The area under a pressure-volume path gives quasi-static expansion work. Equal endpoints can have different heat and work transfers.
By the end: compare paths with the same endpoint states and distinguish isothermal from adiabatic changes.
Turn piston motion into work
A gas pushing a piston does work equal to force times displacement. Pressure is force per area, and piston displacement times area is the change in volume. A quasi-static process changes slowly compared with internal equilibration, so its intermediate states are nearly in equilibrium. Over a small such step, work is pressure times volume change. Add the contributions along the path: on a pressure-volume diagram the result is the area under that path between starting and ending volumes. Expansion does positive work in our convention.
Reversible means that reversing arbitrarily small changes can restore both the system and surroundings without a remaining net change. Slow motion alone does not ensure this if friction is present. The pressure used for boundary work is the opposing external pressure. In a mechanically reversible, quasi-static limit it is arbitrarily close to the gas pressure, so the equilibrium gas curve can calculate the work. For a violent expansion into a vacuum, a continuous equilibrium gas-pressure curve is not the actual path and cannot be substituted for the external pressure.
Build an area from force and distance
Choose the gas alone as the closed system, with the piston outside its boundary. Imagine a piston face of area 0.010 square metres moving outward by 0.10 metres against a constant absolute opposing pressure of 100,000 pascals. The opposing force is pressure times area, or 1,000 newtons. Multiplying that force by the displacement gives 100 joules of work output. The gas loses energy through mechanical transfer even if its boundary admits no heat during this displacement.
The same geometry gives a volume increase of 0.001 cubic metres, which is one litre. Thus 100 kilopascals acting through one litre gives 100 joules. This conversion follows because a pascal is a newton per square metre. Multiplying pressure by volume leaves newtons times metres, the unit of work. On a graph with pressure vertically and volume horizontally, a rectangle's height times width has exactly these units. A picture of area therefore represents an energy, not a force alone.
Piston geometry gives the same work
Schematic piston and swept volume, not equipment dimensions. The stated area and displacement determine the volume change; 100 kPa is absolute opposing pressure.
Now reverse the imagined displacement while keeping the opposing pressure constant. The final volume minus the initial volume becomes negative. Work done by the gas is then minus 100 joules. For an insulated gas with no other transfers, the first law gives an internal energy increase of 100 joules. The negative sign does not mean energy disappeared. It means the mechanical transfer points into the selected system rather than outward. Choosing the piston and gas together would require a different boundary account.
If the opposing pressure changes, one large rectangle no longer gives the exact work. Divide the volume interval into smaller intervals, each with its own nearly constant pressure. Compute pressure times volume increase for every interval and add the results. A falling pressure curve lies below rectangles drawn at each interval's starting pressure and above rectangles drawn at its ending pressure. These two sums bound the work. Narrower intervals tighten the bounds without changing the physical meaning of each little area.
A gas pressure curve needs an extra assumption. Internal equilibrium must be maintained closely enough that a single gas pressure describes each intermediate state. Even then, work crossing a moving boundary depends on the opposing load, not automatically on a pressure measured somewhere inside the gas. In the frictionless reversible limit the pressure difference tends to zero, so the gas curve and opposing pressure give the same limiting area. Friction can leave energy dissipation even when the motion is slow.
A falling curve lies between rectangle sums
Pressure and volume axes are quantitative. The two 5 L rectangles use starting or ending pressures on p = 1000/V. Shading is work area, not gas volume.
Use the rectangle drawing to test a change of geometry before calculating again. If the piston area doubles but its displacement halves, the volume increase stays one litre. At the same opposing pressure, the force doubles and the distance halves, so the work remains 100 joules. If instead the displacement stays fixed, both volume increase and work double. This comparison explains why a pressure-volume description can omit piston shape while retaining the mechanical energy account under the stated boundary assumptions.
Keep the endpoints fixed while changing the route
Start with the sample at 10 L, 100 kPa, 300 K and end at 20 L, 50 kPa, 300 K. An isothermal quasi-static expansion keeps temperature fixed throughout. Pressure falls as volume grows; the work is about 693.15 J. The natural logarithm ln appears when adding the areas under this inverse-volume curve. It is the logarithm to base e, about 2.718; ln 1 is zero and ln 2 is about 0.693.
Alternatively, cool at fixed volume to 50 kPa, then expand at that constant pressure. The vertical segment does no boundary work and the horizontal segment does 500 J. Expanding first at 100 kPa, then cooling at fixed final volume, gives 1,000 J. Every route has the same final internal energy as the start, so net heat input equals work output on each route. The intermediate heat transfers need not all be inward.
