Properties of pure substances

A pure substance is one with a fixed chemical composition throughout. Water is a pure substance, and so is a sealed flask holding ice and liquid water together, because both phases are the same compound. Air, as a gas, behaves as one closely enough to be treated that way. Air with some of it condensed does not, because the liquid that drops out is richer in oxygen than the vapour left behind, and the composition of each part is no longer the composition of the whole. Everything below assumes the substance stays chemically one thing while it changes phase.

Boiling at constant pressure

Take water at \(20\,^\circ\text{C}\) in a cylinder under a piston, with just enough weight on the piston to hold the pressure at \(100\) kPa. The weight does not change, so the pressure does not either, however far the piston travels. Then add heat slowly and watch two instruments: the thermometer, and the position of the piston.

At first the story is dull. The temperature rises steadily and the piston barely moves, because liquid water is very nearly incompressible and expands only slightly on warming. The water in this stretch is a compressed liquid, or equivalently a subcooled liquid: two names for the same thing, the first saying it is at a higher pressure than the pressure at which it would boil at this temperature, the second saying it is at a lower temperature than the temperature at which it would boil at this pressure.

At \(99.6\,^\circ\text{C}\) the first bubble appears. The water is now a saturated liquid — still entirely liquid, but on the point of boiling, with no margin left. Add any heat at all and some of it turns to vapour.

Then comes the part worth stopping on. Heat continues to pour in. The piston rises dramatically, because the vapour being formed occupies more than a thousand times the volume of the liquid it came from. And the thermometer does not move. Not slowly — not at all. It takes \(2257.5\) kJ for every kilogram converted, and throughout that entire transfer the temperature sits at \(99.6\,^\circ\text{C}\). Only when the last drop of liquid has gone, at the state called saturated vapour, does the temperature begin to climb again, and the substance beyond that point is a superheated vapour.

The plateau is the whole reason this topic is difficult, and it is difficult because of a habit of thought rather than any hard mathematics. Temperature is not a measure of energy added. It is a measure of the average kinetic energy of the molecules, and during the plateau the energy going in is not speeding molecules up. It is being spent breaking them apart from one another against their mutual attraction. That is stored energy, real and recoverable — condense the steam and you get all of it back — but it is invisible to a thermometer.

Note also what the pressure was doing during all of this: nothing. Held constant by a weight that never changed. That was not an incidental detail of the apparatus. It is the reason the temperature could not move, and the next section is what happens when you choose a different weight.

The saturation dome

Run the same experiment at a higher pressure and the same three-part story unfolds: a steep rise, a flat plateau, another rise. But the plateau sits at a higher temperature and it is shorter. Higher, because squeezing the substance makes it harder for molecules to escape into the vapour, so more thermal agitation is needed before they can. Shorter, because the vapour formed under greater pressure is denser, so the volume change from liquid to vapour is less dramatic.

The temperature at which the plateau sits is the saturation temperature \(T_{\text{sat}}\), and the pressure at which it happens is the saturation pressure \(P_{\text{sat}}\). They are not two independent facts. They are a single relationship,

which is why the phrase “the boiling point of water” is incomplete without a pressure attached. It is why a pressure cooker works, holding water liquid at \(120\,^\circ\text{C}\) so that food cooks faster, and why on a mountain the same water boils below \(90\,^\circ\text{C}\) and cooks nothing properly at all. The pot is at boiling point in both cases. Boiling point is simply not the same temperature in both cases.

Now repeat the experiment at every pressure, plot each run on the same axes of temperature against specific volume, and mark two points on each: where the plateau starts and where it ends. Joining all the starting points gives the saturated liquid line, and joining all the ending points gives the saturated vapour line. The two curves meet at the top, and together they enclose the region in which liquid and vapour coexist. That enclosed region is the saturation dome, and it partitions the whole diagram into three: the compressed liquid to its left, the superheated vapour to its right, and the mixture beneath.

The two branches meet because the plateau keeps getting shorter as the pressure rises. Push far enough and it shrinks to nothing. For water that happens at \(22.06\) MPa and \(373.95\,^\circ\text{C}\), the critical point, and what happens there deserves to be said plainly rather than passed over. Above it there is no boiling, because there is nothing to boil into. Liquid and vapour have become indistinguishable: the same density, the same refractive index, no meniscus, no surface. Heat a substance past its critical pressure and it goes from something you would call a liquid to something you would call a gas without ever passing through a moment where both existed at once.

Everything after this is bookkeeping about which of the three regions a state is in. The dome is worth being able to sketch from memory, with the state you are working on marked on it, before any table is opened.

Quality

Under the dome there is a difficulty that does not arise anywhere else on the diagram. The state postulate says two independent intensive properties fix the state, and inside the dome pressure and temperature are not independent — the previous section made them one piece of information. Naming both tells you no more than naming either. Something else is needed, and that something is quality:

the fraction of the mass that is vapour. The subscripts are the convention throughout the tables: \(f\) for the saturated liquid, from the German Flüssigkeit, \(g\) for the saturated vapour, and \(fg\) for the difference between them. Quality runs from \(0\) at saturated liquid to \(1\) at saturated vapour, and it is defined only inside the dome. A superheated vapour does not have a quality of \(1.2\); it has no quality at all, and writing one down is a sign that the state was placed in the wrong region.

