Challenge a pressure response without erasing calibration ambiguity

Declare the distribution or system

Original indicated pressure records follow Pind=g[K/V+A/V²], with V expressed in litres, K in kPa L and A in kPa L². The revealed teaching generator uses K=12, while shared pressure gain g and correction A can vary. For the inverse task K is unknown to the analyst and is not independent calibration evidence. If an independently established K=12 were supplied instead, gain would be identifiable as Kfit/12; the nonidentifiability statement would no longer apply. The first two exact-volume records at 1 and 2 litres are calibration data; the 3-litre record is held out for a frozen prediction.

Derive the integral or response

From P1=Kfit+Afit and P2=Kfit/2+Afit/4, obtain Kfit=4P2−P1 and Afit=2P1−4P2. These are indicated coefficients gK and gA, not independently calibrated physical coefficients. An ideal one-parameter fit using only P1 predicts P2=P1/2 and can fail the second observation. The corrected model consumes both first records, so its validation must use the unused third record.

Audit limits and evidence

The third prediction is (8P2−P1)/9. If each calibration reading has a separate deterministic bound epsilon, its direct prediction error is bounded by epsilon. Propagating independent parameter intervals would ignore their shared observations. Add a fresh reading bound separately when judging a residual. These hard bounds are not probabilistic confidence intervals.

Predict a changed premise

More records under the same unknown gain cannot split g from K and A. An independent gain bound 0.98..1.02 implies Kfit/1.02≤K≤Kfit/0.98 when the fitted indicated coefficient is exact. The displayed interval accounts only for that shared gain; include reading errors too for a total physical-parameter interval. A successful third record tests the response family, not every calibration premise.

Worked original case

At A=2,g=1, P1=14 kPa and P2=6.5 kPa. The ideal fit predicts 7 kPa and fails by −0.5 kPa. The corrected fit gives Kfit=12,Afit=2 and predicts P3=38/9≈4.222222 kPa. A gain-only physical K interval is 12/1.02..12/0.98 kPa L.

Investigation

Fit the ideal model to one record, test it on the second, then fit the corrected model to both and freeze a third prediction. Retain the distinction between response-family validation and shared-gain identifiability.

Changed-case practice

Try the changed case before feedback

For P1=16 kPa and P2=7 kPa, predict the unused pressure at 3 litres in kPa.

Numerical result in the requested unitsCompare this result

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.