Real substances

Everything so far has quietly assumed that the working substance obeys $p v = R T$ , and that assumption has to be abandoned before any real machine can be analysed.

Where the ideal gas runs out

The ideal gas model comes from two physical statements: molecules occupy no volume of their own, and they exert no force on one another except during collisions. Both are excellent at low density, because the molecules are then far apart, and both are worthless at high density, because they are then not.

Look at what the model cannot do. Solve $v = R T / p$ for any temperature and any pressure and you always get exactly one answer, a single specific volume that shrinks smoothly as you squeeze. There is no pressure at which the substance suddenly collapses to a thousandth of its volume, so the model has no liquid phase and no condensation. With no condensation there is no boiling, no latent heat, and no critical point at which the distinction between liquid and vapour disappears. A model of matter that cannot boil is not a small idealisation of water. It is a different substance.

The failure lands exactly where engineering lives. A steam power plant boils water at high pressure and condenses it at low pressure, and both processes happen at states the ideal gas model denies exist. A refrigerator does the same with R-134a. Even where the substance is genuinely a vapour, the errors near saturation are large: saturated steam at $10$ MPa has a measured specific volume of $0.018026$ m³/kg, while $R T / p$ with $R = 0.4615$ kJ/kg K and $T_{s a t} = 584.1$ K gives $0.02695$ m³/kg, an overestimate of fifty per cent.

So $p v = R T$ is a limiting law, exact only as density goes to zero, and a real substance has to be described by measurement instead. What follows is the geometry those measurements have.

The p-v-T surface and its shadows

For a pure substance $p$ , $v$ and $T$ are tied by one relation, so the accessible states form a two-dimensional surface in $p$ - $v$ - $T$ space rather than filling the volume. Every equilibrium state of water is a point on one particular surface, measured once and for all.

That surface is awkward to draw, so it is used through its two shadows. Project it onto the $p$ - $T$ plane, looking along the volume axis, and every horizontal phase-change line collapses to a single curve, because pressure and temperature do not change while a substance boils. The result is the phase diagram: three regions, solid, liquid and vapour, separated by the sublimation, fusion and vaporisation lines. All three meet at the triple point, which for water sits at $273.16$ K and $611.7$ Pa, the only condition at which ice, water and steam coexist. That point is so reproducible that it defined the kelvin until $2019$ .

The phase diagram of water on pressure and temperature axes, with the sublimation, fusion and vaporisation lines meeting at the triple point, the vaporisation line ending at the critical point, and the fusion line leaning backwards to the left because ice is less dense than liquid water.

The vaporisation line does not run on forever. It stops at the critical point, at $373.95$ degrees Celsius and $22.06$ MPa for water, first observed for carbon dioxide by Thomas Andrews in $1869$ . Beyond it liquid and vapour are not distinguishable: the meniscus vanishes, the latent heat falls to zero, and a path that loops around the critical point takes a liquid to a vapour with no phase change anywhere along it. The fusion line, by contrast, has no known end.

Water's fusion line leans the wrong way. For almost every substance it slopes up and slightly to the right, so squeezing a liquid freezes it. For water it slopes up and to the left, because ice is less dense than liquid water ( $917$ against $999.8$ kg/m³ near $0$ degrees Celsius), so melting a gram of ice makes it occupy less space, not more. Squeezing ice therefore favours the liquid, and the melting point falls with pressure. The magnitude is small: with a specific volume change on melting of about $9.0 \times 10^{- 5}$ m³/kg and a latent heat of $333.5$ kJ/kg, the slope is roughly $- 7.4 \times 10^{- 8}$ K per pascal, so a hundred atmospheres depresses the melting point by less than a kelvin. Enough to make ice float and to let glaciers creep at their beds, nowhere near enough to explain ice skating, which is a friction and surface-layer effect.

The other projection, onto the $p$ - $v$ plane, keeps what the phase diagram hid. Each phase-change line reopens into a wide region, the saturation dome, bounded on the left by the saturated liquid line and on the right by the saturated vapour line, meeting at the top at the critical point. Inside it, liquid and vapour coexist, and this is the region property tables exist to describe.

