Chapter 3 The Second and Third Laws of Thermodynamics

Chapter 3 The Second and Third Laws of Thermodynamics ~We have seen in the last chapter that the first law of thermodynamics is concerned with the conservation of energy and with the interrelationship of work and heat. ~A second important problem with which thermodynamics deals is whether a chemical or physical change can take place spontaneously. This particular aspect is the concern of the second law of thermodynamics. ~There are several well-known examples of processes which do not violate the first law but which do not occur naturally.

~A cylinder is separated into two compartments by means of a diaphragm (Figure 3.1), with a gas at high pressure on one side and a gas at low pressure on the other. ~If the diaphragm is ruptured, there will be an equalization of pressure. However, the reverse of this process does not occur; if we start with gas at uniform pressure, it is highly unlikely that we will obtain gas at high pressure on one side and gas at low pressure on the other side. ~The first process, in which the gas pressures equalize, is known as a natural or spontaneous process. The reverse process, which does not occur but can only be imagined, is known as an unnatural process. ~Note that the natural, spontaneous process is irreversible in the sense in which we used the term in Section 2.4. We could instead equalize gas pressures by reversibly expanding the gas at the higher pressure, in which case it would perform work. The reverse process could then be carried out, but only at the expense of work that we would have to perform on the system. Such process would not be a spontaneous one. Figure 3.1

Another example of a process that occurs naturally and spontaneously in one direction but not in the other is a mixing process. Suppose with reference to Figure 3.1 that we have oxygen on one side of the partition and nitrogen at the same pressure on the other side. If the partition is removed, the gases will mix. Once two gases have mixed, however, we cannot unmix them simply by waiting for them to do so; we can only unmix them by carrying out a process which involves the performance of work on the system. Unmixing does not involve a violation of the first law, but it would violate the second law. A third example is the equalization of temperature. Suppose that we bring together a hot solid and a cold one. Heat will pass from the hot to the cold solid until the temperatures are equal; this is a spontaneous process. We know from common experience that heat will not flow in the opposite direction, from a cold to a hot body. There would be no violation of the first law if this occurred, but there would be a violation of the second law.

A fourth example relates to a chemical reaction. There are many reactions that will go spontaneously in one direction but not in the other. One of them is the reaction between hydrogen and oxygen to form water: If we simply bring together two parts of hydrogen and one of oxygen, they react together so slowly that there is no observable change. However, we can readily cause reaction to occur essentially to completion, for example by passing a spark through the system. The reaction is accompanied by the evolution of considerable heat (H is negative). There would be no violation of the first law if we were to return this heat to water and reconvert it into hydrogen and oxygen, but in practice this cannot be done unless we employ some process in which we perform work on the system. The reaction from left to right is spontaneous; that from right to left does not occur naturally.

It is obviously a matter of great importance to understand the factors that determine the direction in which a process can occur spontaneously. This amounts to asking what factors determine the position of equilibrium, because a system will move spontaneously toward the state of equilibrium (although it may do so exceedingly slowly). There matters are the concern of the second law of thermodynamics. The second law of thermodynamics can be expressed as follows: It is impossible for an engine to perform work by cooling a portion of matter to a temperature below that of the coldest part of the surroundings. For example, a ship could not be driven by abstracting heat from the surrounding water and converting it into work.

3.1 The Carnot Cycle Carnot presented an interpretation of a particular type of steam engine that had been invented by James Watt. The earliest steam engines had involved a single cylinder with a piston; steam was introduced into the cylinder, causing the piston to move, and the cylinder was then cooled, causing the piston to move in the opposite direction. There was great wastage of heat, and such engines had a very low efficiency, converting only about 1% of the heat into work. Watt’s great innovation was to use two cylinders connected together; one was kept at the temperature of steam, and the other was kept cooled. Early Watt engines had an efficiency of about 8%, and his further innovations led to engines having an efficiency of about 19%. Carnot’s great theoretical contribution was to realize that the production of work by an engine of this type depends on a flow of heat from a higher temperature Th (h standing for hotter) to a lower temperature Tc (c standing for colder). He considered an ideal type of engine, involving an ideal gas, in which all processes occurred reversibly, and showed that such an engine would have the maximum possible efficiency for any engine working between the same two temperatures. Clapeyron restated Carnot’s ideas in the language of calculus, and made use of pressure-volume diagrams, as will now be done in our presentation of the Carnot cycle.

