AcademyFirst-Law Thermodynamics
Academy
Common Thermodynamic Processes
Level 1 - Physics topic page in First-Law Thermodynamics.
Principle
Named thermodynamic processes impose constraints that simplify the first law.
Notation
\(p\)
pressure
\(\mathrm{Pa}\)
\(V\)
volume
\(\mathrm{m^{3}}\)
\(T\)
absolute temperature
\(\mathrm{K}\)
\(Q\)
heat transferred to the system
\(\mathrm{J}\)
\(W\)
work done by the system
\(\mathrm{J}\)
\(\Delta U\)
change in internal energy
\(\mathrm{J}\)
Method
Derivation 1: Apply one constraint at a time
Each named process fixes one variable or one transfer. The first law then tells which remaining energy transfer is required.
Isochoric
\[V=\text{constant}\Rightarrow W=0\]
Isobaric
\[p=\text{constant}\Rightarrow W=p\Delta V\]
Adiabatic
\[Q=0\Rightarrow \Delta U=-W\]
Cyclic
\[\Delta U=0\Rightarrow Q=W\]
Derivation 2: Add the ideal-gas isothermal case
For an ideal gas, internal energy depends only on temperature. Isothermal ideal-gas processes therefore have \(\\Delta U=0\).
Ideal-gas internal energy
\[U=U(T)\]
Isothermal ideal gas
\[\Delta T=0\Rightarrow \Delta U=0\]
First-law result
\[Q=W\]
Rules
These process constraints are used repeatedly in thermal problems.
Isochoric
\[V=\mathrm{constant},\quad W=0\]
Isobaric
\[p=\mathrm{constant},\quad W=p\Delta V\]
Isothermal ideal gas
\[T=\mathrm{constant},\quad \Delta U=0\]
Adiabatic
\[Q=0\]
Cyclic
\[\Delta U=0\]
Examples
Question
A gas is heated at constant volume, receiving
\[500\,\mathrm{J}\]
Find \(W\) and \[\Delta U\]
Answer
At constant volume,
\[W=0\]
The first law gives \[\Delta U=Q-W=500\,\mathrm{J}\]
Checks
- Isochoric means no volume work, not no heat.
- Isothermal does not always mean no heat.
- Adiabatic means no heat transfer, not constant temperature.
- A cyclic process returns all state variables to their starting values.