The Second Law and Cyclic Processes. The second law of thermodynamics may be stated as follows:

This is thought to be generally true, but it cannot be proved. It is a hypothesis. However, the following intuitive development gives the law a reasonable basis. Consider a constant temperature conversion of a system from an initial (i) to a final state (f). This can be done by many different paths, three of which are shown in Figure J-1a. Generally, however, only one of these paths is reversible. All of the others are irreversible. Thus, the reversible path is special, because it is unique. Further, the work required to get the system from the initial to the final state is least along the reversible path. All of this may seem a bit intangible, but it is analogous to travelling between two points, i and f. There are many indirect paths, but only one direct (i.e., great circle) path, as shown in Figure J-1b. The distance travelled is a minimum along the direct path: d_direct < d_indirect.

We now apply the first law to the reversible path and one of the irreversible paths:

Since the initial and final states are the same by both paths, DE_rev = DE_irrev (E is a state function). Subtracting the second equation from the first gives

It therefore follows that q_rev > q_irrev. Thus the heat added to the system is a maximum along the reversible path. But heat is just what we need to discuss entropy.

For an actual constant temperature process,

If the actual process is reversible, DS_surr = -q_rev/T, and DS_univ = 0. However, if the actual process is irreversible, DS_surr = -q_irrev/T. In this case,

This intuitively demonstrates the second law.

Now consider a cyclic process in which a system is taken from state A to state B by path 1, then back to state A by path 2. For such a process the change in any state function is 0. However, the total q (= q₁ + q₂) and w (= w₁ + w₂) are in general not zero, unless path 1 is a reversible path and path 2 is the retracing of path 1. In this case only, q = w = 0. The only cyclic process for which the total heat and work is zero is the reversible one.

For the cyclic process,

Overall, q = -w < 0. The equality holds only for the reversible path. This statement says that by any irreversible cyclic pathway, the total work done on the system must be positive. This is because by path 1, the system will produce less than the maximum amount of work (max obtained by reversible path); by path 2, more than the minimum amount of work (min is for reversible path) will be expended if the path is irreversible. The work obtained in path 1 will not cancel the work expended by path 2, so the total work done on the system is > 0. Thus the heat added to the system, q, is < 0. This means that net heat will be transferred to the surroundings.

The conversion of work to heat in a cyclic irreversible path is illustrated in the P-V diagram of Figure J-1. The work done by the system for path 1 is the area under the plot of P_ext versus V for the path (òP_extdV). The work expended in returning the system to state A along path 2 is the area under the P_ext versus V plot for this path. The difference between these two areas (shaded) is the net work expended for the cyclic path. This work may be considered useless, because it is converted to heat and dissipated in the surroundings.

More Advanced Aspects of Energy and Entropy; Reversible and Irreversible Processes. We begin with some examples.

Example F-1. Calculate heat, work, and the change in internal energy for the reversible isothermal expansion of an ideal gas.

Solution. Reversible processes are of central importance in thermodynamics. It is therefore important that we define and explain the concept carefully. A reversible process is one in which the system and surroundings are in equilibrium throughout the process. In the context of gas expansions, system and surroundings are in equilibrium if they have the same temperature and the same pressure. The word, throughout, is critically important. It means that there is no time during the process when system and surroundings are out of equilibrium. In the expansions considered in Chapter 11, system and surroundings are in equilbrium at the beginning and the end, but not during the process. Neither of these processes is reversible, in the thermodynamic sense of the word. Instead, they are irreversible. If a process is truly reversible, then it can be made to reverse its direction at any time by making an infinitesimal change in some external variable. This provides a useful and easily applied test of reversibility.

To expand a gas reversibly requires that we keep the outside and inside pressures essentially balanced as we carry out the expansion. This can be accomplished with the apparatus shown in Figure 9-4. To the outside of the piston is attached an inflexible rod that is bent in the shape of a squared-off "U". The rod is attached to a string that is passed over a pulley and tied to a pan. Sand is placed on the pan to create an external pressure on the piston. In the figure, the gas is confined in a volume of 1.0 L at a pressure of 3.0 bar. The piston is stationary because the external pressure is the same as the internal pressure. The temperatures of system and surroundings are the same. Thus system and surroundings are in equilibrium at the beginning of the process. The reversible expansion is initiated by removing one grain of sand from the pan. This lowers P_ext by an infinitesimally small amount, and the piston moves out by an infinitesimally small increment. When the piston has stopped moving, a second grain of sand is removed. If this stepwise process is repeated a very large number of times, eventually the gas will have expanded to a final volume of 3.0 L and a final pressure of 1.0 atm. Of course, the process will take a very long time. By making only infinitesimal changes in the external pressure, we insure that system and surroundings are never out of balance. The process therefore satisfies the reversibility criterion.

