SCIENCE · INVESTIGATE AND EXPLAIN

Test a nonideal branch before treating it as equilibrium

The van der Waals equation supplements an ideal gas with an excluded-volume correction and an attraction term. In reduced variables t=T/Tc, v=V/Vc and p=P/Pc, the supplied model is p=8t/(3v−1)−3/v². The domain v>1/3 is essential; extending the expression through its pole is not a physical compression.

Learning outcome: Compare states below, at and above the critical reduced temperature. Retain the excluded-volume domain, pressure sign and slope classification separately, and explain what additional evidence would be required for phase coexistence.

Starting recap

Review definite integrals, probability densities, mean versus mean square, temperature units and the first-law work convention. State the domain and distinguish calibration data from unused validation.

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Try

At t=0.9,v=1, reduced pressure is 0.6 and slope is +0.6, so the homogeneous branch is mechanically unstable. At t=1.1,v=1, pressure is 1.4 and slope −0.6. At t=v=1 the slope is zero, not a tiny rounded nonzero value to invert.

Open the experiment

Predict

What does a positive isothermal pressure-versus-volume slope imply for the homogeneous branch?

Your prediction and reason

Experiment

Compare states below, at and above the critical reduced temperature. Retain the excluded-volume domain, pressure sign and slope classification separately, and explain what additional evidence would be required for phase coexistence.

Differentiate a reduced nonideal isotherm

Worked teaching model · synthetic values · Supplied reduced van der Waals p=8t/(3v−1)−3/v², v>1/3. Positive slope rejects mechanical stability of a homogeneous branch; negative slope is only a local necessary stability check, not a coexistence or global phase determination. No pressurized equipment.

Reduced temperature: 0.9 T/Tc Reduced volume: 1 V/Vc

Reset keeps your prediction, captures and practice history.

Homogeneous van der Waals candidate branches

Graph axes rescale independently; read units and values, not screen angles. 1: supplied isotherm, not coexistence construction (solid). Joined curves show the supplied model, not additional observations.

Equivalent numerical coordinates for Homogeneous van der Waals candidate branches

supplied isotherm, not coexistence construction

reduced volume	reduced pressure
0.5	2.4
0.53125	1.496558
0.5625	0.9912458
0.59375	0.7063047
0.625	0.5485714
0.65625	0.4662717
0.6875	0.4293632
0.71875	0.4198437
0.75	0.4266667
0.78125	0.4429395
0.8125	0.464317
0.84375	0.488049
0.875	0.5124019
0.90625	0.5362966
0.9375	0.5590805
0.96875	0.580379
1	0.6
1.03125	0.6178693
1.0625	0.6339891
1.09375	0.6484093
1.125	0.6612086
1.15625	0.672482
1.1875	0.6823323
1.21875	0.6908644
1.25	0.6981818
1.28125	0.7043845
1.3125	0.7095672
1.34375	0.7138191
1.375	0.7172231
1.40625	0.7198562
1.4375	0.7217891
1.46875	0.723087
1.5	0.7238095
1.53125	0.7240114
1.5625	0.7237424
1.59375	0.723048
1.625	0.7219698
1.65625	0.7205456
1.6875	0.7188097
1.71875	0.7167936
1.75	0.7145258
1.78125	0.7120323
1.8125	0.7093366
1.84375	0.7064605
1.875	0.7034234
1.90625	0.7002433
1.9375	0.6969364
1.96875	0.6935174
2	0.69
2.03125	0.6863963
2.0625	0.6827176
2.09375	0.6789741
2.125	0.675175
2.15625	0.6713288
2.1875	0.6674432
2.21875	0.6635253
2.25	0.6595813
2.28125	0.6556172
2.3125	0.6516382
2.34375	0.6476491
2.375	0.6436542
2.40625	0.6396577
2.4375	0.6356631
2.46875	0.6316736
2.5	0.6276923
2.53125	0.6237218
2.5625	0.6197646
2.59375	0.6158229
2.625	0.6118986
2.65625	0.6079935
2.6875	0.6041094
2.71875	0.6002475
2.75	0.5964092
2.78125	0.5925957
2.8125	0.588808
2.84375	0.5850469
2.875	0.5813133
2.90625	0.577608
2.9375	0.5739314
2.96875	0.5702841
3	0.5666667

At most 82 representative computed coordinates including endpoints; the plotted curve can use a finer grid. A table of a revealed generator is not independent measurement evidence.

Calculated quantities under the declared measurement model
Quantity	Value	Unit
Reduced pressure	0.6	p/pc
Positive-pressure gas premise satisfied	1	flag
Reduced isothermal pressure slope	0.6	dimensionless
Positive-slope unstable branch	1	flag
Critical zero-slope selected state	0	flag
Formal reduced isothermal compressibility	-1.666667	1/pc

Increasing volume raises pressure on this branch: mechanical stability is rejected.

The excluded-volume domain is v>1/3. Some supported formal branches have nonpositive pressure; they do not establish a positive-pressure gas state, even if their local slope is negative. Differentiate p=8t/(3v−1)−3/v² to obtain −24t/(3v−1)²+6/v³. Positive slope implies negative formal compressibility and an unstable homogeneous branch. At t=v=1 the slope vanishes exactly. Determining phase coexistence requires additional thermodynamic construction; the plotted loop is not a literal equilibrium compression path.

Predict before inspecting the values. Save contrasting records with assumptions and one alternative explanation. This virtual display does not certify practical measurement technique.

Increasing volume raises pressure on this branch: mechanical stability is rejected.

Observation to attach to a capture

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

Understand

Declare the distribution or system

Derive the integral or response

At fixed temperature differentiate to obtain dp/dv=−24t/(3v−1)²+6/v³. Isothermal compressibility is proportional to −1/[v(dp/dv)]. A positive pressure slope would give negative compressibility, rejecting mechanical stability of that homogeneous branch. A negative slope is only a local necessary condition; it does not determine which phase has the lowest free energy.

Audit limits and evidence

At t=v=1, pressure is one and the derivative is exactly zero. The model reports that critical boundary without inventing a finite inverse-slope compressibility. Some lower-temperature supported formal branches have nonpositive pressure; these do not establish a positive-pressure gas state even if their slope is negative.

Predict a changed premise

If asked for a formal isothermal work integral, an antiderivative is (8t/3)ln(3v−1)+3/v, in reduced work units. A mathematical area through an unstable loop is not an observed reversible equilibrium path. A full coexistence calculation needs a further construction, such as equality of appropriate thermodynamic potentials, which is not silently supplied by this local derivative.

Worked original case

Investigation

Changed-case practice

Try the changed case before feedback

At t=1.2,v=1, find the reduced isothermal slope dp/dv.

Numerical result in the requested units

This practice is public and unsaved. A matching number does not verify understanding. Retain your explanation, assumptions and changed-case evidence in the separate course workspace.

Sources and boundaries

Original derivations and synthetic cases checked 31 October 2026 against the primary teaching reference. No source exercises or graphics are reproduced. These bounded CPU calculations do not claim actual gas handling, pressure-vessel work, thermal exposure or practical measurement. Public feedback is practice, not independently verified reasoning; retain original work, critique, revision and a newly assigned case in the course evidence workspace.

Apply

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

Which distribution, constitutive law, path or calibration premise supports this inference, and what independent evidence would challenge it?

Your explanation and limitations

Continue

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

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Previous: Integrate distinct work paths between identical end states Next: Solve a finite-body balance with unequal capacities

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