Suppose that 1 mol of an ideal gas is contained in a cylinder with a piston, at an initial pressure of P1,an initial volume of V1and an initial temperature of Th. We refer to this gas as being in state A. Figure 3.2(a) is a pressure-volume diagram indicating the initial state A and paths. Figure 3.2(b) plots the Carnot cycle for 1 mole of an ideal gas using CP/CV=γ=1.40, the value for N2 gas. Figure 3.2b Figure 3.2a Now we will bring about four reversible changes in the system, which will eventually bring it back to the initial state A.

~We first bring about an isothermal expansion AB, the pressure and volume changing to P2 and V2 and the temperature remaining at Th; we can imagine the cylinder to be immersed in a bath of liquid at temperature Th. ~In the second step we bring about an adiabatic expansion (i.e., one in which no heat is allowed to leave or enter the system); we could accomplish this by surrounding the cylinder with insulating material. Since the gas does work during expansion, and no heat is supplied, the temperature must fall; we call the final temperature Tc and the pressure and volume P3 and V3, respectively. Figure 3.2a ~Third, we compress the gas isothermally (at temperature Tc) until the pressure and volume are P4 and V4. ~Finally, the gas is compressed adiabatically until it returns to its original state A (P1,V1,Th). The performance of work on the system, with no heat transfer permitted, raises the temperature from Tc to Th.

Reversible Isothermal Expansion Step A→B is the reversible isothermal expansion at Th. In Section 2.6, we proved that for the isothermal expansion of an ideal gas there is no change of internal energy: We also showed that the work done on the system in an isothermal reversible process is RT ln(Vinitial/Vfinal): Reversible Adiabatic Expansion Step B→C involves surrounding the cylinder with an insulating jacket and allowing the system to expand reversibly and adiabatically to a volume of V3. Since the process is adiabatic, U for an adiabatic process involving 1 mol of gas is

Reversible Isothermal Compression Step C→D involves placing the cylinder in a heat bath at temperature Tc and compressing the gas reversibly until the volume and pressure are V4 and P4, respectively. The state D must lie on the adiabatic that passes through A (Figure 3.2(a)). Since process C → D is isothermal, a positive quantity since V3 >V4 (i.e., we must do work to compress the gas). The work done on the system is a negative amount of heat is absorbed (i.e., heat is actually rejected). Reversible Adiabatic Compression Step D→A. The gas is finally compressed reversibly and adiabatically from D to A. The heat absorbed is zero: The UD→Avalue is

~Table 3.1 gives, as well as the individual contributions, the net contributions for the entire cycle. ~We see that U for the cycle is zero: the contributions for the isotherms are zero, whereas those for the adiabatics are equal and opposite to each other. ~This result that U is zero for the entire cycle is necessary in view of the fact that the internal energy is a state function; in completing the cycle the system is returned to its original state, and the internal energy is therefore unchanged.

B→C For an adiabatic process D→A The net qrev value (see Table 3.1) is A work has been done by the system.

~Note that the net work done by the system, and hence the net heat absorbed, is represented by the area within the Carnot diagram. This is illustrated in Figure 3.3. ~Diagram (a) shows the processes A→B and B→C; both are expansions, and the work done by the system is represented by the area below the lines, which is shaded. ~Diagram (b) shows the processes C→D and D→A, in which work is done on the system in the amount shown by the shaded area. ~The net work done by the system is thus represented by the area in (a) minus the area in (b) and is thus the area within the cycle shown in (c). Figure 3.3 The important thing to note about the Carnot cycle is that we have returned the system to its original state by processes in the course of which a net amount of work has been done by the system. This work has been performed at the expense of heat absorbed, as required by the first law. Since work and heat are not state functions, net work can be done even though the system returns to its original state.

