Phase behaviour: pure substances and binary mixtures

Quick refresher

For a pure substance, gas and liquid exist together only along a line, the vapour pressure curve. At a given temperature the substance boils at one pressure only. As long as both phases are present, the pressure does not change.
The vapour pressure curve ends at the critical point. Above the critical temperature we cannot get a separate liquid phase by any amount of compression. At the critical point itself the properties of gas and liquid coincide.
For a mixture, the two-phase region is not a line but an area. The bubble point and dew point lines bound it, and they meet at the critical point of the mixture. At constant temperature a mixture boils over a range of pressures.
The first gas bubbles are rich in light components, and the last drops of liquid are rich in heavy ones. As the liquid vaporises, the compositions of the phases keep changing.
The cricondenbar and the cricondentherm are the highest pressure and temperature at which two phases are possible. The critical point of the mixture coincides with neither.
Between the critical temperature and the cricondentherm, a reduction of pressure causes not vaporisation but condensation. This is retrograde condensation, and it is the main feature of gas-condensate accumulations.

Key relationships

\ln\frac{p_s}{p_{\text{atm}}} = \ln\frac{p_c}{p_{\text{atm}}}\cdot\frac{1/T_b - 1/T}{1/T_b - 1/T_c}

vapour pressure from two reference points (the normal boiling point and the critical point); temperatures are absolute

p_b = \sum_i x_i\,p_{s,i},\qquad y_i = \frac{x_i\,p_{s,i}}{p_b}

bubble point pressure of an ideal mixture and the composition of the first gas (Raoult's and Dalton's laws)

\frac{1}{p_d} = \sum_i \frac{y_i}{p_{s,i}}

dew point pressure of an ideal mixture

F = C - P + 2

Gibbs phase rule: the number of independent intensive parameters (properties that do not depend on the amount of substance)

Common pitfalls

Assuming that a mixture, like a pure substance, boils at a single pressure. For a mixture, the bubble point and the dew point are different.
Confusing the critical point of a mixture with the cricondenbar and the cricondentherm. Two phases can exist above the critical pressure and above the critical temperature of the mixture.
Assuming that a gas can only expand when the pressure falls. In the retrograde region liquid drops out of it.
Applying the phase diagram of the original mixture to a process in which the composition changes. In the reservoir below the bubble point pressure, gas and liquid move separately, and the envelope of the remaining mixture shifts.
Putting the temperature in degrees Celsius and the gauge pressure into phase-equilibrium formulas. We need absolute temperature (K) and absolute pressure.

After this lesson you can

read the pressure–temperature phase diagram of a pure substance: the vapour pressure curve, the triple point and the critical point
describe what happens to a pure substance and to a mixture when the pressure is reduced at constant temperature, and explain why the bubble point and the dew point of a mixture do not coincide
estimate the vapour pressure from two reference points, and the bubble point and dew point pressures of an ideal mixture (one whose components do not change each other's behaviour) from Raoult's and Dalton's laws
show on the phase envelope of a mixture (the boundary line of its two-phase region) the critical point, the cricondenbar (the highest pressure of the envelope), the cricondentherm (the highest temperature of the envelope), the lines of constant liquid fraction and the region of retrograde condensation
explain retrograde condensation (liquid dropping out of a gas as the pressure falls) and why mixtures of methane with heavy hydrocarbons remain two-phase at pressures far above the critical pressures of their components

Phase, component, equilibrium

Reservoir fluids are mixtures of many substances. Depending on pressure and temperature, they are gas, liquid, or both at once. To understand their behaviour, we start with a pure substance. Then we move on to a mixture of two components. Almost everything that happens to oil and gas is already visible in such mixtures. Composition of reservoir fluids covers the composition of reservoir fluids.

