Ideal and real gas: the gas deviation factor (Z)

Quick refresher

For an ideal gas, $pV = nRT$ . Use only absolute pressure (pressure above a vacuum) and absolute temperature (kelvins), in MPa and K or in psia and °R.
For a real gas, $pV = ZnRT$ . If $Z < 1$ , the attraction between the molecules squeezes the gas more than it would an ideal gas. If $Z > 1$ , the volume of the molecules themselves matters, and this happens at very high pressures.
The law of corresponding states says that $Z$ depends on the reduced temperature and pressure (the temperature and pressure divided by the critical ones). For a mixture we calculate them from the pseudo-critical properties. These are averages weighted by mole fraction, or estimates from the gas specific gravity (the density of the gas relative to air).
The pseudo-critical properties are only a scale for the generalised chart. They are not the true critical point of the mixture.
Hydrogen sulfide and carbon dioxide need the Wichert–Aziz correction. Without it, the $Z$ of a sour gas (a gas that contains hydrogen sulfide) comes out a few per cent too low.
A standard cubic metre is a measure of the amount of gas. At 0.101325 MPa and 20 °C, one kilomole (1000 moles) of ideal gas takes up 24.06 m³, and natural gas at these conditions is almost ideal.

Key relationships

pV = ZnRT

equation of state of a real gas; for an ideal gas $Z = 1$

T_{pr} = \dfrac{T}{T_{pc}},\qquad p_{pr} = \dfrac{p}{p_{pc}}

pseudo-reduced temperature and pressure (absolute)

T_{pc} = \sum_i y_i T_{c,i},\qquad p_{pc} = \sum_i y_i p_{c,i}

pseudo-critical properties of a mixture (Kay's rule)

\rho_g = \dfrac{p\,M}{Z R T}

gas density

Common pitfalls

Using gauge pressure (the reading of a pressure gauge), or temperature in degrees Celsius, in the gas laws.
Confusing the gas deviation factor $Z$ with the gas compressibility $c_g$ . They are different quantities.
Taking the pseudo-critical properties of a mixture for its true critical point.
Estimating $Z$ of a gas with a noticeable content of hydrogen sulfide and carbon dioxide without a correction.
Comparing gas volumes converted to different standard conditions: 20 °C, 15 °C or 60 °F.

After this lesson you can

write the equation of state of an ideal gas (the equation that links pressure, volume and temperature), use absolute pressure and temperature, and use the gas constant in different systems of units
explain why a real gas departs from an ideal one, and why $Z$ first falls below 1 and then rises above 1 as pressure increases
calculate the pseudo-critical and pseudo-reduced pressure and temperature of a gas (the scale values for reading $Z$ from a chart) from its composition or its specific gravity, and find $Z$ from the generalised chart or from an equation that approximates it
account for hydrogen sulfide and carbon dioxide with the Wichert–Aziz correction
calculate the density of a gas and its amount in moles and in standard cubic metres, stating the standard conditions they refer to

Ideal gas

A gas has no shape or volume of its own: it fills the whole vessel it is in. Its molecules move at random, and they create pressure when they hit the walls. In a rarefied gas (one at low pressure) the molecules are far apart. Their own volume is tiny compared with the volume of the vessel, and the forces of attraction between them hardly act. A model that treats the molecules as points is called an ideal gas. The points interact only when they collide and bounce off one another without losing energy (elastic collisions).

Experiments with rarefied gases gave three laws. Together they describe the ideal gas:

Boyle's law (called the Boyle–Mariotte law in Russian-language literature): at constant temperature, the volume of a given mass of gas is inversely proportional to the pressure, $pV = \text{const}$ . If the pressure doubles, the volume halves.
Charles's law (called Gay-Lussac's law in Russian-language literature): at constant pressure, the volume is proportional to the absolute temperature, $V/T = \text{const}$ .
Avogadro's law: equal volumes of any ideal gases at the same pressure and temperature contain the same number of molecules. So a mole (a fixed number of molecules) of any gas takes up the same volume at these conditions.

Equation of state of an ideal gas

pV = nRT

where:

$p$ absolute pressure of the gas, Pa
$V$ volume of the gas, m³
$n$ amount of gas (number of moles), mol
$T$ absolute temperature, $T = t\,[{}^{\circ}\text{C}] + 273.15$ , K
$R$ universal gas constant, 8.314 J/(mol·K), J/(mol·K)

Units: SI (any consistent system of units)

Holds when

The molecules have no volume of their own and do not attract one another. This is a good approximation at low pressure and at a temperature well above the critical one (the highest temperature at which pressure can still turn the gas into a liquid).
Pressure and temperature are absolute.

The equation combines the three laws. The volume of a gas is proportional to the amount of gas and to the absolute temperature, and inversely proportional to the pressure. If the amount of gas does not change, then for a change from state 1 to state 2, $p_1V_1/T_1 = p_2V_2/T_2$ .

