SCIENCE · INVESTIGATE AND EXPLAIN

Audit a closed-cycle heat input before dividing for efficiency

A monatomic ideal gas executes a clockwise pressure-volume rectangle starting at 1 litre and 100 kPa. It heats at fixed volume to pressure 100a, expands at that pressure to b litres, cools at fixed volume to 100 kPa, then compresses back. A kPa litre equals one joule. This specifies quasistatic mechanical paths, not a realizable two-fixed-reservoir heat-transfer schedule.

Learning outcome: Retain a four-row table of delta U, work and signed heat. Add gas entropy changes, then explain what external temperature-contact evidence would be required to calculate total generation.

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

At a=3, b=2, work is 200 J, positive heat input 1050 J, rejected heat 850 J and efficiency 4/21≈0.190476. Energy closes 1050−850=200. The high-temperature corner has six times the low-temperature corner temperature.

Open the experiment

Predict

Which denominator gives the engine thermal efficiency?

Your prediction and reason

Experiment

Retain a four-row table of delta U, work and signed heat. Add gas entropy changes, then explain what external temperature-contact evidence would be required to calculate total generation.

Separate positive heat input from cycle net heat

Worked teaching model · synthetic values · Clockwise rectangle: start (1 litre,100 kPa), heat at fixed volume, expand at high pressure, cool at fixed volume, compress at 100 kPa. Monatomic ideal gas U=3pV/2. Quasistatic mechanical paths; heat-contact implementation is unspecified, not a two-fixed-reservoir reversible engine.

High-to-low pressure ratio: 3 ratio High-to-low volume ratio: 2 ratio

Reset keeps your prediction, captures and practice history.

Ordered clockwise pressure-volume cycle

Graph axes rescale independently; read units and values, not screen angles. 1: start low-left; heat upward, expand right, cool down, compress left (solid). Joined curves show the supplied model, not additional observations.

Equivalent numerical coordinates for Ordered clockwise pressure-volume cycle

start low-left; heat upward, expand right, cool down, compress left

volume (L)	pressure (kPa)
1	100
1	300
2	300
2	100
1	100

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
Net clockwise work	200	J
Positive heat input	1050	J
Rejected heat magnitude	850	J
Energy closure residual	0	J
Thermal efficiency	0.1904762	ratio
Maximum-to-minimum gas temperature ratio	6	ratio

The enclosed work equals net heat, not total heat absorbed on the two heating legs.

Isochoric heat is 3 delta(pV)/2 and isobaric heat is 5p delta V/2. Sum positive legs separately from negative legs. The four gas entropy changes telescope to zero because the gas returns to its state; this does not establish zero entropy generation in unspecified external heat contacts. A range of gas temperatures is not itself a pair of fixed reservoirs.

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

The enclosed work equals net heat, not total heat absorbed on the two heating legs.

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 system and evidence

Derive the state or rate account

Use U=3pV/2. Isochoric heat equals delta U; isobaric heat equals delta U+p delta V=5p delta V/2. The two positive heat terms are 150(a−1) and 250a(b−1) J. The magnitudes rejected on the other legs are 150b(a−1) and 250(b−1) J. Classify the signs before summing.

Audit the constraint and model

The enclosed clockwise work is 100(a−1)(b−1) J. Net heat equals this work because the gas returns to its initial internal energy. Efficiency is W/Qin, not W/Qnet, which would misleadingly equal one for every engine cycle. The four gas entropy changes sum to zero without implying zero entropy production outside the gas.

Predict a changed boundary

At a=2, b=3, predict W=200 J and Qin=1150 J before inspecting the model; efficiency is 4/23. Maximum and minimum gas temperatures differ by factor ab, but those extremes alone do not identify two fixed reservoirs or prove a Carnot implementation. Reversing the path reverses the signed work and changes which device function is being considered.

Worked original case

Investigation

Retain a four-row table of delta U, work and signed heat. Add gas entropy changes, then explain what external temperature-contact evidence would be required to calculate total generation.

Changed-case practice

Try the changed case before feedback

For a=2, b=3, find efficiency as a decimal.

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 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

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?

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: Test which inference survives isolated vacuum expansion Next: Reject a two-reservoir claim using entropy as well as energy

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