Z Factor vs Gas Compressibility Coefficient
Separate dimensionless Z from gas compressibility cg using a synthetic isothermal Z curve, signed pressure derivatives, a worked example and interactive slope controls.
By PetroCalcHub Team | Updated 2026-10-24
Original diagram Interactive example Worked practice 2 technical sources
Related workflow calculators
Engineering context
Gas compressibility factor Z and gas compressibility coefficient cg answer different questions. Z relates a gas state to the ideal-gas equation. The isothermal coefficient cg describes how a fixed amount of gas changes volume with pressure along a constant-temperature, fixed-composition path. Their similar names can hide the difference between a dimensionless state value and a local response with inverse-pressure units.
The example uses a synthetic Z curve near 3,000 psia. It is an arithmetic teaching curve, not measured PVT data or a gas correlation. You can change its local slope while keeping the central Z value fixed at 0.85. Watching cg change at that same point demonstrates why one isolated Z value cannot supply all the information needed for a compressibility calculation.
1 · State value
Z = 0.85
Dimensionless
At 3,000 psia
2 · Local slope
dZ/dP = −0.00002
Per psi
At fixed T and composition
3 · Volume response
cg = 0.000356862745
Per psi
1/P − (1/Z)(dZ/dP)
Change one thing and observe the result
Keep P = 3,000 psia and Z = 0.85. Choose a synthetic local slope on the same-temperature, fixed-composition curve. All three choices pass through the central state.
| Pressure (psia) | Z |
|---|---|
| 2,900 | 0.852 |
| 3,000 | 0.850 |
| 3,100 | 0.848 |
Practical workflow
- 1
State absolute pressure, temperature and composition for the gas state.
- 2
Obtain Z and its local derivative along a compatible isothermal curve.
- 3
Keep the derivative sign and convert its pressure basis before evaluating cg.
- 4
Check the inverse-pressure units and compare with the zero-slope reference 1/P.
Keep the two quantities and their units separate
The real-gas equation is PV = ZnRT for a fixed amount n. Z has no units because it is the ratio PV/(nRT). A value of 0.85 means that ratio is 0.85 at the specified pressure and temperature. It does not mean the gas loses 85% of its volume for a unit pressure increase, and it is not a coefficient expressed per psi.
Isothermal compressibility is cg = −(1/V)(∂V/∂P) at fixed temperature and composition. The minus sign makes cg positive for the usual local behavior in which increasing pressure reduces volume. If pressure is expressed in psi, cg has units 1/psi. A unit label is therefore a quick way to distinguish the two quantities in a table, calculator input or report.
Derive the pressure-slope correction
At fixed amount, temperature and composition, write V = ZnRT/P. Taking a logarithmic derivative gives (1/V)(∂V/∂P) = (1/Z)(∂Z/∂P) − 1/P. Negating that expression gives the gas compressibility relationship below. Both terms have inverse-pressure units, so their difference has the required dimension. Use absolute pressure in the reciprocal term.
The derivative is signed. A falling Z curve has a negative slope, so subtracting its contribution raises cg above 1/P at that point. A rising Z curve lowers cg relative to 1/P. These comparisons hold at the same pressure and Z; they do not say that a negative-slope gas always has a larger coefficient than a different gas at another pressure.
Estimate a local slope from compatible nearby states
For the baseline teaching line, choose Z = 0.852 at 2,900 psia, 0.850 at 3,000 psia and 0.848 at 3,100 psia. All three points represent the same assumed temperature and composition. The centered difference is (0.848 − 0.852)/(3,100 − 2,900) = −0.00002 per psi. It matches the exact slope because this synthetic local curve is linear.
For a curved measured or calculated isotherm, a finite difference is an estimate. Test a suitable pressure spacing, account for measurement noise and compare with a derivative from the underlying model where available. Differencing unrelated temperatures or different compositions follows another path and cannot be inserted as an isothermal derivative. A wide pressure interval may also conceal the local slope at the requested point.
| Pressure (psia) | Z (dimensionless) | Basis |
|---|---|---|
| 2,900 | 0.852 | Same T and composition |
| 3,000 | 0.850 | Central state |
| 3,100 | 0.848 | Same T and composition |
Compare three slopes at the same central Z
At 3,000 psia, 1/P is 0.000333333333 per psi. The negative baseline slope adds 0.000023529412 per psi to that reference, giving cg = 0.000356862745 per psi. A flat local Z curve makes the derivative correction zero. A positive slope of 0.00002 per psi subtracts the same correction and gives 0.000309803922 per psi.
