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Zero entropy

For those who are familiar with the statistical mechanical interpretation of entropy, which asserts that at 0 K substances are nonnally restricted to a single quantum state, and hence have zero entropy, it should be pointed out that the conventional thennodynamic zero of entropy is not quite that, since most elements and compounds are mixtures of isotopic species that in principle should separate at 0 K, but of course do not. The thennodynamic entropies reported in tables ignore the entropy of isotopic mixing, and m some cases ignore other complications as well, e.g. ortho- and para-hydrogen. [Pg.371]

The principle of tire unattainability of absolute zero in no way limits one s ingenuity in trying to obtain lower and lower thennodynamic temperatures. The third law, in its statistical interpretation, essentially asserts that the ground quantum level of a system is ultimately non-degenerate, that some energy difference As must exist between states, so that at equilibrium at 0 K the system is certainly in that non-degenerate ground state with zero entropy. However, the As may be very small and temperatures of the order of As/Zr (where k is the Boltzmaim constant, the gas constant per molecule) may be obtainable. [Pg.373]

Deals with the concept of entropy, which serves as a means of determining whether or not a process is possible. Defines the zero entropy state for any substance in a single, pure quantum state as the absolute zero of temperature. [Pg.209]

If we draw three planes perpendicular to the axes of v, S, U, respectively, these will be the loci of all states which have a constant value of these magnitudes, respectively. The plane v = 0 is evidently fixed, but the planes S = 0, U = 0 can only be fixed by arbitrary choice of the initial states of zero entropy and energy. The origin may therefore be chosen anywhere in the plane of zero volume. [Pg.240]

Within experimental error, S0m (monoclinic) = 5o.m (rhombic), and it seems more probable that they both have zero entropy at 0 K than that both have the same non-zero entropy. [Pg.165]

Experience indicates that the Third Law of Thermodynamics not only predicts that So — 0, but produces a potential to drive a substance to zero entropy at 0 Kelvin. Cooling a gas causes it to successively become more ordered. Phase changes to liquid and solid increase the order. Cooling through equilibrium solid phase transitions invariably results in evolution of heat and a decrease in entropy. A number of solids are disordered at higher temperatures, but the disorder decreases with cooling until perfect order is obtained. Exceptions are... [Pg.177]

Helium is an interesting example of the application of the Third Law. At low temperatures, normal liquid helium converts to a superfluid with zero viscosity. This superfluid persists to 0 Kelvin without solidifying. Figure 4.12 shows how the entropy of He changes with temperature. The conversion from normal to superfluid occurs at what is known as the A transition temperature. Figure 4.12 indicates that at 0 Kelvin, superfluid He with zero viscosity has zero entropy, a condition that is hard to imagine.v... [Pg.178]

SOLUTION (a) Because T = 0, all motion has ceased. We expect the sample to have zero entropy, because there is no disorder in either location or energy. This value is confirmed by the Boltzmann formula because there is only one way of arranging the molecules in the perfect crystal, W = 1. [Pg.398]

To reach W = 1 and S = 0, we must remove as much of this vibrational motion as possible. Recall that temperature is a measure of the amount of thermal energy in a sample, which for a solid is the energy of the atoms or molecules vibrating in their cages. Thermal energy reaches a minimum when T = 0 K. At this temperature, there is only one way to describe the system, so — 1 and — 0. This is formulated as the third law of thermodynamics, which states that a pure, perfect crystal at 0 K has zero entropy. We can state the third law as an equation, Equation perfect crystal T=0 K) 0... [Pg.989]

We have already stated that some defects are related to the entropy of the solid, and that a perfeet solid would violate the second law of thermodynamics. The 2nd law states that zero entropy is only possible at absolute zero temperature. However, most solids exist at temperatures far above absolute zero. Thus, most of the solids that we eneounter are defeet-solids. The defects are usually "point defeets", which are atomlstie... [Pg.71]

In a superconducting metal, some of the electrons are paired into the so-called cooper pairs which are all in the same zero entropy state and do not carry heat. Heat is carried only by unpaired electrons which are in energy states separated from cooper pairs by an energy gap AE(T). The number of unpaired electrons varies as exp(—AE/kBT). Hence ... [Pg.95]

A reversible adiabatic expansion of an ideal gas has a zero entropy change, and an irreversible adiabatic expansion of the same gas from the same initial state to the same final volume has a positive entropy change. This statement may seem to be inconsistent with the statement that 5 is a thermodynamic property. The resolution of the discrepancy is that the two changes do not constitute the same change of state the final temperature of the reversible adiabatic expansion is lower than the final temperature of the irreversible adiabatic expansion (as in path 2 in Fig. 6.7). [Pg.136]

The preceding statement of the third law has been formulated to exclude solutions and glasses from the class of substances that are assumed to have zero entropy at 0 K. Let us examine one example of each exclusion to see that this limitation is essential. [Pg.262]

Likewise, glasses do not have zero entropy at 0 K that is, A5 ,ok is not zero for a transition such as... [Pg.262]

To calculate for this transition, it is necessary to have heat capacity data for both glassy and crystaUine glycerol from near 0 K to the melting point and the heat of fusion of both glass and crystal. Such data [7] lead to a ASm for Equation (11.7) of 19.2 J K mol. Thus, glassy glycerol cannot be assigned zero entropy at OK rather, it possesses a residual entropy of 19.2 J moP. ... [Pg.263]

