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Full chapter index · Practice and worked checks · Thermodynamics subject root
Two ideal gases begin in the same state and finish in the same state. Must they have received the same heat? No. If they follow different pressure–volume paths, they can perform different amounts of work while undergoing the same change in internal energy.
This guide follows one invented monatomic ideal-gas model along three routes. We will calculate work, close the energy accounts, reverse a route and build a complete cycle. The purpose is to distinguish a state change from the transfers that produced it. All numerical paths are paper models, not apparatus designs or operating instructions.
Chapter index
Define the account · Chapters 1–2
1. Define the system and the work convention
2. Specify the same two endpoint states
Compare three paths · Chapters 3–5
3. First route: raise pressure before expanding
4. Second route: expand before raising pressure
5. Third route: increase pressure along a straight line
Repair and extend · Chapters 6–8
6. Repair the mistaken endpoint calculation
7. Reverse a route and keep the signs
8. Join two routes into a cycle
Verify the evidence · Chapters 9–10
9. Keep the second law and useful output separate
10. Verify the model with units and independent checks
Practise and continue · Chapters 11–12
11. Practise with changed endpoints
12. Questions, connected reading and sources
Choose a fixed quantity of gas as a closed system. No matter crosses its boundary. Ignore changes in the gas’s overall kinetic and gravitational potential energy. Assume the only work mode is pressure–volume boundary work.
Use ΔU = Q − W. Heat Q is positive into the gas, and work W is positive when done by the gas on its surroundings. Rearranging gives Q = ΔU + W.
For the paths we calculate, suppose the gas passes through well-defined equilibrium states in an ideal quasistatic, mechanically reversible limit. Its pressure then provides the pressure appropriate to the simple integral W = ∫p dV. This assumption excludes friction and other complications from the model.
In a strongly irreversible process, a plotted gas equilibrium pressure may not exist throughout the change. Boundary forces and the actual external-pressure history need attention. Writing an integral does not justify choosing the wrong pressure inside it.
State A has absolute pressure 100 kilopascals and volume ten litres. State B has absolute pressure 200 kilopascals and volume twenty litres. Both contain the same quantity of the same ideal monatomic gas.
For an ideal gas, pV = nRT. Therefore temperature at B is four times temperature at A: pBVB/(pAVA) = 4, since n and R are fixed. These endpoints do not belong to an isothermal change.
For this monatomic ideal-gas model, U = (3/2)nRT = (3/2)pV. The choice of zero for U does not affect its change. We use this relation only within its stated idealisation.
Check the units: one kilopascal times one litre equals one joule because 1,000 pascals times 0.001 cubic metre is one joule. Thus pAVA = 1,000 joules and pBVB = 4,000 joules.
The endpoint internal energies are 1,500 and 6,000 joules. Every path from A to B in this model has ΔU = 4,500 joules.
CHAPTER 3 OF 12 · Compare three paths
3. First route: raise pressure before expanding
Route H first raises pressure from 100 to 200 kilopascals while keeping volume at ten litres. It then expands from ten to twenty litres at 200 kilopascals.
The fixed-volume leg performs no pressure–volume boundary work because dV = 0. Pressure changes, but the boundary does not sweep out volume.
The expansion leg gives W = pΔV = 200×(20 − 10) = 2,000 joules. Total W for Route H is therefore 2,000 joules. The first law gives Q = 4,500 + 2,000 = 6,500 joules into the gas over the complete route.
The two legs can also be checked separately. On the fixed-volume leg, ΔU = (3/2)×10×(200 − 100) = 1,500 joules, so Q is 1,500. During the constant-pressure expansion, ΔU = (3/2)×200×10 = 3,000 joules and Q = 5,000. Their heat total reproduces 6,500.
CHAPTER 4 OF 12 · Compare three paths
4. Second route: expand before raising pressure
Route L expands from ten to twenty litres at 100 kilopascals, then raises pressure from 100 to 200 kilopascals at fixed volume twenty litres.
Now the only nonzero boundary work occurs at the lower pressure. W = 100×(20 − 10) = 1,000 joules. The endpoint internal-energy change remains 4,500 joules, so Q = 5,500 joules.
