Question

How does Gibbs free energy change relate to work?

Answer 1

The `DeltaG` 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.

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

`dG = dH - d(TS)`

From the definition of enthalpy, `H = U + PV`, where `U` is the internal energy. As a result,

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

From the first law of thermodynamics, `dU = deltaq + deltaw`, where `delta` indicates a path function.

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

Work can be defined as

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

where `"PV"` work defined from the perspective of the system is `deltaw_"PV" = -PdV`. 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_(rev) = TdS`, so:

`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")`