Real-Gas Compressibility Calculations

Introduction

The ideal-gas equation is a model, not an exact description of every gas state. At higher pressures or near condensation, intermolecular attractions and finite molecular size can change the measured pressure-volume relationship. The compressibility factor Z measures the departure: Z = PV/(nRT). A value of one means that particular state matches the ideal equation; a value different from one quantifies the correction.

Core explanation

Define Z = PV/(nRT), with P absolute, V the gas volume, n moles and T in kelvin. Because PV and nRT have the same units, Z is dimensionless. Rearranging gives PV = ZnRT. If measured P, V, T and n are known, compute Z directly. If Z is supplied for the relevant gas composition and state, use V = ZnRT/P or n = PV/(ZRT). The latter shows that using the ideal formula when Z ≠ 1 can bias an amount estimate.

At fixed P, T and n, Z is the real molar volume divided by the ideal molar volume. Thus Z = 0.90 means the measured volume is 90% of the ideal prediction at the same P and T . A value above one means a larger volume than ideal under that comparison. At relatively low pressure, many real gases approach Z ≈ 1 because molecules are far apart. At moderate pressure, attractive interactions can lead to Z below one for some gases; at sufficiently high pressure, excluded molecular volume and repulsive effects can push Z above one. The exact curve depends on gas and temperature, so the sign alone is not a complete molecular explanation.

Suppose 1.00 mol gas at 300 K and 10.0 atm occupies 2.20 L. With R = 0.08206 L atm mol⁻¹ K⁻¹, Z = (10.0)(2.20)/[(1.00)(0.08206)(300)] = 22.0/24.618 = 0.894. The ideal model would predict V ideal = 2.46 L at the same state inputs. The actual specified volume is smaller by about 10.6% of that ideal value. A correction factor determined at one pressure and temperature should not automatically be reused at a substantially different state.

When Z is unknown, measuring only P, V and T does not determine both n and Z. An independent amount measurement, a gas-specific equation of state, or tabulated Z data is required. For a gas mixture, the relevant Z belongs to the mixture at its composition and state; using a pure component's value may be inappropriate. Near phase coexistence, one must also confirm that the sample is a single gas phase rather than treating liquid plus vapour as one gas volume.

OpenStax Chemistry 2e, section 9.6 describes the qualitative competition between attractions and molecular volume and defines Z as a measured-to-ideal molar-volume ratio. The calculation here uses that same comparison without presuming a universal Z curve.

Step-by-step reasoning

1. Confirm the sample is a gas at a specified composition and state, and convert P, V and T to compatible units. 2. If n is known, compute Z = PV/(nRT) and check that all units cancel. 3. If Z is supplied, rearrange PV = ZnRT for the requested variable. 4. Compare Z with one and interpret it only at the stated pressure and temperature. 5. Avoid importing a Z value from a different state or gas without supporting information.

Visual explanation

Draw a horizontal line at Z = 1 on a graph of Z versus pressure. Sketch one possible real-gas curve that begins near one, dips below it and later rises above it. Mark a single measured point, rather than assuming every gas follows that exact path. Beside the graph show two equal-state volume bars, V real = ZV ideal.

Real-world analogy

A map scale predicts a travel distance from straight-line geometry, while the actual route may be shorter or longer because the route follows different constraints. Z acts as a measured-to-model ratio for gas volume at a specified state. The analogy illustrates correction factors, but actual gas departures come from molecular interactions and finite size.

Real-world example

An industrial storage vessel can hold compressed gas at a pressure where ideal-gas estimates of stored moles are inaccurate. Engineers use measured property data or an appropriate equation of state to obtain Z for the gas and conditions, then calculate n = PV/(ZRT). Treating Z as exactly one can misstate inventory and flow.

Why?

Why is Z useful rather than simply saying a gas is “nonideal”? Z gives a numerical deviation at a specified state. It lets a measured volume or pressure be compared directly with the ideal prediction and supports a corrected amount calculation when suitable property data are available.

Common misconception

“A gas has one constant Z.” Compressibility factor varies with pressure, temperature and composition. Another error is to multiply by Z when solving for n: from PV = ZnRT, n = PV/(ZRT), so Z belongs in the denominator for this unknown.

Worked example

A real-gas sample has P = 8.00 atm, V = 4.00 L, T = 320 K and n = 1.10 mol. Using R = 0.08206 L atm mol⁻¹ K⁻¹, Z = PV/(nRT) = 32.0/[(1.10)(0.08206)(320)] = 1.11 to three significant figures. At those specified P, T and n, the real volume is about 11% larger than the ideal-model volume. The result does not prove that all states of this gas have Z above one.

Quick check

1. If Z = 0.80 at a stated P, T and n, what fraction of ideal molar volume is the real molar volume? Answer: V real/V ideal = Z = 0.80, so the real molar volume is 80% of the ideal prediction.

Exam focus

Write the definition Z = PV/(nRT) before rearranging. Keep Z dimensionless, use absolute pressure and Kelvin temperature, and identify the state associated with any tabulated correction. Do not infer a unique molecular mechanism from one Z measurement.

Advanced insight

The slope of Z against pressure near the low-pressure limit relates to virial coefficients, which encode aspects of intermolecular interactions. A measured Z = 1 at one finite pressure does not mean a gas is ideal at every pressure; attraction and finite-size effects can cancel locally.

Summary

Z compares a real gas's PV with nRT at one state. Values below or above one indicate departures from the ideal prediction, and a supplied Z modifies the gas equation to PV = ZnRT. Use gas- and state-specific information rather than assuming a universal correction.

Practice questions

1. For n = 1.00 mol, T = 300 K, P = 10.0 atm and V = 2.20 L, calculate Z using R = 0.08206 L atm mol⁻¹ K⁻¹. Answer: Z = 22.0/24.618 ≈ 0.894. 2. If ideal V is 5.00 L at a stated P and T but Z = 1.10, what is corrected real volume? Answer: V real = ZV ideal = 1.10 × 5.00 = 5.50 L. 3. A sample has PV/(RT) = 2.00 mol and Z = 0.80. What is its corrected amount? Answer: n = PV/(ZRT) = 2.00/0.80 = 2.50 mol.