An Introduction to Thermodynamics phần 9 ppsx

7/10/07 4- 105 How does this K compare with the K from part a? c. If 1 atm of HI at 250 C is placed in a vessel and allowed to reach equilibrium with its o decomposition products at this temperature, what will be the final pressure of HI? 10. The solubility product constant, K , for MgF at room temperature is 7.1 x 10 . Assuming sp 2 -9 activity coefficients are unity, find the solubility of MgF 2 a. in water b. in 0.01 M NaF solution Addition of ions to a dilute solution will decrease the values of activity coefficients. Would the addition of Na SO increase or decrease the solubility of MgF ? 2 4 2 11. Find ∆G for the process of dissolving MgF in water. Is this positive or negative? Explain o 2 why, in terms of what will happen when MgF is added to water. 2 12. The solubility products for barium oxalate at 18 C are given below. o (A) BaC O ·2 H O K = 1.2 x 10 2 4 2 sp -7 (B) BaC O ·½ H2O K = 2.2 x 10 2 4 sp -7 a. Assuming only (A) is present in the solid phase, find the concentration of Ba ion in ++ aqueous solution in equilibrium with this solid. b. Assuming only (B) is present in the solid phase, find the concentration of Ba ion in ++ aqueous solution in equilibrium with this solid. c. Describe what will happen if 1 g of (A) and 1 g of (B) are added to 100 ml of water at 18 C under 1 atm pressure. o d. How many phases will be present at equilibrium? What are they? 13. The dissociation constant of H S in water solution is 9.1 x 10 . 2 -8 H S H + HS 2 + - (The second dissociation, to give S , can be ignored here.) Find the concentration of hydrogen = ion in 0.1 M H S solution. 2 14. The dissociation constant of benzoic acid is 6.30 x 10 at 25 C. Find the concentration of -5 o hydrogen ion a. in 0.001 M benzoic acid solution b. in solution containing 0.001 M benzoic acid plus 0.01 M sodium benzoate. (Benzoic acid dissociates to give one hydrogen ion, BH B + H .) - + 15. For the cell Zn/Zn (a = 0.01)//Ag (a = 0.5)/Ag ++ + a. Write the over-all cell reaction. b. What is E ? o c. What is E ? d. Is the reaction spontaneous? 16. Find the equilibrium constant for the reaction, at 25 C, o 7/10/07 4- 106 2 Fe + Hg 2 Hg + 2 Fe ++ ++ +++ 2 17. The potential of the cell H (1 atm)/HCl(0.01 M),AgCl(s)/Ag(s) 2 is 0.464 V at 25 C. o a. Will the potential be increased or decreased by adding NaCl to a concentration of 0.10 M? b. Will the potential be increased or decreased by alloying the silver with gold? c. Is the silver the anode or cathode? d. As the cell is written, will electrons in the external circuit flow from left to right or from right to left? 18. The Weston normal cell is Cd-Hg/CdSO /Hg SO /Hg, for which the potential is E = 1.0183 4 2 4 [1 - 4.06 x 10 (t - 20) - 9.5 x 10 (t - 20) + 1 x 10 (t - 20) ] V at t C. Find ∆H for this cell at -5 -7 2 -8 3 o 20 C. o 19. The cadmium-calomel cell Cd + Hg Cl Cd + 2 Cl + 2 Hg 2 2 ++ - has been found to have a potential that varies with temperature according to the equation E = 0.6708 - 1.02 x 10 (T - 298) - 2.4 x 10 (T - 298) o -4 -6 2 with T in kelvin and E in V. For the cell reaction at 40 C, calculate o o a. ∆G b. ∆H c. ∆S o o o 20. For the cell Pt, H (1 atm)/HCl/AgCl/Ag 2 in which the electrolyte also contained 0.001 M NaCl, the measured potential, at 25 C, was found o to be 0.724 V. What is the pH of the solution, assuming activities equal to concentrations? 21. Write the reaction to be expected if a block of Mg is attached to a steel ship hull. Will the ship be oxidized or protected from oxidation? 22. Find an oxidizing agent capable of oxidizing Hg to Hg without appreciable formation of 2 ++ Hg . ++ 23. Predict the products of the following reaction: a. HNO + Hg 3 b. HNO + Hg 3 2 ++ c. Hg + Hg ++ 24. In deriving equation 27 it was emphasized that electrochemical potentials are independent of the amount of reaction. (Holding down the button on a flashlight does not increase voltage.) Yet the Nernst equation (see equation 28a) contains n explicitly. How can you reconcile these seemingly contradictory facts? [Answer: 24. Q(a) contains powers of activities proportional to n. Dividing log Q(a) by n removes this apparent dependence on n from the correction term for E . ] 7/10/07 5- 107 5 Heat Engines and Absolute Zero Many innovations of thought and technology contributed to the Industrial Revolution, but certainly a major factor in its development was the replacement of human and animal power by the power of the steam engine and, later, by power from the internal combustion engine. Thermodynamics, too, owes much of its early development to these engines that convert thermal energy to mechanical energy (“heat to work”). It was through his studies of “heat engines” in the early 19 century