System: The amount of substance, within given boundaries, that we are studying. An example is a fluid sample in a laboratory cell.
Component: An individual substance in the system: methane, ethane, carbon dioxide, water. In calculations, a component can also be a simplified group of substances, for example the C7+ fraction.
Phase: A part of a system that is the same throughout (physically homogeneous) and is separated from the rest by a clear boundary: gas, liquid oil, water, paraffin crystals. One phase can contain many components.
Equilibrium: A state in which the measurable properties of the phases do not change with time. As many molecules leave the liquid for the gas as return from it.
Intensive and extensive properties: Intensive properties do not depend on the amount of substance (pressure, temperature, density, specific volume, phase compositions). Extensive properties are proportional to it (volume, mass). We draw phase diagrams in intensive variables. (Specific volume is the volume of one unit of mass.)

Two competing causes govern phase behaviour. Temperature measures the kinetic energy (the energy of motion) of the molecules: the higher it is, the more strongly the molecules tend to fly apart. The forces of attraction between molecules do the opposite and hold them together. These forces are noticeable when the molecules are close. At very close approach they give way to repulsion, which is why a liquid is almost incompressible. Pressure brings the molecules closer together and so helps the attraction. In a gas the molecules are far apart and hardly interact. In a liquid they are densely packed and held together by attraction.

Phase diagram of a pure substance

A phase diagram in pressure–temperature coordinates shows the state of a substance at every combination of $p$ and $T$ . For a pure substance it has three regions (solid, liquid and gas), separated by three lines.

SchematicPhase diagram of a pure substance (schematic): the melting and sublimation curves, the vapour pressure curve, the triple point and the critical point

C

. Path 1–2, at constant temperature, crosses the vapour pressure curve. Path A–B takes the liquid to the gas by going around the critical point.

Vapour pressure curve: The boundary between liquid and gas. On it the two phases are in equilibrium. Above the curve the substance is liquid, and below it the substance is a gas. The pressure on the curve at a given temperature is the vapour pressure $p_s$ .
Critical point $C$: The end of the vapour pressure curve, with coordinates $T_c$ and $p_c$ . More about it below.
Triple point: The only combination of pressure and temperature at which all three phases are in equilibrium.
Sublimation curve: The boundary between solid and gas below the triple point. At low pressure the substance passes from solid to gas without becoming liquid. This is how dry ice (solid $\mathrm{CO_2}$ ) behaves at atmospheric pressure.
Melting curve: The boundary between solid and liquid. For hydrocarbons the melting point rises with pressure, so the line leans to the right. For water, a rare exception, it falls slightly.

Petroleum engineers most often need the vapour pressure curve and its end, the critical point. Solid phases also occur in field practice: deposits of paraffins and asphaltenes, and gas hydrates (ice-like solids made of water and gas). Retrograde condensation, wax, asphaltenes and high-pressure fluids covers them.

Vapour pressure

Vapour pressure rises with temperature very fast, almost exponentially, because on heating more and more molecules get enough energy to break away from the liquid. The point of the curve at atmospheric pressure, 0.101325 MPa, is the normal boiling point $T_b$ .

Below the critical temperature, a substance is liquid at a pressure above $p_s(T)$ and gas at a pressure below $p_s(T)$ . That is why propane in a cylinder at 20 °C stays liquid as long as the pressure above it exceeds about 0.84 MPa. Methane at the same temperature cannot be liquefied at any pressure, because its critical temperature is −82.6 °C.

Vapour pressure from two reference points (Clapeyron–Clausius approximation)

\ln\frac{p_s}{p_{\text{atm}}} = \ln\frac{p_c}{p_{\text{atm}}}\cdot\frac{1/T_b - 1/T}{1/T_b - 1/T_c}

where:

$p_s$ vapour pressure at temperature $T$ , absolute, MPa
$p_{\text{atm}}$ normal atmospheric pressure, 0.101325 MPa, at which the substance boils at $T_b$ , MPa
$T_b$ normal boiling point, K
$T_c,\ p_c$ critical temperature and pressure of the substance, K, MPa
$T$ temperature at which $p_s$ is sought; the formula is intended for $T_b \le T \le T_c$ , K

