Reading Atkins Elements of Physical Chemistry right now, and the book says: "At constant pressure and entropy, an increase in temperature ($\Delta T\gt0$) results in a decrease in $G_\mathrm m$ (molar Gibbs energy) ($\Delta G_\mathrm m\lt0$)." "At constant molar volume and temperature, an increase in pressure ($\Delta p >0$) results in an increase in molar Gibbs energy ($\Delta G_\mathrm m\gt0$)."

They have equations to describe this but I'm not really interested in the mathematical derivations (those are understandable) but rather in a logical, physical, and qualitative explanation of why these relations occur.

Given that $\Delta G$ indicates the maximum non-expansion work derivable from a system, what does this mean for the above relations? Does increasing the pressure of the system require work input ($\Delta G$ is positive) while increasing temperature result in the system doing work/an output of work ($\Delta G$ is negative)?

So, given a system (like a gas or something), why does $G$ change the way it does in response to $T$ and $p$ changes, and what does this physically mean for the system?

I'm a high school student (went through AP Chem) who's just trying to delve deeper into chemistry, so the simplest and most "layman's terms" explanation would probably help.


3 Answers 3


Probably the easiest way to picture it is that, since the Gibbs function is indeed a measure of the potential of the system to transfer energy in certain ways, these transformations affect the energy that can be transferred.

For a certain system at constant pressure and entropy (which therefore cannot absorb thermal energy from its surroundings), an increase in temperature is necessarily linked to an internal transformation - an internal work - that reduces the potential energy available to exert work on its surroundings; that increase in temperature is "using up" part of the potential reversible work as measured by the Gibbs function.

On the other hand, a system at constant volume and temperature which increases its pressure (and which, given these constrains, is therefore receiving some form of energy input from the outside; note that there isn't a pressure-volume tradeoff) will increase the potential reversible work it can exert on its surroundings, as measured by the Gibbs function.

It is important to note the implications of the constant conditions that each of those two cases set, because these constant conditions are telling you which forms of energy transfer are permitted or prohibited in each case.


While the stated problem seeks a non-mathematical answer, it is a good idea to review the equation that is at the source of the problem. For a a homogeneous phase at constant composition we can write:

$$\mathrm dG = V\mathrm dP - S\mathrm dT \tag1$$

This equation describes how the free energy for the system depends on T and P.

A possible source of confusion with the way this problem is formulated is that it compares G at different T or P. Another answer does a good job of explaining how G might be affected by changes in T or P. Still, it remains that $\Delta G$ represents the max amount of of non-$PV$ work that can be performed at constant $T$ and $P$. In that sense, comparing $\Delta G$ computed between two different temperatures or pressures should not be interpreted as the maximum possible non-expansion work that can be performed. It is important to keep this in mind. In chemistry we are most often interested in the effect of changes in $T$ or $P$ on $\Delta G $ for some process such as a phase change or chemical reaction which is otherwise carried out at constant $T$ and $P$. At the risk of redundancy I restate: we are interested in $\Delta G $ specifically because it is intended for the analysis of processes at constant $T$ and $P$.

For such a process from one state to another it is proabably clearer to write

$$\mathrm d \Delta G = \Delta V \mathrm dP - \Delta S \mathrm dT \tag2$$

where we are comparing initial and final states during the process ($\Delta G = G_{final} - G_{initial}$).

Once that is clear, the question remains: why does a decrease in $P$ at constant volumes and temperature, or an increase in $T$ at constant entropies and pressure, result in an increase in $\Delta G$? (note my use of the plural since we are referring to initial and final states of $V$ and $S$)

The short answer is that it doesn't have to!

Or rather, the change in $\Delta G$ depends on whether the difference between the initial and final entropies and volumes is positive or negative. For instance, if the difference in volume is positive - that is, the volume increases when the transformation is carried out - then an increase in pressure will result in a less stable (higher free energy) final state relative to the initial state. In fact, increasing the pressure will require more work to be performed on the system for the process to be completed. Such a response is implicitly described by Le Chatelier's principle.


Total entropy and Gibbs energy:

An alternative view on the Gibbs energy may help to get some insight:

$$\Delta S_{sys} + \Delta S_{surr} = \Delta S_{tot} = -\frac{\Delta G}{T}$$

The second law of thermodynamics tells us that a process is spontaneous only if the total entropy $S_{tot}$ does not decrease. The sign of $\Delta G$ tells us whether $S_{tot}$ increases or decreases. Note that this relation is only valid at constant pressure, and thus Gibbs energy is only really useful for evaluating the feasibility of a process at constant pressure. Let's have a look how $S_{tot}$ changes in the first case.

Increasing temperature at constant pressure and entropy:

It is important to note that the "at constant entropy" refers to $S_{sys}$ only. How do we increase temperature at constant pressure and entropy? We can for instance have an exothermic reaction. One part of the generated heat is used to increase the temperature of the system itself. This would inevitably increase $S_{sys}$, but we can compensate it by transferring the rest of the generated heat to the surroundings. Hence, we increase $S_{surr}$ and in return increases $S_{tot}$:

$$\Delta S_{tot} = \Delta S_{surr} = \int_{T_i}^{T_f}\frac{S_{sys}}{T} dT$$

Increasing pressure at constant volume and temperature:

As mentioned in the beginning, Gibbs energy is not really useful for processes that are not at constant pressure. We can still rationalize the increase of Gibbs energy with increase of pressure by comparing two constant pressure experiments at different pressures, but same initial volume and temperature. Let's look at the following reaction: $$\ce{A_2 <=> 2 A}\quad\text{and}\ \Delta G_1 = 2G_A-G_{A_2}$$

If we now increase the pressure at constant volume and temperature, the Gibbs energy of each component increases by $V\Delta p$ and we end up with: $$\Delta G_2 = \Delta G_1 + V\Delta p$$

Hence, our reaction is less feavored / more disfavored at higher pressure. What energy does $V\Delta p$ correspond to? Our reaction doubles the number of gas molecules. Increasing the number of gas molecules becomes more expensive at higher pressure and $V\Delta p$ is the additional energy we have to pay. It's basically just the manifestation of the Le Chatelier's principle.

  • $\begingroup$ For the last case (V constant) $q_V=\Delta U = \Delta H -V\Delta p = -T \Delta S_{surr}$ $\endgroup$
    – Buck Thorn
    Feb 5, 2021 at 8:58
  • $\begingroup$ @BuckThorn Yeah, you are absolutely right. My second explanation was garbage. I found a better example to rationalize the increase of Gibbs energy with pressure. $\endgroup$
    – Kexanone
    Feb 10, 2021 at 4:23
  • $\begingroup$ "The second law of thermodynamics tells us that a process is spontaneous only if the total entropy π‘†π‘‘π‘œπ‘‘ does not decrease." That doesn't quite capture it. The 2nd law states that π‘†π‘‘π‘œπ‘‘ does not decrease for any process* and that, for spontaneous processes, π‘†π‘‘π‘œπ‘‘ must increase. [*And, for completeness, I need to add: more specifically, for any process involving a sufficiently large number of particles for the 2nd law to apply.] $\endgroup$
    – theorist
    Feb 10, 2021 at 8:14
  • $\begingroup$ "The sign of Δ𝐺 tells us whether π‘†π‘‘π‘œπ‘‘ increases or decreases. Note that this relation is only valid at constant pressure..." For this relationship to hold, you also need to specify constant $T$, and no non-$pV$-work. See my answer here: chemistry.stackexchange.com/questions/124412/… $\endgroup$
    – theorist
    Feb 10, 2021 at 8:15

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