Efficiency of a Reversible Carnot Engine The efficiency of the reversible Carnot engine can be defined as the work done by the system during the cycle, divided by the work that would have been done if all the heat absorbed at the higher temperature had been converted into work. This efficiency is unity (i.e., 100%) only if the lower temperature Tc is zero (i.e., if the heat is rejected at the absolute zero). This result provides a definition of the absolute zero. The efficiency is also the net heat absorbed, qh+ qc, divided by the heat absorbed at the higher temperature, qh: Note that qc is in fact negative. for the reversible engine

Example 3.1 The accompanying diagram represents a reversible Carnot cycle for an ideal gas: a. What is the thermodynamic efficiency of the engine? b. How much heat is absorbed at 500 K? c. How much heat is rejected at 200 K? d. In order for the engine to perform 1.00 kJ of work, how much heat must be absorbed? Solution a. The efficiency = (Th - Tc)/Th =(500-200)/500=0.600 = 60.0%. b. Since the work done is 60.0 J, and the efficiency is 0.600, the heat absorbed at 500 K = 60.0/0.600 = 100 J. c. The heat rejected at 200 K = 100 - 60 = 40 J. d. Since the efficiency is 0.600, the heat required to produce 1000 J of work is 1000/0.600 = 1667 J = 1.67 kJ.

Problem 3.5 Suppose that an iceberg weighing 109 kg were to drift into a part of the ocean where the temperature is 20 C. What is the maximum amount of work that could be generated while the iceberg is melting? Assume the temperature of the iceberg to be 0 C. The latent heat of fusion of ice is 6.025 kJ mol-1. If the process occurred in one day, what would be the power produced? Solution The heat transferred from the water to the ice in melting it is (1012 g/18 g mol-1)6.025 kJ mol-1=3.351011 kJ The fraction of this that can be converted into work is The available work is therefore 0.06823.351011=2.291010 kJ The power, or the rate of doing work, is thus 2.291010/86400=2.65105 kJ s-1=2.65108 W

Carnot’s Theorem ~Carnot’s cycle was discussed previously for an ideal gas. Similar reversible cycles can be performed with other materials, including solids and liquids, and the efficiencies of these cycles determined. ~The importance of the reversible cycle for the ideal gas is that it gives us an extremely simple expression for the efficiency, namely (Th - Tc)/Th. ~We will now show, by a theorem due to Carnot, that the efficiency of all reversible cycles operating between the temperatures Th and Tc is the same, namely (Th - Tc)/Th. This result leads to important quantitative formulations of the second law of thermodynamics. ~Suppose that there are two engines A and B operating reversibly between Th and Tc and that A has a higher efficiency than B. ~By making B use a larger amount of material, it is possible to arrange for it to do exactly the same amount of work in cycle. We can then imagine the engines to be coupled together so that A forces B to work in reverse, so that it rejects heat at Th and absorbs it at Tc. ~During each complete cycle the more efficient engine A draws a quantity of heat qh from the heat reservoir at Th and rejects -qcat Tc. The less efficient engine B, which is being driven backward, rejects q’h at Th and absorbs -q’c at Tc. (If it were operating in its normal manner, it would absorb q’h at Th and reject -q’c at Tc.)

Since the engines were adjusted to perform equal amounts of work in a cycle, engine A only just operates B in reverse, with no extra energy available. The work performed by A is qh + qc, while that performed on B is q’h + q’c, and these quantities are equal: Since engine A is more efficient than B, ~During the operation of the cycle, in which A has forced B to work backward, A has absorbed qh at Th and B has rejected q’h at Th; the combined system A + B has therefore absorbed qh - q’h at Th, and since q’h is greater than qh, this is a negative quantity (i.e., the system has rejected a positive amount of heat at the higher temperature). ~At the lower temperature Tc, A has absorbed qc while B has absorbed -q’c; the combined system has therefore absorbed qc- q’c at this temperature, and this is a positive quantity (it is in fact equal to q’h - qh). ~The A + B system has thus, in performing the cycle, absorbed heat at a lower temperature and rejected it at a higher temperature. It is contrary to experience that heat can flow uphill in this way, during a complete cycle of operations in which the system ends up in its initial state.