Units: absolute pressures in the same units, absolute temperatures in kelvins

Holds when

The formula rests on the Clapeyron–Clausius equation. We treat the vapour as an ideal gas, the volume of the liquid as small compared with the volume of the vapour, and the heat of vaporisation (the heat needed to turn liquid into vapour) as constant. Then $\ln p_s$ depends linearly on $1/T$ .
We draw the straight line through two known points, the normal boiling point and the critical point. The errors of the assumptions partly cancel, and for light hydrocarbons the error is usually no more than a few per cent.
The formula describes only the liquid–gas curve between the triple point and the critical point; above $T_c$ there is no vapour pressure.

$\ln p_s$ falls almost linearly as $1/T$ increases, so the vapour pressure climbs steeply on heating. The lighter the substance, the lower its $T_b$ and $T_c$ , and the further to the left and up its curve lies. At the same temperature a lighter substance has a higher vapour pressure.

CalculatedVapour pressure curves of methane, ethane, propane, n-butane and n-hexane, calculated from the formula above with reference values of

T_b

T_c

and

p_c

. The dots are the critical points. The critical pressures of hydrocarbons do not exceed 5 MPa, which is far below typical reservoir pressures.

Worked examplePressure in a tank of liquefied propane

Given (assumed data)

Normal boiling point of propane: −42.1 °C (231.04 K)
Critical point of propane: 96.7 °C (369.83 K), 4.248 MPa
Tank temperature: −20 °C in winter, 20 °C, 50 °C in summer in the sun

Solution

Slope of the line: $B = \dfrac{\ln(p_c/p_{\text{atm}})}{1/T_b - 1/T_c} = \dfrac{3.736}{0.001624} = 2300$ K.
At 20 °C ( $T = 293.15$ K): $\ln(p_s/p_{\text{atm}}) = B\,(1/T_b - 1/T) = 2.109$ ; $p_s = 0.101325\,\e^{2.109} = 0.835$ MPa.
In the same way at −20 °C: $p_s = 0.242$ MPa; at 50 °C: $p_s = 1.73$ MPa.
For comparison, the measured vapour pressures of propane are about 0.245, 0.836 and 1.713 MPa. The discrepancy is no more than 1.1%.
A pressure gauge shows gauge pressure, which is 0.1 MPa less than absolute pressure. In summer it shows about 1.63 MPa.

Answer: The absolute pressure in the tank is about 0.242 MPa in winter, 0.835 MPa at 20 °C and 1.73 MPa in summer in the sun.

What it means. As long as there is liquid in the tank, the pressure above it equals the vapour pressure. It depends only on temperature, not on how much propane is left, so on a half-empty tank the gauge shows the same as on a full one. The summer pressure is 7 times the winter pressure. Designers of tanks and safety valves must allow for this. The temperatures in this example are assumed; the properties of propane are reference values.

Expansion at constant temperature: the bubble point and the dew point

We study phase behaviour in a PVT cell (from pressure, volume, temperature). This is a cylinder with a piston in a thermostat (a chamber that keeps the temperature constant), and we place the sample in it. By moving the piston, we change the volume and pressure at constant temperature and watch what happens to the sample (Laboratory PVT tests).

Take a pure substance below its critical temperature and start increasing the volume. While the substance is liquid, the pressure falls very fast. A liquid is hardly compressible, so a small increase in volume sharply relieves the pressure. When the pressure reaches the vapour pressure, the first gas bubbles appear. This is the bubble point. As the volume increases further, the liquid vaporises, and the pressure stays the same until the last drop has vaporised. This is the dew point. After it the gas simply expands, and the pressure falls again, but now smoothly.

SchematicIsotherms (lines of constant temperature) in pressure–volume coordinates (schematic). On the left is a pure substance: between the bubble point and the dew point the pressure is constant. On the right is a mixture of two components: as vaporisation proceeds, the pressure falls.

