KELVINS LAW

In many engineering problems, the total cost of an activity (C) depends upon more than one function of a design variable (x) (which may be area, speed, resistance, etc.). If one function causes an increase in cost when X is varied while another causes a decrease, it is possible for an extreme to exist which may represent an optimum value of x (designated x ), if the extreme happens to be a minimum. 

Lord Kelvin considered the optimum cross-sectional area (A) for a copper wire carrying a direct current of magnitude (/). The criterion of selection was minimum yearly total cost of an installation including the loss of power due to resistive loss in the wire and the value of capital invested in copper. 

When a current (I) flows through a wire of resistance (/ ), the voltage drop from one end of the wire to the other (/ ) is from Ch. 10  

The resistance of a long wire R) may be expressed in terms of its specific resistance (p), its length (f), and area ( 4) as follows  

The value of the money invested in copper per year will be (i Q ylA) where Cj is the cost of copper per pound, F is the capital recovery factor (computed from Eq. (13.5) for reasonable values of n and / such as 25 5TS and 10% respectively), yis the specific weight of copper (weight per unit volume), and i and A are the length and area of the conductor. 

---

The Thomson (Kelvin) law is the basis for the description of such phenomena as capillary condensation, nucleation (Chapter IV) and the isothermal mass transfer of substances (see Chapter VII). 

In agreement with the Thomson (Kelvin) law (Gibbs-Freundlich-Ostwald law for solutions, see Chapter 1,3), the increase in chemical potential of substance in small particles in comparison with its value in the bulk phase equals Ap = 2oVJr. This leads to the dependence of the vapor pressure, p, and solubility, c, on particle size, given by ... 

Carnot s research also made a major contribution to the second law of thermodynamics. Since the maximum efficiency of a Carnot engine is given by 1 -T( H, if the engine is to be 100 percent efficient (i.e., Cma = 1), Tc must equal zero. This led William Thomson (Lord Kelvin) to propose in 1848 that Tf must be the absolute zero of the temperature scale later known as the absolute scale or Kelvin scale. ... 

Because Carnot s 1824 manuscript remained unpublished at the time of his death m 1832, it was left to Kelvin and Rudolf Clausius to show how the second law of thermodynamics was implicit in Carnot s work. For this reason Kelvin once referred to Carnot as the profoundest thinker in thermodynamic philosopihy in the first thirty years of the nineteenth century. ... 

Typically, in gas law calculations, temperatures are expressed only to the nearest degree. In that case, the Kelvin temperature can be found by simply adding 273 to the Celsius temperature. 

The second law as it left the hands of Carnot required no explanation. On the caloric theory then prevalent, it was a necessary consequence of a hydrodynamical analogy—the mechanical explanation was in fact, as Carnot s words show, the source of the principle. When the caloric theory was thrown down, the analogy and explanation fell with it, and the reconstruction of Carnot s principle by Clausius and Kelvin resulted in a law of experience. 

The Thermodynamic or Kelvin Temperature Scale Description of the Kelvin temperature scale must wait for the laws of thermodynamics. We will see that the Kelvin temperature is linearly related to the absolute or ideal gas temperature, even though the basic premises leading to the scales are very different, so that... 

The Kelvin-Planck statement of the Second Law also focuses on cyclic devices and limitations. It may be stated as ... 

Like the engine-based statements, Caratheodory s statement invokes limitations. From a given thermodynamic state of the system, there are states that cannot be reached from the initial state by way of any adiabatic process. We will show that this statement is consistent with the Kelvin-Planck statement of the Second Law. 

We wish to show that no points to the leftbb of 2 on the isotherm 62 are accessible from point 1 via any adiabatic path, reversible or irreversible. Suppose we assume that some adiabatic path does exist between 1 and 2. We represent this path as a dotted curve in Figure 2.11a. We then consider the cycle I —>2 —> 1 — 1. The net heat associated with this cycle would be that arising from the last step 1 — 1, since the other two steps are defined to be adiabatic. We have defined the direction 1 — 1 to correspond to an absorption of heat, which we will call qy. From the first law, the net work vv done in the cycle, is given by w = —q, since AU for the cycle is zero. Thus, for this process, iv is negative (and therefore performed by the system), since qy is positive, having been absorbed from the reservoir. The net effect of this cycle, then, is to completely convert heat absorbed at a high temperature reservoir into work. This is a phenomenon forbidden by the Kelvin-Planck statement of the Second Law. Hence, points to the left of 2 cannot be reached from point 1 by way of any adiabatic path. 

As with the first and second laws, the Third Law is based on experimental measurements, not deduction. It is easy, however, to rationalize such a law. In a perfectly ordered3 crystal, every atom is in its proper place in the crystal lattice. At T— 0 Kelvin, all molecules are in their lowest energy state. Such a configuration would have perfect order and since entropy is a measure of the disorder in a system, perfect order would result in an entropy of zero.b Thus, the Third Law gives us an absolute reference point and enables us to assign values to S and not just to AS as we have been restricted to do with U, H, A, and G. 

The Third Law requires that a perfectly crystalline solid of a pure material be present at 0 Kelvin for So to equal zero. Exceptions to the Third Law occur when this is not the case. For example, AgCl(s) and AgBr(s) mix to form a... 

For most substances, the Third Law and statistical calculations of the entropy of the ideal gas are in agreement, but there are exceptions, some of which are summarized in Table 4.2. The difference results from residual entropy, So, left in the solid at 0 Kelvin because of disorder so that St - So calculated from Cp/TdT is less than the St calculated from statistical methods. In carbon monoxide the residual disorder results from a random arrangement of the CO molecules in the solid. Complete order in the solid can be represented schematically (in two-dimensions) by... 

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... 

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... 

In the process represented by equation (4.23), the temperature changes from T to T". What we must investigate is the possibility that T" can equal 0 Kelvin. The Second Law predicts that for this process... 

Figure 10.14a shows such a plot for Kr.10 The straight line below 7 2 = 4 K2 (Ts= 2 K) demonstrates the validity of equation (10.160). A graph similar to the one shown in Figure 10.14a was used in Chapter 4 to extrapolate Cr resultsgg to zero Kelvin when we used the Third Law to obtain absolute entropies. 

Kelvin-Planck statement of Second Law 57 Klotz. I. M. 217. 254. 256 krypton, heat capacity 577-8... 

STRATEGY We expect a positive entropy change because the thermal disorder in a system increases as the temperature is raised. We use Eq. 2, with the heat capacity at constant volume, Cv = nCV m. Find the amount (in moles) of gas molecules by using the ideal gas law, PV = nRT, and the initial conditions remember to express temperature in kelvins. Because the data are liters and kilopascals, use R expressed in those units. As always, avoid rounding errors by delaying the numerical calculation to the last possible stage. 

The full significance of these observations could not be appreciated in advance of the formulation of the second law of thermodynamics by Lord Kelvin and Clausius in the early 1850 s. In a paper published in 1857 that was probably the first to treat the thermodynamics of elastic deformation, Kelvin showed that the quantity of heat Q absorbed during the (reversible) elastic deformation of any body is related in the following manner to the change with temperature in the work — TFei required to produce the deformation ... 

---