P–V Diagrams

A pressure–volume diagram, or P–V diagram, is a graph used to show how the pressure and volume of a gas change during a thermodynamic process.

On a P–V diagram:

• pressure (P) is plotted on the vertical axis
• volume (V) is plotted on the horizontal axis

Each point on the graph represents a particular state of the gas.

By following a path from one point to another, we can determine how the gas changes and, importantly, how much work is done during expansion or compression.

Reading a P–V Diagram

Suppose a gas begins at point A and moves to point B.

If B is to the right of A:

volume increases → expansion

If B is to the left of A:

volume decreases → compression

If B is above A:

pressure increases

If B is below A:

pressure decreases

The direction of the arrow on the path is therefore important.

The same curve travelled in opposite directions represents a different physical process.

State vs. Process

A state describes the condition of the gas at a particular moment.

For example:

P = 200 kPa and V = 0.004 m³

A process describes the change from one state to another.

On a P–V diagram:

• a point represents a state
• a line or curve represents a process
• an arrow shows the direction of the process

A gas can sometimes move between the same initial and final states using different paths.

This becomes important when calculating work.

Constant-Pressure Processes

A process occurring at constant pressure is called an isobaric process.

Because pressure remains constant, it appears as a horizontal line on a P–V diagram.

If the gas moves to the right:

volume increases → gas expands

If it moves to the left:

volume decreases → gas is compressed

For constant pressure, the work done by the gas is:

W = pΔV

where:

ΔV = V₂ − V₁

Therefore:

W = p(V₂ − V₁)

Worked Example: Constant-Pressure Expansion

A gas expands at a constant pressure of 150,000 Pa from:

V₁ = 0.002 m³

to:

V₂ = 0.006 m³

First calculate the change in volume:

ΔV = 0.006 − 0.002

ΔV = 0.004 m³

Now:

W = pΔV

W = 150000 × 0.004

W = 600 J

The gas does 600 J of work on its surroundings.

Area Under a P–V Graph

One of the most important features of a P–V diagram is:

The area under the process curve represents the work done by the gas.

For a constant-pressure expansion, the area forms a rectangle.

Area = height × width

Therefore:

W = pΔV

This gives the same equation we derived from mechanical work.

Why Does Area Represent Work?

Pressure is measured in pascals:

Pa = N/m²

Volume is measured in:

m³

Therefore:

Pa × m³ = (N/m²)(m³)

= N·m

But:

N·m = J

Therefore:

Pa·m³ = J

The area under a pressure-volume graph naturally has units of energy.

Expansion and Compression

The direction of the process determines whether the gas does work or has work done on it.

Expansion

The gas moves from smaller volume to larger volume:

left → right

The gas pushes against its surroundings.

Using our previous First Law convention:

W > 0

The gas does work on the surroundings.

Compression

The gas moves from larger volume to smaller volume:

right → left

The surroundings compress the gas.

Therefore:

W < 0

when W represents work done by the gas.

Work is being done on the gas.

Constant-Volume Processes

A process occurring at constant volume is called an isochoric process or isovolumetric process.

On a P–V diagram, it appears as a vertical line.

Because:

ΔV = 0

the gas does no pressure-volume work.

Therefore:

W = 0

Even though pressure may change substantially, there is no expansion or compression.

Example

A gas is heated inside a rigid sealed container.

Its temperature rises and its pressure increases.

However, the container cannot expand.

Therefore:

ΔV = 0

and:

W = 0

Constant-Temperature Processes

A process occurring at constant temperature is called an isothermal process.

For an ideal gas:

PV = constant

Therefore, as volume increases, pressure decreases.

An isothermal process appears as a curved line on a P–V diagram.

During isothermal expansion:

• volume increases
• pressure decreases
• the gas does work
• the temperature remains constant

For an ideal gas, internal energy depends on temperature. Therefore, during an isothermal process:

ΔU = 0

From the First Law:

ΔU = Q − W

so:

0 = Q − W

Therefore:

Q = W

Energy transferred into the gas by heating equals the work done by the gas.

Adiabatic Processes

An adiabatic process occurs when no energy is transferred by heating between the system and its surroundings.

Therefore:

Q = 0

Using the First Law:

ΔU = Q − W

we get:

ΔU = −W

During adiabatic expansion:

• the gas does work
• internal energy decreases
• temperature decreases

During adiabatic compression:

• work is done on the gas
• internal energy increases
• temperature increases

An adiabatic curve on a P–V diagram is generally steeper than an isothermal curve for the same ideal gas starting from the same state.

