AcademyFirst-Law Thermodynamics
Academy
Paths Between States
Level 1 - Physics topic page in First-Law Thermodynamics.
Principle
Thermodynamic state changes can share endpoints while transferring different heat and work.
Notation
\(A,B\)
initial and final equilibrium states
\(\Delta U\)
change in internal energy
\(\mathrm{J}\)
\(Q\)
heat transferred to the system
\(\mathrm{J}\)
\(W\)
work done by the system
\(\mathrm{J}\)
\(p\)
pressure
\(\mathrm{Pa}\)
\(V\)
volume
\(\mathrm{m^{3}}\)
Method
Derivation 1: Separate state and path quantities
Internal energy is a state function: its change depends only on the endpoints. Heat and work are path quantities: they depend on how the process is carried out.
State-function change
\[\Delta U=U_B-U_A\]
Work path
\[W=\int_A^B p\,dV\]
Different paths
\[W_1\ne W_2\quad\text{can occur for the same }A\to B\]
Derivation 2: Use cycles as a check
A cyclic process returns to its initial state, so every state function returns to its initial value even though heat and work may be nonzero.
Cycle endpoint
\[A\to A\]
Internal-energy change
\[\Delta U_{\mathrm{cycle}}=0\]
Net transfer relation
\[Q_{\mathrm{cycle}}=W_{\mathrm{cycle}}\]
Rules
These relations distinguish endpoint information from path information.
State function
\[\Delta U=U_B-U_A\]
Path work
\[W=\int_A^B p\,dV\]
Cycle energy
\[\Delta U_{\mathrm{cycle}}=0\]
Cycle transfers
\[Q_{\mathrm{cycle}}=W_{\mathrm{cycle}}\]
Examples
Question
A gas goes from state \(A\) to state \(B\) by two different paths. Path 1 has
\[W=300\,\mathrm{J}\]
path 2 has \[W=120\,\mathrm{J}\]
If \[\Delta U=80\,\mathrm{J}\]
find \(Q\) for each path.Answer
Using
\[\Delta U=Q-W\]
\[Q=\Delta U+W\]
Path 1: \[Q=80+300=380\,\mathrm{J}\]
Path 2: \[Q=80+120=200\,\mathrm{J}\]
Checks
- Same endpoints mean the same \(\\Delta U\), not the same \(Q\) or \(W\).
- A cycle can transfer net heat and net work even though \(\\Delta U=0\).
- On a \(p\)-\(V\) graph, work depends on the area under the path.
- Do not write \(\\Delta Q\) or \(\\Delta W\) for path transfers.