For a pure substance, the pressures at the bubble point and at the dew point are the same and equal to $p_s(T)$ . Repeat the experiment at different temperatures. Then two lines appear on the pressure–volume plane: one of bubble points and one of dew points. The two-phase region lies below them, and the lines meet at the critical point. On the pressure–temperature plane the two lines of a pure substance merge into one, the vapour pressure curve.

The critical point and the supercritical fluid

The closer the temperature is to the critical one, the less the gas and the liquid differ along the saturation curve. On heating, the liquid expands, and the gas under high pressure becomes denser. At the critical point their densities and all the other intensive properties become identical. The boundary between the phases disappears, and the interfacial tension (the pull at the surface between the two phases) and the heat of vaporisation fall to zero. The coordinates of this point, the critical temperature $T_c$ and the critical pressure $p_c$ , are constants of the substance (see the table in the lesson on fluid composition).

Above the critical temperature we cannot get a separate liquid phase by any amount of compression. The substance simply becomes denser and denser, without separating into two phases. We call the state at a temperature and pressure above the critical ones supercritical, and we call the substance simply a fluid. In density it can be close to a liquid, and in mobility close to a gas.

A substance can also pass from liquid to gas without boiling at all, by going around the critical point. We compress the liquid above $p_c$ , heat it above $T_c$ , then reduce the pressure and cool it (path A–B in the diagram). At no stage does a phase boundary appear, although at the start the substance was a liquid and at the end a gas. So, far from the saturation curve, “liquid” and “gas” are merely conventional names for one and the same fluid.

More detail:Drying core around the critical point

Core laboratories use this way around the critical point. Suppose a sample has fragile, fibrous clay minerals and we dry it in the ordinary way. The retreating water–air boundary then crushes these fibres with surface tension forces, and the measured permeability of the sample comes out too high. So we first replace the water in the pores with liquids that mix with it, and then with liquid carbon dioxide. We turn the carbon dioxide into gas by going around its critical point (31 °C, 7.4 MPa). No phase boundary appears, and the pore structure is preserved (Core analysis).

A mixture of two components: the phase envelope

We repeat the experiment in the PVT cell with a mixture of two components, a binary mixture. At first everything is the same as for a pure substance: liquid, a sharp fall in pressure, and the first gas bubbles at the bubble point. The pressure there is the bubble point pressure $p_b$ , which for oil we call the saturation pressure. But then, as the liquid vaporises, the pressure keeps falling until it reaches the dew point pressure $p_d$ (the pressure at which condensation begins). There the last drop of liquid disappears (the right-hand panel of the figure above).

The cause lies in the composition of the phases. The first gas bubbles are rich in the more volatile (more easily vaporising) component. So the remaining liquid becomes heavier and heavier, and to keep it vaporising we must reduce the pressure. At the bubble point the liquid still has the composition of the original mixture, while the first gas bubbles are noticeably richer in the light component. At the dew point the picture is the mirror image: the gas has the composition of the mixture, and the last drops of liquid are richer in the heavy component.

For mixtures of similar molecules at moderate pressures, we can estimate these pressures from two classical laws. Raoult's law says that the partial pressure of a component above the liquid equals its mole fraction in the liquid multiplied by the vapour pressure of the pure component. Dalton's law says that the partial pressure of a component in the gas equals its mole fraction in the gas multiplied by the total pressure.

Bubble point pressure and first-gas composition of an ideal mixture

p_b = \sum_i x_i\,p_{s,i}(T),\qquad y_i = \frac{x_i\,p_{s,i}(T)}{p_b}

where:

$p_b$ bubble point pressure of the mixture at temperature $T$ , MPa
$x_i$ mole fraction of component $i$ in the liquid; at the bubble point, in the mixture, —
$p_{s,i}(T)$ vapour pressure of pure component $i$ at $T$ , MPa
$y_i$ mole fraction of the component in the first gas bubbles, —

Units: absolute pressures in the same units

Holds when

An ideal solution: the molecules of the components are close in size and structure (for example propane and butane), and Raoult's law holds for each component.
The gas is ideal, so Dalton's law holds.
The pressures are low, and the temperature is below the critical temperatures of all the components. For reservoir mixtures at high pressures these laws do not work. There we calculate phase equilibrium from equilibrium ratios (the ratio of a component's share in the gas to its share in the liquid) or from equations of state (Vapour–liquid equilibrium, Cubic equations of state).