Follow the energy through the intermediate state
Endpoint temperature does not tell you whether the gas stayed warm throughout a route. On the cool-first route, the intermediate state has volume 10 litres and pressure 50 kilopascals. Its pressure-volume product is 500 joules, half the initial product. Since the amount of ideal gas is fixed, its intermediate temperature is 150 kelvin. This is a different equilibrium state from either endpoint, despite belonging to a route whose initial and final temperatures both equal 300 kelvin.
For the classical monatomic sample, internal energy is three halves of the pressure-volume product. Cooling to that intermediate state therefore changes internal energy from 1,500 to 750 joules. No piston displacement occurs during the constant-volume leg, so boundary work is zero. Heat entering must equal the internal energy change, minus 750 joules. During the next leg, expanding at 50 kilopascals through 10 litres does 500 joules of work while internal energy rises again by 750 joules.
The routes visit different temperatures
State locations use pressure and volume scales. Connecting arrows are omitted; intermediate temperatures follow the fixed amount of ideal gas.
The second leg consequently requires 1,250 joules of heat input. Add 750 joules to replace the internal energy rise and another 500 joules to supply the work output. Summing both legs gives net heat input of 500 joules, even though the total inward heat during the second leg is much larger. An unsigned sum of heat transfers would lose the first leg's outward transfer. The signs in Q and W describe directions across the boundary, not labels for heating equipment.
The expand-first route visits a warmer intermediate state instead. At 20 litres and 100 kilopascals, the pressure-volume product is 2,000 joules and temperature is 600 kelvin. Internal energy reaches 3,000 joules. Expansion adds 1,500 joules to internal energy while delivering 1,000 joules of work, so heat input is 2,500 joules. Cooling at fixed final volume then releases 1,500 joules. Net input is 1,000 joules, twice the net input on the cool-first route between the same endpoints.
These accounts use whole-sample heat capacity, not heat capacity per mole or per kilogram. The sample has constant-volume capacity 5 joules per kelvin and constant-pressure capacity 25/3 joules per kelvin. Warming the cool intermediate state by 150 kelvin at constant pressure takes 1,250 joules, as the direct energy account predicted. The difference between those capacities supplies expansion work. A capacity quoted per mole would first need multiplication by the sample's amount before it could enter this calculation.
Signed transfers for each route
Route and leg
ΔU, J
W out, J
Q in, J
Cool first: fixed V
−750
0
−750
Then expand at 50 kPa
+750
+500
+1,250
Cool-first net
0
+500
+500
Expand first at 100 kPa
+1,500
+1,000
+2,500
Then cool at fixed V
−1,500
0
−1,500
Expand-first net
0
+1,000
+1,000
Exact energy accounts in joules for the classical monatomic sample. Net heat includes heat leaving as a negative transfer.
The temperature labels also reveal where the molecular assumptions matter. The arithmetic assumes a classical monatomic ideal gas with constant capacities throughout each leg. It does not establish that an arbitrary real gas remains dilute, avoids condensation, or has those capacities at the colder intermediate state. Quantum energy spacings can change available energy storage outside a classical range. The route comparison is an exact account inside the stated model, rather than a material-property guarantee for a named gas or a physical apparatus.
Work through the mechanism
Constant opposing pressure
Expansion is positive. In the quasi-static examples, the opposing pressure is arbitrarily close to the gas pressure.
Reversible isothermal ideal-gas expansion
The logarithm adds the changing-pressure contributions under the curve. For the sample, nRT = 1,000 J and doubling gives 1,000 ln 2 ≈ 693.15 J.
Reversible monatomic adiabatic expansion
This assumes an ideal gas with constant monatomic heat capacities and a reversible adiabatic route. The exponent is gamma minus one. Doubling volume from 300 K gives about 188.99 K. It does not apply to irreversible free expansion, and its endpoint differs from the isothermal comparisons below.
Compare paths between the same states
End state: 20 L, 50 kPa, 300 K. ΔU is zero on both paths. Net Q equals W: 693.15 J for the reference; 500 J for the selected path.
Both pressure-volume panels use identical scales and retain the isothermal reference. The shaded area is positive expansion work between A and B; vertical segments do no boundary work. A is 10 L, 100 kPa, 300 K; B is the selected volume at 300 K. Path and volume controls replace the scenario immediately, without simulating physical time.
Close a pressure-volume loop
This rectangular quasi-static cycle joins the expand-first route to the reverse of the cool-first route. It is not a Carnot cycle. Expansion at 100 kPa gives +1,000 J; compression at 50 kPa gives −500 J. Vertical legs do no boundary work. The enclosed area gives 500 J net work, and the completed cycle requires 500 J net heat input because ΔU is zero.
