# Confusion about the use of partial molar Gibbs free energy

#### sponteous

Summary
Is $\Delta G$ useful for situations other than constant temperature and pressure?
If this belongs in classical physics, please move it there. But it seems like the kind of question chemistry people would know so I'm putting it here.

I was reading a textbook on chemical thermodynamics, and it says to raise the partial molar Gibbs free energy of $n$ moles a substance from $\overline{G}_1$ to $\overline{G}_2$ you have to do work $W = n(\overline{G}_2 -\overline{G}_1)$. Sounds reasonable enough. But then they give as an example moving $n$ moles of an ideal gas from a tank at pressure $p_1$ to a higher pressure $p_2$ at constant temperature. This indeed requires compression work $W = nRT(\ln(p_2) - ln(p_1))$, which happens to be equal to $n(\overline{G}_2 -\overline{G}_1)$.

But this seems like kind of a fluke to me. For an isothermal process, it's the change in Helmholtz free energy, $\Delta A$ that measures the work required to bring about the change of state. In this case, $\Delta A = \Delta G$ because we happen to have an ideal gas kept at constant temperature, so the product $pV$ doesn't change. But in general, I thought change in Gibbs free energy was useful only when both temperature and pressure are constant, and you want to know how much non-pV work is required. So it doesn't seem like you'd use it for the case of compressing a gas. Am I wrong?

#### Christopher Grayce

It does seem like a strange example. In general for most simple substances dG = ‒SdT + Vdp + μ dN, so there *are* no processes that can change G of a simple substance at constant T and p, except for the trivial one of adding or subtracting some of the substance (dN != 0). So their example might be rather contrived to avoid the more complicated situation where G is most useful, which is in chemical reactions, meaning you have a mixture with more than one mole number and chemical potential, and G can change at constant T and p by changing the relative mole numbers (and of course those relative changes are constrained by the chemical equations).

#### sponteous

Good points. You are probably right that the author is looking for a simple example. It just seemed to me that they got the right answer for the wrong reason. But I'm still not certain of that, since the material is fairly new to me.

#### Chestermiller

Mentor
In a steady state flow system in contact with a single constant temperature reservoir (not necessarily operating at constant pressure), the decrease in G of the fluid (possibly even comprising a reacting mixture) between inlet and outlet of the control volume provides an upper bound to the shaft work that can be done, for the given inlet and outlet states.

#### sponteous

In a steady state flow system in contact with a single constant temperature reservoir (not necessarily operating at constant pressure), the decrease in G of the fluid (possibly even comprising a reacting mixture) between inlet and outlet of the control volume provides an upper bound to the shaft work that can be done, for the given inlet and outlet states.
Okay, that is pretty much the situation of the example, except in reverse. So if a gas is allowed to decompress from a high pressure tank into a low pressure tank isothermally, it can do some work along the way, like turning a shaft. I just don't get why it is the decrease in G = U-TS+pV, rather than A = U-TS that gives the upper bound to the work it can do.

Edit:
Maybe I'm starting to catch on. I forgot that the gas in the inlet side is "helping" by doing compression work $p_iV_i$ on the parcel of gas that is expanding, and the gas on the outlet side is "hurting" because the parcel is having to do compression work $p_fV_f$ on it. So the net contribution to the cause of turning the shaft is $-(p_fV_f - p_iV_i) = -\Delta(pV)$. The parcel itself does a total of $-\Delta(U-TS)$ in work, and when you add these together you get $-\Delta G$

Last edited:

#### DrDu

$\Delta G= \partial G/\partial \xi |_{T,p}= \partial A/\partial \xi |_{T,V}= \partial H/\partial \xi |_{S,p}= \partial U/\partial \xi |_{S,V}$
where $\xi$ is the reaction number. So Delta G is not only relevant for isothermal and isobaric processes.

#### Chestermiller

Mentor
Okay, that is pretty much the situation of the example, except in reverse. So if a gas is allowed to decompress from a high pressure tank into a low pressure tank isothermally, it can do some work along the way, like turning a shaft. I just don't get why it is the decrease in G = U-TS+pV, rather than A = U-TS that gives the upper bound to the work it can do.