The bubble point pressure is the weighted average of the vapour pressures of the components. The ratio $y_i/x_i = p_{s,i}/p_b$ shows how many times richer in the component the first gas is than the liquid. For the light component it is greater than one, and for the heavy component it is less than one.

Dew point pressure and last-drop composition of an ideal mixture

\frac{1}{p_d} = \sum_i \frac{y_i}{p_{s,i}(T)},\qquad x_i = \frac{y_i\,p_d}{p_{s,i}(T)}

where:

$p_d$ dew point pressure of the mixture at temperature $T$ , MPa
$y_i$ mole fraction of the component in the gas; at the dew point, in the mixture, —
$x_i$ mole fraction of the component in the last drop of liquid, —
$p_{s,i}(T)$ vapour pressure of the pure component, MPa

Units: absolute pressures in the same units

Holds when

The same as for the bubble point pressure.

If the vapour pressures of the components differ, then at one temperature $p_d < p_b$ . So the two-phase state of the mixture occupies a range of pressures, not a single point.

Worked exampleA propane–butane mixture: over what range of pressures does it boil

Given (assumed data)

Mixture: 50 mol% propane and 50 mol% n-butane
Temperature: 20 °C (293.15 K)
Vapour pressure of propane: from the previous example
n-Butane: $T_b = 272.66$ K, $T_c = 425.12$ K, $p_c = 3.796$ MPa

Solution

n-Butane by the two-point formula: slope $B = 2755$ K, $p_s = 0.205$ MPa; propane: $p_s = 0.835$ MPa.
Bubble point pressure: $p_b = 0.5\cdot 0.835 + 0.5\cdot 0.205 = 0.520$ MPa.
Composition of the first bubbles: $y_{\text{propane}} = 0.5\cdot 0.835/0.520 = 0.80$ .
Dew point pressure: $1/p_d = 0.5/0.835 + 0.5/0.205$ , so $p_d = 0.330$ MPa.
Composition of the last drop: $x_{\text{propane}} = 0.5\cdot 0.330/0.835 = 0.20$ .

Answer: At 20 °C the mixture starts to boil at 0.520 MPa and vaporises completely at 0.330 MPa (absolute). The first gas contains 80% propane, the last drop only 20%.

What it means. The mixture boils not at one pressure but over a range from 0.520 to 0.330 MPa, and the compositions of the phases keep changing. Reservoir oil and the production in a separator behave in the same way. The gas released first is richer in light components, and the remaining liquid gradually becomes heavier. Owners of household cylinders with a propane–butane mixture know this too. As gas is drawn off, the liquid in the cylinder loses propane and the pressure in it falls, especially in frost. The data in this example are assumed for illustration; the properties of the substances are reference values.

CalculatedBubble point pressure (straight line) and dew point pressure (curve) of propane–n-butane mixtures at 20 °C as a function of the propane fraction. We calculated them from Raoult's and Dalton's laws, with the vapour pressures from the examples. The horizontal segments join the compositions of the liquid and gas in equilibrium for the 50:50 mixture, at the bubble point and at the dew point.

Repeat the experiment at different temperatures and plot the bubble point and dew point pressures on a pressure–temperature diagram. The result is a loop, the phase envelope. The bubble points form its upper left branch (the bubble point line), and the dew points form its right branch (the dew point line). The branches meet at the critical point of the mixture. Inside the envelope the mixture is two-phase, and outside it is single-phase: above the bubble point line it is liquid, and to the right of and below the dew point line it is gas.

Inside the envelope we draw lines of constant liquid fraction, for example 80, 60, 40 and 20% liquid by volume. All of them converge at the critical point. The envelope of a binary mixture lies between the vapour pressure curves of the pure components. For a mixture rich in the light component it is closer to that component's curve, and for a mixture rich in the heavy component it is closer to the heavy component's curve. Every composition has its own envelope.

