Confusion about the use of partial molar Gibbs free energy

In summary: I forget 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 ####\Delta G
  • #1
sponteous
17
4
TL;DR Summary
Is [itex]\Delta G[/itex] 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 [itex]n[/itex] moles a substance from [itex]\overline{G}_1[/itex] to [itex]\overline{G}_2[/itex] you have to do work [itex]W = n(\overline{G}_2 -\overline{G}_1)[/itex]. Sounds reasonable enough. But then they give as an example moving [itex]n[/itex] moles of an ideal gas from a tank at pressure [itex]p_1[/itex] to a higher pressure [itex]p_2[/itex] at constant temperature. This indeed requires compression work [itex]W = nRT(\ln(p_2) - ln(p_1))[/itex], which happens to be equal to [itex]n(\overline{G}_2 -\overline{G}_1)[/itex].

But this seems like kind of a fluke to me. For an isothermal process, it's the change in Helmholtz free energy, [itex]\Delta A[/itex] that measures the work required to bring about the change of state. In this case, [itex]\Delta A = \Delta G[/itex] because we happen to have an ideal gas kept at constant temperature, so the product [itex]pV[/itex] 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?
 
Chemistry news on Phys.org
  • #2
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).
 
  • Like
Likes sponteous
  • #3
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.
 
  • #4
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.
 
  • Like
Likes sponteous
  • #5
Chestermiller said:
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:
  • #6
##\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.
 
  • Like
Likes sponteous
  • #7
sponteous said:
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.
 
  • Like
Likes sponteous
  • #8
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.
 
  • #9
sponteous said:
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.
 
  • #10
Chestermiller said:
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.
 
  • #11
sponteous said:
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.
 
  • Like
Likes sponteous
  • #12
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.
 
  • Like
Likes Chestermiller

1. What is partial molar Gibbs free energy?

Partial molar Gibbs free energy is a thermodynamic property that describes the change in the Gibbs free energy of a mixture when the amount of one component is changed while holding the amounts of all other components constant.

2. How is partial molar Gibbs free energy different from regular Gibbs free energy?

Regular Gibbs free energy is a thermodynamic property that describes the maximum amount of useful work that can be obtained from a system at constant temperature and pressure. Partial molar Gibbs free energy takes into account the effects of mixing different components in a mixture, while regular Gibbs free energy does not.

3. Why is partial molar Gibbs free energy important in thermodynamics?

Partial molar Gibbs free energy is important because it allows us to understand the behavior of mixtures and the changes in their properties when the amounts of different components are varied. It is also useful in predicting the equilibrium conditions of a mixture.

4. How is partial molar Gibbs free energy calculated?

Partial molar Gibbs free energy can be calculated using the equation: ΔG = ∑niRTln(xi), where ΔG is the change in Gibbs free energy, ni is the number of moles of component i, R is the gas constant, T is the temperature, and xi is the mole fraction of component i in the mixture.

5. What are some practical applications of partial molar Gibbs free energy?

Partial molar Gibbs free energy is used in various fields such as chemistry, biochemistry, and materials science. It is particularly useful in studying phase equilibria and predicting the behavior of mixtures in different conditions. It is also used in the design and optimization of chemical processes and in understanding the properties of biological systems.

Similar threads

Replies
3
Views
765
Replies
12
Views
725
Replies
3
Views
2K
Replies
15
Views
10K
Replies
9
Views
2K
  • Biology and Chemistry Homework Help
Replies
4
Views
1K
  • Chemistry
Replies
4
Views
3K
  • Biology and Chemistry Homework Help
Replies
1
Views
2K
Replies
10
Views
1K
Replies
12
Views
2K
Back
Top