Learn

Learning preferences

Make learning more comfortable. Changes apply immediately.

Enlarges page content, including diagrams and their labels. You can also use your browser’s zoom.

Choose the colours that are easiest for you to distinguish. Applies to lesson diagrams and lab plots. Labels, shapes, hatching and line patterns also distinguish meanings. Source photographs and raw image brightness stay unchanged; mask-only views are available for image overlays.

ExampleComparisonSample link

Saved in this browser on this device. Preferences are separate from progress backups.

SCIENCE · INVESTIGATE AND EXPLAIN

Integrate a temperature-dependent state-energy change

For a fixed-volume system with no nonmechanical work or phase transition, dU=n Cv(T)dT. If heat capacity varies with temperature, the finite energy change is an integral, not one arbitrarily chosen endpoint capacity times deltaT. The packet has n=2 mol, Ti=300 K and molar Cv(T)=A+BT, with A=20 J/(mol K).

Learning outcome: Derive the correction, compare a split interval with the direct integral, and state why constant volume is necessary for the simple Q=delta U account in this packet.

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.

All lessons are open directly. Your saved experiments remain when you leave and return.

Try

For Tf=500 K and B=0.02 J/(mol K²), the integrated change is 11200 J. Freezing capacity at 300 K predicts 10400 J, missing 800 J. The final molar capacity is 30 J/(mol K).

Open the experiment

Predict

For positive B and increasing temperature, the initial-capacity approximation:

Experiment

Derive the correction, compare a split interval with the direct integral, and state why constant volume is necessary for the simple Q=delta U account in this packet.

Integrate heat capacity rather than multiplying one sample

Worked teaching model · synthetic values · Two moles at fixed volume initially 300 K, molar Cv(T)=20 J/(mol K)+B T. No phase transition, other work or leakage. The supplied constitutive law is valid only over 300..650 K.

Reset keeps your prediction, captures and practice history.

State-energy difference from the same reference

internal-energy change (J)temperature (K)0650206500

Graph axes rescale independently; read units and values, not screen angles. 1: integral of two-mole capacity (solid). Joined curves show the supplied model, not additional observations.

Equivalent numerical coordinates for State-energy difference from the same reference

integral of two-mole capacity

temperature (K)internal-energy change (J)
3000
304.375227.8828
308.75456.5313
313.125685.9453
317.5916.125
321.8751147.07
326.251378.781
330.6251611.258
3351844.5
339.3752078.508
343.752313.281
348.1252548.82
352.52785.125
356.8753022.195
361.253260.031
365.6253498.633
3703738
374.3753978.133
378.754219.031
383.1254460.695
387.54703.125
391.8754946.32
396.255190.281
400.6255435.008
4055680.5
409.3755926.758
413.756173.781
418.1256421.57
422.56670.125
426.8756919.445
431.257169.531
435.6257420.383
4407672
444.3757924.383
448.758177.531
453.1258431.445
457.58686.125
461.8758941.57
466.259197.781
470.6259454.758
4759712.5
479.3759971.008
483.7510230.28
488.12510490.32
492.510751.13
496.87511012.7
501.2511275.03
505.62511538.13
51011802
514.37512066.63
518.7512332.03
523.12512598.2
527.512865.13
531.87513132.82
536.2513401.28
540.62513670.51
54513940.5
549.37514211.26
553.7514482.78
558.12514755.07
562.515028.13
566.87515301.95
571.2515576.53
575.62515851.88
58016128
584.37516404.88
588.7516682.53
593.12516960.95
597.517240.13
601.87517520.07
606.2517800.78
610.62518082.26
61518364.5
619.37518647.51
623.7518931.28
628.12519215.82
632.519501.13
636.87519787.2
641.2520074.03
645.62520361.63
65020650

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
Integrated internal-energy increase11200J
Initial-capacity approximation10400J
Missed capacity correction800J
Final molar heat capacity30J/(mol K)
Heat into the fixed-volume system11200J

A state-energy change integrates the capacity along temperature; one endpoint sample is an approximation.

At fixed volume with no other work, dU=n Cv(T)dT and Q=delta U. For Cv=A+BT the integral is n[A deltaT+B(Tf²−Ti²)/2]. Freezing Cv at Ti misses nB(deltaT)²/2 exactly. Reversing the endpoints reverses the state-energy change; splitting the interval does not change its total.

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

A state-energy change integrates the capacity along temperature; one endpoint sample is an approximation.

At fixed volume with no other work, dU=n Cv(T)dT and Q=delta U. For Cv=A+BT the integral is n[A deltaT+B(Tf²−Ti²)/2]. Freezing Cv at Ti misses nB(deltaT)²/2 exactly. Reversing the endpoints reverses the state-energy change; splitting the interval does not change its total.

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

Understand

Declare the distribution or system

For a fixed-volume system with no nonmechanical work or phase transition, dU=n Cv(T)dT. If heat capacity varies with temperature, the finite energy change is an integral, not one arbitrarily chosen endpoint capacity times deltaT. The packet has n=2 mol, Ti=300 K and molar Cv(T)=A+BT, with A=20 J/(mol K).

Derive the integral or response

Integrating gives delta U=n[A(Tf−Ti)+B(Tf²−Ti²)/2]. The factor B has units J/(mol K²), so each term has molar energy units before multiplying by n. At fixed volume with no other work, heat into the system equals this internal-energy increase. That equality would not automatically hold for a volume-changing path.

Audit limits and evidence

The initial-capacity approximation is n(A+BTi)(Tf−Ti). Subtracting it from the exact integral gives nB(deltaT)²/2, positive for positive B. Using the arithmetic mean of initial and final capacities is exact for this linear law, but that trapezoid shortcut is not exact for every nonlinear capacity.

Predict a changed premise

Split a temperature interval at any intermediate temperature and integrate both pieces: the sum must match the original state difference. Reversing the endpoints changes the sign. These operations test state-function consistency; they do not prove that the supplied constitutive law describes an actual substance outside its stated range.

Worked original case

For Tf=500 K and B=0.02 J/(mol K²), the integrated change is 11200 J. Freezing capacity at 300 K predicts 10400 J, missing 800 J. The final molar capacity is 30 J/(mol K).

Investigation

Derive the correction, compare a split interval with the direct integral, and state why constant volume is necessary for the simple Q=delta U account in this packet.

Changed-case practice

Try the changed case before feedback

Use Tf=400 K and B=0.01 J/(mol K²), with the same two moles. Find delta U in joules.

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

For positive B and increasing temperature, the initial-capacity approximation:

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?

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.

Save a science backup or import one

Science work is stored locally in this browser, separately from mathematics and AI. It does not sync automatically. Use this science-only backup; this checkout's older all-subject export does not include science.

A legacy file is available only when the complete snapshot fits the 36 MB limit. For larger snapshots, prepare a complete file set below; never discard notebooks to make a smaller file.

Matching imported notebook drafts replace their current drafts. Omitted lessons and known first attempts remain. Export your current work first if you want to retain both drafts.

Complete file sets and course backups

A file set can hold more work than one science file. Keep its manifest and every part together. Each file is limited to 36 MB. This format needs this version of LearnAI or a compatible newer version; older versions can use only the legacy single-file backup above.

The set includes every saved science course and global continuation at this moment. Later edits need a new export. A conflicting tab exports its own visible work, which may differ from saved work in another tab.

0 file(s) selected. File order and filenames do not determine identity. Missing, changed or conflicting parts block the entire restore; no good part is applied by itself.

Science storage: loading. No account or remote computation is required.