# Fundamental Thermodynamic Relation and Helmholtz Energy

1. May 11, 2013

### jbwebb

I'm confused about the condition for spontaneity for the Helmholtz energy. My textbook (McQuarrie, "Physical Chemistry") derives the conditions as follows. We start with the combined law of thermodynamics:

dU = δq + δw ≤ TdS – PdV since δq/T ≤ dS
dU – TdS + PdV ≤ 0

For a process at constant V and T,

dU – TdS ≤ 0 since dV = 0

which we can rewrite as

d(U – TS) ≤ 0 or dA ≤ 0, where A = U – TS

I'm first confused about the inequality in the combined law. For a reversible process, we get dU – TdS + PdV = 0 because δqrev = TdS and δwrev = –PdV. However, I've read that because this leads to an equation where all the variables are state functions, the equality holds for all processes. So when does the inequality occur?

Secondly, from the combined law dA can be written as dA = –PdV – SdT. If V and T are constant, then how can dA be nonzero? Wouldn't dV = 0 and dT = 0? Wikipedia has a brief explanation (http://en.wikipedia.org/wiki/Helmholtz_free_energy#Minimum_free_energy_and_maximum_work_principles), but it doesn't make much sense to me. Also, why does only the equality of the combined law apply now (i.e. why not dA ≤ –PdV – SdT)?

2. May 11, 2013

### Jano L.

First you need to understand the relation

$$\Delta S \geq \frac{\Delta Q}{T_r}.$$

Then everything will follow.

The relation has this meaning:

Let the system go from equilibrium state 1 to equilibrium state 2 irreversibly, while in thermal contact with reservoir which has constant temperature $T_r$. The system does not need to have temperature during the process.

Because the initial and the final states are equilibrium states, they already determine the change of entropy:

$$\Delta S = S(2) - S(1).$$

Now, Clausius inequality says that this change is greater or equal to absorbed heat $\Delta Q$ divided by the temperature $T_r$ of the $reservoir$.

If the system went from 1 to 2 reversibly, the change would be equal to $\Delta Q / T_r$.

What they mean by "all processes" are not really all processes, but only those that are quasi-static, i.e. those in which the entropy and the volume can be defined at all times (then T and P of the systen are defined as well). That occurs only if the system passes through sequence of equilibrium states. The process may still be irreversible, and it still may be that $dS \geq dQ / T_r$, although $dS = dQ / T$. This happens whenever reservoir heats colder body slowly - it may be quasi-static process, but still it is irreversible, because the heat will never go in the opposite direction.

The equation

$$dA = -SdT - P\,dV$$

follows from the definition of A:

$$A = U - TS$$
and from the equation ("combined law")

$$dU = TdS - P\,dV.$$

All quantities can change, there is no restriction $V,T = const.$ in general. This restriction is imposed for special situations, like when studying reaction which starts and ends at the same T,V.

The equality is used because only quasi-static processes are in mind.

3. May 11, 2013

### jbwebb

Thanks so much for your help! That makes a lot more sense now. Just to make sure I understand you correctly...

The inequality applies to the temperature of the reservoir and not necessarily that of the system. During an irreversible process, if we were able to define a series of equilibrium states as a quasi-static process, then we could still find the infinitesimal change in entropy by the equation $dS = dQ / T$, where T is the temperature of the system. However, if we integrated over all equilibrium states between the initial and final states, then the overall change in entropy would still satisfy $\Delta S \geq \Delta Q/T_r$, where Tr is the temperature of the reservoir (which remains constant during the process). Am I correct in saying that the initial and final temperature of the system will be equal to that of reservoir? So when we're talking about the condition for spontaneity, we're really talking about the temperature of the reservoir in the equation $dA = dU - TdS \leq 0$.

I am still not quite clear about how A can be used as a condition for spontaneity. So V and T are not necessarily constant, but if they are, then a spontaneous process must satisfy $dA \leq 0$, right? How can we reconcile this with $dA = -SdT - PdV$? I guess I don't see how if V and T were to be constant, that $\Delta A$ could be anything other than 0. Is it because T of the system is not really constant in this case, but only is the same in the initial and final equilibrium states? That is, when we say an irreversible process occurs at constant T and V, we really are referencing T of the reservoir.

4. May 11, 2013

### Staff: Mentor

In your original posting, there is a mistake that may be the source of your problems:

δw=-PsurrdV, not -PdV. It is only equal to -PdV for a reversible path, in which case P = Psurr. Therefore, your inequality is incorrect.