It must therefore be concluded that the original postulate is invalid; there cannot be two reversible engines, A and B, operating reversibly between two fixed temperatures and having different efficiencies. Thus, the efficiencies of all reversible engines must be the same as that for the ideal gas reversible engine, namely, ~This conclusion is not a necessary consequence of the first law of thermodynamics. ~If engine A were more efficient than engine B, energy would not have been created or destroyed if engine A drove engine B backward, because the net work done would be equivalent to the heat extracted from the reservoir. ~However, the removal of heat from a reservoir and its conversion into work without any other changes in the system have never been observed. If they could occur, it would be possible, for example, for a ship to propel itself by removing heat from the surrounding water and converting it into work; no fuel would be needed. ~Such a continuous extraction of work from the environment has been called perpetual motion of the second kind. The first law forbids perpetual motion of the first kind, which involves the creation of energy from nothing. The second law forbids the operation of engines in which energy is continuously extracted from an environment which is at the same temperature as the system or at a lower temperature.

The Generalized Cycle : The Concept of Entropy For the efficiency of a reversible Carnot cycle, it follows that This equation applies to any reversible cycle that has distinct isothermal and adiabatic parts, and it can be put into a more general form to apply to any reversible cycle. ~Consider the cycle represented by ABA in Figure 3.4. During the operation of the engine from A to B and back to A, there is heat exchange between the system and its environment at various temperatures. ~The whole cycle may be split up into elements such as ab, shown in the figure. The distance between a and b should be infinitesimally small, but it is enlarged in the diagram. ~During the change from a to b, the pressure, volume, and temperature have all increased, and a quantity of heat has been absorbed. ~Let the temperature corresponding to a be T and that corresponding to b be T + dT. The isothermal corresponding to T + dT is shown as bc and the adiabatic at a as ac; the two intersect at c. The change from a to b may therefore be carried out by means of an adiabatic change ac, followed by an isothermal change cb. Figure 3.4

During the isothermal process an amount of heat dq has been absorbed by the system. If the whole cycle is completed in this manner, quantities of heat dq1, dq2, etc., will have been absorbed by the system during isothermal changes carried out at the temperatures T1, T2, etc. the summation being made round the entire cycle Since the ab elements are all infinitesimal, the cycle consisting of reversible isothermal and adiabatic steps is equivalent to the original cycle A→B→A. It follows that for the original cycle where the symbol denotes integration over a complete cycle. This result that the integral of dqrev/T over the entire cycles is equal to zero is a very important one. We have seen that certain functions are state functions, which means that their value is a true property of the system; the change in state function when we pass from state A to state B is independent of the path, and therefore a state function will not change when we traverse a complete cycle. For example, pressure P, volume V, temperature T, and internal energy U are state functions; thus we can write

It is convenient to write dqrev/T as dS, so that The property S is known as the entropy of the system, and it is a state function. The word entropy was coined in 1854 by Rudolf Julius Emmanuel Clausius; the word literally means to give a direction (Greek en, in; trope, change, transformation). Since entropy is a state function, its value is independent of the path by which the state is reached. Thus if we consider a reversible change from state A to state B and back again (Figure 3.4), it follows that denoting the change of entropy in going from A to B by path (1) denoting the change of entropy in going from B to A by path (2) That is , the change of entropy is the same whatever path is followed.

Problem 3.1 The accompanying diagram represents a reversible Carnot cycle for an ideal gas: a. What is the thermodynamic efficiency of the engine? b. How much heat is rejected at the lower temperature, 200 K, during the isothermal compression? c. What is the entropy increase during the isothermal expansion at 1000 K? d. What is the entropy decrease during the isothermal compression at 200 K? e. What is the overall entropy change for the entire cycle? f. What is the increase in Gibbs energy during the process AB? Solution

Problem 3.4 The following diagram represents a reversible Carnot cycle for an ideal gas: a. What is the thermodynamic efficiency of the engine? b. How much heat is absorbed at 400 K? c. How much heat is rejected at 300 K? d. What is the entropy change in the process AB? e. What is the entropy change in the entire cycle? f. What is the Gibbs energy change in the process AB? g. In order for the engine to perform 2 kJ of work, how much heat must be absorbed? Solution

