0
$\begingroup$

First of all: I know this question has been asked before (Why does the Gibbs free energy only correspond to non-expansion work?), and although I understand the mathematical derivation given there, I struggle with the concept.

If both $\Delta G$ is negative and the change in enthalpy is negative (exothermic), I don't see why we cannot use that heat to do mechanical work? So how is it that it only cooresponds to non-mechanical work?

$\endgroup$
  • $\begingroup$ E, F, G, and H are all derived from $dQ = T dS = dE +pdV$ based on what two independent variables you want to choose, be it (S,V), (S,p), (T,V), or (T,p). So, if you understand what constraints you have chosen then you should directly be able to get to what the particular free energy means. $\endgroup$ – Jon Custer Dec 6 '18 at 22:09
  • $\begingroup$ As Jon Custer explains that is more or less the definition of $\Delta G$. If you are interested in all of the work look at the Helmholtz free energy instead. $\endgroup$ – Buck Thorn Dec 7 '18 at 11:04
2
$\begingroup$

For a closed system in contact with a constant temperature reservoir held at the initial system temperature T, and in contact with a surroundings at an external pressure held at the initial system pressure P, and experiencing a change between two thermodynamic equilibrium states, the change in internal energy is given by: $$\Delta U=Q-W$$If we express the work W as the sum of expansion work and non-expansion work (the latter, say, obtained by generating electricity or turning a shaft), then $$W=P\Delta V+W_{NE}$$And, because there is a single constant temperature bath comprising the surroundings, the heat added to the system is related to the change in entropy between the initial and final thermodynamic equilibrium states of the system by: $$\Delta S=\frac{Q}{T}+S_{gen}$$where $S_{gen}$ is the entropy generated as a result of any irreversibility's that may be present. So, if we combine these equations, we obtain:$$W_{NE}=-\Delta G-TS_{gen}$$where $\Delta G=\Delta U-T\Delta S+P\Delta V$. So, for any process that goes between the same two thermodynamic equilibrium states, the maximum non-expansion work that the system can do on its surroundings is $-\Delta G$ (since $S_{gen}$ can only be positive). The maximum corresponds to a reversible path between the two end states.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.