Work: Since the external pressure never differs from the internal pressure by more than an infinitesimal amount, P_ext = P_int = nRT/V. To calculate the work done on the system we must use equation F-1:

Since the amount of gas, n, and the temperature are constant, they may be taken out from under the integral sign. The remaining integral, òdV/V, is ln V. The work done on the system during a reversible isothermal expansion of an ideal gas is then

For the conditions of this problem,

DE: Since T is constant, and the gas is ideal, DE = DKE = 0.

Heat: Obtain this from the first law: q = DE - w = -w = 3.30 bar-L.

Thus work "flows" out, an equal amount of heat "flows" in, leaving the internal energy of the gas unchanged. Heat taken from the surroundings is used to do work on the surroundings (i.e., the pan, with its ever-decreasing load of sand, is raised against gravity as the gas expands).

Example F-2. Apply the first law to the adiabatic expansion of an ideal gas against a constant external pressure.

Solution. An adiabatic process is one in which no heat flows between system and surroundings. Therefore, q = 0 in the first law expression, and it follows that DE = w. When the gas expands against an external pressure, work is done by the system and therefore w must be negative. It follows that DE < 0. The temperature of the gas therefore decreases. An ideal gas cools off when it expands adiabatically against a constant external pressure. Figure 9-5 shows the initial and final states of an adiabatic expansion. The initial state is the same as in the previous examples: V = 1.0 L, P = 3.0 bar, and T = 298 K. The external pressure is 1.0 bar. The gas expands until its pressure falls to 1.0 bar. The final volume of the gas is less than 3.0 L, the value in the previous expansions, because the temperature decreases during the expansion. It follows that the work done by the system (-w = P_extDV_sys) is less than in Example 9-2, where the gas is kept at constant T throughout the expansion.

With the information available in the problem, we can calculate only the ratio of the final volume to the final temperature:

We cannot calculate either the final volume or the final temperature without further information. Without the final volume we cannot calculate work. However, we can draw the following qualitative conclusions:

Because no heat can enter the system, the system draws on its supply of internal energy in order to push the surroundings back when it expands.

The molecular interpretation of the adiabatic expansion against an external pressure is similar to that for the isothermal expansion, with one important exception. As the molecules collide with and lose energy to the piston, pushing it out against the external pressure, they slow down and the gas cools. However, there is now no heat flow from outside; the kinetic energy of the gas molecules is not restored, as it was in the isothermal process, by the influx of heat. Consequently, the system volume increases less in the adiabatic than in the isothermal expansion, because both the cooling and the expansion of the gas decrease its pressure. For an expansion against constant external pressure, starting from the same initial state, the isothermal process gives a larger DV_sys and produces more work than the adiabatic process.

For the isothermal reversible expansion:

Two things are apparent. First, the surroundings does the minimum work on the system in the reversible expansion. Said an equivalent but more useful way, the system does the maximum work on the surroundings in the reversible process (as discussed earlier, the work done by the system on the surroundings is the negative of the work done by the surroundings on the system). It is not possible for the system to do more work than it does in the reversible process because, to do so, it would have to push back an external pressure that exceeds its own pressure. Under these conditions, of course, the system would contract, not expand. The system does the minimum work on the surroundings -- zero -- in the free expansion, where it pushes nothing back. An intermediate amount of work is obtained from the system when it expands against an external pressure that is constant and less than its own pressure. Similarly, the heat transferred to the system is a maximum in the reversible process, and is less than this maximum for any irreversible process.