For a closed system between two infinitesimally separated equilibrium states,
dU = TdS-PdV (always)

Also, from the definition of A and the above equation, for a closed system between two infinitesimally separated equilibrium states,

dA = -SdT-pdV (always)

So, for V and T constant, dA = 0. Can you think of any infinitesimal processes on a closed system, where, if dT = dV = 0, other than one in which there is absolutely no change from the initial state to the final state. Therefore, dA = 0. So what?

If you have an initial state of a closed system in which, say, you suddenly remove a barrier between two compartments and allow the system to reequilibrate, you will find that the Helmholtz free energy for the final state will be lower than for the initial state.

5. May 11, 2013

### Jano L.

Yes, that is the assumption, except it is better to use total changes $\Delta A$, etc. You put your chemicals into vessel of constant V, everything has temperature $T_r$. That is state 1. You start the reaction, the vessel heats up, you wait until its over, until again everything has temperature $T_r$ (maintained by the heat reservoir - environment). That is state 2. Then you compare the Helmholtz energies and it turns out that

$$\Delta A \leq 0,$$
if the reaction occured spontaneously (by itself, irreversibly).

The equation $dA=−SdT−PdV$ refers to something different. Forget the above example of spontaneity. Now the system is allowed to move only quasi-statically through sequence of equilibrium states, where it has T, V. You take your vessel and change its volume and temperature anyhow. Any changes in T and V are now allowed, and the equation tells you the change in A if you make changes dT and dV in its temperature and volume.

This equation does not need to apply at all for the above question of spontaneity; we can derive $\Delta A \leq 0$ without $dA=−SdT−PdV$ being valid.

However, sometimes the spontaneous process may be so close to quasi-static (slow reaction) that $dA=−SdT−PdV$ applies. Then $T$ can change during the process from $T_r$ to $T_r$ in any possible way, and so can V; they are not constant during the process. In the end, however they have to return to their assumed values $T_r, V$ - that is the assumption of the problem.

6. May 11, 2013

### jbwebb

Ah thanks for the clarification about the pressure. So would the correct statement be

$dU - T_rdS + P_{surr}dV \leq 0$

where the equality holds for reversible processes and the inequality for irreversible processes?

Also, when my textbook says,

"In a system held at constant T and V, the Helmholtz energy will decrease until all the possible spontaneous processes have occurred, at which time the system will be in equilibrium and A will be a minimum. At equilibrium, dA = 0."

what do the "constant T and V" refer to? In this case, is the equality $dA = -SdT - PdV$ not applicable because the initial state is not in equilibrium? Or am I still missing something?

7. May 11, 2013

### jbwebb

Whoops wrote that post before reading your most recent one, Jano! Looking at it now...

8. May 11, 2013

### Jano L.

Chestermiller is right that the work is determined by the external pressure of the reservoir, but that refers rather to situation when the reservoir fixes T, P. In your case, we fix T and V, so in fact we can forget the pressure.

"The constant T,V" refers to the fact that the system is bound to begin and end at the same T,V maintained by the environment and vessel.

9. May 11, 2013

### jbwebb

Oh I think I get it! The equation $dA = -SdT - PdV$ refers specifically to some process between equilibrium states. But this doesn't necessarily apply to spontaneous processes at the infinitesimal level because some cannot be modeled quasi-statically. This isn't a problem because we've already determined that $\Delta A \leq 0$ is a condition for spontaneity at constant $V$ and $T$. For ones that can be modeled quasi-statically, however, you could "hold" the system at constant $V$ and $T$ by ensuring that the final and initial equilibrium states have the same $T_r$ and $V$, even though $T$ and $V$ of the system will vary during the process. This way, when $dA = -SdT - PdV$ can still be applied, $dA$ is not necessarily zero (and less than zero if the process is spontaneous). Is that right?

10. May 11, 2013

### Jano L.

Yes, in our scenario, if the reaction is so slow that $dA = -SdT -PdV$ applies, T and V will vary in some way but eventually return to temperature of the reservoir $T_r$ and original volume (let's denote it $V_0$). Then in principle the change in the free energy can be calculated by integrating $dA$*along the path joining 1, 2 in the space of eq. states:

$$\Delta A = \int_1^2 -SdT-PdV \leq 0.$$

I would like to add that all this works if the reservoir does not get far from equilibrium state; its temperature $T_r$ has to remain constant all the time, so at least it has to be big.

11. May 11, 2013

### jbwebb

Wow, thank you very much! I've learned so much more from this thread in the last few hours than I did pouring over my textbook and searching through the internet this past week. This stuff finally makes sense to me now.

12. May 11, 2013

### Jano L.

That's great!