3.2 Irreversible Processes The treatment of thermodynamically reversible processes is of great importance in connection with the second law. However, in practice we are concerned with thermodynamically irreversible processes, because these are the processes that occur naturally. It is therefore important to consider the relationships that apply to irreversible processes. A simple example of an irreversible process is the transfer of heat from a warmer to a colder body. Suppose that we have two reservoirs, a warm one at a temperature Th and a cooler one at a temperature Tc. We might imagine connecting these together by a metal rod, as shown in Figure 3.5a, and waiting until an amount of heat q has flowed from the hotter to the cooler reservoir. To simplify the argument, let us suppose that the reservoirs are so large that the transfer of heat does not change their temperatures appreciably. Figure 3.5a

In order to calculate the entropy changes in the two reservoirs after this irreversible process has occurred, we must devise a way of transferring the heat reversibly. ~We can make use of an ideal gas to carry out the heat transfer process, as shown in Figure 3.5b. ~The gas is contained in a cylinder with a piston, and we first place it in the warm reservoir, at temperature Th, and expand it reversibly and isothermally until it has taken up heat equal to q. ~The gas is then removed from the hot reservoir, placed in an insulating container, and allowed to expand reversibly and adiabatically until its temperature has fallen to Tc. Finally, the gas is placed in contact with the colder reservoir at Tcand compressed isothermally until it has given up heat equal to q. Figure 3.5b

The entropy changes that have occurred in the two reservoirs and in the gas are shown in Figure 3.5. We see that the two reservoirs, i.e., the surroundings, have experienced a net entropy change of and this is a positive quantity since Th > Tc. The gas, i.e., the system, has experienced an exactly equal and opposite entropy change: This is thus no overall entropy change, as is necessarily the case for reversible changes in an isolated system. On the other hand, for the irreversible change in which the reservoirs are in thermal contact (Figure 3.5a) there is no compensating entropy decrease in the expansion of an ideal gas. The entropy increase in the two reservoirs is the same as for the reversible process, and this is the overall entropy increase. This result, that a spontaneous (and therefore irreversible) process occurs with an overall increase of entropy in the system plus its surroundings, is universally true. The proof of this is based on the fact that the efficiency of a Carnot cycle in which some of the steps are irreversible must be less than that of a purely reversible cycle, since the maximum work is performed by systems that are undergoing reversible processes. Thus, for an irreversible cycle, inequality of Clausius

Figure 3.6 Consider an irreversible change from state A tostate B in an isolated system (Figure 1.4c), as represented by the dashed line in Figure 3.6. Suppose that the conditions are then changed; the system is no longer isolated. It is finally returned to its initial state A by a reversible path represented by the solid line in Figure 3.6. During this reversible process the system is not isolated and can exchange heat and work with the environment. Since the entire cycle A→ B→A is in part reversible, which means that The entropy of the final state B is thus always greater than that of the initial state A if the process A→B occurs irreversibly in an isolated system. Any change that occurs in nature is spontaneous and is therefore accompanied by a net increase in entropy. This conclusion led Clausius to his famous concise statement of the laws of thermodynamics: The energy of the universe is a constant; the entropy of the universe tends always towards a maximum.

3.3 Molecular Interpretation of Entropy ~The logical arguments employed do not require any knowledge of molecules, but many of us find it helpful, in the understanding of thermodynamics, to interpret its principles in the light of molecular structure. ~Any macroscopic property is in reality a consequence of the position and motion of these particles. At any instant we could in principle define the microscopic state of a system, which means that we would specify the position and momentum of each atom. An instant later, even though the system might remain in the same macroscopic state, the microscopic state would be completely different, since at ordinary temperatures molecules are changing their positions at speeds of the order of 103 m s-1. A system at equilibrium thus remains in the same macroscopic state, even though its microscopic state is changing rapidly. ~There is an enormous number of microscopic states that are consistent with any given macroscopic state. Entropy is a measure of how many different microscopic states are consistent with a given macroscopic state. ~When a system moves spontaneously from one state to another, it goes to a state in which there are more microscopic state. We can express this differently by saying that when a spontaneous change takes place, there is an increase in disorder. In other words, entropy is a measure of disorder; an increase in entropy means an increase in disorder.