Entropy and Heat. The search for a quantitative measure of entropy was a long and difficult one. The analysis of steam engines (the original genesis of thermodynamics) was the source of one approach, formulated by Rudolf Clausius in the 1880s. The Clausius formulation expresses entropy in terms of heat and temperature. The second approach, that of Ludwig Boltzmann, was based on considerations of probability and statistics. We have already made the connection between disorder and probability in our discussion in Chapter 10. Although the two quantitative formulations of entropy appear different, it has been shown that they are equivalent in any application to which they may both be applied. We will take up the Clausius formulation of entropy first, and return later to a discussion of the Boltzmann approach.

That there is a connection between disorder and heat is somewhat expected. First, heat, the random thermal motion of molecules, is clearly the most disordered and non-useful form of energy. Second, much of our experience indicates that disorder is produced when heat is added to systems. The solid form of a substance (water, sodium chloride, iron) melts, and the liquid form boils, when heat is added. Gases expand when heated. Atoms form a plasma of nuclei and electrons when heated to the temperatures of stars. These qualitative realizations help us to appreciate the Clausius definition of entropy in equation F-3:

Stated in words, the equation says that the infinitesimal change in entropy that occurs in a system when an infinitesimal part of a process is carried out may be calculated by dividing the infinitesimal amount of heat added to the system by the temperature at which the heat addition occurs. The differential form is required, as it was in the definition of work, to recognize that the temperature may change during a finite process. (The symbol d [rather than the usual "d"] is used with q because heat has what is called an inexact differential. In contrast, entropy has an exact differential, so the usual "d" is used. In general, state functions have exact differentials, and path functions have inexact differentials. We will not deal further with this matter here.) The subscript "rev" on the heat term represents a restriction on the definition. It indicates that entropy must be calculated using the heat added for a reversible path. We return to this in a moment, after we examine the "sense" of equation 11-2. First, the Clausius formulation of entropy recognizes the clear connection between disorder and heat. The more heat we add, the more disorder we produce, at a given temperature. Second, the equation says that the disorder produced by addition of a given amount of heat is less when the system is at high T than when it is at low T. This is also reasonable. In a system at low T (say, near absolute zero) the kinetic energy of random molecular motion is small. At high T (say 1000 K), however, the thermal energy is substantial. Addition of a specified amount of heat to such a system produces a much larger fractional change in disorder at low T than at high T. Equation F-3 thus recognizes that if a system is already highly disordered, a small amount of heat won't make much difference. Finally, entropy is represented in equation 11-2 and in Chapter 11 by an upper case symbol, which we reserve for state functions. Entropy is a state function. This means a change in entropy is independent of path, because it is always calculated as though the process was carried out reversibly. In order to calculate the entropy change for a finite change in the system, equation F-3 must be integrated:

Generally, to perform the integration requires an expression for q in terms of T. There are a number of important processes for which such an expression is available. For isothermal processes, the integration is easily performed to give equation F-5:

This special case formulation is applicable in a number of important situations, as we will see.

The "reversibility" restriction on the heat in the definition of entropy is necessary if entropy is to be a state function. In general, as we have seen, heat is not a state function -- it depends on the path followed between states. However, the heat added to the system is a state function for a particular path; as long as the same path is always followed between states, the same amount of heat will be added every time. Further, if the path is reversible, the heat added to the system is larger than the heat added along any other conceivable path. We "proved" this for the gas expansions in Examples F-1 and F-2, and it is generally true. Thus Clausius defined entropy in terms of the pathway requiring the maximum addition of heat to the system. Figure 11-2 shows three cyclic pathways involving two states, A and B, of a system. For each path, the system begins in state A, is taken to state B by one branch of the path, then is returned to state A by the second branch of the path. The advantage of considering a complete cycle is that the change in any state function of the system must be zero for a cyclic path, because the system ends up in the state that it started in. Thus DE for each cyclic path is zero, and we may focus on the heat, q_total, and the work, w_total, for each path. Regardless of path, the first law of thermodynamics requires that q_total = -w_total. One of the cycles is the reversible path, and is so labelled. The system is taken from A to B along branch "rev a" and is returned to state A along branch "rev b". When the system is taken from state A to state B along this path, system and surroundings are in equilibrium throughout the process. The same is true of the return from B to A. The other cyclic paths are labelled 1 and 2. Because there is only one reversible pathway between two states, paths 1 and 2 must both be irreversible. This means that, although system and surroundings are in equilibrium at state A and at state B, they are not at equilibrium when the system is between states A and B; i.e., while the process is going on. Although Figure 11-2 and the conclusions we will draw from it apply generally to any system, we will assume for the sake of discussion that the A-to-B branches of the processes are the ideal gas expansions described below.