In other unit systems (2)

Metric units

pV = nRT,\qquad R = 0.008314\ \frac{\text{MPa}\cdot\text{m}^3}{\text{kmol}\cdot\text{K}}

$p$ — MPa, $V$ — m³, $n$ — kmol (kilomoles, 1 kmol = 1000 mol), $T$ — K.

$R = 8.314$ kJ/(kmol·K) $= 8.314$ kPa·m³/(kmol·K) $= 0.008314$ MPa·m³/(kmol·K). These are the same number in different units.

Oilfield units

pV = nRT,\qquad R = 10.73\ \frac{\text{psia}\cdot\text{ft}^3}{\text{lb-mol}\cdot{}^{\circ}\text{R}}

$p$ — psia (pounds per square inch, absolute), $V$ — cubic feet, $n$ — pound-moles (1 lb-mol = 453.6 mol), $T$ — degrees Rankine (an absolute scale with Fahrenheit-sized degrees): $T\,[{}^{\circ}\text{R}] = t\,[{}^{\circ}\text{F}] + 459.67$ .

$10.73$ comes from 8.314 J/(mol·K) by changing the units: moles to pound-moles (×453.6), kelvins to degrees Rankine (÷1.8), cubic metres to cubic feet (×35.31) and pascals to psi (÷6895).

Avogadro's law means that at a given pressure and temperature a kilomole of any ideal gas takes up the same volume. This is the molar volume $V_m = RT/p$ . So a gas volume converted to agreed standard conditions (a fixed pressure and temperature) is really a measure of the amount of gas. A standard cubic metre means a definite number of molecules. It is not the volume that the gas takes up in the reservoir or in a pipe.

Conditions to which gas volumes are converted

Conditions: Standard (Russia, Kazakhstan)
Pressure: 0.101325 MPa
Temperature: 20 °C
Molar volume: 24.06 m³/kmol

Conditions: ISO standard (international)
Pressure: 0.101325 MPa
Temperature: 15 °C
Molar volume: 23.64 m³/kmol

Conditions: US oilfield
Pressure: 14.696 psia
Temperature: 60 °F (15.6 °C)
Molar volume: 379.5 ft³/lb-mol

Conditions: Normal
Pressure: 0.101325 MPa
Temperature: 0 °C
Molar volume: 22.41 m³/kmol

Conditions	Pressure	Temperature	Molar volume
Standard (Russia, Kazakhstan)	0.101325 MPa	20 °C	24.06 m³/kmol
ISO standard (international)	0.101325 MPa	15 °C	23.64 m³/kmol
US oilfield	14.696 psia	60 °F (15.6 °C)	379.5 ft³/lb-mol
Normal	0.101325 MPa	0 °C	22.41 m³/kmol

The molar volumes are calculated for an ideal gas. At these conditions natural gas usually differs from an ideal gas by only fractions of a per cent. Gas contracts also use other base pressures, for example 14.65 or 14.73 psia. A volume in “normal cubic metres” is almost 7% smaller than the same gas in standard cubic metres at 20 °C. So always state the base conditions next to the number.

In a mixture of ideal gases, each component behaves as if it were alone in the vessel. The pressures of the components add up to the total pressure (Dalton's law). The volumes that the components would take up at the total pressure and temperature add up to the total volume (Amagat's law). So in a gas the mole fraction of a component equals its volume fraction and its share of the total pressure. The whole mixture is described by one molar mass, $M = \sum y_i M_i$ (with $y_i$ the mole fraction and $M_i$ the molar mass of component $i$ ), and one specific gravity, $\gamma_g = M/28.97$ , which is the molar mass relative to that of air (mole, mass and volume fractions).

Real gas and the gas deviation factor

At reservoir pressures the gas molecules are much closer together, and both assumptions of the ideal gas model fail. First, the molecules attract one another. This attraction helps the outside pressure to squeeze the gas, so the gas takes up less volume than an ideal gas would. Second, the molecules have a volume of their own. When the gas is squeezed very hard, little free space is left between the molecules. The gas then resists further compression more than an ideal gas does, and it takes up more volume.

Equation of state of a real gas

pV = ZnRT,\qquad Z = \frac{V_{\text{real}}}{V_{\text{ideal}}}

where:

$Z$ gas deviation factor at pressure $p$ and temperature $T$ , —
$V_{\text{real}}$ the volume that the gas really occupies, m³
$V_{\text{ideal}}$ the volume of the same amount of ideal gas at the same $p$ and $T$ , which is $nRT/p$ , m³
$p,\ V,\ n,\ R,\ T$ as in the equation of state of an ideal gas; pressure and temperature are absolute, consistent

Units: any consistent system of units

Holds when

$Z$ is not a constant. It depends on the composition of the gas, the pressure and the temperature, so it must be found for each state.
The gas is single-phase (no liquid drops out in the reservoir or in the pipe).

One correction factor in the ideal gas equation covers both effects. If $Z < 1$ , the attraction between the molecules wins, and the gas is squeezed more than the ideal model predicts. If $Z > 1$ , the volume of the molecules themselves wins. When a gas goes from one state to another, $p_1V_1/(Z_1T_1) = p_2V_2/(Z_2T_2)$ .