All three synthetic lines pass through Z = 0.85 at the central pressure, yet they produce different cg values. The explorer changes only that slope and displays the nearby Z points as well as the coefficient. Keep the distinction visible when comparing outputs: the graph's vertical axis is dimensionless Z, while the result card reports cg in 1/psi.
| dZ/dP (1/psi) | cg (1/psi) | Relative to 1/P |
|---|---|---|
| −0.00002 | 0.000356862745 | Higher |
| 0 | 0.000333333333 | Equal |
| +0.00002 | 0.000309803922 | Lower |
Recognize the ideal-gas limit and derivative basis
An ideal gas has Z = 1 throughout the isotherm, so its Z derivative is zero and cg = 1/P. A flat slope at Z = 0.85 also gives cg = 1/P locally, but that one equality does not make the gas ideal. It only says the local Z correction vanishes. State behavior over a range cannot be inferred from one matching coefficient.
A correlation may use reduced pressure Ppr = P/Ppc instead of absolute pressure. At fixed pseudo-critical pressure, convert its slope using dZ/dP = (dZ/dPpr)/Ppc. The reduced-pressure derivative cannot be pasted unchanged into a field expecting 1/psi. Temperature must also remain fixed along the derivative path; changing reduced temperature while stepping pressure introduces another contribution.
Choose the tool that matches the available inputs
Use the derivative-based cg calculator when you already have absolute pressure, Z and a compatible local pressure derivative. The gas Z comparison workflow answers a different task: it evaluates supported Z correlations at reduced conditions. It does not turn an isolated Z output into an isothermal derivative automatically. If a derivative is required, document how the surrounding states or underlying model supplied it.
Check the pressure basis before interpreting a coefficient. A derivative per MPa needs conversion to per psi when using pressure in psia. If the evaluated cg is nonpositive for a proposed ordinary stable gas state, revisit the curve, phase region, units and derivative path rather than replacing the sign or clipping the result. The synthetic controls here stay within a small neighborhood that keeps Z and cg positive.
Preserve the negative slope in a cg calculation
Use P = 3,000 psia, Z = 0.85 and the synthetic fixed-temperature slope dZ/dP = −0.00002 per psi.
- 1. Calculate the local derivative(0.848 − 0.852)/(3,100 − 2,900) = −0.00002 per psi.
The two surrounding points belong to the same synthetic isotherm and composition.
- 2. Evaluate the reciprocal-pressure term1/P = 1/3,000 = 0.000333333333 per psi.
Pressure is absolute and the result has inverse-pressure units.
- 3. Retain the signed Z correction(1/Z)(dZ/dP) = −0.00002/0.85 = −0.000023529412 per psi.
A negative derivative remains negative before the subtraction in the full relationship.
- 4. Subtract the correctioncg = 0.000333333333 − (−0.000023529412) = 0.000356862745 per psi.
The two terms share the same pressure basis and can be subtracted directly.
Result
The gas compressibility coefficient is approximately 3.56862745 × 10⁻⁴ per psi; the Z factor remains 0.85.
Interpretation
The negative local Z slope increases cg above 1/P at this state. A Z value alone would not reveal the magnitude or sign of that correction.
Test your understanding
Keep P and Z fixed but change dZ/dP to +0.00002 per psi. Does cg rise or fall?
Show the worked answer
It falls to 1/3,000 − 0.00002/0.85 = 0.000309803922 per psi. The central Z value stays at 0.85; only the local slope changes.
Method selection guide
Match the calculation method to the physical question and the evidence available.
| Condition | Use | Why |
|---|---|---|
| Z and an isothermal pressure derivative are available | Evaluate the derivative-based cg relationship | The state and its local response supply all three inputs. |
| Only reduced conditions and a gas-correlation choice are available | Evaluate Z first and establish a derivative separately | The current Z comparison reports state values, not cg. |
| Only one measured Z value is available | Obtain compatible local PVT information | Several curves can share the value while having different slopes. |
Before using the result
- Use absolute pressure and identify its unit.
- State a common temperature and gas composition.
- Distinguish dimensionless Z from cg in inverse pressure.
- Preserve the sign of the local Z derivative.
- Convert reduced-pressure or other pressure-unit slopes before use.
- Check the derivative spacing, phase region and local plausibility of the coefficient.
Start with these calculators
These links keep the guide close to the working calculator flow.