2
$\begingroup$

Gibbs Free Energy is defined as

$$G = H - TS$$ and the change in $G,$

$$\Delta G = \Delta H - T\Delta S,$$ (isothermal and isobaric)

Now at constant pressure, $q = \Delta H$

and Entropy is defined as , $\Delta S = \dfrac{q}{T}$ which means that $T\Delta S = q.$

Thus shouldn't Gibbs Free Energy automatically be zero. Or is there something I'm missing, like a specific condition?

$\endgroup$
2
$\begingroup$

The only common situation I'm aware of in which a system can undergo both an isothermal and an isobaric process at the same time is when the system is undergoing a phase change at constant pressure. For instance, we could be considering a system comprised of liquid water and steam in phase equilibrium.

For a quasi-static, isobaric process, $$\Delta H = \Delta (U + pV) = \Delta U + p\Delta V = Q + W + p\Delta V = Q -p\Delta V + p\Delta V = Q,$$ and so the change in enthalpy during the process is exactly equal to the heating of the system. For a quasi-static, isothermal process, $$Q = T\Delta S.$$ Therefore, for a process that is both isothermal and isobaric, $$\Delta G = \Delta (H - TS) = \Delta H - T\Delta S = Q - Q = 0.$$ In other words, yes! - the Gibbs free energy of the whole system is constant during such a process, but note that this is a very special process. It is rare that a system can undergo both a process in which both the temperature and pressure can remain constant at the same time, but such a process$-$*if quasi-static*$-$is a process of constant Gibbs free energy.

So: as long as there is a system containing a single substance in two different phases, and the phases are in thermal, mechanical, and phase equilibrium, then any process occurring at constant pressure also occurs at constant temperature, and so $\Delta G = 0$.

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

The differential of the Gibbs free energy of $n$ species at temperature $T$ and pressure $P$ is $$ \text{d}G = -S \text{d} T + V \text{d}P + \sum_{i=1}^n \mu_i \text{d}N_i. $$ At fixed temperature and pressure, the only change in the Gibbs free energy will come from chemical reactions which change the particle number for some species $N_i$. This is presumably why chemists are so interested in $G$, since any equilibrium process occurring in open air will be at fixed pressure and temperature.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Well, assuming that the system is always in thermal equilibrium with its surroundings and the surroundings' temperature doesn't change; then the process is isothermal. But I don't think that chemical reactions taking place in the open air have to be isothermal. Nonetheless, your point is well-taken (+1). $\endgroup$ – march Nov 12 '15 at 22:09
  • $\begingroup$ @march Yes, anything that releases energy will create a local increase in temperature, but that is by definition out of equilibrium. $\endgroup$ – ZachMcDargh Nov 14 '15 at 0:40

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.