Deck of Cards Analogy ~A deck of 52 cards may be arranged in a particular specified order (suits separate, cards arranged from ace to king ) or in a completely shuffled and disordered state. ~There are many sequences (analogous to microscopic states) that correspond to the shuffled and disordered (macroscopic) state, whereas there is only one microscopic state of the specified order, in which there is less disorder . Thus, the shuffled state has higher entropy than the unshuffled. ~If we start with an ordered deck and shuffle it, the deck moves toward a state of greater randomness or disorder; the entropy increases. The reason the random state is approached when the ordered deck is shuffled is simply that there are many microscopic states consistent with the shuffled condition, and only one consistent with the ordered condition. ~The chance of producing an ordered deck by shuffling a disordered one is obviously very small. This is true for 52 cards; when we are dealing with a much larger number of molecules (e.g., 6.022 × 1023 in a mole), the likelihood of a net decrease in entropy is obviously much more remote.

Entropy Changes ~In the light of these ideas, it is easy to predict what kinds of entropy changes will occur when various processes take place. ~If, for example, we raise the temperature of a gas, the range of molecular speeds becomes more extended; a larger proportion of the molecules have speeds that differ from the most probable value. There is thus more disorder at a high temperature, and the entropy is greater. ~Entropy also increases if a solid melts. The entropy change on melting is the enthalpy of fusion fusH divided by the melting point Tm, and since fusH must be positive, there is always an entropy increase on melting. This is understandable on a molecular basis, since in the solid the molecules occupy fixed sites, while in a liquid there is much less restriction as to position. ~Similarly, for the evaporation of a liquid at the boiling point Tb, the entropy change vapH /Tb must be positive since the latent heat of vaporization vapH must be positive. In the liquid the attractive forces between molecules are much greater than in the vapor, and there is a large increase in disorder in going from the liquid to the vapor state; there are many more microscopic states for the gas compared with the liquid. ~The conversion of a solid into a gas is also accompanied by an entropy increase, for the same reason.

Entropy Changes in Chemical Reactions Entropy changes in chemical reactions can also be understood on a molecular basis. Consider, for example, the process If we convert 1 mol of hydrogen molecules into 2 mol of hydrogen atoms, there is a considerable increase in entropy. The reason for this is that there are more microscopic states (more disorder) associated with the separated hydrogen atoms than with the molecules, in which the atoms are paired together. Again, an analogy is provided by a deck of cards. The hydrogen atoms are like a completely shuffled deck, while the molecular system is like a deck in which aces, twos, etc., are paired. The latter restriction means fewer permissible states and, therefore, a lower entropy. In general, for a gaseous chemical reaction there is an increase of entropy in the system if there an increase in the number of molecules. The dissociation of ammonia, for example, is accompanied by an entropy increase, because we are imposing a smaller restriction on the system by pairing the atoms as N2 and H2, as compared with organizing them as NH3 molecules.

Processes Involving Ions in Solution The situation with reactions in solution is, however, a good deal more complicated. It might be thought, for example that a process of the type occurring in aqueous solution, would be accompanied by an entropy increase, by analogy with the dissociation of H2 into 2H. However, there is now an additional factor, arising from the fact that ions interact with surrounding water molecules, which tend to orient themselves in such a way that there is electrostatic attraction between the ion and the dipolar water molecules. This effect is known as electrostriction or more simply as the binding of water molecules. This electrostriction leads to a considerable reduction in entropy, since the bound water molecules have a restricted freedom of motion. As a result, ionization processes always involve an entropy decrease. An interesting example is provided by the attachment of adenosine triphosphate (ATP) to myosin, a protein that is an important constituent of muscle and that plays an important role in muscular contraction. Myosin is an extended protein that bears a number of positive charges, whereas ATP under normal physiological conditions bears four negative charges. These charges attract and bind water molecules, thus bringing about a lowering of the entropy of the water molecules; when the ATP and myosin molecules come together, there is some charge neutralization and a consequent increase in entropy because of the release of water molecules. This entropy increase associated with the binding of ATP to myosin plays a significant role in connection with the mechanism of muscular contraction.

