A pressure–volume (PV) diagram lets you determine how a gas changes state, identify the thermodynamic process, and calculate work from the path itself. This guide gives you a reusable checklist for reading the axes, interpreting arrows, finding work and heat, and checking answers with the first law of thermodynamics.
Overview
A PV diagram plots pressure, P, on the vertical axis against volume, V, on the horizontal axis. Every point represents an equilibrium state of the system. A path between two points represents a process: the gas changes from an initial state to a final state while pressure and volume may vary.
For this article, use the common physics convention that W is the work done by the gas and Q is heat added to the gas. With that convention, the first law is:
ΔU = Q − W
Equivalently, Q = ΔU + W. If your course defines W as work done on the gas, the signs change, so check the convention before beginning a problem.
The work done by the gas during a process is represented by the signed area under the path:
W = ∫ P dV
Expansion means volume increases, so dV is positive and work done by the gas is positive. Compression means volume decreases, so work done by the gas is negative. A vertical path has no change in volume and therefore zero work, even if the pressure changes substantially.
Checklist by scenario
1. Read the diagram before calculating
- Identify the horizontal and vertical axes. Confirm whether volume is measured in cubic metres, litres, or another unit, and whether pressure is in pascals, kilopascals, or atmospheres.
- Locate the initial and final states. Use the arrow direction rather than assuming that the left-hand point is always the initial state.
- Record the coordinates of important points: Pi, Vi, Pf, and Vf.
- Look at the shape of the path. A horizontal line indicates constant pressure; a vertical line indicates constant volume. A curved path may represent an isothermal or adiabatic process, but the diagram alone may not distinguish them without additional information.
2. Identify the process type
| Process | What stays constant | PV-diagram shape | Useful work expression | Typical clue |
|---|---|---|---|---|
| Isobaric | Pressure | Horizontal line | W = P(Vf − Vi) | Pressure has the same value at every point on the path. |
| Isochoric | Volume | Vertical line | W = 0 | The system is heated or cooled without changing volume. |
| Isothermal | Temperature | Curved hyperbola for an ideal gas | W = nRT ln(Vf/Vi) | For a fixed amount of ideal gas, PV remains constant. |
| Adiabatic | No heat transfer, Q = 0 | Curve, usually steeper than an isotherm | W = −ΔU | The process is insulated or occurs too quickly for significant heat exchange. |
The table gives common idealized descriptions. A curved path should not automatically be labelled isothermal or adiabatic: use the wording of the question, a stated equation, or the relationship between pressure and volume to decide.
3. Find work from the path
For a horizontal isobaric path, calculate the rectangle’s signed area. For example, if a gas expands from 0.20 m3 to 0.50 m3 at a constant pressure of 100 kPa:
W = PΔV = (100,000 Pa)(0.30 m3) = 30,000 J
The result is positive because the path moves to the right. Remember that pressure must be converted to pascals when volume is in cubic metres so that the answer is in joules.
For a curved path, the work is the area under the curve. If the curve has a known equation, integrate it. If the graph is made from straight-line segments, calculate the area of each rectangle or trapezoid and add the signed contributions. A path below another path between the same volumes produces less work during expansion because it encloses less area under the curve.
4. Apply the first law
Once you know work, determine the internal-energy change from the temperature change or from a supplied relation. For an ideal gas, internal energy depends only on temperature, so:
ΔU = nCVΔT
Then use Q = ΔU + W. For an adiabatic process, Q = 0, so ΔU = −W. During adiabatic expansion, work done by the gas is positive and internal energy decreases, which generally means the gas cools. For an isochoric process, W = 0, so all heat transferred changes the internal energy: Q = ΔU.
5. Handle a closed cycle
A closed loop returns the system to its starting state. Therefore, state variables such as internal energy have no net change: ΔUcycle = 0. The net work is the signed area enclosed by the loop, and the first law gives:
Qnet = Wnet
A clockwise cycle usually represents positive net work done by the gas: expansion occurs at higher average pressure than compression. A counterclockwise cycle usually represents negative net work by the gas, meaning net work is done on the gas. Trace the arrows rather than relying only on the visual direction.
What to double-check
- Units: Convert kilopascals to pascals when using cubic metres. The product Pa·m3 equals joules.
- Direction: Rightward motion is expansion and gives positive work by the gas; leftward motion is compression and gives negative work by the gas.
- Area type: Work is the area under the process path, not automatically the area enclosed by two states. For a cycle, use the enclosed area with the correct sign.
- State versus path: Pressure, volume, temperature, and internal energy are state variables. Work and heat depend on the path taken between states.
- Temperature: If using an ideal-gas relationship, use kelvins rather than degrees Celsius.
- Process evidence: Do not infer an adiabatic or isothermal process from curve shape alone. Look for a stated condition such as Q = 0, constant temperature, or a specified equation.
- Sign convention: Write your convention at the top of the solution: “W is work done by the gas” or “W is work done on the gas.”
For a broader approach to setting up physics solutions, see How to Solve Physics Problems: A Step-by-Step Method. If the PV diagram comes from an experiment, uncertainty in pressure, volume, and graph area should also be considered; the physics lab uncertainty guide provides a useful framework.
Common mistakes
Calling every horizontal line an isothermal process: A horizontal line is isobaric because pressure is constant. Temperature may change as volume changes.
Using the change in pressure to calculate work directly: Work depends on P dV. A large pressure change at constant volume still gives zero work.
Forgetting that compression gives negative work by the gas: If the arrow points left, the signed area contributes negatively under the stated convention.
Using the area inside a noncyclic path: For a single process, calculate the area under that process path. The area between two paths matters when comparing work or analysing a cycle.
Mixing first-law conventions: The equations ΔU = Q − W and ΔU = Q + W can both appear in textbooks because they define work differently. The algebra is not the problem; switching conventions halfway through is.
When to revisit
Use this checklist whenever you encounter a new PV graph, especially before solving an exam-style thermodynamics problem or checking a lab calculation. Revisit the process table when a question introduces a new constraint, such as a rigid container, thermal insulation, a constant-temperature bath, or a multistep path.
Before submitting a solution, run this final sequence: read the axes, mark the arrow, label the states, identify the process, convert units, calculate signed work, find ΔU, apply the first law, and check whether the signs make physical sense. For a closed cycle, add one more step: confirm that the system returns to its initial state and compare the calculated net work with the loop’s direction and enclosed area.
As course notation, calculator tools, or assessment requirements change, update your personal physics notes—but keep this underlying reasoning stable. A PV diagram is not just a picture: it is a compact record of states, a path, energy transfer, and the work performed during the process.
