pressure is related to the coefficient of expansion by... [Pg.182]

Because entropy is a state function, the change in entropy of a system is independent of the path between its initial and final states. This independence means that, if we want to calculate the entropy difference between a pair of states joined by an irreversible path, we can look for a reversible path between the same two states and then use Eq. 1 for that path. For example, suppose an ideal gas undergoes free (irreversible) expansion at constant temperature. To calculate the change in entropy, we allow the gas to undergo reversible, isothermal expansion between the same initial and final volumes, calculate the heat absorbed in this process, and use it in Eq.l. Because entropy is a state function, the change in entropy calculated for this reversible path is also the change in entropy for the free expansion between the same two states. [Pg.389]

We choose an isothermal expansion because both temperature and volume affect the entropy of a substance at this stage, we do not want to have to consider changes in both temperature and volume. [Pg.391]

We can show that the thermodynamic and statistical entropies are equivalent by examining the isothermal expansion of an ideal gas. We have seen that the thermodynamic entropy of an ideal gas increases when it expands isothermally (Eq. 3). If we suppose that the number of microstates available to a single molecule is proportional to the volume available to it, we can write W = constant X V. For N molecules, the number of microstates is proportional to the Nth power of the volume ... [Pg.400]

FIGURE 7.9 The energy levels of a particle in a box (a) become closer together as the width of the box is increased, (b) As a result, the number of levels accessible to the particles in the box increases, and the entropy of the system increases accordingly. Die range of thermally accessible levels is shown by the tinted band. The change from part (a) to part (b) is a model of the isothermal expansion of an ideal gas. The total energy of the particles is the same in each case. [Pg.400]

STRATEGY Because entropy is a state function, the change in entropy of the system is the same regardless of the path between the two states, so we can use Eq. 3 to calculate AS for both part (a) and part (b). For the entropy of the surroundings, we need to find the heat transferred to the surroundings. In each case, we can combine the fact that AU = 0 for an isothermal expansion of an ideal gas with AU = w + q and conclude that q = —tv. We then use Eq. 4 in Chapter 6 to calculate the work done in an isothermal, reversible expansion of an ideal gas and Eq. 9 in this chapter to find the total entropy. The changes that we calculate are summarized in Fig. 7.21. [Pg.410]

The thermodynamic linear expansion factor has been related to Flory or thermodynamic interaction parameter, %, and the entropy of dilution parameter, Xs, through the Flory-Fox [10] equations. [Pg.107]

The two coefficients KL and Ks are derived empirically. They are related through the entropy of transition and constrained to reproduce the total enthalpy and entropy increments accompanying the phase transition. Since, the Inden model demands a series expansion in order to calculate the entropy, a simpler related equation by Hillert and Jarl [21] is used in many computer programs. [Pg.47]

As 5 is a thermodynamic property, ASsys is the same in an irreversible isothermal process from the same initial volume Vi to the same final volume V2. However, the change in entropy of the surroundings differs in the two types of processes. First let us consider an extreme case, a free expansion into a vacuum with no work being performed. As the process is isothermal, AU for the perfect gas must be zero consequently, the heat absorbed by the gas Q also is zero ... [Pg.133]

Today we would hesitate to comment on the energy or entropy of the universe, because we have no way to measure these quantities, and we would refer only to the surroundings that are observed to interact with the system. Some cosmological theorists have suggested that the increase in entropy posmlated by the second law is a result of the expansion of the universe [6]. One recent set of astronomical measurements leads to a prediction that the universe will continue to expand, and another predicts that expansion will reach a maximum and reverse [7]. [Pg.148]

The polynomial expansion in equation 5.28 allows the calculation of the high-T enthalpy and entropy of the compound of interest, from standard state values, through the equations... [Pg.238]

When the free energies F of the two crystal structures are identical, the system is at a critical point. The identity of F does not imply identical fimctions (otherwise the two phases would be indistinguishable). Therefore, at the critical point first derivatives of F might differ and therefore enthalpy, volume, and entropy of the two phases would be different. These transformations are first-order phase transitions, according to Ehrenfest [105]. A discontinuous enthalpy imphes heat exchange at the transition temperature, which can easily be measured with DSC experiments. A discontinuous volume is evident under the microscope or, more precisely, with diffraction experiments on single crystals or powders. Some phase transitions are however characterized by continuous first derivatives of the free energy, whereas the second derivatives (specific heat, compressibility, or thermal expansivity, etc.) are discontinuous. These transformations are second-order transitions and are clearly softer. [Pg.59]


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See also in sourсe #XX -- [ Pg.201 ]

See also in sourсe #XX -- [ Pg.820 ]




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