Why thermodynamic work is a path variable
Explain why thermodynamic work depends on the process path and calculate it from the area under a p–V curve.
Offline pages
Saved on this device only. Connect at least once every 30 days to confirm Premium. Sign-out or a change of account removes saved copies.
On this page
You stopped at:
ContinueThe initial and final gas states do not tell you how strongly the gas pushes throughout an expansion. A path specifies pressure at intermediate volumes. Integrating that pressure gives work; changing the path can change the result while leaving both endpoint states fixed.
At a glance
- Prerequisites: piston work and definite integrals as signed areas
- You will learn: calculate work for specified paths; compare routes sharing complete endpoints; explain why a closed path can transfer work
- Key result: for quasistatic pressure–volume work, Wₒᵤₜ = ∫_Vᵢ^VfP(V) dV, Wₒₙ = -Wₒᵤₜ.
- Common trap: treating work as an endpoint property, or assuming every different path must give a different work value.
Work output is positive when the gas delivers energy through work. Work on it has the opposite sign. With unchanged bulk energies, the first law is Δ U = Q + Wₒₙ = Q-Wₒᵤₜ. Here U is internal energy, also written as E elsewhere in the course.
Read the pressure–volume path
Consider a fixed amount and composition of ideal gas between complete equilibrium states i and f. Only piston work is considered here; all paths used for an equilibrium gas-pressure integral are quasistatic. Gas pressure represents the moving-boundary pressure, as in a massless frictionless piston model.
The sketch is qualitative and has no numerical scale. Path A lies above path B over the intermediate volumes, so its expansion area is larger. The numerical routes below are specified separately. In each comparison both routes share the same initial state and the same final state, but the final state need not equal the initial one.
Integrate the route, rather than subtracting states
For an outward boundary displacement, δ Wₒᵤₜ = P dV. On a chosen path, W_(out,i → f) = ∫_Vᵢ^VfP(V) dV. The infinitesimal transfer uses δ Wₒᵤₜ, rather than an exact state differential dW. There is no stored “work content” whose final value minus initial value gives this integral. Internal energy is different: Δ U = Uf-Uᵢ is fixed by the states.
For two monotonic expansions, W_(out,A)-W_(out,B) = ∫_Vᵢ^Vf[PA(V)-PB(V)] dV. If A has higher pressure throughout, its output is greater. If the curves cross, compare the signed areas between them. Extra area in one part can cancel a deficit in another, so different paths can give equal work.
Use matching units: Pa m³ = J or kPa L = J. Expansion at positive pressure gives positive output; compression reverses the integration limits. A shrinking volume interval makes the work tend to zero for bounded pressure, but a closed loop can return to its original volume after both expansion and compression and still give nonzero work.
A fast irreversible route can lack a single equilibrium pressure at intermediate states. Joining its endpoints by an invented gas-pressure curve does not establish its work. Use the actual boundary-pressure and piston-energy model instead. The OpenStax work treatment compares pressure–volume integrals for different quasistatic routes.
Two routes with different work
Worked example 1
Two paths between the same gas states
Problem
A fixed amount of ideal gas changes quasistatically through a massless frictionless piston from Pᵢ = 2.0 × 10⁵ Pa, Vᵢ = 1.0 × 10⁻³ m³ to Pf = 1.0 × 10⁵ Pa, Vf = 3.0 × 10⁻³ m³. On path A it first expands at Pᵢ, then its pressure falls at fixed Vf. On path B the pressure decreases linearly with volume between the two states. Find the work output along each path and show why the endpoints alone do not determine work.