Quality, and why two properties are still needed

Inside the dome an isotherm on the $p$ - $v$ diagram is a horizontal line. Boiling water at $100$ kPa stays at $99.61$ degrees Celsius from the first bubble to the last drop, and adding heat only converts more liquid to vapour. So within the dome, pressure and temperature are not independent: knowing one fixes the other. Give me $100$ kPa and $99.61$ degrees Celsius and I still cannot tell you the volume or the enthalpy, because I do not know how much of the mass has boiled.

The state postulate is not violated, it is being misread. It demands two independent properties, and here $p$ and $T$ are one property wearing two hats. The second must distinguish states along the horizontal line, and the natural choice is the quality

x = \frac{m_{g}}{m_{g} + m_{f}}

the fraction of the total mass that is vapour. It runs from $0$ on the saturated liquid line to $1$ on the saturated vapour line, and it is meaningless outside the dome.

Because volume is extensive, the mixture's specific volume is the mass-weighted average of the two saturated values $v_{f}$ and $v_{g}$ . The total volume is $m_{f} v_{f} + m_{g} v_{g}$ , and dividing by the total mass gives $v = (1 - x) v_{f} + x v_{g}$ , which rearranges to the form worth memorising:

v = v_{f} + x (v_{g} - v_{f})

The same argument works unchanged for internal energy and enthalpy, since those are extensive too, so $u = u_{f} + x u_{f g}$ and $h = h_{f} + x h_{f g}$ , where the subscript $f g$ denotes the difference across the dome. This one relation, plus a table of $v_{f}$ and $v_{g}$ , replaces the equation of state entirely inside the saturation region.

Example. A mixture of liquid water and steam at $100$ kPa has a quality of $0.35$ . What is its specific volume, given $v_{f} = 0.001043$ m³/kg and $v_{g} = 1.694$ m³/kg?

Apply the mixing rule directly: $v = v_{f} + x (v_{g} - v_{f}) = 0.001043 + 0.35 \times (1.694 - 0.001043) = 0.001043 + 0.35 \times 1.692957 = 0.5936$ m³/kg.

Now you. The same mixture at $100$ kPa has a quality of $0.80$ . What is its specific volume, in m³/kg?

Answer

$v = 0.001043 + 0.80 \times 1.692957 = 0.001043 + 1.354366 = 1.355$ m³/kg.

Reading the tables

Property tables come in two families. Saturated tables list, for each saturation state, the temperature, the corresponding pressure, and the values of $v_{f}$ , $v_{g}$ and the internal energies and enthalpies alongside. The identical data is printed twice, once indexed by round temperatures and once by round pressures, purely so that whichever you are given is a row rather than an interpolation. Superheated tables cover states outside the dome, where $p$ and $T$ are independent again, and are laid out as a block per pressure with temperature running down the rows.

For water at $100$ kPa the saturated row reads $T_{s a t} = 99.61$ degrees Celsius, $v_{f} = 0.001043$ m³/kg and $v_{g} = 1.694$ m³/kg. Notice the ratio: the vapour occupies $1624$ times the volume of the liquid it came from, which is why a boiler is a pressure vessel and a leak is violent.

Between rows, interpolate linearly, which for a smooth property over a small interval is accurate to a fraction of a per cent. Superheated water at $100$ kPa is tabulated at $1.6959$ m³/kg for $100$ degrees Celsius and $1.9367$ m³/kg for $150$ degrees Celsius. For $120$ degrees Celsius the fraction of the way across is $(120 - 100) / (150 - 100) = 0.4$ , so

v \approx 1.6959 + 0.4 (1.9367 - 1.6959) = 1.7922 m^{3} / kg

The ideal gas value, $R T / p = 0.4615 \times 393.15 / 100 = 1.814$ m³/kg, is $1.2$ per cent high, which is the residual non-ideality of steam a few degrees above its own boiling point.

Interpolation is a convenience, not a physical claim, and it fails wherever the property is not nearly linear: near the critical point, or between widely spaced pressure blocks. Never interpolate across a phase boundary.

A rigid vessel of water

Take a rigid, sealed vessel of volume $0.100$ m³ containing $1.00$ kg of water at $100$ kPa. What is inside it?