CalculatedGas deviation factor of natural gas against pseudo-reduced pressure, at four pseudo-reduced temperatures

T_{pr}

. It is calculated from an equation that approximates the generalised chart (advanced section). The dot is the gas of the lesson example:

T_{pr} = 1.70

, between the curves for 1.5 and 2.0.

At zero pressure every gas is ideal, $Z = 1$ . As pressure rises, the attraction shows first, and $Z$ falls. For natural gas at reservoir temperatures the lowest $Z$ is usually 0.7–0.9, at pressures of 12–18 MPa. With more compression, the volume of the molecules themselves takes over. $Z$ rises and, at pressures of about 30–40 MPa, becomes greater than 1.

The most accurate way to get $Z$ is to measure it in a PVT laboratory (a lab that tests how the pressure, volume and temperature of a fluid are linked). A gas sample is compressed in a cell at reservoir temperature, and its measured volume is compared with the volume of an ideal gas (Laboratory PVT tests). More often, though, $Z$ is calculated, from the composition or from the specific gravity of the gas.

The law of corresponding states

The $Z$ isotherms (curves at constant temperature) of different gases have similar shapes but different scales. Methane has a critical temperature of −82.6 °C. At 20 °C and 5 MPa it is still a gas, with $Z \approx 0.9$ . Ethane has a critical temperature of 32.2 °C, and at 20 °C it already turns into a liquid at 3.8 MPa. But if we measure temperature and pressure as fractions of the critical values, the differences almost disappear. This is the law of corresponding states: at the same reduced temperature and pressure, different substances have almost the same $Z$ .

Reduced temperature and pressure

T_r = \frac{T}{T_c},\qquad p_r = \frac{p}{p_c}

where:

$T_r,\ p_r$ reduced temperature and pressure, —
$T,\ p$ absolute temperature and pressure of the gas, K, MPa
$T_c,\ p_c$ critical temperature and pressure of the substance (critical point), K, MPa

Units: dimensionless ratios; temperature only in kelvins or degrees Rankine

Holds when

The law works best for simple non-polar molecules (those with evenly spread electric charge: methane, nitrogen, argon). For large, asymmetric and polar molecules (heavy hydrocarbons, hydrogen sulfide, water) the deviations are larger.

The reduced properties show how far a substance is from its critical point. Equal $T_r$ and $p_r$ mean the same “position” relative to the critical point, and so similar values of $Z$ .

For a mixture there is nothing to put into these formulas. The true critical point of a mixture is set by its phase envelope (the curve that bounds the pressures and temperatures where gas and liquid exist together). No simple rule links it to the critical points of the components (critical point of a mixture). So for a mixture we use stand-in pseudo-critical values of temperature and pressure. They are averages of the critical values of the components, weighted by mole fraction (Kay's rule). They are not a real critical point. They are just a scale that lets the generalised chart work for mixtures too.

Pseudo-critical properties of a mixture (Kay's rule)

T_{pc} = \sum_i y_i\,T_{c,i},\qquad p_{pc} = \sum_i y_i\,p_{c,i}

where:

$T_{pc},\ p_{pc}$ pseudo-critical temperature and pressure of the gas, K, MPa
$y_i$ mole fraction of component $i$ in the gas, fraction
$T_{c,i},\ p_{c,i}$ critical temperature and pressure of the component (table of component properties), K, MPa

Units: K and MPa (or °R and psia)

Holds when

The composition of the gas is fully known, $\sum y_i = 1$ . For the C7+ fraction (the heavy end of the mixture, treated as one component) the critical properties are estimated from correlations that use its molar mass and density.
There are few non-hydrocarbon components. Hydrogen sulfide and carbon dioxide need a correction (below).

Pseudo-critical properties always lie between the critical properties of the components. The true critical point of a mixture can lie outside these limits.

Pseudo-reduced temperature and pressure

T_{pr} = \frac{T}{T_{pc}},\qquad p_{pr} = \frac{p}{p_{pc}}

where:

$T_{pr},\ p_{pr}$ pseudo-reduced temperature and pressure, —
$T,\ p$ absolute temperature and pressure of the gas (for example, reservoir values), K, MPa

Units: dimensionless ratios

Holds when

$T$ and $p$ are absolute and in the same units as $T_{pc}$ and $p_{pc}$ .

With this pair of numbers we find $Z$ on the generalised chart or from an equation that approximates the chart.

The generalised chart of the gas deviation factor (the Standing–Katz chart) was built in the early 1940s. It used measurements on “sweet” natural gases (gases without hydrogen sulfide) that were mostly methane. It covers $T_{pr}$ from 1.05 to 3 and $p_{pr}$ up to 15. Today people rarely read it by hand. Software uses equations that approximate it (one is in the advanced section). For complex fluids it uses equations of state, which are more general equations linking pressure, volume and temperature (Cubic equations of state).