SchematicPhase envelope of a mixture (schematic): the bubble point and dew point lines, the critical point

C

, the cricondenbar, the cricondentherm, the lines of constant liquid fraction and the region of retrograde condensation. Path 1–2–3–4 is a pressure reduction at a temperature between the critical temperature and the cricondentherm.

The critical point of a mixture is still the point where the properties of liquid and gas coincide, but it is no longer the top of the two-phase region. The highest pressure at which two phases are still possible is the cricondenbar, and the highest temperature is the cricondentherm. The critical point lies on the envelope, away from them: for gas mixtures usually to the left of and below the cricondenbar, and for oils to the right of it.

The critical temperature of a binary mixture lies between the critical temperatures of the components. But the critical pressure can be much higher than the critical pressures of both. This is especially strong for mixtures of methane with heavy hydrocarbons. A mixture of methane and decane stays two-phase at pressures several times higher than the 4.6 MPa of methane and the 2.1 MPa of decane. So reservoir oil with a lot of dissolved methane can have a bubble point pressure of tens of megapascals, although none of its components alone could be in two phases at such a pressure.

Retrograde condensation

Ordinary experience suggests that when the pressure falls, liquid vaporises and gas expands. But mixtures have a region where everything is the other way round. Take a gas at a temperature between the critical temperature of the mixture and the cricondentherm, and reduce the pressure (path 1–2–3–4 in the diagram). At point 2 the gas crosses the dew point line, and liquid begins to drop out of it. As the pressure falls further, more liquid forms, and the liquid fraction reaches a maximum (near point 3). Then the liquid vaporises again, and at point 4, on the lower branch of the dew point line, the mixture is single-phase once more. The dropout of liquid as the pressure falls is retrograde condensation. “Retrograde” means “reverse”: it is the opposite of ordinary behaviour.

The explanation lies in the properties of a dense gas. Under high pressure a gas comes close to a liquid in density, and it dissolves heavy molecules like a liquid solvent. When the pressure is reduced, the gas becomes thinner, and its dissolving power falls. The heavy components then come out as drops. At still lower pressures ordinary vaporisation begins to predominate, and the drops pass back into the gas.

On the phase diagram the region of retrograde condensation has the dew point line above it. Below it is the line that joins the points of greatest liquid fraction on the isotherms (lines of constant temperature). In temperature, the region lies between the critical point and the cricondentherm.

For a reservoir this means the following. Suppose the reservoir temperature lies between the critical temperature of the mixture and the cricondentherm. Then the fluid is a gas under initial conditions. But when the reservoir pressure falls below the dew point pressure, condensate drops out in the pores. Capillary forces hold the drops, and most of this liquid never reaches the wells. The most valuable heavy components stay in the reservoir. Types of reservoir fluids covers how to deal with this.

Terms and symbols

Phase: A part of a system that is the same throughout, separated from the rest by a boundary.
Vapour pressure $p_s$: The pressure at which the liquid and gas of a pure substance are in equilibrium at a given temperature.
Critical point: The state in which the properties of liquid and gas coincide. For a pure substance, it is the end of the vapour pressure curve ( $T_c$ , $p_c$ ).
Bubble point: The state in which the first gas bubbles come out of the liquid as the pressure falls. For oil, the pressure at this point is called the saturation pressure.
Dew point: The state in which the first drops of liquid drop out of the gas (or, as the pressure falls in the two-phase region, the last ones disappear).
Phase envelope: The boundary of the two-phase region of a mixture on the pressure–temperature diagram. It is made of the bubble point and dew point lines, which meet at the critical point.
Cricondenbar and cricondentherm: The highest pressure and the highest temperature at which a mixture can exist in two phases.
Retrograde condensation: The dropout of liquid from gas as the pressure falls. It occurs at temperatures between the critical temperature of the mixture and the cricondentherm.

