How does Gibbs free energy change relate to work?

May 16, 2018

The $\Delta G$ for a reversible process is equal to the maximum non-PV work that can be performed at constant temperature and pressure on a conservative system.

Consider the differential relationship between the Gibbs' free energy, enthalpy, and entropy:

$\mathrm{dG} = \mathrm{dH} - d \left(T S\right)$

From the definition of enthalpy, $H = U + P V$, where $U$ is the internal energy. As a result,

$\mathrm{dG} = \mathrm{dU} + d \left(P V\right) - d \left(T S\right)$

From the first law of thermodynamics, $\mathrm{dU} = \delta q + \delta w$, where $\delta$ indicates a path function.

$\mathrm{dG} = \delta q + \delta w + P \mathrm{dV} + V \mathrm{dP} - T \mathrm{dS} - S \mathrm{dT}$

Work can be defined as

$\delta w = \delta {w}_{\text{PV" + deltaw_"non-PV}}$,

where $\text{PV}$ work defined from the perspective of the system is $\delta {w}_{\text{PV}} = - P \mathrm{dV}$. Non-PV work can be, e.g. electrical work (think electrochemistry).

From this, assuming that the process performed is reversible (in thermal equilibrium the whole way through), ${q}_{r e v} = T \mathrm{dS}$, so:

color(green)(dG) = overbrace(cancel(TdS))^(q_(rev)) + deltaw_"non-PV" overbrace(- cancel(PdV))^(w_(rev,"PV")) + cancel(PdV) + VdP - cancel(TdS) - SdT

$= \textcolor{g r e e n}{- S \mathrm{dT} + V \mathrm{dP} + \delta {w}_{\text{non-PV}}}$

In the end, we find that at constant temperature and pressure, the Gibbs' free energy corresponds to the maximum non-compression and non-expansion work that can be performed:

$\textcolor{b l u e}{\mathrm{dG} = \delta {w}_{\text{non-PV", " const T & P}}}$