SCIENCE · INVESTIGATE AND EXPLAIN

Distinguish weak corrections from cancelling gas effects

Use a controlled nonideal model to test competing attraction and finite-size explanations.

Learning outcome: Explain pressure departures using separate particle assumptions and challenge ideality claims with changed conditions.

Starting recap

At the same molar volume v and absolute temperature T, an ideal gas has P=RT/v. Z=Pv/(RT) equals one for that baseline.

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Try

At one selected condition two substantial nonideal corrections cancel, giving Z=1. A second volume reveals the difference from an ideal model.

Open the experiment

Predict

Predict whether Z=1 proves particles have no attraction or size. Predict how dilution and heating affect each correction separately.

Your prediction and reason

Experiment

Start at 300 K and v/b=2. Capture the two nonzero terms, then compare v/b=4 and 20. At fixed v/b=4 compare 600 K and size-only models. Use the fixed graph and particle assumptions to justify the changes.

Separate competing departures from ideal pressure

Constructed nonideal comparison · stipulated parameters · Original van der Waals teaching parameters, not fitted species data: b=40 mL/mol, a=2R(300 K)b. Compare homogeneous stable branches at 300/600 K with v/b≥1.5. No phase-equilibrium, liquid fraction, critical-region accuracy or actual material identification is inferred.

Controlled hypothetical model: Both corrections at 300 K case Molar volume divided by excluded-volume parameter: 2 v/b

Reset keeps your prediction, captures and practice history.

Agreement at one condition can hide two large corrections

Both corrections at 300 K; molar volume v = 80.00000 mL/mol = 2b. The ideal pressure at this same v and T is 31.17923 MPa; the combined teaching model gives 31.17923 MPa. These are calculated values for stipulated parameters, not observed pressures of a named gas.

Contribution	Change to Z	Interpretation at fixed v and T
Ideal baseline	1	Point particles with no intermolecular force
Finite-size correction	+1.00000	Less accessible volume raises pressure relative to ideal
Attraction correction	-1.00000	Mean attraction lowers pressure relative to size-only model
Combined Z	1.00000	Corrections cancel here; neither must vanish

Change volume to test a single-point explanation. At 300 K, v=2b gives Z=1 even though the positive and negative corrections both have magnitude one. At v=4b, Z=5/6. Increasing temperature to 600 K reduces the attraction term relative to RT while retaining the same b correction at fixed v. Thus “higher temperature means exactly ideal” is also too strong.

In the dilute limit v/b→∞ both corrections vanish. At the finite endpoint 20b they are small but not zero. Real gases can also condense at lower temperature; this activity does not solve phase coexistence or extend its homogeneous branch into that regime.

Optional equation and parameter audit

The van der Waals teaching equation is P=RT/(v−b)−a/v² for molar volume v in m³/mol. Here R=8.314462618 J/(mol K), b=0.00004 m³/mol and a=0.19954710283200003 Pa m⁶/mol². Multiplying by v/(RT) gives Z=1+b/(v−b)−a/(RTv). The constants were deliberately constructed to give a cancellation example; they are not experimental fit parameters for a real species.

All authored states have v>b, positive pressure and negative isothermal dP/dv. The full model critical temperature 8a/(27Rb) is 177.77778 K, below both temperatures used here. These checks exclude a mechanical-instability loop in this chosen domain; they do not validate real-material accuracy. The model’s excluded-volume parameter is not a literal measured container volume occupied by the drawn circles.

OpenStax nonideal-gas mechanisms and equation

Z = 1 + 1.00000 − 1.00000 = 1.00000.

Z=Pv/(RT) compares pressure at the same molar volume and temperature. A value near one can reflect weak corrections or cancellation; one observation alone does not distinguish them.

Observation to attach to a capture

Captures retain the hypothetical comparison and molar-volume ratio. Replay reconstructs separate finite-size and attraction corrections; it is not a measured equation of state or validated real-material prediction.

Understand

One matching pressure can hide two effects

Start with the 300 K model at molar volume v=2b. It gives the same pressure as the ideal gas law. Before reading the ledger, predict whether that proves particle size and intermolecular attraction are negligible. Then change only the volume to 4b.