The first move is always the same: compute the specific volume, which the geometry hands you directly, $v = V / m = 0.100 / 1.00 = 0.100$ m³/kg, then compare it with the saturation values at the given pressure. Since $v_{f} = 0.001043 < 0.100 < 1.694 = v_{g}$ , the state lies inside the dome and the vessel holds a saturated mixture at $99.61$ degrees Celsius. Had $v$ come out below $v_{f}$ it would be compressed liquid; above $v_{g}$ , superheated vapour. That single comparison settles the phase.

The quality follows from the mixing rule, solved for $x$ :

x = \frac{v - v_{f}}{v_{g} - v_{f}} = \frac{0.100 - 0.001043}{1.694 - 0.001043} = \frac{0.098957}{1.692957} = 0.0585

So $58.5$ grams of the water is vapour and $941.5$ grams is liquid. The volumes tell the opposite story: the vapour fills $0.0585 \times 1.694 = 0.0991$ m³ and the liquid only $0.9415 \times 0.001043 = 0.00098$ m³. Ninety-nine per cent of the vessel is steam while six per cent of the contents is, which is why quality can never be guessed by eye from a sight glass.

Example. A rigid, sealed vessel of $0.250$ m³ holds $2.00$ kg of water at $100$ kPa. What phase is inside, and how much of the mass is vapour?

The specific volume is $v = V / m = 0.250 / 2.00 = 0.125$ m³/kg. Since $v_{f} = 0.001043 < 0.125 < 1.694 = v_{g}$ , this is a saturated mixture at $99.61$ degrees Celsius. Solving the mixing rule for quality, $x = (0.125 - 0.001043) / (1.694 - 0.001043) = 0.123957 / 1.692957 = 0.0732$ , so the vapour mass is $0.0732 \times 2.00 = 0.146$ kg.

Now you. A rigid, sealed vessel of $0.050$ m³ holds $1.00$ kg of water at $100$ kPa. What phase is inside, and how many grams of it are vapour?

Answer

$v = 0.050 / 1.00 = 0.050$ m³/kg, which lies between $v_{f}$ and $v_{g}$ , so it is a saturated mixture at $99.61$ degrees Celsius. Then $x = (0.050 - 0.001043) / 1.692957 = 0.0289$ , so $28.9$ grams of the water is vapour.

Now heat the vessel. Rigid means constant $V$ , sealed means constant $m$ , so $v$ stays pinned at $0.100$ m³/kg and the state moves vertically up a line of constant specific volume on the $p$ - $v$ diagram. Pressure and temperature both rise together along the saturation curve, and since our fixed $v$ is far to the right of the critical specific volume of water ( $0.003106$ m³/kg), the vertical line exits the dome through the saturated vapour side. The quality therefore increases, the liquid level falls, and the last drop evaporates where $v_{g} = 0.100$ m³/kg, which the tables place at about $2.0$ MPa and $212.4$ degrees Celsius. Beyond that the vessel holds superheated steam and its pressure climbs steeply.

The direction of that result depends entirely on which side of the critical volume you start. Fill the same vessel with $50$ kg of water instead, giving $v = 0.002$ m³/kg, and the vertical line exits through the saturated liquid side: the liquid level rises and the vessel fills completely with compressed liquid, which is how a sealed, over-filled system bursts.

The compressibility factor

Tables are exact but they are also a different book for every substance. A compact alternative starts by asking how badly a gas violates $p v = R T$ , and calls the answer the compressibility factor

Z = \frac{p v}{R T}

By construction $Z = 1$ for an ideal gas, $Z < 1$ where attraction pulls molecules closer than ideal, and $Z > 1$ where their own volume keeps them apart. Saturated steam at $100$ kPa has $Z = 1.694 / 1.720 = 0.985$ ; the same steam at $10$ MPa has $Z = 0.018026 / 0.02695 = 0.669$ .

The useful discovery, made by van der Waals himself, is that $Z$ is not a private fact about each gas. Plot $Z$ against the reduced pressure $p_{R} = p / p_{c r}$ and the reduced temperature $T_{R} = T / T_{c r}$ , measuring each state as a fraction of that substance's own critical values, and the data for nitrogen, methane, carbon dioxide, water and a dozen others collapse onto very nearly a single chart. This is the principle of corresponding states: two substances at the same reduced conditions are in mechanically similar states. The generalised chart built from it is good to a few per cent for most gases, and worst for strongly polar or hydrogen-bonded ones, water and ammonia included.

