SCIENCE · INVESTIGATE AND EXPLAIN

Interpret nonideal gas evidence without a single-cause shortcut

Separate two model corrections and test the domain before interpreting compressibility.

Learning outcome: Compare300 and 500 K at Vm1, then vary volume at fixed temperature . Retain each correction, total pressure and Z; verify Vm>b and the declared above-critical boundary.

Starting recap

Review amount dimensions, chemical formulas and charge, signed differences, ratio bounds and the distinction between a theoretical constraint and measured conversion.

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Try

At Vm=1 L/mol,T=350 K, ideal pressure is 2909.9 kPa; excluded volume adds153.153 and attraction subtracts200, giving2863.053 kPa and Z≈.98390.

Open the experiment

Predict

Does Z close to 1 prove that intermolecular interactions are absent?

Your prediction and reason

Experiment

Compare300 and 500 K at Vm1, then vary volume at fixed temperature . Retain each correction, total pressure and Z; verify Vm>b and the declared above-critical boundary.

Compare attractive and excluded-volume corrections

Worked teaching model · synthetic values · Fictional van der Waals gas with a=200 kPa L²/mol² and b=0.050 L/mol: P=RT/(Vm−b)−a/Vm², R=8.314. Selected T exceeds this model critical temperature8a/(27Rb)≈142.6 K and Vm exceeds b. Positive-pressure single-phase comparison; not a real gas property fit or pressure-system design.

Molar volume: 1 L/mol Temperature: 350 K

Reset keeps your prediction, captures and practice history.

Axes rescale. Connected samples show a declared parameter relationship, not elapsed reaction time. Dashed comparison, when present, is the unbiased gas pressure or ideal-gas Z=1 reference described in the model.

Numerical diagram coordinates

molar volume (L/mol)	compressibility Z
0.5	0.9736493
0.625	0.9769871
0.75	0.9797874
0.875	0.9820565
1	0.9839007
1.125	0.9854175
1.25	0.986682
1.375	0.9877498
1.5	0.9886622
1.625	0.9894501
1.75	0.990137
1.875	0.9907408
2	0.9912756
2.125	0.9917524
2.25	0.9921802
2.375	0.9925661
2.5	0.9929158
2.625	0.9932343
2.75	0.9935255
2.875	0.9937927
3	0.9940389

Dashed reference
molar volume (L/mol)	compressibility Z
0.5	1
0.625	1
0.75	1
0.875	1
1	1
1.125	1
1.25	1
1.375	1
1.5	1
1.625	1
1.75	1
1.875	1
2	1
2.125	1
2.25	1
2.375	1
2.5	1
2.625	1
2.75	1
2.875	1
3	1

Calculated evidence under the stated assumptions
Quantity	Value	Unit
Ideal pressure	2909.9	kPa
Excluded-volume pressure correction	153.1526	kPa
Attractive pressure correction	-200	kPa
Model pressure	2863.053	kPa
Compressibility factor	0.9839007	ratio
Model critical temperature	142.5529	K
Available molar volume	0.95	L/mol

Attraction lowers and excluded volume raises the pressure relative to the ideal model at fixed T and Vm.

The two corrections compete; Z near1 can reflect cancellation rather than absence of interactions. The bounded domain stays above the stipulated critical temperature with positive pressure and Vm>b. This fictional equation is not an empirical fit to a named gas or a description of liquid–vapor coexistence.

Retain the reaction/measurement boundary with every number. This mathematical packet is not a chemical procedure or evidence of actual laboratory technique.

Attraction lowers and excluded volume raises the pressure relative to the ideal model at fixed T and Vm.

Observation to attach to a capture

Captures preserve recipes and replay their calculated results. They are synthetic model evidence, not physical measurements.

Understand

Define a dimensionless diagnostic

Compressibility factor is Z=PVm/(RT), where Vm is molar volume. An ideal gas has Z=1. Values above or below one indicate departure from that ideal relation, but a single Z value does not uniquely identify a microscopic cause. Calibration error, composition error or a compensating physical effect can also affect an inference.

Audit the supplied equation

The fictional van der Waals model uses P=RT/(Vm−b)−a/Vm², with a=200 kPa L²/mol², b=.050 L/mol and R=8.314 kPa L/(mol K). Relative to ideal RT/Vm, the excluded-volume part raises pressure and the attractive part lowers it at fixed T and Vm. Their sum, not either term alone, determines the comparison.

Work an original state

At Vm=1 L/mol and T=350 K, the ideal term is 2909.9 kPa. Excluded volume adds2909.9/.95−2909.9≈153.153 kPa, while attraction subtracts200. The final pressure is 2863.053 kPa and Z≈.98390. The result is close to one even though both corrections are nonzero and partly cancel.

Test temperature and domain

At the same volume but500 K, the excluded-volume increment grows with temperature while the stipulated attractive correction remains−200. Their balance can produce Z above one. The allowed domain has Vm greater than b, positive calculated pressure and temperatures above the model critical temperature 8a/(27Rb)≈142.6 K. Those checks avoid interpreting a singular denominator or a liquid–vapor coexistence region as a simple uniform gas.

Interpret without overclaiming

The equation parameters are fictional and not fitted properties of a named gas. A near-ideal measured state would not prove that attractions and finite-size effects vanish; it might reflect cancellation. Independent measurements at multiple temperatures and densities could test competing models. This bounded activity does not specify safe operating pressures or authorize a pressure-system experiment. Retain the domain checks and the alternative explanations alongside Z.

Sources and evidence boundary

Conceptual sources checked 30 October 2026: Nonideal-gas behavior. All prose, cases, code and diagrams are original. The supplied models are not measured structures, a performed laboratory investigation or certification of independent mastery. No chemical handling, heating, pressure vessel or cryogenic experiment is requested. Retain predictions, contrasting evidence and a changed-case explanation.

Apply

Public practice checks a specific claim. It does not establish independent mastery.

Why is an apparently good Z value not a unique fit of a real gas mechanism?

Your explanation and limitations

Continue

Completion requires a saved prediction, at least one capture and a reflection. Presence checks do not grade your reasoning.

Skipping retains experiments and awards no mastery.

Previous: Compare molecular speed without confusing it with bulk transport Next: Separate internal-energy change from heat and boundary work

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