How to find Z of a natural gas

1
Pseudo-critical properties
Use Kay's rule if you know the composition. If you know only the specific gravity of the gas, use a correlation (next section).
2
Correction for sour gases
If the gas contains hydrogen sulfide or carbon dioxide, correct the pseudo-critical properties by the Wichert–Aziz method.
3
Pseudo-reduced properties
Find $T_{pr}$ and $p_{pr}$ from the absolute pressure and temperature at which you need $Z$ .
4
The factor Z
Read it from the generalised chart or from an equation that approximates it. Check that $T_{pr}$ and $p_{pr}$ are not outside the limits for which the chart was built.

Pseudo-critical properties from specific gravity

We do not always have a complete gas analysis. Often we know only the specific gravity of the gas relative to air, $\gamma_g = M/28.97$ , measured at the separator. For hydrocarbon gases with a small amount of nitrogen, carbon dioxide and hydrogen sulfide, the pseudo-critical properties are closely tied to this number. The heavier the gas, the higher its pseudo-critical temperature and the slightly lower its pseudo-critical pressure.

Pseudo-critical properties of natural gas from specific gravity (Standing's approximation)

T_{pc} = 168 + 325\,\gamma_g - 12.5\,\gamma_g^2,\qquad p_{pc} = 677 + 15\,\gamma_g - 37.5\,\gamma_g^2

where:

$T_{pc}$ pseudo-critical temperature, °R
$p_{pc}$ pseudo-critical pressure, psia
$\gamma_g$ specific gravity of the gas relative to air, —

Units: oilfield: $T_{pc}$ in degrees Rankine, $p_{pc}$ in psia

Holds when

Natural gases that are dry and lean (with little heavy hydrocarbon) and have no more than a few per cent of nitrogen, carbon dioxide and hydrogen sulfide.
This is an approximation of a graphical correlation. For gas-condensate mixtures (gases that drop liquid when the pressure falls) another curve is used (advanced section). For wet gases (gases rich in heavier hydrocarbons), later correlations are more accurate.
When there is an appreciable amount of sour gases, we account for them separately, with a correction, as in the next section.

The correlation replaces a composition analysis when you need a quick calculation or do not know the composition. For a gas of unusual composition, with a lot of nitrogen, carbon dioxide or heavy hydrocarbons, check the result with a calculation from the composition.

In other unit systems (1)

Metric units

T_{pc}\,[\text{K}] = \frac{168 + 325\,\gamma_g - 12.5\,\gamma_g^2}{1.8},\quad p_{pc}\,[\text{MPa}] = 0.006895\,(677 + 15\,\gamma_g - 37.5\,\gamma_g^2)

$T_{pc}$ — K, $p_{pc}$ — MPa.

1.8 is the number of degrees Rankine in one kelvin; 1 psi = 0.006895 MPa.

Worked exampleGas deviation factor and density of a reservoir gas

Given (assumed data)

Gas specific gravity $\gamma_g$: 0.65; non-hydrocarbon components less than 2%
Reservoir pressure $p$: 20.0 MPa (absolute)
Reservoir temperature $t$: 80 °C

Solution

Pseudo-critical properties: $T_{pc} = (168 + 325\cdot 0.65 - 12.5\cdot 0.65^2)/1.8 = 374.0/1.8 = 207.8$ K; $p_{pc} = 0.006895\cdot 670.9 = 4.626$ MPa.
Absolute temperature $T = 80 + 273.15 = 353.15$ K. Pseudo-reduced properties: $T_{pr} = 353.15/207.8 = 1.70$ , $p_{pr} = 20.0/4.626 = 4.32$ .
From the generalised chart (or from the equation that approximates it, in the advanced section): $Z = 0.869$ .
Molar mass $M = 0.65\cdot 28.97 = 18.83$ g/mol. Gas density in the reservoir: $\rho_g = \dfrac{pM}{ZRT} = \dfrac{20{,}000\cdot 18.83}{0.869\cdot 8.314\cdot 353.15} = 148$ kg/m³ (pressure in kPa). For an ideal gas it would have come out as 128 kg/m³.
In 1 m³ of pores filled with this gas, $n = pV/(ZRT) = 7.84$ kmol. At standard conditions (20 °C) this is $7.84\cdot 24.06 = 189$ m³. A calculation for an ideal gas would have given 164 m³.

Answer:

Z = 0.869

. The gas density in the reservoir is 148 kg/m³. Each cubic metre of pores holds about 189 m³ of gas converted to standard conditions.

What it means. The gas is near the minimum of the isotherm. The attraction between the molecules makes it $1/Z = 1.15$ times denser than an ideal gas. A calculation for an ideal gas would underestimate the amount of gas in the reservoir, and with it the gas in place, by 13%. The last number in the answer is 1 divided by the gas formation volume factor, which is the volume that one standard cubic metre of gas takes up in the reservoir (Gas formation volume factor, compressibility and viscosity). The data in this example are assumed for illustration.

