Entropy

Almost everyone meets entropy twice: once in a popular science book, where it is disorder and the heat death of the universe, and once in a thermodynamics course, where it is a column in a steam table. Reconciling the two is nobody’s job and the attempt wastes a great deal of time. This article builds the second one and comes back to the first at the end, honestly, in the last section. The construction is short, and it starts with an integral that turns out to be zero.

Where entropy comes from

Everything in the second law so far has been a prohibition. Heat will not climb a temperature gradient on its own. No engine can beat a reversible one working between the same two reservoirs. These are useful statements and they are all of the form this cannot happen. What they do not give is a quantity — something you can compute for a proposed process and compare against a limit, the way you compute an energy balance and compare it against zero. The first law has such a quantity, and the whole of this topic is the second law acquiring one.

Start with the reversible cycle you already know. A Carnot engine draws \(Q_H\) from a reservoir at \(T_H\) and rejects \(Q_L\) to one at \(T_L\), and the result that makes the thermodynamic temperature scale work is that the two heats stand in the ratio of the two temperatures:

Adopt the usual sign convention, in which heat into the system counts positive and heat out counts negative, and that equation says something more suggestive: the two contributions \(Q/T\) cancel exactly when they are added round the loop. Any reversible cycle at all can be approximated as closely as you like by a mesh of small Carnot cycles, so the same cancellation survives in the limit, and the general statement is that

For an irreversible cycle it is not zero but negative, which is the Clausius inequality, \(\oint \delta Q/T \le 0\), and that sign is returned to in the next section. Here it is the equality that does the work, because of a piece of reasoning that is entirely about bookkeeping and not at all about physics.

If the integral of some quantity round every closed loop is zero, then the integral between any two states is the same along every path. Take two reversible paths from state 1 to state 2. Run out along the first and back along the second and you have a cycle, whose total is zero, so the outward integral along the first path equals the outward integral along the second. The route has cancelled itself. A quantity whose change between two states does not depend on how you got there is precisely what is meant by a property, so there is a property here whether or not anybody was looking for one. Clausius named it entropy and defined it by the differential:

Per unit mass this is \(s\), in kJ/kg·K, and it is the same lower-case convention used for every other specific property. Note what has happened: entropy was not measured and then explained. It was defined into existence, because an integral vanished and something had to be the thing whose change that integral was.

The subscript on \(\delta Q\) is the part that gets dropped and should not be. Entropy is defined along a reversible path. That does not restrict which processes you can analyse, because entropy is now a property and the change between two given states is fixed by the states alone. It means that when the real process is irreversible, you evaluate \(\Delta S\) along an imagined reversible path between the same end states, and the answer is right for the real one. What you may not do is compute \(\int \delta Q/T\) along the irreversible path and call the result the entropy change. It will come out too small, and the deficit has a name.

The increase of entropy principle

Write the deficit down and the second law becomes an equation you can balance. For any process undergone by any closed system,

The sum is entropy transferred: every heat interaction carries entropy across the boundary alongside it, in the amount \(Q_k\) divided by the absolute temperature \(T_k\) of the boundary where that heat crosses. Work carries none, which is the sharpest distinction the subject draws between the two — a paddle wheel and a flame can put identical energy into the same fluid, and only the flame brings entropy with it. The remaining term, \(S_{\text{gen}}\), is entropy generated inside the boundary by irreversibility: friction, unrestrained expansion, mixing, chemical reaction, electrical resistance, and heat conducted across a finite temperature difference. It is zero for a reversible process, positive for a real one, and never negative.

Two things about \(S_{\text{gen}}\) repay being said plainly. It is not a property: it depends on how the process was run, not on where it started and finished, so it has no value at a state and never appears in a table. And it is the measure the second law was missing. A process with \(S_{\text{gen}} = 0\) is reversible, one with \(S_{\text{gen}} > 0\) is possible but wasteful, and one for which the arithmetic returns \(S_{\text{gen}} < 0\) is impossible. That last case is the useful one. A negative generation is not an error to be rounded away; it is the calculation reporting that the process you described cannot occur.

Draw the boundary so that no heat crosses it — an isolated system — and the transfer term disappears, leaving \(\Delta S_{\text{iso}} = S_{\text{gen}} \ge 0\). Since any system can be isolated by taking in enough of its surroundings, the same holds for a system together with everything it interacts with:

Here is where the popular statement does damage. “Entropy always increases” is true of the total and false of a system, and students who have absorbed the slogan will refuse to write down a negative \(\Delta S\) even when it is correct. Put a can of drink in a refrigerator. It cools, its entropy falls, and there is nothing controversial about that. The refrigerator dumps rather more entropy into the kitchen than the can lost, because it also had to reject the work it consumed, and the total goes up. Every real cooling, freezing, condensing or crystallising process lowers the entropy of something. It is the sum that is constrained.