On the expansion leg, ΔU = (3/2)×100×10 = 1,500 joules and Q = 2,500. On the fixed-volume leg, ΔU = (3/2)×20×100 = 3,000 joules and Q = 3,000. Their sum is 5,500.
The final gas state is the same as for Route H. The surroundings’ energy-transfer history differs. Route H supplies an extra 1,000 joules of heat, and the gas delivers an extra 1,000 joules of work.
No energy disappears. The difference is exactly what conservation requires when ΔU is held fixed but W changes.
CHAPTER 5 OF 12 · Compare three paths
5. Third route: increase pressure along a straight line
Route D joins A and B with a straight line on a pressure–volume graph. Pressure rises linearly from 100 to 200 kilopascals as volume rises from ten to twenty litres.
For this specified linear relation, work is the trapezoid area: W = [(100 + 200)/2]×10 = 1,500 joules. Therefore Q = 4,500 + 1,500 = 6,000 joules.
The arithmetic mean pressure is exact here because the path is linear in volume. It is not a universal average to use for any curve between these endpoints. An isothermal ideal-gas path, for example, has a different shape and cannot connect these particular endpoints at one constant temperature.
| Invented A-to-B path | Work done by gas | Internal-energy increase | Net heat into gas |
|---|---|---|---|
| Raise pressure, then expand | 2,000 J | 4,500 J | 6,500 J |
| Expand, then raise pressure | 1,000 J | 4,500 J | 5,500 J |
| Straight pressure–volume line | 1,500 J | 4,500 J | 6,000 J |
The middle result is a property of this chosen line, not evidence that every physically possible route lies between the other two work values.
A common error is to calculate work as Δ(pV) = 4,000 − 1,000 = 3,000 joules. That quantity changes a product of state variables. It is not the same as integrating p dV.
For differentiable paths, d(pV) = p dV + V dp. The extra V dp term shows why the two operations differ. A change in pressure at fixed volume changes pV while performing no pressure–volume boundary work in our model.
Another error uses final pressure times total volume change regardless of the route. That happens to give Route H’s work because its expansion is at final pressure. It fails for Route L and the diagonal route.
The repair is concrete: mark every volume-changing leg and identify the pressure along that leg. Then add the signed areas. OpenStax’s work, heat and internal-energy chapter develops the underlying relationship. Our three-path comparison supplies an original numerical test of whether the distinction has become usable.
Follow Route H backwards from B to A. The gas first compresses at 200 kilopascals from twenty to ten litres, then loses pressure at fixed volume ten litres.
Work done by the gas is W = 200×(10 − 20) = −2,000 joules. The negative sign means the surroundings perform net pressure–volume work on the gas.
Internal energy changes by −4,500 joules because the endpoints are reversed. Therefore Q = −4,500 − 2,000 = −6,500 joules. Heat leaves the gas overall.
Check it verbally: work input adds 2,000 joules, heat loss removes 6,500, so stored internal energy falls by 4,500. That explanation protects the sign convention better than memorising that compression “uses a minus”.
Reversing the ideal path reverses its signed integrals. It does not establish that an actual frictional or uncontrolled process can be undone without leaving changes in the surroundings. The ideal comparison and a real operating history are different evidence claims.
Go from A to B using Route H, then return from B to A along reversed Route L. The gas returns to its starting state, so total ΔU is zero.
Net work is 2,000 − 1,000 = 1,000 joules. Net heat is 6,500 − 5,500 = 1,000 joules. The first law closes: zero = 1,000 − 1,000.
On the pressure–volume plot, the clockwise rectangular cycle encloses area (200 − 100)×(20 − 10) = 1,000 joules. Traversing the same loop anticlockwise reverses the sign of the net work.
This is a cycle result, not an instruction to equate work with the enclosed area for an unfinished open path. For a path that does not return, work is the signed area under that path with the appropriate limits.
Nor does zero ΔU mean no heat or work crossed the boundary. It means the stored-state change over the complete cycle is zero. Large transfers can occur while their net effect on storage cancels.
