SCIENCE · INVESTIGATE AND EXPLAIN

Separate internal-energy change from heat and boundary work

Choose a system and sign convention before combining transfers.

Learning outcome: Compare expansion, no volume change and compression with the same heat. Then construct two different heat/work combinations giving the same internal-energy change.

Starting recap

Review signed energy transfers, dimensional consistency, reaction coefficients, state functions and phase/structural evidence boundaries.

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Try

With300 J heat into the system and expansion1 L against100 kPa, work on the system is−100 J and internal energy rises200 J.

Open the experiment

Predict

Can a system absorb heat yet have a smaller internal-energy increase because it does work on its surroundings?

Your prediction and reason

Experiment

Compare expansion, no volume change and compression with the same heat. Then construct two different heat/work combinations giving the same internal-energy change.

Separate energy change from heat and boundary work

Worked teaching model · synthetic values · Closed system with constant external pressure100 kPa, only pressure-volume work w=−PextΔV, chemical sign convention positive work into system.1 kPa L=1 J. Constant external pressure is not a claim of a reversible path or fixed temperature.

Heat into system: 300 J Volume change: 1 L

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Axes rescale. On the phase map, short dashes show solid–gas coexistence, long dashes liquid–gas and the solid line solid–liquid; the dot is the selected state. In the Hess diagram, the dashed line is the independent direct packet. A heating plot follows supplied energy, not elapsed time.

Numerical diagram coordinates

heat into system (J)	internal energy change (J)
-500	-600
-400	-500
-300	-400
-200	-300
-100	-200
0	-100
100	0
200	100
300	200
400	300
500	400

Calculated evidence under the stated assumptions
Quantity	Value	Unit
Heat into system	300	J
Work on system	-100	J
Internal energy change	200	J
Expansion work delivered by system	100	J

Internal energy is a state function; heat and work describe transfers across a chosen boundary.

Expansion gives negative work on the system, compression positive. At constant volume only-PV-work gives ΔU=q. At constant pressure with only PV work and appropriate equilibrium endpoints, heat relates to ΔH instead. A sign convention does not make heat a substance stored inside a system.

Retain the system, phase and structural assumptions with every number. Supplied or calculated evidence is not actual calorimetry, chemical handling or X-ray operation.

Internal energy is a state function; heat and work describe transfers across a chosen boundary.

Observation to attach to a capture

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

Understand

Fix the boundary and signs

Internal energy belongs to a specified system. Heat and work describe energy crossing its boundary, not substances stored inside it. Use the chemical convention ΔU=q+w, with heat into the system positive and work done on it positive. The surroundings have the complementary transfer signs when the combined boundary is appropriately closed.

Calculate pressure-volume work

For expansion against a constant external pressure, w=−PextΔV. The negative sign expresses energy leaving the system as it pushes its surroundings. Compression has negative ΔV and positive work on the system. One kPa multiplied by one liter equals one joule, so100 kPa and1 L expansion give−100 J without a hidden factor of1000.

Reconcile a complete ledger

If q=300 J and ΔV=1 L, ΔU=300−100=200 J. If the volume instead stays fixed, only-PV work vanishes and ΔU=300 J. With1 L compression, ΔU=400 J. The same heat input does not imply the same state change when other transfers differ.

Exhibit path dependence

An energy increase of200 J can arise from300 J heat and−100 J work, or from100 J heat and+100 J work. Equal ΔU does not require equal q and w. This is the distinction between a state change and the transfers used to accomplish it; the ledger alone does not specify every physical endpoint variable.

Connect to calorimetry carefully

At constant volume and with only pressure-volume work, heat equals ΔU. At constant pressure, with only PV work and appropriate equilibrium endpoints, heat is associated with ΔH. Other work modes or an incomplete boundary can invalidate those shortcuts. The selected constant external pressure is not a claim that every step is reversible or isothermal. Retain the sign convention, boundary and omitted work modes before using a temperature measurement to infer chemical energy.

Sources and evidence boundary

Conceptual sources checked 30 October 2026: Energy and enthalpy. All prose, cases, code and diagrams are original. The supplied models are not measured structures, a performed laboratory investigation or certification of independent mastery. No chemical handling, heating, pressure vessel or cryogenic experiment is requested. Retain predictions, contrasting evidence and a changed-case explanation.

Apply

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

Which conditions are required before identifying measured heat with ΔU or ΔH?

Your explanation and limitations

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Previous: Interpret nonideal gas evidence without a single-cause shortcut Next: Challenge a calorimetric inference with a missing heat path

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