SCIENCE · INVESTIGATE AND EXPLAIN

Integrate distinct work paths between identical end states

Fix the same initial and final equilibrium states of a monatomic ideal gas: initial volume V0=1 litre and pressure p0, final volume rV0 and pressure p0/r. Both endpoints have the same pV and hence the same temperature. This does not force every intervening path to be isothermal.

Learning outcome: Compute both path areas before revealing the ledger, close the two paths into a conceptual cycle and assign work/heat signs. Explain the reservoir and quasistatic assumptions rather than claiming a performed engine.

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.

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Try

At p0=200 kPa and r=3, isothermal work is 200ln3≈219.722458 J. Straight-path work is 200×2×(1+1/3)/2≈266.666667 J. Their difference is about 46.944209 J; delta U is zero for both paths.

Open the experiment

Predict

What is fixed by the two endpoint states even when the path changes?

Your prediction and reason

Experiment

Compute both path areas before revealing the ledger, close the two paths into a conceptual cycle and assign work/heat signs. Explain the reservoir and quasistatic assumptions rather than claiming a performed engine.

Compare paths with the same equilibrium endpoints

Worked teaching model · synthetic values · Monatomic ideal gas with initial volume 1 litre, final volume r litres and final pressure p0/r. Compare quasistatic isothermal and straight-line pressure-volume paths. Heat reservoirs needed to realize each path are not simulated; work positive by the gas.

Final-to-initial volume ratio: 3 ratio Initial pressure: 200 kPa

Reset keeps your prediction, captures and practice history.

Same endpoint states, different path integrals

Graph axes rescale independently; read units and values, not screen angles. 1: isothermal (solid).2: straight pressure path (dashed). Joined curves show the supplied model, not additional observations.

Equivalent numerical coordinates for Same endpoint states, different path integrals

isothermal

volume (L)	pressure (kPa)
1	200
1.025	195.122
1.05	190.4762
1.075	186.0465
1.1	181.8182
1.125	177.7778
1.15	173.913
1.175	170.2128
1.2	166.6667
1.225	163.2653
1.25	160
1.275	156.8627
1.3	153.8462
1.325	150.9434
1.35	148.1481
1.375	145.4545
1.4	142.8571
1.425	140.3509
1.45	137.931
1.475	135.5932
1.5	133.3333
1.525	131.1475
1.55	129.0323
1.575	126.9841
1.6	125
1.625	123.0769
1.65	121.2121
1.675	119.403
1.7	117.6471
1.725	115.942
1.75	114.2857
1.775	112.6761
1.8	111.1111
1.825	109.589
1.85	108.1081
1.875	106.6667
1.9	105.2632
1.925	103.8961
1.95	102.5641
1.975	101.2658
2	100
2.025	98.76543
2.05	97.56098
2.075	96.38554
2.1	95.2381
2.125	94.11765
2.15	93.02326
2.175	91.95402
2.2	90.90909
2.225	89.88764
2.25	88.88889
2.275	87.91209
2.3	86.95652
2.325	86.02151
2.35	85.10638
2.375	84.21053
2.4	83.33333
2.425	82.47423
2.45	81.63265
2.475	80.80808
2.5	80
2.525	79.20792
2.55	78.43137
2.575	77.6699
2.6	76.92308
2.625	76.19048
2.65	75.4717
2.675	74.76636
2.7	74.07407
2.725	73.3945
2.75	72.72727
2.775	72.07207
2.8	71.42857
2.825	70.79646
2.85	70.17544
2.875	69.56522
2.9	68.96552
2.925	68.37607
2.95	67.79661
2.975	67.22689
3	66.66667

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.

straight pressure path

volume (L)	pressure (kPa)
1	200
1.025	198.3333
1.05	196.6667
1.075	195
1.1	193.3333
1.125	191.6667
1.15	190
1.175	188.3333
1.2	186.6667
1.225	185
1.25	183.3333
1.275	181.6667
1.3	180
1.325	178.3333
1.35	176.6667
1.375	175
1.4	173.3333
1.425	171.6667
1.45	170
1.475	168.3333
1.5	166.6667
1.525	165
1.55	163.3333
1.575	161.6667
1.6	160
1.625	158.3333
1.65	156.6667
1.675	155
1.7	153.3333
1.725	151.6667
1.75	150
1.775	148.3333
1.8	146.6667
1.825	145
1.85	143.3333
1.875	141.6667
1.9	140
1.925	138.3333
1.95	136.6667
1.975	135
2	133.3333
2.025	131.6667
2.05	130
2.075	128.3333
2.1	126.6667
2.125	125
2.15	123.3333
2.175	121.6667
2.2	120
2.225	118.3333
2.25	116.6667
2.275	115
2.3	113.3333
2.325	111.6667
2.35	110
2.375	108.3333
2.4	106.6667
2.425	105
2.45	103.3333
2.475	101.6667
2.5	100
2.525	98.33333
2.55	96.66667
2.575	95
2.6	93.33333
2.625	91.66667
2.65	90
2.675	88.33333
2.7	86.66667
2.725	85
2.75	83.33333
2.775	81.66667
2.8	80
2.825	78.33333
2.85	76.66667
2.875	75
2.9	73.33333
2.925	71.66667
2.95	70
2.975	68.33333
3	66.66667

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
Quantity	Value	Unit
Final pressure	66.66667	kPa
Isothermal work by gas	219.7225	J
Straight-line-path work by gas	266.6667	J
Internal-energy change for either path	0	J
Isothermal heat into gas	219.7225	J
Straight-line heat into gas	266.6667	J
Path-work difference	46.94421	J

Heat and work depend on the path even when the internal-energy change is fixed.

One kPa litre is one joule. For the isotherm integrate p0V0/V to obtain p0V0 ln r. For a straight pressure path use the trapezoid area (p0+pf)(Vf−V0)/2. Both endpoints have equal pV, so a monatomic ideal gas has delta U=3 delta(pV)/2=0. The first law Q=delta U+W makes each heat equal its own work. Intermediate temperatures need not be equal between paths.

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

Heat and work depend on the path even when the internal-energy change is fixed.

Observation to attach to a capture

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

Understand

Declare the distribution or system

Derive the integral or response

For a quasistatic isotherm p(V)=p0V0/V, work by the gas is the integral p dV=p0V0 ln r. For a straight pressure-versus-volume path joining the endpoints, integrate the affine pressure to obtain (p0+pf)(Vf−V0)/2. One kPa litre equals one joule, which makes the displayed numerical conversion transparent.

Audit limits and evidence

The monatomic ideal-gas relation U=3pV/2 gives zero internal-energy change between these particular endpoints. The first law with work positive by the system gives Q=delta U+W. Each path therefore needs heat equal to its own work; equal endpoint temperatures do not imply equal heat transfer.

Predict a changed premise

The straight path generally passes through different intermediate temperatures and requires an appropriate external pressure and heat-reservoir schedule to remain quasistatic. The controls compare mathematical paths, not a complete physical engine. Traversing a path backward reverses its work; closing the two paths into a cycle produces a signed enclosed area.

Worked original case

At p0=200 kPa and r=3, isothermal work is 200ln3≈219.722458 J. Straight-path work is 200×2×(1+1/3)/2≈266.666667 J. Their difference is about 46.944209 J; delta U is zero for both paths.

Investigation

Changed-case practice

Try the changed case before feedback

With p0=100 kPa and r=2, find straight-path work by the gas in joules.

Numerical result in the requested units

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

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?

Your explanation and limitations

Continue

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Previous: Integrate a temperature-dependent state-energy change Next: Test a nonideal branch before treating it as equilibrium

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