Contraction of Muscle or Rubber The entropy change that occurs on the contraction of a muscle is also of interest. A stretched strip of muscle, or a stretched piece of rubber, contracts spontaneously. When stretched, muscle or rubber is in a state of lower entropy than in the contracted state. Both muscle or rubber consist of very long molecules. If a long molecule is stretched as far as possible without breaking bonds, there are few conformations available to it. However, if the ends of the molecule are brought closer together, the molecules can then assume a large number of conformations, and the entropy will therefore be higher. In 1913 Archibald Vivian Hill made accurate measurements of the heat produced when a muscle contracts and found it to be extremely small. The same is true of a piece of rubber. When muscle or rubber contracts, there is therefore very little entropy change in the surroundings, so that the overall entropy change is essentially that of the material itself, which is positive. We can therefore understand why muscular contraction, or the contraction of a stretched piece of rubber, occurs spontaneously. Processes of this kind that are largely controlled by the entropy change in the system are referred to as entropic processes.

Problem 3.18 Predict the signs of the entropy changes in the following reactions when they occur in aqueous solution. a.Hydrolysis of urea: H2NCONH2 + H2OCO2 + 2NH3 b.H++OH-H2O c.CH3COOHCH3COO-+H+ d.CH2BrCOOCH3 + S2O32-CH2(S2O3-)COOCH3 + Br- Solution a. Positive (increase in number of molecules) b. Positive (decrease in electrostriction) c. Negative (increase in electrostriction) d. Positive (decrease in electrostriction)

3.4 The Calculation of Entropy Changes We have seen that entropy is a state function, which means that an entropy change SA→Bwhen a system changes from state A tostate B is independent of the path. This entropy change is given by for the transition A → B by a reversible path. That is, whereas SA→Bis independent of the path, the integral may be equal to or less thanthe entropychange: It is equal to the entropy change if the process is reversible and less than the entropy change if the process is irreversible. It follows that if a system changes from state A to state B by an irreversible process, we cannot calculate the entropy change from the heat transfers that occur. Instead we must contrive a way of bringing about the change by purely reversible processes. Some examples of how this is done will now be considered.

Changes of State of Aggregation ~If we keep the pressure constant, a solid will melt at a fixed temperature, the melting point Tm at which solid and liquid are at equilibrium. As long as both solid and liquid are present, heat can be added to the system without changing the temperature; the heat absorbed is known as the latent heat of melting (or fusion) fusH of the solid. ~Since the change occurs at constant pressure, this heat is an enthalpy change and is the difference in enthalpy between liquid and solid. Thus, ~It is easy to heat a solid sufficiently slowly at its melting point that the equilibrium between liquid and solid is hardly disturbed. The process is therefore reversible, since it follows a path of successive equilibrium states, and the latent heat of melting is thus a reversible heat. Since the temperature remains constant, the integral becomes simply the heat divided by the temperature:

The entropy of melting or fusion is thus For example, fusH for ice is 6.02 kJ mol-1 and the melting point is 273.15 K, so that The entropy of vaporization can be dealt with in the same way, since when a liquid is vaporized without any rise in temperature, the equilibrium between liquid and vapor remains undisturbed. Thus for water at 100 C, The same procedure can be used for a transition from one allotropic form to another, provided that the process occurs at a temperature and pressure at which the two forms are in equilibrium. Gray tin and white tin, for example, are in equilibrium at 1 atm pressure and 286.0K, and trsH = 2.09 kJ mol-1. The entropy change is thus

Ideal Gases A particularly simple process is the isothermal expansion of an ideal gas. Suppose that an ideal gas changes its volume from V1 to V2 at constant temperature. In order to calculate the entropy change we must consider the reversible expansion; since entropy is a state function, S is the same however the isothermal expansion from V1 to V2 occurs. We have seen that if n mol of an ideal gas undergoes a reversible isothermal expansion, at temperature T, from volume V1 to volume V2, the heat absorbed is Since the temperature is constant, S is simply the reversible heat absorbed divided by the temperature:

If a volume change occurs in an ideal gas with a change in temperature, we proceed as follows. Suppose that the volume changes from V1 to V2 and that the temperature changes from T1 to T2. Again, we imagine a reversible change, knowing that S will be the same whether the change is reversible or not: Integration If CV,m is independent of temperature If the temperature of a gas is changed by a substantial amount, it will be necessary to take into account the variation of the heat capacity with the temperature. The procedure is best illustrated by an example.

