Audit a complete thermal cycle before claiming useful work

Establish the model and evidence

A cyclic device returns its working gas to the starting state, so its net internal-energy change is zero. That statement does not mean every stage has zero ΔU. Compute heat and work stage by stage, then add the signed values. Forgetting one cooling or compression segment can create a fictitious device that produces work without a complete energy account.

The rectangular PV path begins at low pressure and small volume, heats at fixed volume, expands at high pressure, cools at fixed volume and compresses at low pressure. The clockwise enclosed area is net work delivered. Vertical legs contribute no volume work but generally do exchange heat. Heat absorbed during warming must not be confused with net heat over the whole cycle.

The stage ledger uses the work-on convention: ΔU=Q+Won. Summing it gives zero=Qnet+Won,net. Efficiency, shown as an optional synthesis quantity, divides net delivered work by total positive heat input, not by net heat input. The latter would misleadingly give unity for every ideal closed cycle with positive work.

The supplied hot and cold reservoirs bracket all modeled gas temperatures. Numerical reversible-efficiency comparisons are enrichment beyond the framework’s qualitative second-law boundary. The required physical conclusion is that a cyclic device cannot take heat from one reservoir and convert all of it to work without another effect. Real engine construction, friction losses and operation are not modeled or requested.

Worked evidence

At pressure and volume ratios 1.6, enclosed area is(160−100) kPa×(4.8−3.0) L=108 J. Stage heats are 270, 720,−432 and−450 J; total input 990 J, rejection 882 J and net 108 J. Efficiency is 108/990≈10.91%, not 100%.

Predict and investigate

Predict whether doubling a cycle’s heat input alone is enough to infer twice its useful work.

Capture ratios 1.6/1.6 and reconstruct all four stage ledgers. Then change only pressure ratio and compare enclosed area, total positive heat and net work. Explain why zero net ΔU cannot be applied separately to every stage.

Transfer to changed conditions

A proposed cycle absorbs 500 J from one reservoir, rejects no heat and delivers 500 J of work. Does satisfying the first law make it possible?

Compare the worked transfer

The energy balance alone is consistent, but the single-reservoir, all-heat-to-work cyclic claim violates the qualitative second law. A complete explanation needs both laws.

Explain the boundary

Identify the denominator of efficiency and explain why net heat is the wrong one. Give an energy-balanced proposal that the qualitative second law would still reject.

Sources and authorship

Physical relationship reference. Current AP Physics 2 framework, checked 30 October 2026. The numerical cases, text and computed representations are original. All browser evidence is synthetic or calculated, not a record of completed practical work. No hazardous apparatus is required.