# I Where is the work done coming from in Helmholtz free energy

1. Jul 29, 2017

### tze liu

Sorry,i want to ask a question here

the note said the volume is "fixed" here.

if the volume if fixed,how comes the work done(because no change of volume) here
i totally get lost here

thank

#### Attached Files:

File size:
71.8 KB
Views:
139
2. Jul 29, 2017

### stevendaryl

Staff Emeritus
That description is a little confusing, and the Wikipedia article on the topic is no better.

The article goes on to say that $\Delta A$ can be nonzero because the full definition is $dA = -S dt - P dV + \mu dN$, where $N$ is the number of particles, and $\mu$ is the chemical potential. So changing the number of particles can also change $A$. That's true, but doesn't really address the seeming contradiction, as far as I can see.

But here's the way that I think of the various kinds of "free energy". In the first place, imagine that you have two systems that are thermally isolated (no heat in or out) but share the same volume, as in figure 1:

Figure 1: Thermally isolated systems sharing a volume

If system $S_1$ expands by pushing the partition to the right, then it will do work on system $S_2$. The amount of work done is given by:

$\Delta W = P \Delta V_1$

We can use the definition of internal energy to relate the work to the heat going in or out:

$\Delta U = \Delta Q - \Delta W = T \Delta S - P \Delta V$

(Sometimes this is written with a plus sign in front of $\Delta W[/tex], but that depends on whether you're talking about the work done by the system or the work done on the system. If the changes are reversible, as we're assuming, the two are just the negatives of each other). If the systems are thermally isolated, then [itex]\Delta Q = T \Delta S = 0$, which means that $\Delta W = - \Delta U$

So the work done by the system is just equal to the loss of internal energy of the system. So in this scenario, the internal energy is a measure of the ability of the system to do work.

Note: The above discussion is for a reversible change. In an irreversible change, the work done by the system can be less than $-\Delta U$. So we really get:

$\Delta W \leq - \Delta U$

Now, let's relax the constraint that the two systems are thermally isolated, and instead assume that they are free to exchange heat with a surrounding medium, as in Figure 2:

Figure 2: Systems may exchange heat with surrounding medium, but temperature is kept constant

In this case, it's no longer necessarily true that $\Delta Q = 0$ when system 1 expands. As it expands, it can absorb heat from the surrounding heat bath, making $\Delta Q > 0$. But assuming the heat bath is much larger than either of the two systems, the temperature will remain constant during any reversible change of system 1. So we can use a different thermodynamic identity:

$\Delta A = \Delta U - \Delta (S T) = - P \Delta V - S \Delta T$

In this case, the heat bath is keeping the temperature constant, so $\Delta T = 0$. So we have:

$\Delta A = -P \Delta V = -\Delta W$

So in this case, it is $A$ that gives a measure of the potential of the system to do work, rather than $U$.

Again, this is for a reversible change. For an irreversible change, $-\Delta A$ is an upper bound on the amount of work done by the system:

$\Delta W \leq - \Delta A$

The discussion in Wikipedia is misleading, because of course, a system can't do any work if the volume is not allowed to change. I think that the issue is resolved by assuming that there is more than one system involved. The volume of the composite system doesn't change, but the volumes of the smaller subsystems can change, and Helmholtz free energy is a measure of the capacity of the subsystem to do work.

3. Jul 29, 2017

### tze liu

is that your assumption?

4. Jul 29, 2017

### Staff: Mentor

You are correct that no P-V work is done. So, if the only focus is P-V work, and the P-V work is zero, then the inequality is satisfied. But there are other kinds of work that can be done on the system (or the system can do on the surroundings), and, if these are operative, then the inequality applies. For a more detailed derivation and description of all this, see Denbigh, Principles of Chemical Equilibrium.

5. Jul 29, 2017

### tze liu

""But there are other kinds of work that can be done on the system""
thank
do u means those work done doesn't change the volume of the system?

6. Jul 29, 2017

### Staff: Mentor

Yes

7. Jul 29, 2017

### tze liu

when I study thermodynamic,

i never heard that the work done doesn't change the volume of the system

So what is such "work done" looks like and how does this "work done" act on the system?

do u have any example?thank

8. Jul 29, 2017

### Staff: Mentor

Electrochemical work, stirring work

9. Jul 29, 2017

### tze liu

why those work doesn't change the volume?

10. Jul 29, 2017

### Staff: Mentor

When you stir a liquid in a rigid closed container, does the volume of the container or the liquid change?

11. Jul 29, 2017

### tze liu

oh i see

the Work here is the free mechanical work.
but not contributed to the change of volume.

thank

12. Jul 29, 2017

### Staff: Mentor

Don't forget also the possibility of electrochemical work like a battery.