Show full solution
Calculate the rectangular area for path A
Method
Add the work of the isobaric expansion and the constant-volume pressure change.Reason
Only a volume change contributes to ∫ P dV; the vertical segment contributes zero.Working
Answer
W_(out,A) = Pᵢ(Vf-Vᵢ) + 0 = (2.0 × 10⁵)(2.0 × 10⁻³) = 400 J.Integrate the linear pressure path
Method
Use the average of the endpoint pressures to find the trapezoidal area.Reason
For a linear P(V), the integral equals mean pressure times the volume interval. In general, if P = aV + b, it is (a/2)(Vf²-Vᵢ²) + b(Vf-Vᵢ).Working
Answer
W_(out,B) = ((Pᵢ + Pf)/2)(Vf-Vᵢ) = (1.5 × 10⁵)(2.0 × 10⁻³) = 300 J.Compare work for identical endpoints
Working
Answer
Path A gives 400 J and path B gives 300 J of work output, despite identical initial and final states. Their areas differ, establishing path dependence. Different pressure profiles can sometimes have equal areas, so a different path does not necessarily give a different work value.
Test whether different paths must give different work
Guided practice 2
Different pressure profiles with equal work
Problem
Two quasistatic expansion paths for the same fixed ideal-gas sample start at (V,P) = (1.0 L,200 kPa) and end at (3.0 L,200 kPa), with only piston work. Path A stays at 200 kPa. Path B consists of straight-line segments joining these points in order:
| Volume (L) | Pressure (kPa) |
|---|---|
| 1.0 | 200 |
| 1.5 | 300 |
| 2.0 | 200 |
| 2.5 | 100 |
| 3.0 | 200 |
Find path B’s output from 1.0 to 2.0 L, from 2.0 to 3.0 L, and over the whole expansion. Compare its total with path A. Does the comparison make work a state function?
Add the straight-line areas
Hints
Hint 1
Use each interval’s width
Each straight segment spans 0.50 L. Multiply that width by its own mean pressure, then sum the segments.Show solution step by step
Integrate the four segments
Method
Use the trapezoid rule exactly for each straight-line segment.Reason
The endpoint mean is the volume-average pressure on a linear segment.Working
Answer
The first two segments give 125 + 125 = 250 J. The next two give 75 + 75 = 150 J. Thus W_(out,B) = 250 + 150 = 400 J. For path A, W_(out,A) = 200(3.0-1.0) = 400 J.
Interpret the equality
Working
Answer
Relative to A, B adds 50 J of area in the first half and loses 50 J in the second. Their work happens to agree. This single equality does not make work a state function: the preceding example already gives unequal outputs between identical endpoints. Path dependence allows equality for some pairs; it does not require unequal work for every pair of different paths.
Return by a different route
Check your understanding 1
Use the first worked example’s endpoints: (Pᵢ,Vᵢ) = (200 kPa,1.0 L) and (Pf,Vf) = (100 kPa,3.0 L). Go outward by path A (expand at 200 kPa, then lower pressure at fixed final volume). Return by reversing path B, the straight pressure–volume line between those states. The full path is quasistatic and only piston work occurs. Find net output and net work on the gas. Does returning to the initial volume require zero work?
Show answer
Outward work is 400 J. Reversing the linear path changes its output from + 300 to -300 J. Thus W_(out,net) = 400-300 = 100 J and W_(on,net) = -100 J.
The state return gives Δ U = 0, not zero work. The enclosed loop area is 100 J, with expansion above compression. Check that your explanation uses the actual traversal direction, includes both paths and separates state change from transferred work. This mechanical reversal does not by itself establish thermodynamic reversibility.
Practice set
Check your understanding 2
A gas expands quasistatically at 2.0 × 10⁵ Pa from 1.0 × 10⁻³ m³ to 3.0 × 10⁻³ m³. Find work output and work on the gas for the first law.
Show hint
Show answer
Check your understanding 3
Two quasistatic monotonic expansions span the same volume interval. Path 1 has higher pressure at every intermediate volume than path 2. Which produces more output? What changes if both paths are instead traversed as compressions?
Show hint
Show answer
Check your understanding 4
Must every complete quasistatic cycle give zero net output because the final state equals the initial state?
Show hint
Show answer
Before you move on
Work depends on the pressure acting while volume changes, not just on the endpoints. Compare signed areas under the actual paths, including compression signs. Some different routes have equal work; closed routes can still deliver net output because no stored work property is being restored.
Spotted an error? Report a correction
Syllabus and review details
No official syllabus alignment is listed for this lesson.