Compressibility Factor Z and Deviation from Ideal Gas Behaviour
The ideal gas law PV = nRT works well for gases at low pressure and high temperature. At high pressure or low temperature, real gases disagree with it. IIT-JAM, GATE and CSIR-NET papers test this through one number, the compressibility factor Z. If you can read Z, you can say at once whether attraction or repulsion dominates in a given gas.
The formulas
Van der Waals: (P + a n²/V²)(V − nb) = nRT
Low-pressure form: Z ≈ 1 + B/Vm, with B = b − a/(RT)
Boyle temperature: TB = a / (R b)
What each term means
- Vm – molar volume, V/n.
- a – the attraction term. A larger a means stronger intermolecular attraction. It lowers the pressure that the walls feel.
- b – the excluded volume per mole, linked to the size of molecules. It makes the gas harder to compress.
- B – the second virial coefficient. For the van der Waals gas, B = b − a/RT.
- Z < 1 – the gas is more compressible than an ideal gas; attraction wins.
- Z > 1 – the gas is less compressible than an ideal gas; repulsion (size) wins.
Constants a and b differ slightly between textbooks and data tables, and some books give a in atm L² mol⁻² and others in bar L² mol⁻². In the examples below we use the values stated in the question: a = 3.59 atm L² mol⁻², b = 0.0427 L mol⁻¹ (a CO₂-like gas), R = 0.082057 L atm mol⁻¹ K⁻¹. In an exam, always use the constants given in the paper.
Worked examples
Example 1: Z from measured data. 2.00 mol of a gas occupies 5.00 L at 300 K and shows a pressure of 8.00 atm. Find Z.
nRT = 2.00 × 0.082057 × 300 = 49.234 L atm.
PV = 8.00 × 5.00 = 40.00 L atm.
Z = 40.00 / 49.234 = 0.812.
Since Z < 1, attraction dominates. The ideal gas law would have predicted P = 49.234/5.00 = 9.85 atm, so the real pressure is lower, as attraction predicts.
Example 2: Z from the van der Waals equation. Find Z for 1.00 mol at 300 K in 1.00 L, 0.500 L and 10.0 L, using a = 3.59 and b = 0.0427.
RT = 0.082057 × 300 = 24.617 L atm/mol. For each V, P = RT/(V − b) − a/V².
V = 1.00 L: P = 24.617/0.9573 − 3.59/1.00 = 25.715 − 3.590 = 22.125 atm. Ideal P = 24.617 atm. Z = 22.125 × 1.00 / 24.617 = 0.899.
V = 0.500 L: P = 24.617/0.4573 − 3.59/0.25 = 53.832 − 14.360 = 39.471 atm. Ideal P = 49.234 atm. Z = 39.471 × 0.500 / 24.617 = 0.802.
V = 10.0 L: P = 24.617/9.9573 − 3.59/100 = 2.4723 − 0.0359 = 2.4364 atm. Ideal P = 2.4617 atm. Z = 2.4364 × 10.0 / 24.617 = 0.990.
As the gas expands, Z moves towards 1: the gas behaves more and more ideally.
Example 3: the virial shortcut. Check the V = 10.0 L result with Z ≈ 1 + B/Vm.
B = b − a/RT = 0.0427 − 3.59/24.617 = 0.0427 − 0.14584 = −0.10314 L/mol.
Z ≈ 1 + (−0.10314)/10.0 = 0.9897, matching 0.990 from Example 2. At 1.00 L the shortcut gives 0.897, close to the exact 0.899, but it is only a low-density approximation and drifts at smaller volumes (0.5 L: 0.794, against 0.802).
Example 4: Boyle temperature. For this gas, TB = a/(Rb) = 3.59 / (0.082057 × 0.0427) = 3.59 / 0.0035040 = 1.02 × 10³ K (about 1025 K).
At TB the second virial coefficient B is zero, so the gas follows PV = nRT closely over a wide range of low pressures. Below TB (like 300 K here) B is negative and Z dips below 1 at moderate pressure. Above TB, B is positive and Z is greater than 1.
Reading a Z versus P graph
| Feature of the curve | What it tells you |
|---|---|
| Starts at Z = 1 when P → 0 | All gases become ideal at very low pressure. |
| Dips below 1 first, then rises | Attraction dominates at moderate P; size (b) dominates at high P. |
| Rises straight from 1 (Z > 1 always) | Temperature is above TB for that gas, or a is very small (typical for light gases such as H₂ and He at room temperature). |
| Deepest dip at lower T | Attraction matters more when molecules move slowly. |
Common errors
- Mixing units: a in bar L² mol⁻² with R in L atm. Keep P, V and R in a matching set.
- Using V instead of V − nb inside the pressure expression.
- Saying “Z > 1 means attraction”. It is the opposite: Z < 1 is attraction, Z > 1 is repulsion.
- Applying Z ≈ 1 + B/Vm at high density, where higher virial terms matter.
- Forgetting that Z is dimensionless: a unit in your answer is a red flag.
Exam relevance
| Exam | Typical use |
|---|---|
| IIT-JAM | Compute Z, compare real and ideal pressure, identify attraction or repulsion. Check the official syllabus for the exact scope. |
| GATE / CSIR-NET / CUET-PG | van der Waals constants, Boyle temperature, virial form and the Z–P curve. Check the current notification for the syllabus. |
The calculator below uses the ideal gas law. Use it to find the ideal pressure, then divide the real measured pressure by it to get Z (since Z = Preal / Pideal at the same n, V and T).
Compute the ideal-gas value first; then compare it with the real measurement to get Z.
Open the Ideal Gas Law (PV = nRT) Calculator →Previous-year CSIR-NET questions and timed quizzes to check what you just read.
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