Recognizing Common Processes

Being able to recognize these shapes makes P–V diagrams much easier to interpret.

Work Along a Curved Path

If pressure changes as the gas expands, we cannot generally use one pressure value in:

W = pΔV

Instead, the work corresponds to the area under the curve.

In more advanced mathematics this is written:

W = ∫P dV

The integral means that we add many very small pressure × volume-change contributions across the entire process.

At this level, the important idea is:

larger area under the P–V curve → more work done by the gas

Estimating Work from a Graph

Suppose pressure decreases approximately linearly from 300 kPa to 100 kPa while volume increases from 0.002 m³ to 0.006 m³.

Because the graph forms a trapezoid, we can use average pressure:

Average pressure = (300000 + 100000) / 2

Average pressure = 200000 Pa

Volume change:

ΔV = 0.006 − 0.002

ΔV = 0.004 m³

Work:

W = average pressure × ΔV

W = 200000 × 0.004

W = 800 J

The gas does approximately 800 J of work.

Comparing Different Thermodynamic Paths

Suppose a gas moves from the same initial state A to the same final state B using two different paths.

Path 1 has a higher pressure during most of the expansion.

Path 2 has a lower pressure.

Because work is the area under the P–V curve:

Path 1 has the greater area → more work is done.

This demonstrates an important thermodynamic idea:

Work depends on the path taken between two states.

Internal Energy Is Different

Although work depends on the path, the change in internal energy depends only on the initial and final states.

Suppose a gas moves from state A to state B.

Whether it follows Path 1 or Path 2:

ΔU is the same

provided the initial and final states are identical.

However:

W may be different

and therefore, from:

ΔU = Q − W

the amount of energy transferred by heating must also be different.

This distinction between state quantities and path-dependent quantities is fundamental in thermodynamics.

State Functions and Path Functions

Internal energy is a state function.

Its change depends only on:

• initial state
• final state

It does not depend on the route taken.

Heat transfer and work are path dependent.

Their values depend on how the system moves between the two states.

A useful analogy is climbing a mountain.

Your change in altitude depends only on your starting and finishing elevations.

But the distance you walk depends on the route you take.

Similarly:

ΔU → depends on endpoints

Q and W → depend on path

Multi-Step Paths

Some P–V diagrams contain several separate processes.

Suppose a gas follows:

A → B → C

From A to B, it expands at constant pressure.

From B to C, its pressure decreases at constant volume.

A → B

Horizontal movement means:

constant pressure

Volume increases, so the gas does work.

B → C

Vertical movement means:

constant volume

Therefore:

W = 0

The total work for the complete process is therefore just the area under the horizontal A → B section.

Worked Multi-Step Example

A gas begins at:

A: P = 200 kPa, V = 0.002 m³

It expands at constant pressure to:

B: P = 200 kPa, V = 0.005 m³

It then drops in pressure at constant volume to:

C: P = 100 kPa, V = 0.005 m³

Work from A to B

Convert pressure:

200 kPa = 200000 Pa

Volume change:

ΔV = 0.005 − 0.002

ΔV = 0.003 m³

Work:

W = 200000 × 0.003

W = 600 J

Work from B to C

Volume does not change.

Therefore:

W = 0

Total Work

W_total = 600 + 0

W_total = 600 J

Cyclic Processes

Sometimes a system eventually returns to its original state.

This is called a cycle.

On a P–V diagram, the path forms a closed loop.

Because the gas returns to its original state:

ΔU = 0 for the complete cycle

However, the gas may still do net work.

The area enclosed by the loop represents the net work done during the cycle.

This idea is extremely important in understanding:

• engines
• refrigerators
• heat pumps
• power stations

Clockwise and Counterclockwise Cycles

For the sign convention used in these notes:

Clockwise Cycle

Expansion generally occurs at higher pressure than compression.

The gas does more work during expansion than is done on it during compression.

Therefore:

net work is done by the gas

Counterclockwise Cycle

Compression occurs at higher pressure.

Therefore:

net work is done on the gas

The direction of the loop therefore provides useful information about the overall energy transfer.

Connecting P–V Diagrams to the First Law

Recall:

ΔU = Q − W

A P–V diagram gives us information about W.

Once we determine the work and know either Q or ΔU, we can calculate the remaining quantity.

Example

A gas expands and the area under its P–V curve represents 700 J of work.