Edit:
Maybe I'm starting to catch on. I forgot that the gas in the inlet side is "helping" by doing compression work $p_iV_i$ on the parcel of gas that is expanding, and the gas on the outlet side is "hurting" because the parcel is having to do compression work $p_fV_f$ on it. So the net contribution to the cause of turning the shaft is $-(p_fV_f - p_iV_i) = -\Delta(pV)$. The parcel itself does a total of $-\Delta(U-TS)$ in work, and when you add these together you get $-\Delta G$
I was thinking of it more like this:

For a steady flow process, the open system versions of the first and 2nd laws (applied to a control volume) give $$\dot{Q}-\dot{W}_s+\sum{\dot{m}_{in}h_{in}}-\sum{\dot{m}_{out}h_{out}}=0$$and$$\sum{\dot{m}_{out}s_{out}}-\sum{\dot{m}_{in}s_{in}}=\frac{\dot{Q}}{T}+\dot{\sigma}$$assuming that the system in the control volume is in contact with a single constant temperature reservoir at temperature T, and where $\dot{W}_s$ is the rate of doing shaft work, $\dot{m}_{in}$ is the rate of mass flow in inlet streams to the control volume, $\dot{m}_{out}$ is the rate of mass flow in outlet streams from the control volume, h is specific enthalpy, s is specific entropy, and $\dot{\sigma}$ is the rate of entropy generation within the control volume. If we combine these two equations by eliminating the rate of heat flow between the reservoir and the fluid in the control volume, we obtain:$$\dot{W}_s=\sum{\dot{m}_{in}(h_{in}-Ts_{in})}-\sum{\dot{m}_{out}(h_{out}-Ts_{out})}-T\dot{\sigma}$$or
$$\dot{W}_s=-\dot{m}\Delta g-T\dot{\sigma}$$where $$\dot{m}=\sum{\dot{m}}_{in}=\sum{\dot{m}_{out}}$$and $$\Delta g=\frac{\sum{\dot{m}_{out}(h_{out}-Ts_{out})}-\sum{\dot{m}_{in}(h_{in}-Ts_{in})}}{\dot{m}}$$Note that, since the rate of entropy generation $\dot{\sigma}$ is positive definite, the maximum rate of doing shaft work possible (for the same inlet and outlet states) is $-\dot{m}\Delta g$, the rate of decrease in Gibbs free energy in passing through the control volume.

#### sponteous

Thank you for the derivation. That's a good idea, using enthalpies in the first law statement. That way the accounting of the various expansion work contributions is automatically taken care of. I hadn't seen that version before, but that is definitely useful. Thanks.

#### Chestermiller

Mentor
Thank you for the derivation. That's a good idea, using enthalpies in the first law statement. That way the accounting of the various expansion work contributions is automatically taken care of. I hadn't seen that version before, but that is definitely useful. Thanks.
Please do yourself a favor and look up the derivation of the open system (control volume) version of the first law of thermodynamics. This is in all thermo books that I am familiar with, such as Smith and van Ness, Moran et al, Denbigh, etc.

#### sponteous

Please do yourself a favor and look up the derivation of the open system (control volume) version of the first law of thermodynamics. This is in all thermo books that I am familiar with, such as Smith and van Ness, Moran et al, Denbigh, etc.
Okay I will. I'm pretty sure I see how to derive it already, but it can't hurt to see the official way. Thanks again!

By the way, in Classical Physics someone posted a question: Maximum work done by a body in an external medium that hasn't been adequately answered yet. It has to do with a confusing passage in Landau & Lifshitz' famous book on stat. mech. Maybe you would know about it.

#### Chestermiller

Mentor
Okay I will. I'm pretty sure I see how to derive it already, but it can't hurt to see the official way. Thanks again!

By the way, in Classical Physics someone posted a question: Maximum work done by a body in an external medium that hasn't been adequately answered yet. It has to do with a confusing passage in Landau & Lifshitz' famous book on stat. mech. Maybe you would know about it.
I looked it over, but their writing style drives me crazy. I'm not really able to fully understand the situation they are describing. But I think it might be something similar to what Denbigh describes in discussing the physical interpretation of the change in Helmholtz free energy. Sorry I can't help more.

#### sponteous

No problem, thanks for taking a look. I suspect the lack of clarity has to do with it being translated from Russian. Their little book on mechanics is extremely clear. Maybe it just has to do with the subject matter being more difficult to talk about precisely.

"Confusion about the use of partial molar Gibbs free energy"

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