Processes on state diagrams

A state diagram displays a process path

Pressure-volume area is work because expansion pushes a boundary through distance. Two paths connecting the same states can have different work even though the internal-energy change is identical.

Turn boundary motion into p dV

For a piston of area A moving dx, boundary force is F = pA and displacement work is dW = Fdx = pA dx = p dV because dV = A dx. Integrating gives W = ∫p dV. At constant pressure, W = p(V₂ − V₁); for a cycle, the signed enclosed area is net work.

A gas expands at constant p = 200 kPa from 1.0 to 3.0 L. W = 200,000(0.002) = 400 J. The unit check is Pa·m³ = J.

Quasistatic is not the same as slow only

The p dV expression uses a well-defined boundary pressure, usually an approximately quasistatic path. A violently expanding gas may not have uniform internal pressure, so a single p(V) curve is not available.

Work it: Compare work for a 1 L to 3 L expansion at 100 kPa constant pressure and along p(V) = 300 kPa·L/V with V in litres. Which path does more work?