CHAPTER 9 OF 12 · Verify the evidence
9. Keep the second law and useful output separate
An energy balance is necessary for a proposed process, but it does not establish a complete engine design or its physical implementation. The heat exchanges need compatible thermal surroundings, and irreversibility has to be addressed in a fuller model.
Our quasistatic paths can be discussed as ideal controlled comparisons. They do not identify a pair of fixed-temperature reservoirs or establish a Carnot efficiency. Substituting arbitrary temperatures into a two-reservoir formula would add unsupported conditions.
The calculated W is total gas boundary work under the stated pressure convention. It is not automatically useful shaft output after atmospheric displacement, friction and auxiliary demands. When comparing an actual machine claim, ask which work crosses which boundary.
For that next layer, follow finite bodies, entropy generation and heat flow and the broader thermodynamics science guide. They keep conservation and direction connected without treating them as the same test.
CHAPTER 10 OF 12 · Verify the evidence
10. Verify the model with units and independent checks
Plot volume horizontally and absolute pressure vertically. Label A and B, then draw each route separately. A vertical leg has zero horizontal width and contributes zero boundary-work area.
For each path, maintain columns for ΔU, W and Q. Check Q − W = ΔU both leg by leg and over the complete route. A total can accidentally be correct even when two leg errors cancel.
Keep pressure units consistent. Gauge pressure describes pressure relative to a reference; it is not interchangeable with absolute pressure in the ideal-gas equation. Replacing one by the other changes both the state interpretation and the total-work account.
A spreadsheet can integrate small volume increments along the diagonal and reproduce 1,500 joules. That checks arithmetic under the specified linear path. It does not verify an actual gas’s equation of state, frictionless motion or measured pressure history. Separate the correctness of the calculation from evidence that the model fits a physical system.
Check one: a new rectangle
For the same ideal monatomic model, take A at 80 kilopascals and five litres, and B at 160 kilopascals and fifteen litres. The pV change is 2,400 − 400 = 2,000 joules, so ΔU = 3,000 joules.
Expanding after raising pressure gives W = 160×10 = 1,600 joules and Q = 4,600. Expanding before raising pressure gives W = 800 and Q = 3,800. The straight line gives W = 1,200 and Q = 4,200 joules.
Check two: close the new cycle
Use the high-pressure outward route and reverse the low-pressure route. Net work is 800 joules, equal to the rectangle’s area 80×10. Net heat is also 800, while ΔU is zero.
Check three: identify missing information
An account gives only the two endpoint states and asks for heat supplied. You can calculate ΔU with the supplied gas model, but Q also needs work or an adequate process description. State what is missing rather than choosing a familiar path silently.
CHAPTER 12 OF 12 · Practise and continue
12. Questions, connected reading and sources
Why is internal-energy change the same on every route?
Internal energy is a state function. For the same system and endpoints its change is fixed. Heat and work describe transfers along a process, so their separate values can differ.
Does fixed volume mean no energy transfer?
It means zero pressure–volume boundary work. Heat or other work modes can still transfer energy. This article excluded other work modes explicitly.
Can I use average endpoint pressure for every curve?
No. It gives the exact integral for our linear pressure–volume path. A different curve needs its own integral or justified numerical approximation.
Which guide comes next?
Return to The Gold Standard of Thermodynamics for systems, signs and energy accounting. The two existing science treatments—heat, work and energy and energy, entropy and conversion limits—provide broader conceptual routes. For a compression application, read bicycle pumps, pressure, volume and work.
For other energy-account questions, follow changing heat capacities, mixing when the container also warms and unequal operating mixes in a heat-pump comparison. For direction beyond the first law, use finite bodies and entropy generation. Thermal engineering connects physical models with performance evidence.
The Gold Standard of Education and How X Works extend the habit of identifying what remains the same when the process changes.
OpenStax’s work, heat and internal-energy chapter supports pressure–volume work and the monatomic ideal-gas energy model. Its first-law chapter and thermodynamic-processes chapter supply the wider distinctions. The endpoints, three-route comparison, leg accounts and transfer exercises are original numerical teaching examples.
Sources checked 21 October 2026. The worked values are educational models with their assumptions stated in the article.