Route to 20 L, 50 kPa
Work output
Net heat input
Isothermal
693.15 J
693.15 J
Cool first, then expand
500 J
500 J
Expand first, then cool
1,000 J
1,000 J
All three routes start and end at 300 K, so ΔU is zero. Work uses quasi-static boundary pressures.
Use the path controls to test a prediction
Select the double-volume preset and the cool-first route in the existing path comparison. Before reading its work result, identify the horizontal leg's height as 50 kilopascals and its width as 10 litres. Their product predicts 500 joules. Switch to expand-first without changing final volume. The endpoints remain fixed, but the horizontal leg rises to 100 kilopascals. Its rectangle now predicts 1,000 joules. The reference curve remains between these two routes because its pressure falls continuously between their horizontal pressures.
Keep double volume selected and divide the isothermal reference into two intervals on paper, from 10 to 15 litres and from 15 to 20 litres. Along this curve, pressure equals 1,000 divided by volume in litres, in kilopascals. The three boundary pressures are therefore 100, about 66.67, and 50 kilopascals. Starting-pressure rectangles give an upper estimate of about 833.33 joules. Ending-pressure rectangles give a lower estimate of about 583.33 joules. The displayed 693.15 joules lies inside those bounds.
Splitting into four equal intervals tightens the estimates. At 10, 12.5, 15, 17.5, and 20 litres, the pressures are 100, 80, about 66.67, about 57.14, and 50 kilopascals. Each rectangle is 2.5 litres wide. The starting-pressure sum is about 759.52 joules and the ending-pressure sum is about 634.52 joules. These bounds explain the logarithm. It adds the limiting sequence of changing-pressure contributions instead of replacing the whole route with a single pressure.
Next choose triple volume while keeping the same route selection. Predict the final pressure before reading it. The initial pressure-volume product is 1,000 joules, so 30 litres at 300 kelvin requires about 33.33 kilopascals. The low rectangle has width 20 litres and gives about 666.67 joules. The high rectangle gives 2,000 joules. The reference gives about 1,098.61 joules. Its extra work for the second ten litres is smaller than for the first ten because the opposing pressure is lower.
Narrower intervals tighten the work bounds
Horizontal positions use a common energy scale, with a truncated axis starting at 500 J. These are area-sum bounds, not uncertainty in the exact formula.
This diminishing extra work does not mean the gas has less internal energy at the triple-volume endpoint. For this fixed amount of ideal gas, internal energy depends only on temperature, which remains 300 kelvin at both endpoints. The reference needs heat input equal to each work contribution throughout its isothermal expansion. The other routes can store and release energy on intermediate legs. The controls change the proposed route account immediately, not a physical elapsed time or a rate of heat supply.
Finally compare the two-volume and three-volume settings while watching the axes, not just the numerical work. The horizontal distance grows, and the final pressure falls. In the low route these changes partly oppose each other, while the high route retains its original pressure. That is why their work values do not grow in the same proportion. You can now predict the ordering for any displayed volume ratio above one. Low-route work stays below the reference, which stays below the high-route work.
Equal volume increments give unequal work
Bar lengths share a zero-based joule scale. Each bar is reversible isothermal work at 300 K over a 10 L interval.
Do not equate constant temperature with no heat
Isothermal means temperature stays constant. Adiabatic means no heat crosses the boundary. During a reversible isothermal expansion of an ideal gas, heat replaces the energy leaving as work. During an insulated expansion against a load, work comes from internal energy, so a classical ideal gas cools. For a reversible monatomic adiabatic path, pressure falls faster than on an isotherm. Doubling volume from the initial 300 K state gives about 188.99 K on that reversible adiabatic route, rather than the 300 K endpoint of the isothermal routes shown below.
An insulated ideal gas freely expanding into an evacuated chamber is a counterexample to the claim that every adiabatic expansion cools. The rigid outer boundary does no work and no heat enters, so initial and final U and T are equal. The rapid intermediate state need not have one temperature. A closed cycle returns to its starting U; its net work is the signed area enclosed by a quasi-static pressure-volume loop, not the total area under every leg added as positive.
Separate a process condition from a route law
Adiabatic describes a boundary condition, zero heat transfer. It does not specify opposing pressure or guarantee a reversible curve. For the reversible monatomic example, doubling volume lowers temperature from 300 to about 188.99 kelvin. With whole-sample constant-volume capacity 5 joules per kelvin, internal energy falls by about 555.06 joules. Since Q is zero, positive work output must equal that loss. This gas reaches a colder endpoint than any of the three 300 kelvin routes in the path comparison.