We now consider the return branches, which must be compressions. To compress the gas, it is clearly necessary to have an external pressure which is at least equal to the pressure of the gas in the system, otherwise the gas will not compress. If the external pressure is equal to the internal pressure (actually infinitesimally larger), the compression is reversible. The work and heat along this branch are equal and opposite to those for the reversible expansion. The path followed in the reversible expansion is exactly retraced in the reversible compression, and the total heat and work for the cyclic path are zero: q_revtotal = -w_revtotal = 0. When both branches are done reversibly, no work is required and no heat flows! It is very significant that the reversible branch from A to B is the only one that may be exactly retraced in returning from B to A. Neither 1a nor 2a (nor any other conceivable path from A to B) may be so retraced. This is in fact the true meaning of the word "reversible". We now show the non-retraceability of paths 1a and 2a. Along path 1a the gas expands against P_ext = 1 atm. At all times during the expansion except at the end, the gas pressure exceeds 1 atm. When the gas pressure drops to 1 atm, expansion stops. Clearly, this path cannot be retraced in returning to A, because 1 atm of external pressure will not compress the gas! To compress the gas, we must make P_ext larger than the internal pressure of the gas. The least subtle way to carry out the compression is to make the external pressure equal to 3 bar. Compression stops when P_int = 3 atm, which is state A. However, the work done on the system in doing this is 6 atm-L, and the heat added to the system is -6 atm-L (i.e., heat flows to the surroundings). For cyclic path 1 with the least subtle return branch, q_1total = (2 - 6) atm-L = -4 atm-L, and w_1total = 4 atm-L. Irreversible cycle 1 therefore takes useful work from the surroundings and converts it to non-useful heat! The most subtle way to return from B to A is via the reversible path, "rev b". This will require the least work of any conceivable return path. However, this work still exceeds the -2 atm-L done on the system during the expansion! Cyclic path 1 therefore involves a minimum w_1total of 1.3 atm-L, and a maximum q_1total of -1.3 atm-L. It is therefore unavoidable that cyclic path 1 takes useful work from the surroundings and converts it to non-useful heat in the surroundings. You can apply the same reasoning to cycle 2 and find that the situation is even worse, because the system does no work in expanding, but a minimum of 3.3 atm-L of work must be done to recompress it.

We can sum up this rather intricate line of reasoning in either of 2 statements:

It is clear that this is true. Both cycles 1 and 2 involve the net conversion of work (useful) to heat (non-useful). The reversible cycle involves no net work or heat, therefore no degradation in the useful energy. The implication of this is that there is no such thing as a cyclic process which converts non-useful heat to useful work! Thus the second statement:

These statements are both alternative ways of stating the 2nd Law of Thermodynamics.

There is one final important conclusion to be drawn from the cyclic paths of Figure 11-2. For all three paths (and for any other cyclic path), DS_sys = 0. This must be true because entropy is a state function. For path "rev", it is also true that DS_surr = 0, because no heat is added to the surroundings in this cycle. However, for both (irreversible) paths 1 and 2, heat is added to the surroundings. Therefore DS_surr > 0, and DS_univ > 0 for all irreversible paths. This makes more general our original statement of the 2nd law:

Although we have not demonstrated it, these statements are true for non-cyclic processes as well. What then would we say about a process with DS_univ < 0? Based on the second law of thermodynamics, we would say that such a process is impossible. There has never been a process observed which had the effect of decreasing the entropy of the universe. Often these are processes which we would very much like to occur. Some examples of impossible processes follow:

11-2 Entropy Calculations. Equation F-4 may be applied to any isothermal process for which we can envision a reversible path. Two such processes which are important in chemistry and which have been discussed in other contexts are phase changes and gas expansions.

