# Why does the Gibbs free energy only correspond to non-expansion work?

It has been defined as the energy available for work other than expansion work. Why can't it be used for expansion work

The association of the Gibbs free energy with “additional”, or “non-expansion” work, is simply a mathematical result of its definition: $G = H - TS = U + pV - TS$.

From the First Law, for a closed system, we have $\mathrm{d}U = \delta q + \delta w$.

For a reversible change, the Second Law tells us that $\mathrm{d}S = \delta q_\text{rev}/T$; so $\delta q_\text{rev} = T\,\mathrm{d}S$.

We can split $\mathrm{d}w$ into two parts, expansion work and additional work: expansion work is given by the formula $\mathrm{d}w_{\text{exp}} = -p\,\mathrm{d}V$, and for now we can just label additional work as $\mathrm{d}w_{\text{add}}$. In this context, additional simply means anything that isn’t expansion work. Putting all this together, we have

$$\mathrm{d}U = T\,\mathrm{d}S - p\,\mathrm{d}V + \delta w_{\text{add}}$$

and from the definition of the Gibbs free energy in the first paragraph,

\begin{align} \mathrm{d}G &= \mathrm{d}(U + pV - TS)\\ &= \mathrm{d}U + (p\,\mathrm{d}V + V\,\mathrm{d}p) - (T\,\mathrm{d}S + S\,\mathrm{d}T)\\ &= T\,\mathrm{d}S - p\,\mathrm{d}V + \delta w_{\text{add}} + p\,\mathrm{d}V + V\,\mathrm{d}p - T\,\mathrm{d}S - S\,\mathrm{d}T\\ &= V\,\mathrm{d}p - S\,\mathrm{d}T + \delta w_{\text{add}} \end{align}

Now, if a closed system is kept at constant pressure and temperature, $\mathrm{d}p = 0$ and $\mathrm{d}T = 0$. Therefore $\mathrm{d}G = \delta w_{\text{add}}$.

The simplest application of this interpretation of $G$ is in electrochemistry, where one could move $|\mathrm{d}n|$ moles of electrons through a certain potential difference $E$; the “additional work” done in this case is given by $-FE\,|\mathrm{d}n|$ (you could consult a physics text for details). This gives us $\mathrm{d}G = -FE\,|\mathrm{d}n|$, and (after several more steps) the familiar equation $\Delta_\mathrm{r} G = -zFE$.