The activity uses deliberately constructed parameters, not data fitted to a real gas. Its purpose is to test an inference: a net result can hide opposing contributions. A real material would need measured evidence across suitable conditions to evaluate the model’s accuracy.

Connect each assumption to a pressure change

The ideal model treats particles as points without intermolecular attractions. Finite-size particles exclude some configurations from other particles. In this simple model, reducing accessible volume raises pressure at the same molar volume and temperature. Attractions instead lower pressure relative to the corresponding size-only model. Near conditions that lead to condensation, attractions can be especially important; this activity does not model that phase change.

The three particle pictures show separate assumptions. Their circles, counts and dashed interaction are qualitative, not measured molecular diameters, forces or trajectories. The excluded-volume parameter b should not be read directly as the literal volume of the drawn particles.

Compare pressure with the ideal baseline using the dimensionless ratio Z=Pv/(RT), where v is volume per mole. Z below one means lower pressure than ideal at the same v and T; above one means higher pressure. The chart’s horizontal axis is molar volume divided by b, not pressure or time.

Make a changed-condition comparison

At 300 K and 2b, the ledger is Z=1+1−1=1. Both corrections are substantial. At 4b the finite-size term becomes 1/3 while attraction contributes −1/2, giving Z=5/6≈0.83333. This second observation distinguishes the constructed model from a truly ideal prediction.

At the same 4b but 600 K, the size correction stays 1/3 and the attraction contribution becomes −1/4. Now Z=13/12≈1.08333. The attraction parameter has not disappeared: its contribution relative to thermal energy is smaller. Heating does not necessarily make every pressure exactly ideal, especially when finite-size effects remain.

Switch to the size-only comparison at 300 K. At 4b it gives Z=4/3. Explain the difference using both the pressure ledger and the particle assumptions, rather than saying simply that one curve is “more real.” All three are model comparisons with different stated assumptions.

At 20b the full 300 K model gives Z=1+1/19−1/10≈0.95263. It is approaching one as the sample becomes more dilute, but this finite endpoint is not the exact dilute limit. Capture the cancellation case and a changed condition with your explanation.

Optional quantitative model

The calculations use P=RT/(v−b)−a/v², the van der Waals equation. Here v is in m³/mol, b=40 mL/mol=0.000040 m³/mol, R=8.314462618 J/(mol K), and a=2R(300 K)b in Pa m⁶/mol². The constants were selected to make the cancellation case exact; they are not reference parameters for a named substance.

Multiplying by v/(RT) gives Z=1+b/(v−b)−a/(RTv). This isolates the positive finite-size and negative attraction contributions. The qualitative mechanism and changed-condition reasoning are the core lesson; deriving this equation is optional depth rather than a claim about required AP computation.

Only volumes 1.5b–20b and temperatures 300/600 K are used. Pressure remains positive and decreases as volume increases at each temperature. Both temperatures exceed this hypothetical model’s critical temperature, approximately 177.78 K. No unstable loop, condensation fraction or phase-coexistence calculation is presented. Those domain checks establish internal consistency, not accuracy for any real gas.

Transfer and sources

A fictional observed Z is one, while an independently proposed model assigns corrections +0.30 and −0.30. Identify a change of volume or temperature that would help distinguish cancellation from negligible effects. State what additional observations would be needed before choosing between models. Agreement at one point alone does not identify microscopic causes.

AP Chemistry CED, 3.6.A.1: attractions and particle volume as causes of nonideal behavior.
OpenStax nonideal-gas behavior: physical mechanisms, compressibility factor and van der Waals representation.
Return to ideal gas pressure and volume for the baseline or molecular speed distributions for temperature and microscopic-motion context.

Apply

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

Explain why one pressure observation cannot distinguish weak interactions from cancellation. Propose a changed-condition check and distinguish this constructed example from empirical validation or a condensation prediction.

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 distributions and threshold fractions Next: Repair a solution diagram using atoms, charge and concentration

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