Insulated free expansion uses the same zero-heat condition but has no opposing pressure to push through the newly accessible space. For the rigid combined container, external boundary displacement is zero as well. Thus work output and heat input both vanish. Internal energy stays at 1,500 joules and final equilibrium temperature remains 300 kelvin in this ideal-gas model. The initial and final volumes can match the reversible isothermal example even though the actual energy transfers match neither that reference nor the reversible adiabatic expansion.
During free expansion the gas need not pass through uniform equilibrium states. Drawing an isotherm between its endpoints is useful as a reference calculation, but that curve is not a measured history of its pressure. The reference can later calculate the gas entropy difference because entropy is a state function. Actual heat transfer cannot do that job by itself here. It is zero, while the gas entropy increases. The irreversible change creates entropy within the isolated total rather than importing it as heat.
Zero heat does not select a unique endpoint
Schematic boundary and energy-flow comparison. Both routes double occupied volume. Only the reversible adiabatic route does work and cools in this ideal-gas model.
Now follow the rectangular cycle already shown. Expansion at 100 kilopascals through 10 litres supplies 1,000 joules. Compression at 50 kilopascals through minus 10 litres contributes minus 500 joules. The vertical legs contribute no boundary work. Net work is therefore 500 joules, equal to the enclosed rectangle of height 50 kilopascals and width 10 litres. The area beneath the compression leg must be subtracted. Adding both unsigned areas would falsely report 1,500 joules for the completed cycle.
Reversing the direction around that same rectangle reverses the work sign. The system then receives net mechanical work of 500 joules and must release net heat of 500 joules to return to its initial internal energy. A closed loop guarantees zero change in every state function after a cycle, but it does not guarantee zero net transfers. Nor does a clockwise loop establish a Carnot engine, reversible heat exchange with two reservoirs, a cycle duration, or an efficiency for a real device.
As a final diagram check, imagine attaching an insulated label to the entire rectangular cycle without changing any state or work value. That claim fails the first law. Returning to the initial internal energy requires net Q equal to the nonzero net W, whereas perfect insulation would require Q equal to zero. The contradiction concerns the cycle, even if one leg is adiabatic. Naming a process condition is meaningful only when its transfer account agrees with all the stated endpoints and boundaries.
Direction sets the signed loop area
Quantitative pressure-volume geometry. Arrows reverse while the enclosed area remains 50 kPa × 10 L = 500 J. These are not Carnot cycles.
Exercises
Check both stages
For the cool-first route to 20 L, calculate the heat in each stage. The initial monatomic U is 1,500 J.
Hint
At the intermediate state, p = 50 kPa and V = 10 L. Use U = 3pV/2.
Worked answer
Stage
ΔU
W
Q
Cool at fixed V
−750 J
0 J
−750 J
Expand at fixed p
+750 J
+500 J
+1,250 J
Both stages
0 J
+500 J
+500 J
Each row obeys Q = ΔU + W.
Cooling at 10 L lowers U to 750 J, does no boundary work, and releases 750 J as heat. Expansion at 50 kPa restores U to 1,500 J and does 500 J of work, requiring 1,250 J of heat input. Net heat is 1,250 minus 750, or 500 J. Net heat is not the same as total inward heat summed without the outward transfer.
Reject a false work path
The same ideal gas expands into a vacuum from 10 L to 20 L inside a rigid insulated container. Someone uses the isothermal curve to claim 693 J of work. What is wrong?
Hint
Identify the opposing pressure and the actual outer boundary motion.
Worked answer
The rigid outer container does not move. Removing the internal partition allows free expansion without boundary work. The markers represent the same amount of gas, and intermediate states need not be uniform.
Process
Q
W
Final T
Reversible isothermal expansion
693.15 J
693.15 J
300 K
Insulated free expansion
0 J
0 J
300 K
Same endpoints do not license using the same pressure path.
The opposing pressure is zero and the outer container is rigid, so boundary work is zero. Heat input is also zero. Internal energy, and therefore the equilibrium temperature of this ideal gas, are unchanged. The initial and final states match the reversible isothermal example, but the actual irreversible route does not. The isothermal curve’s area is not its work.
Sources and further reading
These optional references support the stated laws and model assumptions. The numerical examples and diagrams are authored teaching models, not experimental measurements or engineering specifications. Selected relevant passages were reviewed, not entire textbooks or a comprehensive research literature.