Figure 11-3a shows two ways in which the melting of an ice cube can be carried out. At left, the ice cube is submerged in a container of water which is at 50^oC. Since the temperature of the water is so much above the melting point of ice (0^oC), the ice cube will spontaneously melt. In order to reverse the melting process, the temperature of the water bath would have to be lowered to just slightly below 0^oC. This is a finite, not an infinitesimal, change in an external variable. We therefore conclude that the melting process at left is irreversible. At right, the ice cube is submerged in water at 0^oC, the normal melting point of water. By adding an infinitesimal amount of heat to the water bath, melting can be initiated. It will occur very slowly, of course, because the temperature differential between cube and bath is so small. To reverse the melting process at any point, we need only lower the bath temperature by an infinitesimal amount again. The melting process at right is reversible. On the basis of this discussion, we conclude that, in general, phase changes carried out at the normal phase change temperature are reversible.

Example F-3. What is the entropy change when 3.4 g of ice melts reversibly at 0^oC under a constant pressure of 1 bar? It is found by calorimetry that 1130 J of heat are absorbed by the system during this process.

Solution. Because the process is reversible and isothermal, we may apply equation F-6.

The equality of q and DH for the process follows because pressure is constant.

For any reversible phase change, DS_{phase change} = DH_{phase change}/T_{phase change}.

Figure 11-3b shows two ways in which a gas may be expanded isothermally from 2.0 atm and 1.0 L to 0.8 atm and 2.5 L. The top expansion is carried out against a constant external pressure of 0.8 atm. At any point during the expansion except at the end, the internal pressure exceeds 0.8 atm. In order to reverse the expansion, then, it is necessary to increase the external pressure by a finite amount. The top expansion is therefore irreversible. The heat involved in this process cannot be used in equation 11-4 to calculate the entropy change of the expansion. Instead, the heat involved in the bottom process must be used. Here the external pressure balances the internal pressure at each point in the process. The expansion may be turned into a compression at any point by making an infinitesimal change in P_ext. The bottom expansion is reversible.

Example F-4. Calculate DS_sys for the reversible isothermal expansion of an ideal gas from 1.0 L to 2.5 L at 298 K.

Solution. We apply equation 11-4.

The number of moles of gas was obtained by applying the ideal gas law to the initial pressure, volume, and temperature conditions.

Example F-5. What is DS_sys for the top gas expansion in Figure 11-3b?

Solution. This is much easier than it may seem. Since the initial and final states of the system are exactly the same as for the reversible expansion at right in the figure, the entropy change is the same as well. Entropy is a state function. Thus

The take-home message from this example follows. DS_sys for the isothermal expansion of a gas from a particular V_i to a particular V_f is the same no matter how the expansion is carried out: DS_sys = nRln V_f /V_i.

Example F-6. Consider the apparatus in Figure 11-4. It consists of two cylinders of equal diameter, on the right and left, each fitted with a piston. The cylinders are joined by valves through a tank of volume 1.00 L which is joined to another tank of volume 1.00 L. Initially an ideal gas is in the left cylinder and the lower small tank at a pressure of 1.00 atm, and the temperature is 298 K. Calculate the quantities indicated for each process.

Solution. We can make our work as easy as possible by recognizing at the outset that since the system is an ideal gas and T is constant for all processes, DE = 0 for each of the three processes, and q = -w.

a. Reversible compression from V_i = 10.0 L to V_f = 1.0 L. The gas volume is reduced by a factor of 10, so the pressure increases to 10.0 atm.

To express q and w in Joules, multiply by the ratio of R in Joules to R in atm-L:

Since the compressions is reversible, the heat just calculated is the reversible heat needed to calculate the entropy change. Further, the process is isothermal. Thus

Heat flows from system to surroundings in this process. Therefore the heat added to the surroundings, necessary to calculate the entropy change of the surroundings, is 2330 J:

Note that, as expected for a reversible expansion, DS_univ = DS_sys + DS_surr = 0 J/K.

b. Closing valve 1 isolates the gas from the left cylinder/piston. When valve 2 is subsequently opened, the gas will undergo a free expansion to occupy both 1.0 L boxes. When volume is doubled, pressure is halved. The final pressure of the gas is thus 5.0 atm. No work is done and therefore no heat flows in the isothermal free expansion:

The entropy change for the system is calculated by imagining the expansion to be done reversibly:

Moles of gas are calculated by the ideal gas law:

During the isothermal free expansion, no heat flows to or from the surroundings. Therefore

The entropy change for the universe in this expansion is

The free expansion is irreversible.

c. When valve 2 is closed, the gas in the top 1.00 L box is isolated, and is therefore not important in determining what happens when valve 3 is opened. We therefore focus on the 0.202 moles of gas exerting 5.00 atm pressure in the bottom 1.00 L box. When valve 3 is opened, this gas will expand, pushing back the right piston against the constant external pressure of 1 atm, until the gas pressure is 1.00 atm. Since the gas pressure drops by a factor of 5 during expansion, volume increases by the same factor to 5.00 L.