that the young Frenchman Sadi Carnot discovered the principle now best th described in terms of entropy, and even today the heat engine is considered important as a means of demonstrating the principles of classical thermodynamics. But the principles of the heat engine are also of great practical importance in modern technology and in current research. New devices are constantly being sought to convert thermal energy into electrical power. One way of stating the second law is that mechanical energy (as work) can be totally converted to thermal energy (or “heat”) but thermal energy cannot be totally converted to work (i.e., mechanical energy) in a cyclic process. Thermodynamics provides an understanding of the possible effectiveness, and the inherent limitations, of all devices to convert thermal energy into mechanical energy. The theory of heat engines is also important to a discussion of absolute zero and the third law of thermodynamics. Heat Engines and Heat Pumps The essential ingredients of a steam engine are a boiler and a source of heat for it; a cylinder — into which the steam from the boiler can expand — fitted with a piston against which the gas will push, performing work (and thereby losing energy and becoming cooled); and means for further cooling and/or venting the expanded steam so that the piston can be returned to its initial position. In the operation of the engine, heat is put into the engine at some elevated temperature, the engine itself does work on its surroundings (transmitted by the shaft of the piston), and a portion of the thermal energy taken in is discharged, as heat, at a lower temperature (into the cooling water). The energy transfers are represented in Figure 1. Figure 5.1. A “heat engine” takes thermal energy from a reservoir at a higher temperature, T , does work on the surroundings, and dumps excess thermal energy at T . H C (1) - input output efficiency U Q W == 7/10/07 5- 108 Theoretical heat engines always operate on a completely reversible, closed cycle. Real heat engines seldom, if ever, operate on a fully closed cycle. Expanded steam, for example, is typically at least partially released to the atmosphere, rather than being condensed to liquid and recycled. That has little effect on the overall efficiency, so we retain the cyclic model. The efficiency of a heat engine is defined to be the output, (-) W, divided by the input, Q , U which is the thermal energy put into the engine at the upper temperature, T . U For a complete cycle, Q + Q + W = ∆E = 0, and Q is positive, Q , which is the thermal energy U L U L (or “heat”) absorbed by the engine at the lower temperature, is negative, and W, the work done on the heat engine, is negative. Thus, in terms of all positive values (-W) = Q - (- Q ) U L or |W| = |Q | - |Q | U L Therefore Q > (- W) and the efficiency is less than one (or less than 100%). U CARNOT’S THEOREM. Between 1698 and 1782, a series of British inventors designed steam engines, originally to remove water that flooded the coal mines but eventually to initiate the Industrial Revolution that was largely responsible for the creation of the British Empire. Toward the end of the century, Lazare Carnot, a French mathematician, engineer, general, and administrator under Napoleon, studied the design of steam engines and concluded that efficiency was decreased by any “jerkiness” (sudden acceleration) of the engine. His son, Sadi, followed in his footsteps as an officer, until his career was terminated by Napoleon’s collapse. Sadi then retired, at half pay, at the age of 24 (in 1820) to study physics, economics, and steam engines. In 1824 he wrote a small pamphlet, Reflections on the Motive Power of Fire, that laid out the theory of heat engines, including a derivation of the maximum possible efficiency of any such device. Consider two heat engines, A and B, operating between the same two temperatures. We wish to avoid any jerkiness in their operation, so we may imagine that engine B, at least, operates reversibly. We put no restrictions on engine A, so we may start with the trial assumption that the efficiency of A is greater than that of B. The two engines can then be coupled together as shown in Figure 2. Engine A drives engine B in reverse, and because (-W)/Q is greater for engine A, we U set (Q ) = - (Q ) (by adjusting the length of the stroke). Then (W) > (-W) , so there is more U A U B B A than enough mechanical energy available to drive engine B. By allowing some input of energy at the lower temperature, from the cooling water, we now have a pair of engines that will give us perpetual motion (of the second kind). Engine A drives engine B with excess work available to carry out some other task. But we know this is impossible, both