A number makes the point better than the sentence does. Let 1000 kJ pass from a reservoir at 800 K to one at 300 K, with the two of them inside a single boundary so nothing else is involved. A reservoir is large enough that its temperature does not shift, so its entropy change is simply \(Q/T\). The source loses \(1000/800 = 1.25\) kJ/K. The sink gains \(1000/300 = 3.33\) kJ/K. The same energy, divided by a smaller temperature, buys more entropy on arrival than it cost on departure, and the difference of 2.08 kJ/K did not come from anywhere: nothing crossed the outer boundary. It was generated, by the act of letting heat fall through a temperature difference of 500 degrees with no engine in the gap. That gap is the lost opportunity, and \(S_{\text{gen}}\) is its size.

One point of technique. In the transfer term, \(T_k\) is the temperature at the boundary where the heat actually crosses, not the average temperature of the system. Choosing the boundary far enough out that it sits at the surroundings temperature is a standard trick, because it moves the irreversibility of the temperature drop inside the system and lets you find the total generation in one balance rather than two.

T–s diagrams

Rearranging the definition gives \(\delta Q_{\text{rev}} = T\,dS\), and integrating,

which is the reason entropy earns a diagram of its own. Plot temperature against entropy and the heat transferred in a reversible process is the area under the process line, exactly as the area under a process on a P–v plot is the boundary work. The second law stops being a rule to be remembered and becomes a picture that can be measured with a ruler.

The vertical axis is absolute temperature, and this is not a formality. An area is measured from the axis, so the axis has to sit at true zero for the area to mean the heat. A T–s diagram drawn in Celsius has areas that are wrong by \(273.15\,\Delta s\), and the same warning applies to every entropy calculation in this article: temperatures go in in kelvin. The only place a Celsius difference survives is inside a \(\Delta T\) where the offset cancels, and inside a logarithm it emphatically does not.

Two processes now have shapes worth memorising. A reversible adiabatic process transfers no heat, so \(dS = 0\) and the process is a vertical line — it is isentropic, and the two words are used interchangeably even though the equivalence needs both halves of the condition. An adiabatic process that is irreversible generates entropy and moves to the right; an isothermal process is a horizontal line, along which \(Q = T\,\Delta S\) comes straight out with no integration at all.

Put two of each together in the right order and the Carnot cycle is a rectangle. Heat in along the top isotherm at \(T_H\), isentropic expansion down the right-hand side, heat out along the bottom isotherm at \(T_L\), isentropic compression back up the left. Take \(T_H = 800\) K, \(T_L = 300\) K and an entropy change of 1.25 kJ/kg·K along the isotherms. Heat in is \(800 \times 1.25 = 1000\) kJ/kg, the full rectangle down to absolute zero. Heat out is \(300 \times 1.25 = 375\) kJ/kg, the strip below. The net work is the difference, 625 kJ/kg, and it is the enclosed rectangle. The efficiency is one area divided by the other,

and the \(\Delta s\) cancelling top and bottom is the entire reason a Carnot efficiency depends on nothing but the two temperatures. The result was proved in the previous topic by a contradiction argument about impossible engines. Here it is two sides of a rectangle.

One caution about what may be drawn. An irreversible process passes through states that are not in equilibrium and therefore have no single temperature or entropy, so strictly it cannot be drawn as a line at all. Where one is sketched — a real turbine expansion leaning to the right of the vertical, say — it is a dashed indication of where the end state ends up, not a path. The area under it is not the heat transferred, and reading one off is a mistake that survives all the way into exam scripts.

Evaluating entropy change

For a real substance the honest answer is that you look it up. Entropy sits in the property tables in exactly the same columns as internal energy and enthalpy, with \(s_f\), \(s_{fg}\) and \(s_g\) in the saturated tables and a value per cell in the superheated blocks. Inside the dome the lever rule applies unchanged, because entropy is extensive and averages by mass like everything else:

For water at 100 kPa, \(s_f = 1.3028\) and \(s_{fg} = 6.0562\) kJ/kg·K, so a mixture of quality 0.8 has \(s = 1.3028 + 0.8(6.0562) = 6.148\) kJ/kg·K. A compressed liquid, as before, is taken as saturated liquid at the same temperature, \(s \approx s_f(T)\), and the approximation is better for entropy than it is for enthalpy because there is no \(Pv\) term to neglect.