Sour gases: the Wichert–Aziz correction

Hydrogen sulfide and carbon dioxide behave differently from hydrocarbons. The hydrogen sulfide molecule is polar: one end is slightly positive and the other slightly negative. In the straight (linear) carbon dioxide molecule the electric charge is also uneven: the oxygen atoms carry a partial negative charge and the carbon carries a positive one. Both molecules attract their neighbours more strongly than methane does. This is why their critical temperatures are high: 373 and 304 K, against 191 K for methane. The generalised chart was built from gases with almost none of these components. Kay's rule averages the critical temperatures by mole fraction, so it exaggerates the contribution of the sour components in a mixture that is mostly methane. The pseudo-critical temperature comes out too high. As a result, for a sour gas Kay's rule gives a $Z$ that is too low, by a few per cent or more, depending on how much of the sour components there is.

Wichert and Aziz (the correction was published in the early 1970s) did not change the chart itself. They proposed to correct the pseudo-critical properties instead. These are reduced by an amount $\varepsilon$ that depends on the fractions of hydrogen sulfide and carbon dioxide. After that, we find $Z$ in the usual way.

Wichert–Aziz correction to the pseudo-critical properties

T'_{pc} = T_{pc} - \varepsilon,\qquad p'_{pc} = \frac{p_{pc}\,T'_{pc}}{T_{pc} + B\,(1 - B)\,\varepsilon}

where:

$T'_{pc},\ p'_{pc}$ corrected pseudo-critical temperature and pressure; we calculate $T_{pr}$ and $p_{pr}$ from them, K, MPa
$T_{pc},\ p_{pc}$ pseudo-critical properties from Kay's rule (or from a correlation), K, MPa
$\varepsilon$ correction term (formula below), K
$B$ mole fraction of hydrogen sulfide, fraction

Units: temperatures and $\varepsilon$ in the same units (K or °R), pressure in any

Holds when

The correction was built from experiments with gases that contained up to about 54 mol% (mole per cent) of carbon dioxide and up to 74 mol% of hydrogen sulfide.
If the gas contains a lot of nitrogen or heavy hydrocarbons, an accurate calculation needs a laboratory analysis or an equation of state.

The correction always lowers the pseudo-critical temperature. The reduced temperature then rises, and the gas “moves away” from the critical point. At typical reservoir conditions, $Z$ increases after the correction.

Wichert–Aziz correction term

\varepsilon = 120\,\left(A^{0.9} - A^{1.6}\right) + 15\,\left(B^{0.5} - B^{4}\right)

where:

$\varepsilon$ correction term, °R
$A$ total mole fraction of hydrogen sulfide and carbon dioxide, $y_{\mathrm{H_2S}} + y_{\mathrm{CO_2}}$ , fraction
$B$ mole fraction of hydrogen sulfide, $y_{\mathrm{H_2S}}$ , fraction

Units: $\varepsilon$ in degrees Rankine; for kelvins, divide the result by 1.8

Holds when

The fractions are mole fractions, written as numbers between 0 and 1, not as percentages.

For kelvins, divide the coefficients by 1.8: use 66.7 instead of 120 and 8.33 instead of 15. With 10–20 mol% of sour components, $\varepsilon$ is about 7–15 K. Hydrogen sulfide gives a larger correction than the same amount of carbon dioxide.

Assumed sour gas of the example and critical properties of its components

Component: Nitrogen $\mathrm{N_2}$
$y_i$: 0.010
$T_{c,i}$ , K: 126.2
$p_{c,i}$ , MPa: 3.40
$M_i$ , g/mol: 28.01

Component: Carbon dioxide $\mathrm{CO_2}$
$y_i$: 0.050
$T_{c,i}$ , K: 304.2
$p_{c,i}$ , MPa: 7.38
$M_i$ , g/mol: 44.01

Component: Hydrogen sulfide $\mathrm{H_2S}$
$y_i$: 0.120
$T_{c,i}$ , K: 373.2
$p_{c,i}$ , MPa: 9.00
$M_i$ , g/mol: 34.08

Component: Methane $\mathrm{C_1}$
$y_i$: 0.710
$T_{c,i}$ , K: 190.6
$p_{c,i}$ , MPa: 4.60
$M_i$ , g/mol: 16.04

Component: Ethane $\mathrm{C_2}$
$y_i$: 0.060
$T_{c,i}$ , K: 305.4
$p_{c,i}$ , MPa: 4.87
$M_i$ , g/mol: 30.07

Component: Propane $\mathrm{C_3}$
$y_i$: 0.035
$T_{c,i}$ , K: 369.9
$p_{c,i}$ , MPa: 4.25
$M_i$ , g/mol: 44.10

Component: n-Butane
$y_i$: 0.015
$T_{c,i}$ , K: 425.2
$p_{c,i}$ , MPa: 3.80
$M_i$ , g/mol: 58.12

Component	$y_i$	$T_{c,i}$ , K	$p_{c,i}$ , MPa	$M_i$ , g/mol
Nitrogen $\mathrm{N_2}$	0.010	126.2	3.40	28.01
Carbon dioxide $\mathrm{CO_2}$	0.050	304.2	7.38	44.01
Hydrogen sulfide $\mathrm{H_2S}$	0.120	373.2	9.00	34.08
Methane $\mathrm{C_1}$	0.710	190.6	4.60	16.04
Ethane $\mathrm{C_2}$	0.060	305.4	4.87	30.07
Propane $\mathrm{C_3}$	0.035	369.9	4.25	44.10
n-Butane	0.015	425.2	3.80	58.12

The composition is assumed. The critical properties are reference values, in kelvins, as in the table of the lesson on the composition of fluids.