from long experience and from our understanding of the second law of thermodynamics. The conclusion of this thought experiment is that the efficiency of engine A cannot exceed the efficiency of any reversible heat engine, such as B. This conclusion, reached by Sadi Carnot, is known as Carnot’s theorem: (2a) ln 1 2 111 V V nRTQW ==− 7/10/07 5- 109 Figure 5.2. Two heat engines may be coupled, such that the reversible engine (left) provides thermal energy at the higher temperature to the other engine, which in turn provides the work to drive the first engine. The efficiencies of any two reversible heat engines, operating between the same two temperatures, are the same, and this is the maximum efficiency for any heat engine operating between these two temperatures. From Carnot’s theorem we can see that the nature of the “working fluid” — whether steam, ideal gas, mercury, or any other vapor, liquid, or vapor-liquid combination — cannot change the efficiency, nor can the nature of the cyclic process alter the efficiency, so long as the steps are reversible and thermal energy transfers occur only at the upper and lower temperatures. The efficiency will, in fact, be decreased by non-reversible heat transfers or transfers at intermediate temperatures and, to a lesser extent, by friction, as we will see below, but these are differences associated with the departure from reversibility. It is sufficient, therefore, to choose a very simple cyclic process, operating with an ideal gas. The result will be applicable to any real device, apart from corrections for irreversibility. CARNOT’S CYCLE. The cycle most conveniently treated is called the “Carnot cycle”. It consists of an isothermal reversible expansion at the upper temperature, an adiabatic reversible expansion from the upper temperature to the lower temperature, an isothermal reversible compression at the lower temperature, and an adiabatic reversible compression from the lower to the upper temperature. More briefly, it alternates isothermal and adiabatic steps, with every step reversible. The important equations of the Carnot cycle are obtained quite easily for an ideal gas. Step 1: at T , ∆E = 0; V 6 V : 1 1 1 2 Step 2: Adiabatic, Q = 0; T 6 T ; V 6 V : a 1 2 2 3 W = ∆E = n C (T - T ) (2b) a a V 2 1 and in more detail, dE = nC dT= w = - PdV= - nRT dV/V so C ln T /T = - R ln V /V a V a V 2 1 3 2 (2c) ln 3 4 222 V V nRTQW ==− (3) 2 2 1 1 T Q T Q −= ( ) ln 1 2 12 ∑ −== V V TTnRWW i (4) input output 11 21 1 T T T TT Q W ∆ = − = − = 7/10/07 5- 110 Step 3: at T , ∆E = 0; V 6 V : 2 2 3 4 Step 4: Adiabatic, Q = 0; T 6 T ; V 6 V : b 2 1 4 1 W = ∆E = n C (T - T ) (2d) b b V 1 2 and in more detail, dE = nC dT= w = - PdV= - nRT dV/V so C ln T /T = - R ln V /V b V b V 1 2 1 4 Then, because ln T /T = - ln T /T , it follows that ln V /V = - ln V /V and thus 1 2 2 1 3 2 1 4 V /V = V /V , so 2 1 3 4 and The efficiency is SHORTCUT: We may ignore the details of the ideal-gas expansions and compressions and, in fact, remove the restriction to an ideal gas, by emphasizing the changes in entropy. In the initial reversible isothermal step at T , ∆E = 0 and 1 1 - W = Q = T ∆S 1 1 1 1 For the reversible adiabatic step (T to T ), Q is zero and therefore ∆S is zero. In the reversible 1 2 isothermal step at T , ∆E = 0 and 2 2 W = - Q = - T ∆S 2 2 2 2 and in the reversible adiabatic step returning from T to T , Q is zero and ∆S is zero. 2 1 Now we note that for the over-all process, ∆S = ∆S + ∆S = 0 1 2 because the engine has undergone a complete cycle. It follows that (4a) - efficiency 11 T T Q W ∆ == %4.21 373 80 - eff 11 == ∆ == T T Q W %6565.0 873 570 eff === 7/10/07 5- 111 ∆S = - ∆S 1 2 Also, because ∆E = 0 for the complete cycle, - W = Q = Q + Q = (T - T ) ∆S 1 2 1 2 1 Let ∆S be represented by the symbol κ (Greek kappa). The important equations are then 1 Q = κ T (5a) 1 1 Q = - κ T (5b) 2 2 - W = κ (T - T ) = κ ∆T (5c) 1 2 with Q the amount of thermal energy transferred to the engine at T , Q the amount transferred to 1 1 2 the engine at T , and W the total amount of work done on the engine in a complete cycle. These 2 three equations must apply to any reversible heat engine. Only the value of κ will vary from one engine to another, for a given choice of T and T . 