11 min readConcept sheetSection A objective (2 marks, match the columns); Section B very short answer (2 marks): isothermal against adiabatic, or work as the area under a P-V graph

Thermodynamic processes and work from a P-V graph

A process changes P, V or T. Isothermal keeps T fixed; adiabatic lets no heat in or out. Work done equals the area under the P-V graph.

Work in expansionW = PΔV at constant pressure; in any process, W = area under the P-V (indicator) diagram

Isothermal processTemperature stays constant; the walls conduct heat; the change is very slow

Adiabatic processNo heat enters or leaves (ΔQ = 0); the walls insulate; the change is quick

Cyclic processThe system returns to its starting state; net work = area enclosed by the loop

Process, path and indicator diagram

A system goes through a thermodynamic process when it moves from one equilibrium state to another and P, V or T change. For example, a gas heated in a cylinder pushes its piston out. The graph of a process is called its path. A graph of pressure against volume is called an indicator diagram, or P-V diagram. Real changes are never perfectly in equilibrium, so we study quasi-static processes. These go so slowly that the system is almost in equilibrium at every stage.

Types of thermodynamic process

Process	What stays constant	How it is done	Example
Isothermal	Temperature (T)	Walls conduct heat well, and the change is made very slowly, so heat flows in or out to keep T fixed	Gas slowly compressed in a metal cylinder kept in a water bath
Adiabatic	No heat flows in or out (ΔQ = 0)	Walls insulate perfectly, or the change is very fast	Air compressed quickly in a bicycle pump; the barrel warms
Isobaric	Pressure (P)	Heated while open to the air, or under a free piston	Water heated in an open vessel
Isochoric	Volume (V)	Heated in a rigid, sealed container, so no work is done	Gas heated in a closed steel cylinder
Cyclic	The final state is the same as the first	A series of processes that ends where it began	The working of any engine

Reversible vs irreversible processes

ReversibleEvery stage is an equilibrium state, and the process can be run backwards along the same path to the start. It must be very slow and free of friction. It is an ideal that real processes only approach. Near examples: a gas compressed very slowly; masses added to a spring one by one and then removed.

IrreversibleIt cannot be retraced along the same path, and its middle stages are not equilibrium states. All natural processes are irreversible. Examples: heat produced by friction, sugar dissolving in water, iron rusting in air.

Work done from a P-V graph

A gas at pressure P pushes a piston of area A.

1
The force on the piston is F = PA.
2
If the piston moves out a small distance Δx, the work done by the gas is ΔW = F × Δx = PAΔx.
3
AΔx is the small increase in volume ΔV, so ΔW = PΔV.
4
On the P-V graph, PΔV is the area of a thin strip under the curve.
5
Adding all the strips from V₁ to V₂: total work = area under the curve between V₁ and V₂.
6
In expansion the gas does work, which is positive. In compression work is done on the gas, which is negative.
7★
So, work depends on the path: different curves between the same two states have different areas. In a cycle, the net work = the area enclosed by the loop.

★ key step

How to draw an indicator diagram

1
Draw axes: pressure P up the side and volume V along the bottom.
2
Mark V₁ and V₂ on the V-axis. Mark the starting point A (P₁, V₁) and the end point B (P₂, V₂).
3
Join A to B with a smooth curve sloping down. Put an arrow on it to show the direction of change.
4
Drop dotted lines from A and B to the V-axis. Shade the region under the curve.
5
Draw one thin strip of width ΔV and height P inside it, labelled PΔV.
6
Write work done = shaded area. For comparison, an adiabatic curve from A falls more steeply than an isothermal one.

Worked example: work in expansion and compression

Problem

A gas at a steady pressure of 2.0 × 10⁵ Pa expands from 2.0 L to 5.0 L. (a) Find the work done by the gas. (b) What is the work if it is pushed back to 2.0 L at the same pressure? (1 L = 10⁻³ m³)

Working

ΔV = 5.0 L − 2.0 L = 3.0 L = 3.0 × 10⁻³ m³.
(a) W = PΔV = 2.0 × 10⁵ Pa × 3.0 × 10⁻³ m³ = 600 J.
(b) Now ΔV = −3.0 × 10⁻³ m³, so W = −600 J: the work is done on the gas.

Answer

(a) +600 J, done by the gas; (b) −600 J, that is 600 J done on the gas

In real life

Pump a bicycle tyre fast and the barrel warms up: quick compression is nearly adiabatic, so the work heats the air. Water boiling in an open pot is an isobaric process, at the steady pressure of the air.

Exam angle

2 marks: “A gas is compressed once slowly in a thin metal cylinder standing in water, and once quickly in an insulated cylinder. Name the two processes and give two differences between them.” Slow, in the metal cylinder: isothermal; heat flows out, so T stays constant. Quick, insulated: adiabatic; no heat flows (ΔQ = 0), so the work done on the gas raises T.

2 marks: “Show that the work done by a gas in a small expansion is PΔV. What does the area under a P-V graph represent?” The force PA moves the piston Δx, so ΔW = PAΔx = PΔV. The area under the P-V curve is the work done in the process.

2 marks: “Distinguish between reversible and irreversible processes, with one example of each.” Reversible: can be retraced through equilibrium states; ideal and very slow (slow compression of a gas). Irreversible: cannot be retraced; all natural processes (rusting of iron, heat from friction).

Remember it

Iso-THERM-al: same temperature. A-dia-BATIC: heat cannot pass. Iso-BAR-ic: same pressure. Iso-CHOR-ic: same volume (space).