Worked exampleGas deviation factor of a sour gas with and without the correction

Given (assumed data)

Gas composition: as in the table above: 12 mol% hydrogen sulfide, 5 mol% carbon dioxide
Reservoir pressure $p$: 30.0 MPa (absolute)
Reservoir temperature $t$: 100 °C ( $T = 373.15$ K)

Solution

Molar mass $M = \sum y_i M_i = 22.18$ g/mol, $\gamma_g = 0.766$ . By Kay's rule: $T_{pc} = \sum y_i T_{c,i} = 234.2$ K, $p_{pc} = \sum y_i p_{c,i} = 5.247$ MPa.
Correction: $A = 0.12 + 0.05 = 0.17$ , $B = 0.12$ ; $\varepsilon = 66.7\,(0.17^{0.9} - 0.17^{1.6}) + 8.33\,(0.12^{0.5} - 0.12^{4}) = 12.5$ K.
Corrected properties: $T'_{pc} = 234.2 - 12.5 = 221.7$ K; $p'_{pc} = \dfrac{5.247\cdot 221.7}{234.2 + 0.12\cdot 0.88\cdot 12.5} = 4.939$ MPa.
With the correction: $T_{pr} = 373.15/221.7 = 1.68$ , $p_{pr} = 30.0/4.939 = 6.07$ , and from the chart $Z = 0.918$ .
Without the correction: $T_{pr} = 1.59$ , $p_{pr} = 5.72$ , $Z = 0.876$ , which is 4.6% less.

Answer: With the Wichert–Aziz correction,

Z = 0.918

. Without it,

Z = 0.876

What it means. The amount of gas in the pores is inversely proportional to $Z$ . So a calculation without the correction would overestimate the gas in place by 4.8%. For the sour gases of the Pre-Caspian Basin this correction is only the minimum. In projects, engineers calculate their properties with equations of state tuned to laboratory experiments. The partial pressure of hydrogen sulfide in the reservoir here is 3.6 MPa (its mole fraction times the total pressure). So the materials and safety measures are chosen as for an environment containing hydrogen sulfide (Composition of reservoir fluids). The data in this example are assumed for illustration.

Terms and symbols

Ideal gas: A model of a gas made of point molecules that do not attract one another. It obeys the equation $pV = nRT$ .
Gas deviation factor $Z$: The volume of a real gas divided by the volume of an ideal gas at the same pressure, temperature and amount of gas.
Reduced properties: The temperature and pressure divided by the critical ones: $T_r = T/T_c$ , $p_r = p/p_c$ .
Pseudo-critical properties: The stand-in critical temperature and pressure of a mixture. They are averages by the mole fractions of the components, or estimates from the gas specific gravity.
Law of corresponding states: At the same reduced temperature and pressure, different gases have close values of $Z$ .
Molar volume: The volume of one kilomole of gas. For an ideal gas at 0.101325 MPa and 20 °C it is 24.06 m³.
Standard conditions: The pressure and temperature to which gas volumes are converted. In Russia and Kazakhstan they are 0.101325 MPa and 20 °C.

Common mistakes and misconceptions

Wrong

“ $Z = 0.9$ means that the gas is 10% more compressible than an ideal gas.”

In fact

$Z$ compares volumes, not compressibilities. The gas takes up 90% of the volume of an ideal gas, so the same volume holds 11% more of it. Compressibility $c_g$ is a different quantity. We find it from the slope of $Z$ against pressure (Gas formation volume factor, compressibility and viscosity).

Wrong

“For a mixture, let's take the critical point from the phase diagram.”

In fact

The generalised chart takes the pseudo-critical properties, which are averages by mole fraction. The true critical point of a mixture can lie far from them.

Wrong

“Hydrogen sulfide is just one more component, Kay's rule will cope.”

In fact

For sour gases Kay's rule underestimates $Z$ by a few per cent. You need the Wichert–Aziz correction or a calculation with an equation of state.

Wrong

“The temperature in the formula is 80, the pressure is the pressure gauge reading.”

In fact

Use absolute values only: 353.15 K, and the pressure with the atmosphere (0.101 MPa) added.

In depth: advanced detail

Where the dip of the isotherm comes from: the van der Waals equation

The simplest equation of state that includes both effects is the van der Waals equation: $\left(p + a/V_m^2\right)(V_m - b) = RT$ , where $V_m$ is the molar volume. The constant $a$ describes the attraction between molecules, and $b$ describes their own volume. Solving it for $Z = pV_m/(RT)$ , we get:

Gas deviation factor of a van der Waals gas

Z = \frac{V_m}{V_m - b} - \frac{a}{R\,T\,V_m}

where:

$V_m$ molar volume of the gas, m³/kmol
$a$ constant of the attraction between molecules, MPa·m⁶/kmol²
$b$ the volume of the molecules themselves in one kilomole, m³/kmol

Units: SI or any consistent units

Holds when

This is a qualitative model. It is not accurate enough for natural gas in numbers, so in practice more advanced equations of state are used (Cubic equations of state).