The rule of thumb then becomes concrete. $Z$ is within one per cent of unity when $p_{R}$ is below about $0.1$ at any temperature, or when $T_{R}$ is above roughly $2$ and the pressure is not extreme. Room-temperature air is safe on both counts, with $T_{R} = 298 / 132.5 = 2.25$ and, at atmospheric pressure, $p_{R} = 0.1 / 3.77 = 0.027$ . Steam near its own saturation line satisfies neither.

Example. Superheated steam at $100$ kPa and $150$ degrees Celsius is tabulated at $v = 1.9367$ m³/kg. Compute $Z$ and decide whether the ideal-gas model is safe here.

With $R = 0.4615$ kJ/kg K and $T = 423.15$ K, the ideal value is $R T / p = 0.4615 \times 423.15 / 100 = 1.9528$ m³/kg, so $Z = p v / R T = 1.9367 / 1.9528 = 0.992$ . The error is under one per cent: fifty degrees of superheat is enough to make the ideal gas model acceptable for this steam.

Now you. Superheated steam at $100$ kPa and $100$ degrees Celsius is tabulated at $v = 1.6959$ m³/kg. Compute $Z$ and decide whether the ideal-gas model is safe here.

Answer

$R T / p = 0.4615 \times 373.15 / 100 = 1.7221$ m³/kg, so $Z = 1.6959 / 1.7221 = 0.985$ . The error is $1.5$ per cent, so this close to the saturation line the ideal gas model is already marginal, and it only gets worse at higher pressure.

Van der Waals, and what one equation can buy

In his $1873$ doctoral thesis, Johannes Diderik van der Waals asked what minimal repair to $p v = R T$ would let a gas condense, and made two corrections on physical grounds. First, molecules have size, so the volume available for them to move in is not $v$ but $v - b$ , where $b$ is roughly the volume the molecules themselves occupy. Second, a molecule about to strike the wall is pulled back by its neighbours behind it, so the measured pressure is less than the ideal one. That pull scales with the density of the pullers and with the density of the pulled, so it goes as $(1 / v)^{2}$ . Together:

(p + \frac{a}{v^{2}}) (v - b) = R T

The two constants are fixed by a neat observation: at the critical point the critical isotherm has both a zero slope and an inflection, so $\partial p / \partial v$ and $\partial^{2} p / \partial v^{2}$ both vanish there. Imposing that gives $a = 27 R^{2} T_{c r}^{2} / 64 p_{c r}$ and $b = R T_{c r} / 8 p_{c r}$ , so the constants come from two measured numbers per substance and nothing is fitted to the bulk of the data.

Qualitatively the payoff is large. Below the critical temperature the equation is a cubic in $v$ with three real roots at a given pressure, the smallest a liquid-like volume and the largest a vapour-like one, so a single algebraic expression now contains both phases and predicts condensation. That is genuinely more than the ideal gas can do.

Quantitatively it is poor. The same critical-point conditions force the equation to predict a critical compressibility factor of exactly $3 / 8 = 0.375$ for every substance, whereas the measured values cluster near $0.27$ to $0.29$ for simple gases (nitrogen is $0.289$ ) and reach down to $0.229$ for water. An equation thirty per cent wrong about the point it was calibrated on is not trusted for design work, and it is not used for any.

There is one further embarrassment. The middle root of the cubic makes the subcritical isotherm double back, so that over a range of volumes $\partial p / \partial v$ is positive: compressing the fluid would lower its pressure, which is mechanically unstable and does not happen. James Clerk Maxwell repaired it in $1875$ by replacing the loop with a horizontal line drawn so that the two areas it cuts off are equal, a construction later shown to follow from equality of the Gibbs function in the two phases. That flat line is the saturation pressure, and the dome is recovered from theory. The equation explains why the dome exists; the tables say where it lies.

Nothing in this lesson said which direction a process runs. The tables would happily let a lukewarm vessel spontaneously separate into ice and steam, and the First Law would not object either. Finding the principle that forbids it is the work of the next lesson.