The Memory Trick
💡 IIAP — Each Process Holds One Thing Fixed
Four standard thermodynamic processes each fix one variable while letting the others change, and each simplifies the First Law (ΔU = Q − W) in a distinct way. Isothermal holds Temperature fixed. Isobaric holds Pressure fixed. Adiabatic means no heat exchange (Q=0). Isochoric (or isovolumetric) holds Volume fixed.
Why It Works
Fixing one variable eliminates one term from the general First Law equation, making each process's math dramatically simpler — recognizing which process a problem describes immediately tells you which term of ΔU = Q − W drops out or simplifies.
Step by Step
Working Through Each Process
1
Isothermal — constant T
For an ideal gas, internal energy U depends only on temperature — so if T is constant, ΔU = 0, meaning Q = W exactly. Follows PV = constant (Boyle's Law) on a PV diagram, tracing a hyperbola.
A gas expanding very slowly in contact with a large heat reservoir stays at essentially constant temperature throughout — a classic isothermal process.
2
Isobaric — constant P
Work is simply W = PΔV, and heat follows Q = nCpΔT (using the constant-pressure heat capacity). On a PV diagram, this traces a horizontal line.
A gas heated in a cylinder with a freely-moving piston expands at constant pressure (set by the piston's weight), a classic isobaric process.
3
Adiabatic (Q=0) and Isochoric (constant V)
Adiabatic: no heat exchange at all, so ΔU = −W entirely; follows TV^(γ−1) = constant, tracing a STEEPER hyperbola than isothermal on a PV diagram. Faster processes tend to be more adiabatic (less time for heat exchange). Isochoric: no volume change means W = 0, so ΔU = Q = nCvΔT entirely; traces a vertical line on a PV diagram.
Rapid compression (like in a diesel engine cylinder) is approximately adiabatic, since it happens too quickly for significant heat exchange with surroundings.
🏥 Worked Example
A gas undergoes an isochoric (constant volume) process, absorbing 500 J of heat. How much work does it do, and what is its change in internal energy?
1
Recall the isochoric simplification: constant volume means no work is done, W = 0.
2
Apply the First Law: ΔU = Q − W = 500 − 0 = 500 J.
3
Conclusion: all 500 J of absorbed heat goes directly into increasing internal energy, since with no volume change, there's no way for the gas to do (or have done to it) any pressure-volume work.
📌 Exam Application
Exams test correctly identifying which of the four processes a scenario describes, applying the corresponding simplified First Law equation, and recognizing each process's characteristic shape on a PV diagram.
⚠️ Most Common Thermodynamic Processes Mistakes
The most common trap is confusing adiabatic (Q=0, no heat exchange) with isothermal (T=0 change, constant temperature) — they are NOT the same thing; an adiabatic process can and typically does change temperature significantly, precisely because no heat is exchanged to compensate for the work being done.
✓ Quick Self-Test
1) What does IIAP stand for? Isothermal, Isobaric, Adiabatic, Isochoric. 2) What is held constant in an isothermal process, and what does this imply about ΔU for an ideal gas? Temperature is constant; ΔU = 0, so Q = W exactly. 3) What defines an adiabatic process, and what is the simplified First Law for it? No heat exchange (Q=0); ΔU = −W. 4) What is held constant in an isochoric process, and what does this imply about work done? Volume is constant; W = 0, so ΔU = Q. 5) On a PV diagram, which of the four processes traces a vertical line, and which traces a horizontal line? Isochoric traces a vertical line (constant V); isobaric traces a horizontal line (constant P).
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