Common mistakes and misconceptions

Wrong

“At constant temperature a mixture boils at one pressure, like water.”

In fact

For a mixture, the pressure in the two-phase region changes from the bubble point pressure to the dew point pressure. In the lesson example it goes from 0.52 to 0.33 MPa at 20 °C.

Wrong

“Above the critical temperature of a mixture, liquid cannot appear.”

In fact

For a mixture that is true only above the cricondentherm. Between the critical temperature and the cricondentherm, liquid drops out of the gas as the pressure falls. This is retrograde condensation.

Wrong

“The critical pressure of a mixture is the average of the critical pressures of the components.”

In fact

For mixtures of methane with heavy hydrocarbons, the critical pressure and the cricondenbar can be several times higher than the critical pressures of all the components.

Wrong

“If the vapour pressure of propane at 20 °C is 0.84 MPa, then the pressure in a cylinder is 0.84 MPa only while it is full.”

In fact

As long as liquid remains in the cylinder, the pressure above it equals the vapour pressure and depends only on temperature. We cannot use the pressure to judge how much propane is left.

Wrong

“Since the fluid in the reservoir is a gas, it behaves like a light, rarefied gas.”

In fact

At reservoir pressures a gas is compressed hundreds of times, and a rich gas comes close to a light liquid in density. This is the dense phase. It dissolves heavy hydrocarbons, and they can drop out as condensate when the pressure falls.

In depth: advanced detail

Gibbs phase rule

Why do two phases of a pure substance exist together along a line, but those of a mixture in a whole region? The Gibbs phase rule gives the answer. It counts how many intensive parameters we can change independently without disturbing the equilibrium.

Gibbs phase rule

F = C - P + 2

where:

$F$ number of degrees of freedom — independent intensive parameters (pressure, temperature, phase compositions), —
$C$ number of components, —
$P$ number of phases in equilibrium, —

Units: integers

Holds when

The system is in equilibrium, there are no chemical reactions, and every component can be present in every phase.

For a pure substance ( $C = 1$ ) with two phases, $F = 1$ . Once the temperature is fixed, the pressure is fixed too, so liquid–gas equilibrium is a line. At the triple point ( $P = 3$ ) $F = 0$ , which is a point. For a binary mixture with two phases, $F = 2$ . At a given temperature we can still vary the pressure, and the compositions of the phases change with it. So the two-phase state of a mixture of a given composition occupies a whole region of the pressure–temperature diagram.

Where the vapour pressure formula comes from

The Clapeyron–Clausius equation links the slope of the vapour pressure curve to the heat of vaporisation:

Clapeyron–Clausius equation

\frac{\dd p_s}{\dd T} = \frac{\Delta H_v}{T\,\Delta v}\quad\Rightarrow\quad \frac{\dd \ln p_s}{\dd (1/T)} = -\frac{\Delta H_v}{R}

where:

$\Delta H_v$ molar heat of vaporisation (the heat needed to vaporise one mole), J/mol
$\Delta v$ difference of the molar volumes of gas and liquid, m³/mol
$R$ universal gas constant, 8.314 J/(mol·K), J/(mol·K)
$T$ absolute temperature, K

Units: SI

Holds when

The second form follows from the first if we take $\Delta v \approx v_{\text{g}} = RT/p_s$ . This means we neglect the volume of the liquid and treat the vapour as an ideal gas.

At a constant heat of vaporisation, $\ln p_s$ depends linearly on $1/T$ , with slope $-\Delta H_v/R$ . The two-point formula draws this line through the normal boiling point and the critical point. So its slope $B$ is an “effective” heat of vaporisation: $\Delta H_v \approx R B$ . For propane $R B = 8.314\cdot 2300 \approx 19.1$ kJ/mol, while the measured heat of vaporisation at the normal boiling point is about 19.0 kJ/mol. More precisely, $\dfrac{\dd \ln p_s}{\dd (1/T)} = -\dfrac{\Delta H_v}{R\,\Delta Z}$ , where $\Delta Z$ is the difference between the gas deviation factors $Z$ of the gas and the liquid. Near the critical point both the heat of vaporisation and $\Delta Z$ tend to zero, but their ratio changes little. So $\ln p_s$ remains an almost linear function of $1/T$ right up to the critical point.