The use of it is the lever rule. The total volume is the liquid volume plus the vapour volume, so dividing through by the total mass,

which says the specific volume sits a fraction \(x\) of the way along the chord that runs across the dome at that temperature. Read it forwards to get \(v\) from the quality; read it backwards, as \(x = (v - v_f)/v_{fg}\), to get the quality from a volume. And the argument used nothing about volume in particular. Any property that is extensive and therefore averages by mass obeys the same rule:

One fraction places every property of a mixture at once, which is why quality is the second coordinate rather than some more obvious quantity.

The error to guard against is reading \(x\) as a volume fraction. It is a mass fraction, and the two are nowhere near each other. At \(100\) kPa, a half-and-half mixture by mass has \(v_f = 0.001043\) and \(v_g = 1.6941\) m³/kg, so the vapour occupies a little over \(99.9\%\) of the volume. A tank that looks empty of liquid may be half liquid by mass. Intuition built by looking at the thing is not worth much here; the arithmetic is.

Enthalpy and the tables

One more property has to be introduced before the tables make sense, and it arrives in a slightly unsatisfying way — as a combination that keeps appearing, given a name because writing it out grew tiresome:

Enthalpy. Its full justification belongs to the accounting of heat and work and to flowing systems, where \(Pv\) turns out to be exactly the work needed to push a unit mass of fluid across a boundary, so that \(h\) is the energy a stream carries plus the work it does arriving. For now it is enough that it is a property, because \(u\), \(P\) and \(v\) are all properties, and that it is tabulated.

The definition is easy to verify against a table and worth doing once. For saturated water at \(100\) kPa, the liquid has \(u_f = 417.40\) and \(h_f = 417.51\) kJ/kg, a difference of \(0.11\) — because \(Pv\) for a liquid of specific volume \(0.001043\) is almost nothing. The vapour has \(u_g = 2505.6\) and \(h_g = 2675.0\), a difference of \(169.4\), which is \(100 \times 1.6941\) to the last digit. The definition is not doing anything mysterious.

The layout of the tables follows directly from the geometry of the dome. Saturated tables give one row per state of saturation, indexed by temperature in one table and by pressure in another, purely for convenience of lookup — the two tables hold the same information. Each row carries \(T_{\text{sat}}\), \(P_{\text{sat}}\), and then \(v_f\) and \(v_g\), \(u_f\), \(u_{fg}\) and \(u_g\), and the same triples for \(h\) and \(s\). One row, because under the dome pressure and temperature are one fact.

Superheated tables cannot be laid out that way, because outside the dome pressure and temperature are independent again and every combination is a distinct state. So they are blocks: one block per pressure, rows by temperature within it. Compressed liquid tables exist too, and are short, for reasons the next section explains.

Values between rows are found by linear interpolation, and it is worth being deliberate about it, because interpolation is where careless arithmetic quietly costs marks. Go in as far as you went along:

At \(200\) kPa, superheated steam at \(215\,^\circ\text{C}\) lies \((215 - 200)/(250 - 200) = 0.30\) of the way from the \(200\,^\circ\text{C}\) row to the \(250\,^\circ\text{C}\) row, so every property is taken three tenths of the way across. Two cautions. Interpolate the property you want, not something you will later square or invert, and never interpolate across the saturation line: the properties change slope discontinuously there, and a straight line drawn from a superheated row to a saturated one describes nothing physical.

Which table to use

Almost every mistake made with property tables is made before the table is opened, in deciding which region the state is in. It is worth reducing to a fixed procedure and doing it the same way every time.

Given a pressure and a temperature, look up \(T_{\text{sat}}\) at that pressure and compare. If the temperature is below it, the state is a compressed liquid and sits to the left of the dome. If it is above it, the state is a superheated vapour and sits to the right. If the two are equal, the state is somewhere on the chord beneath the dome, and \(P\) and \(T\) together have told you one thing rather than two; a quality or a specific property is needed before anything can be read off.

Given a pressure and a specific volume instead — or an internal energy, or an enthalpy — the comparison runs the same way against the saturated values at that pressure. Below \(v_f\) the state is compressed liquid, above \(v_g\) it is superheated, and between them it is a mixture whose quality follows from the lever rule. This is why a quality that comes out negative or greater than one is useful rather than embarrassing: the arithmetic is reporting that the state was never under the dome, and the answer is to change tables, not to round the quality to the nearest end.

Compressed liquid data are scarce because liquids barely care about pressure. Squeezing a liquid at fixed temperature changes its properties very little, while warming it changes them a great deal, so to good accuracy a compressed liquid can be treated as a saturated liquid at the same temperature:

The temperature is what to enter with, and using the pressure instead is the standard error. Where a better figure for enthalpy is wanted, the correction \(h \approx h_f(T) + v_f(P - P_{\text{sat}})\) restores the pressure term that the approximation dropped, and it matters only at high pressures.

One last warning, because it is the one that catches people who have just met the tables and are about to meet the ideal gas equation. Steam is not an ideal gas at the conditions any of these problems use. Water vapour near saturation is thoroughly non-ideal, and \(Pv = RT\) applied to it can be wrong by tens of per cent. When the substance is water and the state is anywhere near the dome, the tables are not the slow route to the answer. They are the only route.