OK can be computed from heat capacity measurements [8] for each crystalline form from near 0 K to the transition temperature (368.6 K) and the heat of transition. The result is zero within experimental error. Hence, both rhombic and monoclinic sulfur ate assigned zero entropy at 0 K. [Pg.263]

The equality sign is reached for a reversible process, entailing overall zero entropy production. Concomitantly, efficiency will be below Camot in the presence of dissipative entropy producing fluxes. While this statement is strictly speaking correct, we will show below that it can be misleading. The delicate... [Pg.189]

So far, we have been able to calculate only changes in the entropy of a substance. Can we determine the absolute value of the entropy of a substance We have seen that it is not possible to determine absolute values of the enthalpy. However, entropy is a measure of disorder, and it is possible to imagine a perfectly orderly state of matter with no disorder at all, corresponding to zero entropy an absolute zero of entropy. This idea is summarized by the third law of thermodynamics ... [Pg.456]

Let s calculate the entropy of a tiny solid made up of four diatomic molecules of a binary compound such as carbon monoxide, CO. Suppose the four molecules have formed a perfectly ordered crystal in which all molecules are aligned with their C atoms on the left. Because T = 0, all motion has ceased (Fig. 7.5). We expect the sample to have zero entropy, because there is no disorder in either location or energy. This value is confirmed by the Boltzmann formula because there is only one way of arranging the molecules in the perfect crystal, W = l and S = k In 1 =0. Now suppose thar the molecules can lie with their C atoms on either side yet still have the same total energy (Fig. 7.6). The total number of ways of arranging the four molecules is... [Pg.457]

The third law (also called the Nernst heat theorem) states that all perfectly ordered crystalline substances have zero entropy at absolute zero temp... [Pg.701]

In the few years that have elapsed since these observations, a great deal of information has accumulated on the properties and rate coefficients of the hydrated electron. Some of its characteristic properties are shown in Table 8. The subscript f refers to formation from a hypothetical standard state for electrons having zero entropy, enthalpy and free energy and in a vacuum at the bulk electrostatic po-... [Pg.82]

To find the equilibrium distribution, we consider a system of fixed volume10 and number of particles in contact with a thermal reservoir at temperature T. By the second law, at equilibrium there must be a zero entropy change of the universe for all infinitesmal changes in the system, because Auniv is a maximum at equilibrium. In particular, for the change involved in increasing the number of molecules in the jih state (with e = e7) of the system by 1, while decreasing the number of molecules in the lowest state of the system (with e = 0)... [Pg.138]

An alternative way to clarify the nature of this state is to test its stability with respect to a metal-insulator transition. This has received a lot of theoretical attention recently. The JT singlet ground state makes these compounds free from the tendency towards a magnetic instability observed in so many Mott insulators. In fact, their ground state does not break any symmetry and Capone et al. explained [43] that it then has a zero entropy, which makes a direct connection with a metal impossible (it would violate the Luttinger theorem). These authors predict that the only way to go from the insulator to the metal would be through an exotic superconducting phase or a first-order transition. [Pg.190]

At absolute zero entropy of all pure elements and compounds is zero. [Pg.12]

When w = 1, however, the expression (1.7) reduces to zero, showing that the ordered state has zero entropy. This is as we should expect there is only one arrangement of the atoms, all the a s being on one lattice, all the b s on the other, so that then is no randomness at all. [Pg.296]

Equation (3.340) yields a paraboloid-like change of dissipation with respect to forces A, andX2, as seen in Figure 3.4. The system tends to minimize the entropy and eventually reaches zero entropy production if there are no restrictions on the forces. On the other hand, if we externally fix the value of one of the forces, for example, A 2 = X20, then the system will tend toward the stationary state characterized by the minimum entropy production at X2 = X20. The system will move along the parabola of Figure 3.4 and stop at point [Pg.147]

Studies related to entropy changes revealed that reduction in temperature leads to decrease in entropy change for all processes. It was therefore, postulated that for a process occurring at absolute zero temperature the entropy change would be zero. This has led to a basis from which absolute values of entropy can be determined, taking entropy at absolute zero of temperature to be zero. Thus, unlike i/and F whose changes can be accurately measured but not the absolute value, the absolute value of entropy can indeed be measured. We take, for calculation purposes, enthalpy of elements in a defined standard state to be zero, but that assumption is only for convenience, no molecule or atom can have zero heat content at ambient conditions. On the other hand, a fully ordered (crystalline) solid at absolute zero temperature will have zero entropy. [Pg.57]


See other pages where Zero entropy is mentioned: [Pg.397]    [Pg.781]    [Pg.803]    [Pg.99]    [Pg.484]    [Pg.108]    [Pg.18]    [Pg.266]    [Pg.273]    [Pg.100]    [Pg.379]    [Pg.287]    [Pg.452]    [Pg.38]    [Pg.298]    [Pg.1]    [Pg.90]    [Pg.356]    [Pg.359]    [Pg.355]    [Pg.54]    [Pg.262]    [Pg.266]   
See also in sourсe #XX -- [ Pg.431 ]

See also in sourсe #XX -- [ Pg.832 ]




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