A P-V diagram is a graph with pressure on the vertical axis and volume on the horizontal axis that represents a gas's thermodynamic process or cycle, where the area under the curve equals the magnitude of work done on or by the gas (W = -PΔV for constant pressure).
A P-V diagram (pressure-volume diagram) plots a gas's pressure against its volume as the gas goes through a thermodynamic process. Each point on the graph is one state of the gas, and the path between points shows how the gas got from one state to another. Compress a gas and the point moves left. Let it expand and the point moves right. Heat it at constant volume and the point moves straight up.
The diagram's real power is geometric. The area under the curve equals the magnitude of the work involved in the process. The CED defines work done on a system by a constant external pressure as W = -PΔV, so when volume decreases (compression), W is positive and energy flows into the gas. When volume increases (expansion), the gas does work on its surroundings and W on the gas is negative. A P-V diagram turns the first law of thermodynamics into a picture you can read instead of a pile of equations.
P-V diagrams live in Topic 9.4 (The First Law of Thermodynamics) in Unit 9. They directly support learning objective 9.4.B, describing the behavior of a system using thermodynamic processes, and they feed into 9.4.A because the diagram lets you track changes in internal energy. Here's the core move. The first law, ΔU = Q + W, has three quantities, and the diagram hands you one of them for free. Read the work from the area under the curve, figure out ΔU from how temperature changed (U = 3/2 nRT for an ideal monatomic gas, and PV tells you about T through the ideal gas law), and solve for Q. Almost every Unit 9 quantitative problem runs through this graph, which is why it shows up constantly on both multiple choice and free response.
Keep studying AP® Physics 2 Unit 9
ΔU = Q + W (Unit 9)
The P-V diagram is basically the first law drawn as a picture. The area under the path gives you W, and the start and end points (through PV and temperature) tell you ΔU, so the diagram lets you solve for Q without measuring heat directly.
Thermodynamic cycle (Unit 9)
A cycle is a closed loop on a P-V diagram, and the gas ends exactly where it started. Since temperature returns to its initial value, ΔU = 0 for the full cycle, and the area enclosed by the loop equals the net work for one trip around.
Work done by a gas (Unit 9)
Same area under the curve, opposite sign convention. When a gas expands, it does positive work on its surroundings, which means the work done ON the gas is negative. Getting this sign straight is half the battle in first law problems.
Constant volume process (Unit 9)
On a P-V diagram, a constant volume (isochoric) process is a vertical line. No change in volume means zero area under the path, so W = 0 and every bit of heat added goes straight into internal energy (ΔU = Q).
Expect P-V diagrams in multiple choice as ranking or comparison tasks. You might be shown several paths between the same two states and asked to rank the work done, the heat added, or the change in internal energy. The key insight is that ΔU depends only on the endpoints, but W and Q depend on the path taken. On the free response side, P-V diagrams are a classic setup for quantitative reasoning and graph-sketching. You may be given a process (say, a piston compressing a gas at constant pressure) and asked to sketch the path, calculate work from the area, and apply ΔU = Q + W to find heat transfer. Always check the direction of the arrow on the path. Rightward (expansion) means the gas does work on its surroundings, and leftward (compression) means work is done on the gas.
Both graphs use area to find work, but they answer different questions. A force-displacement graph gives work done by a single force on an object. A P-V diagram gives work for a gas, with the twist of the negative sign convention. W = -PΔV is the work done ON the gas, so an expanding gas (area to the right) has negative work done on it, while a compressed gas has positive work done on it. On a force-displacement graph there's no built-in sign flip like that, which is why so many wrong answers come from importing the wrong sign habits.
A P-V diagram plots pressure against volume, and each point on it represents one state of the gas.
The area under the curve equals the magnitude of the work, with W = -PΔV giving the work done on the gas for a constant pressure process.
Compression (volume decreases) means positive work is done on the gas, while expansion means the gas does work on its surroundings.
A vertical line on a P-V diagram is a constant volume process, and since the area under it is zero, no work is done.
The change in internal energy depends only on the starting and ending states, but work and heat depend on the path taken between them.
For a closed loop (a thermodynamic cycle), ΔU = 0 and the net work equals the area enclosed by the loop.
It's a graph with pressure on the vertical axis and volume on the horizontal axis that shows how a gas changes state during a thermodynamic process. The area under the path equals the magnitude of the work done on or by the gas, which makes it the main tool for first law problems in Unit 9.
No. The area gives the magnitude of work, but the sign depends on direction. If the gas expands (path moves right), the work done on the gas is negative because W = -PΔV. If the gas is compressed (path moves left), work done on the gas is positive.
Both use area under the curve, but a P-V diagram has a built-in sign convention from W = -PΔV. Expansion means negative work done on the gas even though the area is a positive number, while a force-displacement graph has no such automatic sign flip.
Because the gas returns to its exact starting point, its pressure, volume, and therefore temperature are unchanged. Since internal energy of an ideal gas depends only on temperature (U = 3/2 nRT for monatomic), no net change in T means no net change in U, so Q = -W for the full cycle.
A vertical line is a constant volume (isochoric) process. Since ΔV = 0, the area under the path is zero and no work is done, so the first law reduces to ΔU = Q. All heat added goes directly into internal energy.
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