1 2 APPLICATIONS OF HEAT ENGINES AND HEAT PUMPS. The maximum possible efficiency for a heat engine producing net work (κ > 0, W < 0) is that of a reversible engine. From equation 4 this is For example, a steam engine operating between 100 C and 20 C has a maximum possible o o efficiency of Steam engines can be designed to operate with superheated steam, with a resultant improvement in possible performance. Operating between 600 C (a typical federal limit) and 30 C, the o o maximum possible efficiency would be which still represents a loss of more than a third of the energy, and even this value is too optimistic. PRACTICAL LIMITATION OF CURZON AND AHLBORN. It has always been apparent that the Carnot cycle puts an unrealistic demand on the system, because it assumes thermal energy transfer at T 1 and at T . Such a transfer, without a thermal gradient, would be infinitely slow. Although such 2 an arrangement would give maximum efficiency, it would give zero power output because it could (6) eff 1 1 21 Q W T TT − = − = (7) - input output (warming) efficiency 11 T T W Q ∆ === (7a) input output (cooling) efficiency 22 T T W Q ∆ === 7/10/07 5- 112 never finish its first cycle. Curzon and Ahlborn have shown that optimum power output occurs when the demands of rate of thermal energy transfer are balanced against the engine efficiency. When this limitation is included, the actual “best” efficiency becomes This is a substantially lower efficiency than predicted by Carnot, but it agrees quite well with the actual efficiencies obtained from real heat engines. (This shows that friction, usually blamed for poor performance, is relatively inconsequential in well-designed heat engines.) By the Curzon and Ahlborn formula, the efficiency of an engine operating between 20 C and o 100 C would drop to 1 - %(293/373) = 11%. Moving to the higher temperatures of 600 C and o o 30 C would give an efficiency of 1 - %(303/873) = 41%. o REVERSING A HEAT ENGINE. Our proof of Carnot’s theorem relied on the assumption that it is possible to reverse a heat engine, which would mean taking in thermal energy at a low temperature from the surroundings (e.g., the cooling water or from outdoors) and releasing high- temperature thermal energy to the surroundings, as a consequence of our having done work on the engine. Under these circumstances, our “efficiency” becomes the amount of thermal energy transferred (at T or at T , depending on our intent) divided by the amount of work we must do on 1 2 the engine. To transfer thermal energy to the reservoir at T , 1 whereas the efficiency for removing thermal energy from the reservoir at T would be 2 Suppose, for example, we wish to move thermal energy from outside, where it is freezing (0 C) to inside where it is 25 C. Then the calculated efficiency is 298/25 = 11.92 = 1192%. o o Whereas we could buy 1 kWh of electrical energy and get 1 kWh of thermal energy put into the room from an electric heater (which is necessarily 100% efficient), if we could put that same power into a reversible Carnot engine we could get 11.92 kWh of thermal energy into the room, most of it (10.92 kWh ) coming from the 0 C air outside. Is this possible? o The answer has to be yes and no. Yes, we can build reversible heat engines, which for the application described would be called heat pumps. Yes, the heat pumps would extract thermal energy from outside and dump large amounts of thermal energy inside. The same type of reversed heat engine can also pull thermal energy from the inside of a freezer or refrigerator and dump it (3a) 2 1 2 1 Q Q T T − = 7/10/07 5- 113 into the room. But no, for the same reasons discussed by Curzon and Ahlborn, we cannot get very close to 1100% efficiency. Actual measured efficiencies typically range between 3 and 5 (or 300% and 500%), which is still quite good, but not as spectacular as Carnot would indicate. Many people believe an “efficiency” should inherently be less than, or at most equal to, 100%. The output/input (i.e., efficiency > 1) of a heat pump or refrigerator is therefore typically labeled as a “coefficient of performance”. The Third Law of Thermodynamics As we have seen, the first law of thermodynamics deals with the conservation of energy as it is transformed and transferred between the system and surroundings. Although temperature is important in measuring energy changes, it enters only as temperature differences, for which the scale is quite arbitrary. The second law of thermodynamics predicts which of the many conceivable transformations and transferrals of energy can actually occur, given sufficient time. In the definition of the entropy function, an “absolute” temperature appears, which recurs in the equations for phase equilibrium, in the expressions for the efficiencies of heat engines, and in other applications of the second law. The third law of thermodynamics is comparatively recent, having been proposed early in the 20 century. Several forms have been suggested for this th additional postulate, but all are intimately involved with the “absolute” temperature scale, and particularly with the absolute zero of temperature. DETERMINATION OF THE ABSOLUTE TEMPERATURE SCALE. It has not been possible to achieve the absolute