Work and the Joule Experiment

Work Depends on How You Get There

For a fluid the elementary work done on the system is $\delta W = -P\,dV$, so a finite process gives

$$W = -\int_{V_1}^{V_2} P\,dV$$

and the integral needs the whole path, not just the endpoints. Different systems have their own forms, $\delta W = \sigma\,dA$ for a surface, $\mathcal{E}\,dq$ for a cell, $-\mathbf{m}\cdot d\mathbf{B}$ for a magnetic moment, but all share the pattern intensive variable times change in extensive variable.

Comparing two paths between the same endpoints makes the path dependence concrete:

Same start, same finish, different answers. That is why $\delta W$ is written with a $\delta$ rather than a $d$, it is an inexact differential with no function behind it.

Joule's experiment is what rescues energy from this. He performed adiabatic work on water in three different ways — a falling weight turning a paddle, electrical heating, mechanical compression — and found that the work needed to move between two given states was always the same, regardless of method. If adiabatic work is path-independent, it defines a state function:

$$\Delta U = W_{adiabatic}$$

Once $U$ exists, heat is defined as the difference for a non-adiabatic path: $Q = \Delta U - W$. Heat is not a separate substance but a residual, whatever energy crossed the boundary that the work terms do not account for.

Common pitfall: reading free expansion into a vacuum as "the gas did work by expanding". There is nothing to push against, so $P_{ext} = 0$ and $W = 0$ exactly. The gas expands, but does no work and, for an ideal gas, does not cool either.