13. Jul 29, 2017

### tze liu

is this sentense incorrect?

even that the system is thermal isolated and the volume is fixed
there are STILL SOME electrochemical work or other type of mechanical work here
so the delta U is not zero?

#### Attached Files:

File size:
74.3 KB
Views:
111
14. Jul 29, 2017

### Staff: Mentor

You can have a current-carrying wire coming into the container and another wire exiting the container, with the current driven by a battery situated inside the container (elecrochemical). In this case, the internal energy of the system in the container changes (i.e., the battery is running down). The current can drive a motor outside the container. So the system is doing work.

15. Jul 29, 2017

### Lord Jestocost

This is always confusing. The essential point is the following:

Calculate the free energy of a system at a given temperature T in state #1: F#1
Calculate the free energy of the system at a given temperature T in state #2: F#2
(to do such calculations you generally need statistical physics)

The work which is done when going from state #1 to #2 in course of a reversible, isothermal process (at temperature T ) is then

Wrev = F#2 - F#1

E.g., in case F#2 < F#1, ABS(F#2 - F#1) is the maximum work you can extract during this isothermal process.

EDIT: In case of irreversible processes you have (work done positive, work extracted negative sign):

ΔW ≥ ΔF

Last edited by a moderator: Jul 29, 2017
16. Jul 31, 2017

### tze liu

By the way,Is that true the reverior and the system both are at the same constant temperature in my case?

17. Jul 31, 2017

### Staff: Mentor

No. If you had checked out Denbigh like I suggested, you would have found that T is the temperature of the reservoir and also the temperature of the system at the interface with the reservoir. It is also the initial and final temperature of the system in the initial and final equilibrium states. But, during the process, it is not the temperature of the system (which is non-uniform) except at the interface with the reservoir.

Chet

18. Jul 31, 2017

### tze liu

however, there is dF=d(U-TS) in my equation.
how can they use the temperature in the reservoir but not using the temperature inside the system?

there are two T in this case
the T in the reservoir and the other T in my system which can be described by F free energy.
why the equation dF=d(U-TS) ignore the temperature change inside the system.

19. Jul 31, 2017

### Staff: Mentor

If the temperature is not uniform within your system during the process, then F (which is a function of equilibrium state) is undefined. If you want to define F within your system when it is not at equilibrium, then you have to integrate the local free energy per unit mass over the mass of your system.

Last edited: Jul 31, 2017
20. Jul 31, 2017

### tze liu

it seems
if the temperature is constant in my process, i assume this temperature is T2

and the environment temperature is T1 ( T1 cannot be T2,if so, then there is no heat transfer)

why it should be dF=d(U-T1S) but not dF=d(U-T2S) ?

i dont know why they use the environment temperature in the equation but not using the system's temperature( assume the system has constant temperature)

21. Jul 31, 2017

### Staff: Mentor

As I said, see Denbigh.

22. Jul 31, 2017

### Staff: Mentor

Do you really think that, in an irreversible process, the temperature within the system is uniform?

From the Clausius inequality applied to this problem, we have: $$\Delta S=\frac{Q}{T}+\sigma$$ where Q is the heat transferred from the reservoir to the system, T is both the initial and final temperature of the system as well as the temperature of the reservoir (at the heat transfer boundary with the system), and $\sigma$ is the entropy generated within the system during the process (always positive). So we have: $$Q=T\Delta S-T\sigma$$
If we substitute this into the first law of thermodynamics, we obtain:
$$\Delta U=T\Delta S-T\sigma+W$$where W is the work done by the surroundings on the system. From this, it follows that $$W=\Delta F+T\sigma$$Since $\sigma$ is always positive, it follows that $$W\gt \Delta F$$

Last edited: Jul 31, 2017
23. Jul 31, 2017

### Staff: Mentor

24. Jul 31, 2017

### Staff: Mentor

Here is the direct quote from Denbigh: "Consider the special case (a) that the only heat transferred to the system is from a heat reservoir which remains at the constant temperature T; (b) that the initial and final temperatures, $T_1$ and $T_2$, of the system are equal, and are equal to the temperature T of the reservoir."

Note that this description fails to mention the important fact that the temperature of the system is not uniform during the process, or that it is equal to T only at the boundary with the reservoir (except in the initial and final states).

25. Jul 31, 2017

### tze liu

Oh I see.than you very much .There are still problems here.if the temperature at the beginning are equal to the reverior,how comes the heat transfer between them at the beginning?(because it violates the zero law)