zero of temperature and there is good reason to believe that it cannot be reached, although temperatures routinely achieved are now in the general range of nanokelvin, 10 K. -9 Measurements at very low temperatures are important, but identification of absolute zero is critical to studies even at room temperature. For example, embedded in Carnot’s equations is a definition of thermodynamic temperature, measured in terms of thermal energy transfer. From equation 3, If we compare measurements at the freezing point of water and the boiling point of water, the measured value of this ratio is 1.366, from which we find T = 2.73 ∆T and T = 3.73 ∆T. Then if 2 1 we define ∆T to be 100 K, we establish the kelvin scale with T = 273 K and T = 373 K. 2 1 Alternatively, it is found preferable to set one fixed point, taken as the triple point of water at 273.16 K. The boiling point is then measured to be 100 K above the freezing point of water. NERNST’S HEAT THEOREM. The entropy change for a chemical reaction at a temperature T can 2 be found if the entropy change at some other temperature, T , is known and the heat capacities of 1 reactants and products are known. The procedure is analogous to Kirchhoff’s law (equation 17, Chapter 1). For a constant pressure, ( ) ( ) (8) productsreactants 2 1 1 2 1 T 2 dT T C SdT T C S T T P T P ∫∫ +∆+=∆ 7/10/07 5- 114 From measurements of entropies of reaction near room temperature, combined with measurements of heat capacities over a range of temperatures, Nernst observed that calculated entropies of reaction seemed to go to zero as the temperature approached zero. This observation provided the basis for a generalization known as the third law of thermodynamics. There are several ways of stating the third law, and many of the implications are best understood from the standpoint of quantum statistical mechanics. For present purposes the following statement will suffice: Every substance in its lowest energy state, at absolute zero, has the same entropy. An equivalent statement would be the entropy change for every process (chemical or physical change) becomes zero when the process occurs between substances in their lowest equilibrium states at absolute zero. We cannot demonstrate this by carrying out reactions at 0 K. In the first place it is impossible to achieve absolute zero, and in the second place, the rates of chemical reactions and other processes decrease so rapidly with decreasing temperature that few, if any, chemical reactions could proceed at temperatures even near 0 K. It is necessary, therefore, to rely on calculated entropies of reaction. It is now known that such calculated values include many “apparent exceptions” to the third law, which generally prove to be the best supporting argument for the law. From a knowledge of the structure of a substance, together with an understanding of statistical mechanics, it is possible to predict which substances are likely to be “apparent exceptions” to the third law and to calculate the exact amount of the discrepancy to be expected. The excellent agreement obtained between theory and experiment on the “exceptions” is evidence of the validity of the original postulate. Assume, for example, that a substance has an energy state that differs only slightly in energy (or enthalpy) from the lowest energy state, but that this upper energy state has an appreciable entropy difference from the lowest state. Labeling the upper state with an asterisk and the lowest state with a subscript zero, H* - H is positive but very small, whereas S* - S is positive and not o o vanishingly small. Then, at a temperature greater than 0 K, T(S* - S ) may be greater than H* - o H , so the free-energy change for the transition to the upper state will be negative and the system o will exist predominantly in the upper state. Only for vanishingly small temperatures will the enthalpy term be able to overcome the entropy term and make the lowest energy state most stable, but at such low temperatures the time required for the change of state may be so great that the lowest energy state is never achieved experimentally. Entropy is a measure of disorder in a system and its evaluation depends upon a counting operation, so the entropy differences between states are often expressible as a function of small integers. Energies, however, may be arbitrarily close. Energy states that are very close together are said to be “degenerate” or “nearly degenerate.” An example of near degeneracy of energy levels is provided by carbon monoxide, CO, which can pack in a solid crystal in the perfectly ordered arrangement CO CO CO or in disordered arrangements, such as CO CO OC CO, in which some molecules are turned around. Although these have only slightly different energies, [...]