The entropy change of the system is calculated in the usual way:

The heat added to the surroundings during the expansion is the negative of the heat added to the system, or -405 J. Thus

The entropy change for the universe in this expansion is

The expansion is irreversible.

In all three expansions in the preceding example, we have calculated the entropy change of the system in the agreed-upon manner: we have used the heat for a reversible path. However, to calculate DS_surr, we have used the actual heat put into the surroundings in the process. Why is it that we have to imagine a reversible path for the system, but not for the surroundings? We justify this seeming contradiction as follows.

In general, we expect that

This equation is consistent with the definition of entropy due to Clausius. If we now restrict our attention to isothermal processes, as we have done above in calculating DS_sys, this simplifies to

We now make a statement which is true because the surroundings is essentially infinite: from the standpoint of the surroundings, all processes are reversible. This is true because the surroundings is so large; any amount of heat added to it is for all practical purposes infinitesimal. Thus the actual heat added to the surroundings during a process is q_rev from the standpoint of the surroundings, and

This equation rationalizes our use of the actual process heat in calculating the entropy change of the surroundings.

The implications of this approach are a bit disturbing. An isothermal gas expansion from V_i to V_f can be carried out in many ways (e.g., reversibly, freely, or against a constant P_ext). No matter how we do it the entropy change for the system is the same because we have defined entropy to be a state function for the system. However, the entropy change for the surroundings clearly very much depends on the choice of process. Entropy is not a state function -- is not conserved -- for the surroundings, or for the universe. It is therefore a very different quantity from energy, which is conserved for the universe in every process.

The Boltzmann Interpretation of Entropy. The Clausius interpretation of entropy is a pragmatic one, based on engine analysis. Ludwig Boltzmann took a different approach, based on the idea that it is reasonable to draw a parallel between the amount of disorder in a system in a particular state and the number of possible ways the system may be arranged in that state. In a very general way, this can be stated mathematically as

where W = the number of arrangements possible for a given state of the system, and f is some to-be-determined function. We agree at the outset that a state with only one possible arrangement has no disorder -- that is S = 0 -- and that S should increase smoothly as W increases. This is our first constraint on the form of the function, f. To arrive at a second constraint on the form of f, we consider a simple example, illustrated in Figure 11-5. Shown is a row of 10 buckets placed side by side. A ping pong ball is tossed at the row of buckets and lands in one of them (we assume for simplicity that there are no misses -- the ball always goes in one of the buckets). The state of the resulting system is then described by saying that there is one ball in the set of 10 buckets. How many arrangements are possible for this state? Clearly, since the ball can be in any one of the buckets, W = 10. There is therefore some disorder associated with this state. An observer asked to view the row of buckets containing 1 ball would be uncertain as to where the ball is. His uncertainty is a measure of the disorder.

We now double the amount of material in the system by tossing a second ball, which also lands in one of the buckets. The state of the system is described by saying that there are 2 balls in the set of 10 buckets. How many arrangements are possible? The first ball can be in any of 10 locations. For each of these possibilities, there are 10 possible locations for the second ball. The total number of possible arrangements for the state is thus 10 x 10 = 10². By similar reasoning, if n balls in succession were tossed, the number of arrangements would be 10ⁿ. In general, for a state consisting of N balls placed in L locations, the number of arrangements is given by F-8:

The entropy of the state is therefore S = f(L^N). We now impose the second constraint on the form of the function f: the nature of f must be such that S doubles when the amount of material in the system (N) doubles. In equation form,

The function that satisfies both of the imposed constraints is the logarithm. Thus

where C is a constant of proportionality. Equation F-10 is, in essence, the Boltzmann interpretation of entropy.

