What is the physical significance of Helmholtz Free Energy?

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Faith Wise · Accepted Answer

For a closed system undergoing a thermodynamic process between two thermodynamic equilibrium states, the first law of thermodynamics tells us that

Δ U = q - w

If the process takes place with the system solely in contact with a single constant temperature reservoir at temperature T that is also the initial (and final) temperature of the system (so that all the heat transfer q between the system and its surroundings takes place at the reservoir temperature T), the second low of thermodynamics tells us that

Δ S = \frac{q}{T} + σ

where

σ

is the amount of entropy generated within the system during the process. If we solve this second law equation for q and substitute it into the first law relationship, we obtain:

Δ U = T Δ S - σ T - w

or

w = - Δ A - σ T

where A is the Helmholtz Free Energy. This equation tells us that the maximum amount of work that can be extracted from the system between the initial and final thermodynamics equilibrium states under these constraints is minus the change in the Helmholtz free energy for the system between these end states.

What is the physical significance of Helmholtz Free Energy?

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