# How does Gibbs free energy change relate to work?

##### 1 Answer

The

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

#dG = dH - d(TS)#

From the definition of enthalpy,

#dG = dU + d(PV) - d(TS)#

From the first law of thermodynamics,

#dG = deltaq + deltaw + PdV + VdP - TdS - SdT#

**Work** can be defined as

#deltaw = deltaw_"PV" + deltaw_"non-PV"# ,

where *from the perspective of the* ** system** is

**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),

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

#= color(green)(-SdT + VdP + deltaw_"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:

#color(blue)(dG = deltaw_"non-PV", " const T & P")#