The first term is greater than 1: it comes from the volume of the molecules themselves. The second term is negative: it comes from the attraction. While the gas is rarefied ( $V_m \gg b$ ), $Z \approx 1 + (b - a/RT)/V_m$ . At a low temperature the attraction is stronger, and $Z$ falls as pressure rises. Under strong compression $V_m$ comes close to $b$ , the first term grows quickly, and $Z$ rises above 1. The higher the temperature, the weaker the second term. Above the so-called Boyle temperature there is no dip at all. In this model it is $a/(Rb)$ ; for real gases it is roughly 2.5–2.7 times the critical temperature. This is why the curve for $T_{pr} = 3$ on the chart hardly falls below 1.

The generalised chart as an equation

For calculations on a computer, the generalised chart is replaced by equations fitted to its points. One of the most widely used is the Dranchuk–Abou-Kassem equation (mid-1970s), which has eleven constants. The chart in this lesson and its examples were calculated with it.

Dranchuk–Abou-Kassem equation

Z = 1 + \left(A_1 + \frac{A_2}{T_{pr}} + \frac{A_3}{T_{pr}^3} + \frac{A_4}{T_{pr}^4} + \frac{A_5}{T_{pr}^5}\right)\rho_r + \left(A_6 + \frac{A_7}{T_{pr}} + \frac{A_8}{T_{pr}^2}\right)\rho_r^2 - A_9\left(\frac{A_7}{T_{pr}} + \frac{A_8}{T_{pr}^2}\right)\rho_r^5 + A_{10}\left(1 + A_{11}\rho_r^2\right)\frac{\rho_r^2}{T_{pr}^3}\,\e^{-A_{11}\rho_r^2}

where:

$\rho_r$ reduced gas density, $\rho_r = 0.27\,p_{pr}/(Z\,T_{pr})$ (0.27 is the critical gas deviation factor assumed in the approximation), —
$T_{pr},\ p_{pr}$ pseudo-reduced temperature and pressure, —
$A_1 \ldots A_{11}$ constants of the approximation (table below), —

Units: dimensionless

Holds when

It applies for $1.0 < T_{pr} \le 3.0$ and $0.2 \le p_{pr} < 30$ . For $T_{pr}$ close to 1 and $p_{pr} > 1$ the accuracy is noticeably worse.
It approximates the same generalised chart, so it has the same limits: sour gases need corrected pseudo-critical properties.

$Z$ appears on both sides of the equation (through $\rho_r$ ), so we solve it by iteration (repeated trial and correction). Set $Z = 1$ , find $\rho_r$ , then refine $Z$ . This is usually done by Newton's method applied to $\rho_r$ . Within the range of the chart, the equation reproduces it with an average error of about half a per cent.

Constants of the Dranchuk–Abou-Kassem equation

: $A_1$
Value: 0.3265
: $A_7$
Value: −0.7361

: $A_2$
Value: −1.0700
: $A_8$
Value: 0.1844

: $A_3$
Value: −0.5339
: $A_9$
Value: 0.1056

: $A_4$
Value: 0.01569
: $A_{10}$
Value: 0.6134

: $A_5$
Value: −0.05165
: $A_{11}$
Value: 0.7210

: $A_6$
Value: 0.5475

Value

	Value		Value
$A_1$	0.3265	$A_7$	−0.7361
$A_2$	−1.0700	$A_8$	0.1844
$A_3$	−0.5339	$A_9$	0.1056
$A_4$	0.01569	$A_{10}$	0.6134
$A_5$	−0.05165	$A_{11}$	0.7210
$A_6$	0.5475

Wet gases and gas-condensate mixtures

Take the reservoir mixture of a gas-condensate accumulation, recombined from samples of gas and condensate (Types of reservoir fluids). Its specific gravity is higher, and its pseudo-critical properties follow a different relation to it. For such mixtures we use a separate curve of the same graphical correlation:

Pseudo-critical properties of a gas-condensate mixture from specific gravity (Standing's approximation)

T_{pc} = 187 + 330\,\gamma_g - 71.5\,\gamma_g^2,\qquad p_{pc} = 706 - 51.7\,\gamma_g - 11.1\,\gamma_g^2

where:

$T_{pc},\ p_{pc}$ pseudo-critical temperature and pressure, °R, psia
$\gamma_g$ specific gravity of the reservoir mixture relative to air, —

Units: oilfield: °R and psia (divide by 1.8 for K, multiply by 0.006895 for MPa)

Holds when

The mixture in the reservoir is single-phase (the pressure is above the dew point pressure, the pressure at which the first drops of liquid form).
There are few non-hydrocarbon components. For rich mixtures it is more accurate to calculate from the composition, with C7+ pseudo-components (lumped groups of heavy components), or to use an equation of state.

