Internal energy constant-volume

Instead of defining a perfect gas as we have done, by Boyle s law and Joule s law, we may prefer to assume that a thermodynamic temperature scale is known, and that the perfect gas satisfies the general gas law PV = const. X T. Then we can at once use the relation (6.2) to calculate the change of internal energy with volume at constant temperature, and find it to be zero. That is, we show directly by thermodynamics that Joule s lawr follows from the gas law, if that is stated in terms of the thermodynamic temperature. 

Three pounds of wet steam containing 15% moisture and initially at a pressure of 400 psia (2758.0 kPa) expands at constant pressure (P = C) to 600°F (315.6°C). Determine the initial temperature Ti, enthalpy Hx, internal energy E, volume Vi, entropy S), final enthalpy H2, final internal energy... 

Five pounds (2.3 kg) of wet steam initially at 120 psia (827.4 kPa) with 30% moisture is heated at constant volume ( V = C) to a final temperature of 1000°F (537.8°C). Determine the initial temperature 7). enthalpy Hi, internal energy E, volume V, final pressure P2, final enthalpy H2, final internal energy E2, final volume V2, heat added <2i, work output W, change in internal energy A E, change in volume A V, and change in entropy AS. 

An isolated system is a system that does not exchange work dW = 0, heat dQ = 0, or matter dN = 0 with its surroundings. Consequently, the total internal energy and volume remain constant ... 

Tins gives us the relation between the change of internal energy with volume and the change of entropy with volume in the case of any substance whatsoever which is undergoing a reversible change at constant temperature... 

An ideal gas constrained to remain at constant volume and T, is also a system at constant internal energy and volume, since U is only a function of temperature for the ideal gas. Consequently, at equilibrium, the entropy should be a maximum. 

In the previous section we used the result dS = 0 to identify the equilibrium state of an initially nonuniform system constrained to remain at constant mass, internal energy, and volume. In this section we explore the information content of the stability criterion... 

We begin the discussion of intrinsic stability by considering further the example of Figure 7.1-1 of the last section, equilibrium in a pure fluid at constant mass (actually, we will use number of moles), internal energy, and volume. Using the (imaginary) subdivision of the system into two subsystems, and writing the extensive properties N, U, V, and S as sums of these properties for each subsystem, we were able to show in Sec. 7.1 that the condition... 

As the first application of these criteria, consider the problem of identifying the state of equilibrium in a closed, nonreacting multicomponent system at constant internal energy and volume. To be specific, suppose N moles of species 1, moles of species 2, and so on are put into an adiabatic container that will be maintained at constant volume, and that these species are only partially soluble in one another, but do not chemically react. What we would like to be able to do is to predict the composition of each of the phases present at equilibrium. (A more difficult but solvable problem is to also predict the number of phases that will be present. This problem is briefly considered in Chapter 11.) In the analysis that follows, we develop the equation that will be used in Chapters 10, 11, and 12 to compute the equilibrium compositions. 

Here 5°, ]J°, and V° are the molar entropy, internal energy, and volume of pure component i in some reference state, and Cv.i is its constant-volume heat capac-ity. 

Furthermore, many such partial derivatives can be constructed that cannot be determined experimentally. (Example Can you construct an experiment in which the entropy remains constant That can sometimes be extremely difficult to guarantee.) Equations like 4.18 and 4.19 eliminate the need to do that They tell us mathematically that the change in internal energy with respect to volume at constant entropy equals the negative of the pressure, for example. There is no need to measure internal energy versus volume. All we need to measure is the pressure. 

The vaiiabiUty of circumstances, under which the heat capacity can be measured, extends as soon as the thermodynamical states of a system are no longer unambiguously characterized by means of the internal energy, the volume and the numbers of moles alone (or other variables which can be gained from these variables by Legendre-transformations) (Section II. 1.5). Thus the variaUes p and V, for example, are not sufficient to describe in an unique way the mechanical state of an elastic solid under the influence of external forces (Section II.1.5.1). On the contrary, the pressure has to be replaced by the six components of the stress tensor and the volume by the six conqxments j/fp of the strain tensor (multiplied by a suitable factor which makes these variables extensive variables). The analc e of Cp is then the heat capacity at constant strain [Eq.(37)], the analogue of Cp tiie heat capacity at constant stress [Eq.(38)]. Eq.(39) takes the place of the relation (25a). Butaccord-ing to the particular physical situation heat capacities can also be measured which correspond to a mixed set of variables, like constant y . t , y , t , y . 

A Closed System of Constant Internal Energy and Volume... 

We conclude, therefore, that if a system is allowed to reach equilibrium - at constant mass, internal energy, and volume - its entropy assumes the maximum possible value under these constraints. Consequently, (dS/dt) = 0 or just ... 

Thus, when a closed system reaches equilibrium, constrained to constant internal energy and volume ... 

Since the solvent molecules, the polymer segments, and the lattice sites are all assumed to be equal in volume, reaction (8.A) impUes constant volume conditions. Under these conditions, AU is needed and what we have called Aw might be better viewed as the contribution to the internal energy of a pairwise interaction AUp jj., where the subscript reminds us that this is the contribution of a single pair formation by reaction A. 

The integrated terms are simply the specific heat of the unit mass of adsorbent and its associated adsorbate. The specific heat at constant volume has been used for the adsorbate since, theoretically, there is no expansion of the adsorbate volume and the heat required to raise the temperature is the change in internal energy. In practice there will be some expansion and a pessimistically high estimate could use the specific heat at constant pressure The specific heat of the adsorbed phase is in any case difficult to estimate and it is common to approximate it to that of saturated liquid adsorbate at the same temperature. 

Clearly, Aid is equal to the heat transferred in a constant pressure process. Often, because biochemical reactions normally occur in liquids or solids rather than in gases, volume changes are small and enthalpy and internal energy are often essentially equal. 

Suppose now that we have an ensemble of N non-interacting particles in a thermally insulated enclosure of constant volume. This statement means that the number of particles, the internal energy and the volume are constant and so we are dealing with a microcanonical ensemble. Suppose that each of the particles has quantum states with energies given by i, 2,... and that, at equilibrium there are Ni particles in quantum state Su particles in quantum state 2, and so on. 

One pound-mole of an ideal gas is compressed at a constant pressure of 1 atm in a piston-like device from an initial volume of 1.5 ft to a final volume of 0.5 ft. The internal energy is known to decrease by 20 Btu. How much heat was transferred to or from the gas ... 

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