During the process, 1,100 J is transferred to the gas by heating.

Therefore:

ΔU = Q − W

ΔU = 1100 − 700

ΔU = +400 J

The internal energy increases by 400 J.

Interpreting Physical Changes

A P–V diagram is more than a mathematical graph. Each movement corresponds to something physically happening to the gas.

Moving right

Gas expands.

Moving left

Gas is compressed.

Moving upward

Pressure increases.

Moving downward

Pressure decreases.

Horizontal path

Constant pressure.

Vertical path

Constant volume.

Curved path

Pressure and volume are both changing.

By combining this information with the First Law and the ideal gas model, we can build a much more complete description of what is happening.

Units on P–V Diagrams

Always inspect the axis units carefully.

Pressure may be shown in:

• Pa
• kPa
• MPa

Remember:

1 kPa = 1000 Pa

1 MPa = 1,000,000 Pa

Volume may be shown in:

• m³
• L
• cm³

Remember:

1 L = 1 × 10⁻³ m³

When calculating work in joules using pressure and volume:

use Pa and m³

because:

Pa·m³ = J

Common Mistakes

Thinking the slope represents work

Work is related to the area under the curve, not the slope.

Using the area beside the curve

For expansion or compression, work corresponds to the area between the process path and the volume axis over the relevant volume interval.

Calculating work for a vertical line

A vertical process has:

ΔV = 0

so:

W = 0

Ignoring the direction of the process

The same path can represent expansion or compression depending on the arrow direction.

Assuming work depends only on the endpoints

Work depends on the path taken.

Forgetting unit conversions

Using kPa and m³ without accounting for the prefix can produce an answer off by a factor of 1,000.

Confusing work with internal energy

The area under a P–V curve represents work, not the total internal energy of the gas.

Key Terms

P–V diagram — a graph showing pressure against volume for a thermodynamic system.

Thermodynamic state — the condition of a system described by variables such as pressure, volume, and temperature.

Thermodynamic process — a change from one state to another.

Isobaric process — a process occurring at constant pressure.

Isochoric process — a process occurring at constant volume.

Isothermal process — a process occurring at constant temperature.

Adiabatic process — a process in which no energy is transferred by heating.

Expansion — an increase in the volume of a gas.

Compression — a decrease in the volume of a gas.

Work — energy transferred mechanically; on a P–V diagram it is represented by the area under a process curve.

State function — a quantity whose change depends only on the initial and final states.

Path dependent — describing a quantity whose value depends on the particular process followed.

Thermodynamic cycle — a sequence of processes that returns a system to its original state.

Key Takeaways

• A P–V diagram plots pressure vertically and volume horizontally.
• A point represents a thermodynamic state.
• A line or curve represents a thermodynamic process.
• Movement to the right represents expansion.
• Movement to the left represents compression.
• A horizontal line represents constant pressure.
• A vertical line represents constant volume.
• The area under a P–V curve represents work done by the gas.
• A constant-volume process has W = 0.
• Different paths between the same states can involve different amounts of work.
• Internal-energy change depends only on the initial and final states.
• Work and heat transfer depend on the path.
• For a complete cycle, the area enclosed by the loop represents net work.
• P–V diagrams connect graphical analysis with the First Law of Thermodynamics.

Check Your Understanding

1. Which quantity is plotted on the horizontal axis of a P–V diagram?
2. What does movement from left to right represent physically?
3. What type of process appears as a horizontal line?
4. What type of process appears as a vertical line?
5. Explain why no pressure-volume work is done during a constant-volume process.
6. A gas expands at 120,000 Pa from 0.002 m³ to 0.007 m³. Calculate the work done.
7. A gas expands at constant pressure of 250 kPa from 3.0 L to 7.0 L. Calculate the work done in joules.
8. A gas expands while its pressure decreases linearly from 400 kPa to 200 kPa. Its volume increases from 0.002 m³ to 0.006 m³. Calculate the work using the area of the trapezoid.
9. Two paths connect the same initial and final states. Path A lies entirely above Path B on a P–V diagram during expansion. Which path involves more work done by the gas? Explain.
10. A gas follows a vertical path from P = 100 kPa to P = 300 kPa. How much pressure-volume work is done?
11. A P–V diagram shows that a gas does 900 J of work while receiving 1,400 J by heating. Calculate its change in internal energy.
12. Explain why two processes connecting exactly the same initial and final states can have the same ΔU but different values of Q and W.