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SCIENCE · INVESTIGATE AND EXPLAIN

Test which inference survives isolated vacuum expansion

An insulated gas expands into an initially evacuated part of a rigid enclosure. The external boundary does no work and no heat crosses it, so delta U=0. The actual expansion is not a sequence of equilibrium piston states. A pressure-volume curve drawn through intermediate equilibrium states must not be mistaken for its actual work path.

Learning outcome: Derive the model-consistency identity, retain ideal and nonideal endpoint predictions and explain why integrating equilibrium p dV along the displayed endpoint family would assign the wrong actual work.

Starting recap

Review logarithmic integrals, equilibrium states, absolute temperature and the first-law work convention. State the domain and distinguish calibration data from unused validation.

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Try

For A=200 J L and r=3, T2=3500/9≈388.888889 K. The entropy change is 12 ln(35/36)+8 ln 3≈8.450848 J/K. The pressure is 8T2/3−200/9≈1014.814815 kPa; the endpoint energy equals its initial value.

Open the experiment

Predict

Which statement follows for both stipulated models in isolated vacuum expansion?

Experiment

Derive the model-consistency identity, retain ideal and nonideal endpoint predictions and explain why integrating equilibrium p dV along the displayed endpoint family would assign the wrong actual work.

Test an ideal inference against a consistent nonideal model

Worked teaching model · synthetic values · Insulated expansion into a vacuum from V1=1 litre and T1=400 K. Supplied consistent model U=12T−A/V joule and p=8T/V−A/V² kPa for V in litres. No boundary work or heat. Endpoints lie in a positive-pressure, locally stable supported region; this is not a full phase diagram.

Reset keeps your prediction, captures and practice history.

Energy-conserving endpoint family, not intermediate equilibrium states

final temperature (K)final volume ratio044000

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

Equivalent numerical coordinates for Energy-conserving endpoint family, not intermediate equilibrium states

supplied nonideal model

final volume ratiofinal temperature (K)
1400
1.0375399.3976
1.075398.8372
1.1125398.3146
1.15397.8261
1.1875397.3684
1.225396.9388
1.2625396.5347
1.3396.1538
1.3375395.7944
1.375395.4545
1.4125395.1327
1.45394.8276
1.4875394.5378
1.525394.2623
1.5625394
1.6393.75
1.6375393.5115
1.675393.2836
1.7125393.0657
1.75392.8571
1.7875392.6573
1.825392.4658
1.8625392.2819
1.9392.1053
1.9375391.9355
1.975391.7722
2.0125391.6149
2.05391.4634
2.0875391.3174
2.125391.1765
2.1625391.0405
2.2390.9091
2.2375390.7821
2.275390.6593
2.3125390.5405
2.35390.4255
2.3875390.3141
2.425390.2062
2.4625390.1015
2.5390
2.5375389.9015
2.575389.8058
2.6125389.7129
2.65389.6226
2.6875389.5349
2.725389.4495
2.7625389.3665
2.8389.2857
2.8375389.207
2.875389.1304
2.9125389.0558
2.95388.9831
2.9875388.9121
3.025388.843
3.0625388.7755
3.1388.7097
3.1375388.6454
3.175388.5827
3.2125388.5214
3.25388.4615
3.2875388.403
3.325388.3459
3.3625388.29
3.4388.2353
3.4375388.1818
3.475388.1295
3.5125388.0783
3.55388.0282
3.5875387.9791
3.625387.931
3.6625387.884
3.7387.8378
3.7375387.7926
3.775387.7483
3.8125387.7049
3.85387.6623
3.8875387.6206
3.925387.5796
3.9625387.5394
4387.5

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
QuantityValueUnit
Nonideal final temperature388.8889K
Ideal-model final temperature400K
State entropy change8.450848J/K
Ideal-model entropy change8.788898J/K
Energy residual3.126388e-13J
Final pressure1014.815kPa

Zero heat and zero work imply constant internal energy, not universally constant temperature.

For U=12T−A/V, conservation gives T2=400−(A/12)(1−1/r). The pressure law satisfies (∂U/∂V)T=T(∂p/∂T)V−p=A/V², so dS=12 dT/T+8 dV/V is consistent. The ideal A=0 limit is isothermal; all supported endpoint entropy changes are positive. No quasistatic pressure path is assigned to the actual vacuum expansion.

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

Zero heat and zero work imply constant internal energy, not universally constant temperature.

For U=12T−A/V, conservation gives T2=400−(A/12)(1−1/r). The pressure law satisfies (∂U/∂V)T=T(∂p/∂T)V−p=A/V², so dS=12 dT/T+8 dV/V is consistent. The ideal A=0 limit is isothermal; all supported endpoint entropy changes are positive. No quasistatic pressure path is assigned to the actual vacuum expansion.

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

Understand

Declare the system and evidence

An insulated gas expands into an initially evacuated part of a rigid enclosure. The external boundary does no work and no heat crosses it, so delta U=0. The actual expansion is not a sequence of equilibrium piston states. A pressure-volume curve drawn through intermediate equilibrium states must not be mistaken for its actual work path.

Derive the state or rate account

For an ideal gas U=CvT, constant U implies unchanged temperature. Our supplied alternative has U=12T−A/V joule and p=8T/V−A/V² kPa, with V in litres. These laws are thermodynamically consistent because T(∂p/∂T)V−p=A/V²=(∂U/∂V)T. An arbitrary added pressure correction without a compatible internal energy would not support this inference.

Audit the constraint and model

From initial V1=1 litre, T1=400 K to V2=r litres, conservation gives T2=400−(A/12)(1−1/r). Attraction allows cooling while total internal energy stays fixed. Combining dU+p dV cancels the attraction terms and gives dS=12 dT/T+8 dV/V. Thus delta S=12 ln(T2/400)+8 ln(r), positive throughout the supported range.

Predict a changed boundary

At A=0 the ideal limit is recovered exactly. At A>0, zero heat does not mean zero entropy change and zero internal-energy change does not mean zero temperature change. The supported positive-pressure endpoints are locally stable for this supplied model; no global coexistence or molecular measurement is claimed.

Worked original case

For A=200 J L and r=3, T2=3500/9≈388.888889 K. The entropy change is 12 ln(35/36)+8 ln 3≈8.450848 J/K. The pressure is 8T2/3−200/9≈1014.814815 kPa; the endpoint energy equals its initial value.

Investigation

Derive the model-consistency identity, retain ideal and nonideal endpoint predictions and explain why integrating equilibrium p dV along the displayed endpoint family would assign the wrong actual work.

Changed-case practice

Try the changed case before feedback

Use A=300 J L and final ratio r=2. Find T2 in kelvin.

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 diagrams are reproduced. All thermal operations are mathematical proposals or supplied records, not physical heating, pressure-vessel or exposure instructions. Public feedback and retained prose do not certify independent reasoning or actual practical participation. Preserve first work, critique, revision and a new assigned response in the separate course workspace.

Apply

Which statement follows for both stipulated models in isolated vacuum expansion?

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

Which system boundary, reference path, constitutive law or independently calibrated record supports this entropy inference?

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