Where there is no table, the two \(T\,ds\) relations take over. Apply the first law to a reversible process in a closed system, with \(\delta Q = T\,ds\) and \(\delta W = P\,dv\), and then substitute \(h = u + Pv\) in the result:

Both were derived along a reversible path, and both relate properties only. That makes them valid for any process whatever, reversible or not, which is the same argument that let entropy be evaluated on an imagined path in the first section. It is worth being clear that this is a licence about evaluation, not about physics: the relations tell you the entropy change, they do not tell you that the heat transfer was \(T\,ds\).

For an ideal gas, substitute \(du = c_v\,dT\), \(dh = c_p\,dT\) and \(Pv = RT\), and each relation integrates to a pair of logarithms. With the specific heats held constant,

Use whichever form matches the data you were given; they agree, because \(c_p - c_v = R\) and \(P v = RT\) connect them. Take air from 300 K and 100 kPa to 600 K and 400 kPa, with \(c_p = 1.005\) and \(R = 0.287\) kJ/kg·K. Then \(\Delta s = 1.005\ln 2 - 0.287 \ln 4 = 0.697 - 0.398 = 0.299\) kJ/kg·K. The temperature term raised it, the pressure term pulled it back, and the two very nearly cancelled. Set the whole thing to zero and rearrange and you have the isentropic relation, \(T_2/T_1 = (P_2/P_1)^{(k-1)/k}\), which is not a new result but this one solved for a special case. Where the temperature range is wide enough that constant specific heats will not do, the tabulated function \(s^{\circ}(T)\) replaces the first logarithm and the pressure term is unchanged.

For an incompressible substance — any liquid or solid, to a good approximation — \(dv = 0\), so the first \(T\,ds\) relation loses its work term entirely, and the two specific heats collapse into one:

Pressure does not appear. Squeezing a liquid at constant temperature changes its entropy by nothing worth writing down, which has a consequence worth carrying: an isentropic process in an incompressible substance is an isothermal one. The pump in a steam cycle raises the pressure of the feedwater by a hundredfold and barely warms it, and that is not an idealisation being sloppy — it is what this equation says.

What entropy is not

Entropy is not disorder, in any sense of the word that would be recognised outside a physics department, and the bedroom analogy — tidy room, low entropy; messy room, high entropy; tidying requires effort — is worse than useless. Tidiness is not a thermodynamic state. Two identical volumes of the same gas at the same temperature and pressure have the same entropy to the last decimal place, whether you imagine the molecules neatly ranked or scattered, because the entropy is fixed by the state and the arrangement of individual molecules is not part of it. The analogy predicts nothing, gives no number, and leaves people expecting a quantity that behaves like a judgement about housekeeping.

What the statistical reading actually claims is narrower and much more specific. Boltzmann’s relation is

where \(k = 1.380649 \times 10^{-23}\) J/K and \(W\) is the number of distinct microscopic states — quantum states of the whole assembly, positions and momenta and energy levels together — that are consistent with the macroscopic state you have specified by its energy, volume and composition. Not the number of ways to arrange objects on a shelf. The number of ways the system’s energy can be distributed over the levels available to it while the measured properties stay exactly what they are. Adding heat opens up higher levels, which multiplies the number of ways, and \(W\) climbs very steeply indeed. The logarithm is there so that entropy adds when two systems are placed side by side, since the counts multiply.

That is a real statement and it explains real things — why entropy is extensive, why it rises with temperature, why mixing two gases generates some. It also has a consequence that finishes off the disorder story on its own terms. Water freezing in a freezer becomes more ordered by any everyday standard, and its entropy falls, and the process happens anyway, because the latent heat carried out to the kitchen raises the entropy there by more. Order appearing locally is the ordinary case, not the exception, and no amount of talk about disorder will tell you which way that particular contest goes. The arithmetic of the previous sections will.

There is a reference point worth knowing, since entropy values look absolute in a way that internal energy does not. The third law says that the entropy of a pure crystalline substance at absolute zero is zero, which fixes a genuine origin for the scale rather than an arbitrary one. In practice the steam tables do not use it: they set \(s_f = 0\) for saturated liquid water at the triple point, 0.01 °C, purely as a convention, exactly as they set \(u_f = 0\) there. Since every problem asks for a difference, the choice of origin cancels — provided both states are read from the same table, which is the one way this can go wrong.

So the working picture is small and it is enough. Entropy is a property, with a value at every equilibrium state, tabulated for real substances and computable from two logarithms for an ideal gas. It is carried across a boundary by heat and never by work. It is created by irreversibility and destroyed by nothing, so the total for a system and its surroundings can only rise. Whether that also explains the direction of time is a question this course does not need answered, and the answer, if there is one, will not change the value of \(s_g\) at 100 kPa.