The final step is to find out what the constant, C, is. To do this, we tie together the Boltzmann and Clausius definitions of entropy by comparing the expression for DS based on 11-11 with DS for the expansion of an ideal gas from V_i to V_f:

Application of F-11 to the gas expansion is not difficult. The amount of material in the system is the number of gas molecules, N. When the gas expands, the number of locations available for the molecules increases from L_i to L_f. The change in entropy on expansion is therefore

Equating the two expressions for DS and replacing the number of moles of gas by N/N_o (N_o is Avogadro's number),

Division by N and replacement of R/N_o by k, Boltzmann's Constant, gives

It seems reasonable to assume the number of locations for gas molecules, L, to be directly proportional to the available volume, V. Then L_f/L_i = V_f/V_i, and we conclude that the proportionality constant, C, is Boltzmann's constant. The final form of the Boltzmann interpretation of entropy is F-12:

This equation serves as Boltzmann's epitaph; it is inscribed on his tomb in Vienna.

Finally, if a state "a" has more possible arrangements than a second state "b", then state "a" is more probable than state "b". The Boltzmann interpretation of entropy thus explicitly recognizes the connection between entropy and probability that we developed intuitively in Chapter 10. According to this interpretation, the second law of thermodynamics says that the universe seeks states of higher probability.

Example F-7. Consider a system consisting of a pair of dice. The system has 11 states, one for each combination of numbers (2 through 12) showing on the top faces of the dice. Which state is most probable? Which is least probable? Discuss the entropies of the most and least probable states.

Solution. We must figure out the number of ways to obtain each number combination:

The most probable state is 7. States 2 and 12 have only 1 arrangement each, so are both least probable states. State 7 has highest entropy. States 2 and 12, since they have only one arrangement, both have S = 0.

The entropy associated with number combinations of dice may be thought of as follows. A blindfolded man is told that the dice show a total of 2, and is asked what number is showing on each die. He can, with certainty, state that each die shows 1. In contrast, when told that the dice show a total of 7, he can only guess at the numbers on individual die. He is uncertain. The extent of uncertainty is a measure of the entropy of the state.

Comments on the Temperature Dependence of DG_R^o and K_eq. Consider a reaction for which the following statements are true:

The van't Hoff relation and Le Chatelier's principle tell us that because DH_R^o < 0, K_eq for the reaction should decrease with increasing T, and the reaction should shift toward reactants. At the same time, the equation DG_R^o = DH_R^o - TDS_R^o tells us that as T increases, the TDS_R^o term becomes dominant, and implies that products should become more favored as T increases. How do we rationalize these apparently conflicting conclusions?

We accomplish this in terms of the equations

From these equations, we conclude that the following are consequential to an increase in T:

The first and last consequences, that DG_R^o becomes more negative but at the same time K_eq decreases, seem to violate physical intuition. The key, however, is equation (3). As long as T increases at a faster rate than -DG_R^o increases, -DG_R^o can increase while at the same time K_eq decreases.

In summary, we make the following statements about the effects of an increase in T on a reaction for which DH_R^o < 0 and DS_R^o > 0:

The equilibrium constant is a function of partial pressures of the participants. A good starting point, then, is the manner in which the free energy of a pure gaseous substance depends on its pressure. We consider a small, reversible change in G for 1 mole of the substance:

Differentiation of the TS term is accomplished with the chain rule. Constraining the temperature to be constant requires dT = 0. Then

Substitute for dH using the definition of enthalpy; this gives

Then substitute for dE by differentiating the First Law:

The expression appears to be growing more complex. However, we can now take advantage of the restriction of reversibility. This means that system and surroundings are in equilibrium throughout. Thus P_ext = P (the pressure of the system -- the gaseous substance). Further, dq = dq_rev, which is of course TdS. Some cancellation of terms now yields a much simplified expression:

For a single pure substance, G depends in a very simple way on P. Equation 12-16 can be readily integrated for an ideal gas, for which V = nRT/P. The result is F-16.

Here G^o is the standard free energy of the substance at the standard condition of 1.0 bar; G is the free energy at arbitrary pressure P; and the other terms are familiar. Applying F-16 to each pure substance in the general reaction,

gives the extremely important equation F-18.

The familiar pressure expression under the logarithm sign is Q, the reaction quotient.

At equilibrium, when the chemical reaction occurs reversibly, DG = 0 and Q = K_eq. Making these substitutions and rearranging gives the desired relationship between the two spontaneity measures:

This is a fundamental result of thermodynamics.