AcademyFirst-Law Thermodynamics
Academy
Work During Volume Changes
Level 1 - Physics topic page in First-Law Thermodynamics.
Principle
Pressure-volume work is the energy transferred when a system boundary moves against pressure.
Notation
\(W\)
work done by the system
\(\mathrm{J}\)
\(p\)
gas pressure at the boundary
\(\mathrm{Pa}\)
\(V\)
system volume
\(\mathrm{m^{3}}\)
\(\Delta V\)
change in volume
\(\mathrm{m^{3}}\)
\(A\)
piston area
\(\mathrm{m^{2}}\)
\(x\)
piston displacement
\(\mathrm{m}\)
Method
Derivation 1: Build work from a moving piston
A gas pushing a piston exerts force \(F=pA\). If the piston moves outward a small distance \(dx\), the gas does positive work.
Pressure force
\[F=pA\]
Small displacement work
\[dW=F\,dx=pA\,dx\]
Volume change
\[dV=A\,dx\]
Boundary work
\[dW=p\,dV\]
Derivation 2: Add the work along the process
If pressure changes during the process, the total work is the signed area under the path on a \(p\)-\(V\) graph.
General work
\[W=\int_{V_i}^{V_f}p\,dV\]
Constant pressure
\[W=p(V_f-V_i)=p\Delta V\]
Constant volume
\[\Delta V=0\Rightarrow W=0\]
Rules
These are the compact work relations for volume-changing processes.
Boundary work
\[dW=p\,dV\]
Path work
\[W=\int_{V_i}^{V_f}p\,dV\]
Isobaric work
\[W=p\Delta V\]
Isochoric work
\[W=0\]
Examples
Question
A gas expands at constant pressure
\[2.0\times10^5\,\mathrm{Pa}\]
from \[0.030\,\mathrm{m^3}\]
to \[0.050\,\mathrm{m^3}\]
Find work done by the gas.Answer
\[W=p\Delta V=(2.0\times10^5)(0.020)=4.0\times10^3\,\mathrm{J}\]
The sign is positive because the gas expands.Checks
- Expansion gives \(W>0\) with the section sign convention.
- Compression gives \(W\<0\).
- Work is the area under the actual path, not just the endpoint difference.
- Constant-volume processes do no pressure-volume work.