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SCIENCE · INVESTIGATE AND EXPLAIN

Compare route-dependent heat with endpoint-dependent internal energy

Choose the sign convention before balancing thermal energy: heat into the gas is positive and work done by the gas is positive, so ΔU=Q−W. For a quasistatic constant-pressure expansion, W=pΔV; a fixed-volume segment has zero boundary work. On a pressure-volume diagram, the area under the path represents this boundary work.

Learning outcome: Draw both routes and a closed clockwise cycle for two selected endpoints. Keep ΔU, W and Q distinct and explain why equal endpoint temperatures would not generally fix the work or heat.

Starting recap

Declare temperature and time references, system boundaries and model regimes. Use absolute temperatures for ratios; distinguish state, path, oscillator phase and wave propagation.

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Try

For final 150 kPa and 2 L, ΔU=300 J. The high-pressure route does 150 J work and absorbs 450 J heat; the low-pressure route does 100 J work and absorbs 400 J. The clockwise rectangle area is 50 J.

Open the experiment

Predict

For two paths between the same equilibrium states of the same ideal gas, the quantity guaranteed equal is:

Experiment

Draw both routes and a closed clockwise cycle for two selected endpoints. Keep ΔU, W and Q distinct and explain why equal endpoint temperatures would not generally fix the work or heat.

Compare heat and work along two paths between the same states

Worked teaching model · synthetic values · Monatomic ideal gas starts at V=1 L and p=100 kPa. Compare heating at fixed initial volume then expansion at final pressure, with expansion at initial pressure then heating at fixed final volume. Quasistatic boundary work and U=3pV/2; no apparatus.

Reset keeps your prediction, captures and practice history.

Two quasistatic routes between shared endpoints

absolute pressure (kPa)volume (L)032000

Unlike graph quantities rescale independently; read units and values, not screen angles. 1: raise pressure then expand (solid).2: expand then raise pressure (dashed). Joined curves show the supplied model, not additional observations.

Equivalent numerical coordinates for Two quasistatic routes between shared endpoints

raise pressure then expand

volume (L)absolute pressure (kPa)
1100
1200
3200

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.

expand then raise pressure

volume (L)absolute pressure (kPa)
1100
3100
3200

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
Endpoint internal-energy change750J
Work by gas on high-pressure expansion path400J
Work by gas on low-pressure expansion path200J
Heat into gas on high-pressure path1150J
Heat into gas on low-pressure path950J
Clockwise rectangle cycle work200J

Internal energy depends on endpoints; heat and boundary work depend on the route.

One kPa litre equals one joule. With work by the gas positive, Delta U=Q−W. Isochoric segments do no boundary work, while a constant-pressure expansion does p Delta V. A clockwise closed rectangle does positive net work but has zero net Delta U. Isothermal ideal-gas change has Delta U=0, not necessarily Q=0; adiabatic means Q=0, not necessarily constant temperature.

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

Internal energy depends on endpoints; heat and boundary work depend on the route.

One kPa litre equals one joule. With work by the gas positive, Delta U=Q−W. Isochoric segments do no boundary work, while a constant-pressure expansion does p Delta V. A clockwise closed rectangle does positive net work but has zero net Delta U. Isothermal ideal-gas change has Delta U=0, not necessarily Q=0; adiabatic means Q=0, not necessarily constant temperature.

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

Understand

Declare the state, system and observation

Choose the sign convention before balancing thermal energy: heat into the gas is positive and work done by the gas is positive, so ΔU=Q−W. For a quasistatic constant-pressure expansion, W=pΔV; a fixed-volume segment has zero boundary work. On a pressure-volume diagram, the area under the path represents this boundary work.

Build the algebraic model

The monatomic ideal gas starts at 100 kPa and 1 L. Compare two paths to 200 kPa and 3 L. One first raises pressure at fixed volume and then expands at 200 kPa; the other expands at 100 kPa before raising pressure. Since one kPa litre equals one joule, their works are respectively 200×2=400 J and 100×2=200 J.

Check the original evidence packet

For the supplied monatomic model U=3pV/2, both paths have ΔU=1.5(200×3−100×1)=750 J. Therefore heat inputs are 1150 and 950 J. The difference in heat equals the difference in work while the state change is identical. A clockwise rectangle combining the high-pressure expansion with the low-pressure return does net work 200 J and returns to the original internal energy.

Challenge the domain and transfer

Isothermal means constant temperature; for an ideal gas that gives zero internal-energy change, not necessarily zero heat transfer. Adiabatic means zero heat transfer, not necessarily zero temperature change. An adiabatic expansion that does positive work lowers internal energy under this sign convention. Isochoric and isobaric identify fixed volume and pressure instead. These distinctions are conditions on a process; calling a material an ideal gas does not automatically make every process isothermal, adiabatic or reversible.

Worked original case

For final 150 kPa and 2 L, ΔU=300 J. The high-pressure route does 150 J work and absorbs 450 J heat; the low-pressure route does 100 J work and absorbs 400 J. The clockwise rectangle area is 50 J.

Bounded investigation

Draw both routes and a closed clockwise cycle for two selected endpoints. Keep ΔU, W and Q distinct and explain why equal endpoint temperatures would not generally fix the work or heat.

Changed-case numerical practice

Try the changed case before feedback

An adiabatic gas expansion does 80 J of work on its surroundings. Under ΔU=Q−W, find internal-energy change in J.

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 limits

Original content checked 30 October 2026 against OpenStax College Physics 2e. Packets are supplied or generated teaching evidence, not measured apparatus. No heating, pressure vessels, physical oscillators, sound exposure or construction is requested. Public checks and saved reports do not establish supervised practical skill, equipment safety or independent mastery.

Apply

For two paths between the same equilibrium states of the same ideal gas, the quantity guaranteed equal is:

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Which information belongs to the state and which requires a path?

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