The Four Named Processes

You’ll be able to

Identify isobaric, isochoric, isothermal and adiabatic processes on a PV diagram
State which term of the first law vanishes in each process
Predict the direction of heat flow and temperature change for each

Each process kills one term

The four named processes each hold one thing constant, and each therefore makes one term of ΔU = Q + W disappear or simplify. Knowing which term drops out is the entire content of these problems. Isobaric: constant pressure, so W = −PΔV is easy to compute. Isochoric: constant volume, so W = 0. Isothermal: constant temperature, so ΔU = 0 and Q = −W. Adiabatic: no heat exchange, so Q = 0 and ΔU = W.

The four processes

isobaric (P const): W_on = −PΔV · isochoric (V const): W = 0, so ΔU = Q · isothermal (T const): ΔU = 0, so Q = −W · adiabatic (Q = 0): ΔU = W

For an ideal gas ΔU depends only on temperature, which is why constant T means constant U.

Why isothermal and adiabatic look similar but differ

Both appear as curves sloping down to the right on a PV diagram, but the adiabatic is steeper. On an isothermal path, heat flows in during expansion to keep the temperature fixed. On an adiabatic path no heat can enter, so the expanding gas must draw the energy from its own internal store — it cools. That extra temperature drop is what makes the pressure fall faster, and it is why an adiabatic curve cuts across isotherms rather than following one.

Adiabatic in practice

A process is effectively adiabatic when it happens too fast for heat to flow or in a well-insulated container. A bicycle pump warming as you compress air, a spray can cooling as it empties, and the compression stroke in an engine are all approximately adiabatic. Note the pattern: adiabatic compression heats the gas and adiabatic expansion cools it, which follows directly from ΔU = W with Q = 0.

Worked example

An ideal gas is compressed adiabatically, with 500 J of work done on it. Find the change in internal energy, the heat exchanged, and state what happens to the temperature.

1.Adiabatic means Q = 0 by definition.
2.First law: ΔU = Q + W_on = 0 + 500 J.
3.ΔU = +500 J.
4.For an ideal gas, internal energy depends only on temperature, so a rise in U means a rise in T.

Answer: ΔU = +500 J, Q = 0, and the temperature rises. Adiabatic compression always heats the gas, because the work done on it has nowhere to go except into internal energy.

Watch out

Isothermal does not mean no heat flows — it means heat flows exactly fast enough to keep the temperature fixed. Adiabatic means no heat flows and the temperature therefore changes. Confusing these two reverses both Q and ΔU.

Checkpoint

An ideal gas expands isothermally. Which is true?

Checkpoint

A gas is heated at constant volume. The heat added:

Checkpoint

On a PV diagram, an adiabatic expansion is steeper than an isothermal expansion from the same starting point because:

On the exam

Write down which term is zero before doing any arithmetic — "adiabatic, so Q = 0" or "isochoric, so W = 0". That single line makes the first law solvable and is often awarded on its own.

Answer the 3 checkpoints as you read.