Connections to other topics

Composition of reservoir fluids — what components reservoir mixtures are made of and the critical parameters of the components.
Types of reservoir fluids — how the position of the reservoir temperature relative to the phase envelope decides the fluid type.
Vapour–liquid equilibrium — quantitative calculation of bubble point and dew point pressures and phase compositions of real mixtures.
Cubic equations of state — how the phase envelopes of reservoir mixtures are calculated from equations of state.
Laboratory PVT tests — how the bubble point pressure and the dew point are measured in a PVT cell.
Retrograde condensation, wax, asphaltenes and high-pressure fluids — solid phases — paraffins, asphaltenes, hydrates — and retrograde condensation in the reservoir.

Check yourself

1
A pure substance below its critical temperature is expanded at constant temperature in a PVT cell, starting from the liquid state. What happens to the pressure between the bubble point and the dew point?
1. A.It falls
2. B.It rises
3. C.It stays constant
4. D.It first rises, then falls
Show the answer
Correct answer: C. It stays constant
For a pure substance, liquid and gas can be in equilibrium at a given temperature only at the vapour pressure. When the volume grows, liquid vaporises to fill it, and the pressure does not change until the last drop has vaporised.
2
What is the highest temperature at which a mixture can exist in two phases called?
1. A.Cricondentherm
2. B.Dew point temperature
3. C.Cricondenbar
4. D.Critical temperature
Show the answer
Correct answer: A. Cricondentherm
The cricondentherm is the rightmost point of the phase envelope. The critical temperature of the mixture is lower. Between the two lies the region of retrograde condensation.
3
Why is propane in a cylinder liquid at 20 °C, while methane at the same temperature cannot be liquefied by any pressure?
Show the answer
20 °C is below the critical temperature of propane (96.7 °C). At a pressure above the vapour pressure, about 0.84 MPa, propane is liquid. The critical temperature of methane is −82.6 °C, so at 20 °C methane is supercritical. Compression only makes it denser, and it never forms a separate liquid phase.
4
An ideal mixture of 30 mol% propane and 70 mol% n-butane is at 20 °C. The vapour pressures are 0.835 and 0.205 MPa. What is the bubble point pressure?
1. A.0.84 MPa
2. B.0.52 MPa
3. C.0.26 MPa
4. D.0.39 MPa
Show the answer
Correct answer: D. 0.39 MPa
$p_b = 0.3\cdot 0.835 + 0.7\cdot 0.205 = 0.394$ MPa. The first gas contains $0.3\cdot 0.835/0.394 \approx 0.64$ propane, which is twice as much as the liquid.
5
A gas in a reservoir at a temperature between the critical temperature and the cricondentherm is at a pressure above the dew point pressure. What will happen as the reservoir pressure falls?
Show the answer
While the pressure is above the dew point, the gas is single-phase. Below it, retrograde condensation begins. Liquid drops out in the pores, its fraction grows to a maximum, and as the pressure falls further the liquid partly vaporises. The condensate that has dropped out is mostly immobile (it does not flow), and it is lost to production unless we maintain the pressure.
6
Why can reservoir oil have a bubble point pressure of 25 MPa if the critical pressures of all its hydrocarbon components are below 5 MPa?
Show the answer
The critical pressure of a mixture is not the average of the critical pressures of its components. For mixtures of methane with heavy hydrocarbons, the two-phase region extends to pressures many times the critical pressures of the components. A lot of methane dissolves in heavy hydrocarbons only under high pressure, and the more methane is dissolved, the higher the pressure at which it starts to come out. So the bubble point pressure of an oil with a lot of dissolved methane is tens of megapascals.

Updated: 19 October 2026