Example 3.2 A mole of hydrogen gas is heated from 300 K to 1000 K at constant volume. The gas may be treated as ideal with Calculate the entropy change. Solution V constant and n = 1

Problem 3.11 One mole of an ideal gas, with CV,m = 3R/2, is heated (a) at constant pressure and (b) at constant volume, from 298 K to 353 K. Calculate S for the system in each case. Solution a. b.

Problem 3.22 At 0C 20 g of ice are added to 50 g of water at 30 C in a vessel that has a water equivalent of 20 g. Calculate the entropy changes in the system and in the surroundings. The heat of fusion of ice at 0 C is 6.02 kJ mol-1, and the specific heat capacities of water and ice may be taken as constant at 4.184 and 2.094 J K-1 g-1, respectively, and independent of temperature. Solution Heat balance equation Reversible processes a. Cool 70 g of water to 0 C

b. Melt 20 g of ice to 0 C c. Heat 90 g of water to 5.57 C

Entropy of Mixing Suppose that we have two ideal gases at equal pressures, separated by a partition as shown in Figure 3.7. If the partition is removed, the gases will mix with no change of temperature. If we want to calculate the entropy change, we must imagine the mixing to occur reversibly. This can be done by allowing the first gas to expand reversibly from its initial volume V1 to the final volume V1 + V2; the entropy change is Similarly, for the second gas Figure 3.7 the pressures and temperatures are the same The mole fractions x1 and x2 of the two gases in the final mixture are

The total entropy change S1 + S2 is thus The entropy change per mole of mixture is thus For any number of gases, initially at the same pressure, the entropy change per mole of mixture is If the initial pressures of two ideal gases are not the same, the entropy increase is also applies to the mixing of ideal solutions Note that the mixing of two gases each at volume V1 + V2, to give a mixture of volume V1 + V2, involves no change in entropy.

Problem 3.29 A vessel is divided by a partition into two compartments. One side contains 5 mol O2 at 1 atm pressure; the other, 5 mol N2 at 1 atm pressure. Calculate entropy change when the partition is removed. Solution For O2 For N2

Example 3.3 Exactly one liter of a 0.100 M solution of a substance A is added to 3.00 liters of a 0.050 Msolution of a substance B. Assume ideal behavior and calculate the entropy of mixing. Solution 0.100 mol of substance A is present and the volume increases by a factor of 4: 0.150 mol of B is present and the volume increases by a factor of 4/3: The net S is therefore

Informational or Configurational Entropy ~This treatment of entropy of mixing brings out an important point that is often misunderstood. ~Consider a simple example relating to the mixing of ideal gases. Suppose that we have half a mole of an ideal gas, separated by a partition from another half mole at the same volume, pressure, and temperature; if we remove the partition, what is the entropy change? The important point to appreciate is that this cannot be answered unless we have further information. ~If the gases are identical, there is obviously no entropy change. There is, however, an entropy of mixing if the gases are different. As just explained, we calculate an entropy of mixing by expanding each gas reversibly and isothermally to twice its volume; the heat absorbed in each case is 1/2RTln2 so that the total entropy change is Rln2, which is equal to 5.76 J K-1. ~There is no entropy change when we mix the two expanded gases at the same volume; the entropy of mixing Rln2 is really an informational or configurational entropy, and its value depends on the information we possess.

~For the shuffling of 52 cards, we can say nothing about the entropy change unless we have certain information. ~If the cards have been printed so that all are identical, there will be no entropy change on shuffling. If, however, the cards have been printed normally, there is an increase in entropy when an ordered deck is shuffled. ~This is entirely a configurational entropy, arising from the fact that for an ordered deck we accept only one arrangement of cards out of the large number (52!, approximately 8.07×1067) of possible arrangements. The calculated configurational entropy change for shuffling is 52kBln52! = 1.12 ×10-19 J K-1. This, of course, is much smaller than the change for the mixing of two gases, because the number of cards is much smaller than the number of molecules.

Chapter 3 The Second and Third Laws of Thermodynamics