... a substance, with heat capacity CP, from the temperature T2 to the temperature T2 - dT, and ∆T is the difference in 3 See E.M Loebl, J Chem Ed 37, 361-363 ( 196 0), and R Fowler and E.A Guggenheim, Statistical Thermodynamics, Cambridge, 193 9, pp 224-227 7/10/07 5- 116 temperature between the substance and the surroundings If the ratio C P/T 2 becomes infinite, as might be expected, when T2 goes to zero,... 2.4 197 .66 213.78 237.8 223.1 202.8 130.68 198 .70 186 .90 173.78 206.6 188.83 232.7 205.8 116.1 260.6 64.7 82.6 78 .9 32.7 27.0 89. 6 57.2 191 .01 HNO3 NH3 Na NaCl NaOH O2 O3 P PH3 S SF6 SO2 CBr4 CCl4 CF4 CH4 C2H2 C2H4 C2H6 CH3OH C2H5OH C6H6 g g g c c c g g red white g rhombic g g g g lq g g g g g g lq g lq lq 304.4 266 .9 192 .7 51.3 72.1 64.5 205.15 238 .9 22.8 41.1 210.2 32.05 291 .5 248.2 358.1 3 09. 4 214.4... any given temperature change of the substance would indeed be infinite It has been found, experimentally, however, that CP also goes to zero as the temperature approaches 0 K, and in fact the ratio CP /T2 becomes zero, rather than infinite Therefore, we cannot argue that the second law prohibits us from reaching absolute zero If CP/T2 does go to zero as T2 goes to zero, as observed experimentally and... T Rearranging, we obtain o o S B − S A = ∫ CP ( A) T1 0 T2 dT dT − ∫ C P (B) 0 T T We may assume T1 $ T2; if this is not satisfied we need only relabel the phases A and B to make it so Because the transformation from A, at T 1, to B, at T2, is “isentropic” (no entropy change), it should be possible, in principle, to achieve an adiabatic, reversible transformation, A(T1) -6 B(T2) We are free to choose... equation 8 can be applied to find entropies at the absolute zero (The integral is not defined unless CP/T remains finite.) Then it becomes possible to show that the third law of thermodynamics prohibits us from reaching absolute zero The proof is as follows Let A and B represent any two phases that differ in some respect but which may be interconverted For example, A and B might represent reactants and products... entropies can be added or subtracted to find entropies of reaction, which can be combined with enthalpies of reaction to calculate free energies of reaction, and hence equilibrium constants Entropies can also be calculated, employing equations of statistical mechanics, if the molecular properties are known from spectroscopic measurements Al AlCl3 AlF3 Al2O3 State c c c c (α) Compound Table 1 STANDARD ENTROPIES,... (9) T→0 for any isothermal process involving only phases in internal equilibrium or involving any phase in a pseudoequilibrium state, provided the process does not disturb this pseudoequilibrium The importance of the third law, for classical thermodynamics, is that it provides a natural reference level2 from which entropies can be measured and tabulated These “absolute” entropies are available in handbooks,... 200 .9 2 19. 3 2 29. 2 281.6 160.7 281.6 160.7 173.4 ATTAINMENT OF ABSOLUTE ZERO.3 As we have seen, the most efficient refrigeration system is a reversible heat pump The work required to withdraw an amount of heat Q = - Q 2 from a lowtemperature reservoir at T2 is, from equation 7a, W = (∆T/T2 )Q2 As T2 goes to zero, therefore, the amount of work required to remove a given amount of thermal energy from any... as Avogadro’s number 2 This reference level is often called zero entropy, but that name is somewhat misleading No absolute value for entropy can be predicted from classical thermodynamics, and there is no way in which an absolute value of entropy could be meaningfully employed The so-called “zero of entropy” is usually not the lowest entropy state attainable, for it ignores the spin effects mentioned... any substance at the temperature T2 will become infinite It thus appears that the second law prohibits us from reaching the absolute zero A more meaningful equation for our purposes is obtained by finding the work required for a given change in temperature Replacing q2 with CP dT, the differential form of equation 7a becomes w= CP ∆TdT T2 (10) in which w is the amount of work required to change the . of statistical mechanics, it is possible to predict which substances are likely to be “apparent exceptions” to the third law and to calculate the exact amount of the discrepancy to be expected becoming cooled); and means for further cooling and/or venting the expanded steam so that the piston can be returned to its initial position. In the operation of the engine, heat is put into the engine. 361-363 ( 196 0), and R. Fowler and E.A. 3 Guggenheim, Statistical Thermodynamics, Cambridge, 193 9, pp. 224-227. 7/10/07 5- 116 Br lq 152.2 g 304.4 2 g 245.47 HNO g 266 .9 3 C graphite 5.74 NH g 192 .7 3