Below the dew point pressure there are two phases in the reservoir (gas and liquid), and one $Z$ for “the gas” is no longer enough. We calculate the properties of each phase with an equation of state. For material balance (an accounting of the fluid produced against the pressure drop) we use the so-called two-phase $Z$ , measured in the laboratory.

Connections to other topics

Composition of reservoir fluids — mole fractions, molar mass and specific gravity of a gas, critical properties of the components.
Phase behaviour: pure substances and binary mixtures — the critical point of a pure substance and of a mixture, from which the reduced properties are measured.
Gas formation volume factor, compressibility and viscosity — how to get the formation volume factor, compressibility and viscosity of a gas from $Z$ .
Black oil model: solution gas–oil ratio and formation volume factors — the gas formation volume factor in the black oil model, and how to convert gas volumes.
Cubic equations of state — cubic equations of state, the next step after the generalised chart.

Check yourself

1
In which of these states is natural gas closest to ideal?
1. A.0.5 MPa and 120 °C
2. B.20 MPa and 120 °C
3. C.20 MPa and 20 °C
4. D.60 MPa and 20 °C
Show the answer
Correct answer: A. 0.5 MPa and 120 °C
At low pressure the molecules are far apart, and a high temperature weakens the effect of the attraction. So $Z$ differs from 1 by only fractions of a per cent. At 20 MPa, especially at a low temperature, the gas departs strongly from ideal. At 60 MPa the volume of the molecules themselves wins, and $Z > 1$ .
2
For the gas in a reservoir $Z = 0.85$ . What does this mean?
1. A.The gas is 15% more compressible than an ideal gas
2. B.The gas takes up 85% of the volume of an ideal gas. A given volume holds about 18% more of it than an ideal-gas calculation gives
3. C.The gas takes up a volume 15% larger than an ideal gas
4. D.The gas is 15% lighter than air
Show the answer
Correct answer: B. The gas takes up 85% of the volume of an ideal gas. A given volume holds about 18% more of it than an ideal-gas calculation gives
From $V = ZnRT/p$ , the volume of the real gas is 0.85 of the ideal one. In a given volume, $n = pV/(ZRT)$ , which is $1/0.85 = 1.18$ times more than for an ideal gas. Compressibility is a different quantity, and the specific gravity relative to air has no link to $Z$ .
3
A mixture of 90 mol% methane and 10 mol% ethane is at 12 MPa (absolute pressure) and 40 °C. Find the pseudo-critical and pseudo-reduced properties ( $T_c$ and $p_c$ of methane are 190.6 K and 4.60 MPa, of ethane 305.4 K and 4.87 MPa).
Show the answer
$T_{pc} = 0.9\cdot 190.6 + 0.1\cdot 305.4 = 202.1$ K; $p_{pc} = 0.9\cdot 4.60 + 0.1\cdot 4.87 = 4.63$ MPa. $T_{pr} = 313.15/202.1 = 1.55$ , $p_{pr} = 12/4.63 = 2.59$ . From the generalised chart $Z \approx 0.81$ .
4
How does the Wichert–Aziz correction change the calculation for a sour gas?
1. A.It lowers the pseudo-critical temperature and pressure. The reduced temperature rises, and $Z$ at reservoir conditions usually increases
2. B.It raises the pseudo-critical temperature, and $Z$ decreases
3. C.It changes only the molar mass of the gas
4. D.It removes hydrogen sulfide and carbon dioxide from the composition
Show the answer
Correct answer: A. It lowers the pseudo-critical temperature and pressure. The reduced temperature rises, and $Z$ at reservoir conditions usually increases
The correction term $\varepsilon$ is always positive, and we subtract it from the pseudo-critical temperature. The pseudo-critical pressure also decreases. The gas is then “farther” from the critical point, and $Z$ , which Kay's rule gave too low, increases. This is what happens in the lesson example (from 0.876 to 0.918).
5
The gas specific gravity is 0.70. Find its density at standard conditions (0.101325 MPa, 20 °C) and in the reservoir at 15 MPa (absolute pressure), 70 °C and $Z = 0.85$ .
Show the answer
$M = 0.70\cdot 28.97 = 20.28$ g/mol. At standard conditions, $\rho = M/V_m = 20.28/24.06 = 0.843$ kg/m³ (the gas is practically ideal there). In the reservoir, $\rho = pM/(ZRT) = 15{,}000\cdot 20.28/(0.85\cdot 8.314\cdot 343.15) \approx 125$ kg/m³, which is almost 150 times greater.
6
How do the pseudo-critical properties of a mixture differ from its true critical point?
Show the answer
The pseudo-critical properties are averages of the critical properties of the components, taken by mole fraction. They are a scale that lets the generalised chart of $Z$ work for mixtures. The true critical point of a mixture is the point on its phase envelope where the properties of liquid and gas become the same. It depends on the composition in a more complicated way, and it can lie outside the range of the critical properties of the components. So the pseudo-critical pressure cannot be used, for example, as an estimate of the pressure above which the mixture is single-phase.

Updated: 1 October 2026