# How to derive the pressure dependency for the Gibbs free energy?

$$\mathrm{d}G=\left(\frac {\partial G}{\partial p}\right)_T+\left(\frac {\partial G}{\partial T}\right)_p$$ The above is an exact differential. Firstly what does this mean physically? (I hear about the fact that a change in G is independent of the "path" but I don't understand this argument because I don't know what is meant by this path. Also how would something be dependent on the path?)

It can also be shown that $\mathrm{d}G$ can be written: $$\mathrm{d}G=V\mathrm{d}p-S\mathrm{d}T$$

So by comparing coefficients: $$\left(\frac {\partial G}{\partial p}\right)_T=V$$ It is useful to integrate this to see how $G$ varies with $p$: $$\int_{p_1}^{p_2}\left(\frac {\partial G}{\partial p}\right)_T\mathrm{d}p=\int_{p_1}^{p_2}V\mathrm{d}p$$ $V$ can be expressed in terms of $p$ using the equation $V=\frac{n\mathcal{R}T}p$ Where $n\mathcal{R}T$ are constants. The above equation becomes: $$\int_{p_1}^{p_2}\left(\frac {\partial G}{\partial p}\right)_T\mathrm{d}p=n\mathcal{R}T\int_{p_1}^{p_2}\frac{\mathrm{d}p}p$$ Integrating the LHS of the above equation is the crux of my question (I don't understand how that can be integrated). I am not sure whether the following step is correct mathematically, please explain why this is correct/incorrect. I cancel out the $\mathrm{d}p$'s on the LHS: $$\int_{p_1}^{p_2}\mathrm{d}G=n\mathcal{R}T\int_{p_1}^{p_2}\frac{\mathrm{d}p}p$$ Then the result is: $$G(p_2)-G(p_1)=n\mathcal{R}T\ln\left(\frac{p_2}{p_1}\right)$$ Or... $$\Delta G=n\mathcal{R}T\ln\left(\frac{p_2}{p_1}\right)$$ This result is correct but I don't understand the left hand side. I've been told that the reason it's possible to write "$G(p_2)-G(p_1)$" or "$\Delta G$" is because $G$ is a state function/exact differential but, again, I don't understand this. If $G$ was not a state function/exact differential how would the result change?

This question has a lot of parts! I hope I manage to answer most but probably edits will be required.

$$\mathrm{d}G=\left(\frac {\partial G}{\partial p}\right)_T+\left(\frac {\partial G}{\partial T}\right)_p$$

What does this mean physically?
This is simply an expression that shows the contributions to the change in the Gibb's free energy, which varies with both pressure and temperature (i.e., $G$ is a function of $p$ and $T$). However, it is possible to separate the exact differential of $\mathrm{d}G$ into two terms using the mathematical expression of partial derivatives, first term is the variation of $G$ with pressure at constant temperature, and the second term is the variation of $G$ with temperature at constant pressure.

What does it mean for $\Delta G$ to be independent of the path? (same question: what does it mean for $\Delta G$ to be a state function?)
You can calculate a change in $G$, or equivalently, $\Delta G$, for any reaction. So for example for a reaction like a combustion process for glucose

$$\ce{C6H12O6~(s) + 3 O2~(g) -> 6 CO2~(g) + 6 H2O~(g),}$$

you can calculate $\Delta G$ for this reaction at any particular temperature/pressure that you would like. And in fact, if you don't do the combustion directly but somehow do it via completely dissociated gaseous atoms

$$\ce{C6H12O6~(s) + 3 O2~(g) -> 6 C~(g) + 18 O~(g) + 12 H~(g) -> 6 CO2~(g) + 6 H2O~(g)}$$

You will still end up with the same $\Delta G$ for the overall reaction. Obviously if you calculate the $\Delta G$ for each individual step/arrow in the gaseous atoms case it will be very different, but they will combine to give the same overall $\Delta G$ as the direct combustion. In other words, $\Delta G$ won't be able to tell you whether you have done the combustion directly or done it laboriously via gaseous atoms, it just tells you what is the difference between the two states, first state being glucose + oxygen, second state being carbon dioxide + water. $\Delta G$ for glucose that you have eaten and combined with oxygen you breathe in and breathe out carbon dioxide + water would still the same $\Delta G$ as for combustion, even though now it is all done through cells, Kreb's cycle and a lot of biochemistry.

How would something be dependent on the path?
As you can see from my example above, the amount of effort you need to make $$\ce{C6H12O6~(s) + 3 O2~(g) -> 6 CO2~(g) + 6 H2O~(g)}$$ happen can differ a lot, and for each of the cases, what you "get back" from doing that reaction is different, such as heat released when burning glucose vs keeping your cellular biochemistry going with respiring glucose. In physics/physical chemistry, this is formally "work done". Work is dependent on path.

I am a little unsure whether you went from

$$\mathrm{d}G=V\mathrm{d}p-S\mathrm{d}T \tag1$$

to

$$\left(\frac {\partial G}{\partial p}\right)_T=V \tag2$$

in the right way, so I am going to go through it again.

At constant $T$, $\mathrm{d}T=0$. So $(1)$ becomes

$$\mathrm{d}G=V\mathrm{d}p \tag3$$

which after rearranging is equivalent to $(2)$.

The integration then should go from $(3)$ to give

$$\int_{p_1}^{p_2}\mathrm{d}G=\int_{p_1}^{p_2}V\mathrm{d}p \tag4$$

The reason why what you wrote look a little weird

$$\int_{p_1}^{p_2}\left(\frac {\partial G}{\partial p}\right)_T\mathrm{d}p=\int_{p_1}^{p_2}V\mathrm{d}p$$

is that you usually don't just cancel partial differentials like you did. Whereas $(3)$ is an exact differential, which you integrate over in $(4)$, and that is OK. Unfortunately I am not sure whether what you wrote is wrong, it is just something I would not write myself.

The rest of the derivation is fine.

Final question. I believe you are fine with mathematically going from $\int_{p_1}^{p_2}\mathrm{d}G$ to $G(p_2)-G(p_1)$, however, if you think about this result physically, $\Delta G$ does not depend on any other $p_3, p_4, p_5...p_n$ that the system might pass through on the way from $p_1$ to $p_2$. If $G$ is not a state function, then you would expect to have to consider at the very least all these different points of $p$ that you might have gone through, and different ways of doing the reaction passing through different $p$ would change $G$ even if the initial and final $p$ are the same. In that case, an integral like what we have written here would be a very poor way of finding $\Delta G$ because we are missing the contribution from all the intervening $p_n$, and we would have to use some different mathematical tools. In fact, I believe we would have failed to derive the master equation $(1)$ much earlier on, so the rest of the derivation would simply not follow.

• Welcome to Chemistry.SE! To acquaint yourself with this page, take the tour and visit the help center. Furthermore this tutorial shows you how math and chemical formulae can be nicely formatted on this site. – Philipp Dec 12 '14 at 1:45
• Concerning your problem with "cancelling partial differentials" have a look here. Of course, you are right that it has nothing to do with cancelling like in normal products and that this is mathematically much more involved but nevertheless it's a good mnemonic because it usually can simply be treated just like that. – Philipp Dec 12 '14 at 1:51
• Thanks! I didn't realise you can use mhchem, should probably have tested it. Thanks for fixing my answer. Also thanks for clarifying that you can cancel partial differentials (and finding a German link, though I should probably clarify I have only recently moved to Germany!) – selkie222 Dec 12 '14 at 8:00