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### Topic: Describing evaporation without entropy — what about Helmholtz?  (Read 4793 times)

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#### MackTuesday

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##### Describing evaporation without entropy — what about Helmholtz?
« on: November 06, 2015, 10:25:25 PM »
I have in mind a rigid, impervious, heat-insulating vessel containing a monatomic liquid and vacuum. The vessel has an insulating forcefield inside, exactly at the surface of the liquid. It can be toggled.

Step 1: The liquid is in thermal equilibrium. The field is switched off for a very short, predetermined time. Some portion of the molecules at the surface exceed some threshold of kinetic energy, enough to escape into the vacuum.

Step 2: Now the liquid is no longer in thermal equilibrium because some of its fastest molecules have escaped. The same can be said for the vapor, except it lacks slow ones. We wait until both regain equilibrium -- they will do so separately because of the forcefield -- and then turn the field back on.

We return to step 1. This time there is some condensation as well as evaporation.

We repeat steps 1 and 2 again and again. At each iteration we can always predict the proportion of liquid that will evaporate, and the proportion of vapor that will condense, based purely on the temperatures of the liquid and vapor, the volume of the liquid, the pressure of the vapor, and the length of time the field is off. We can do this without considering changes in entropy.

But the Helmholtz energy is what determines the direction of this process, and its calculation requires knowledge of the change in entropy.

So is there a problem with my model?

#### Corribus

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##### Re: Describing evaporation without entropy — what about Helmholtz?
« Reply #1 on: November 07, 2015, 12:28:14 AM »
Step 2: Now the liquid is no longer in thermal equilibrium because some of its fastest molecules have escaped.
If the force field is turned back on after they escape, so the liquid is an isolated state again, why is it no longer in thermal equilibrium?
(Or, depending on how you want to define things - was the system ever in thermal equilibrium, if molecules escaped when you lowered the force field?)
What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman

#### MackTuesday

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##### Re: Describing evaporation without entropy — what about Helmholtz?
« Reply #2 on: November 07, 2015, 11:14:58 AM »
Step 2: Now the liquid is no longer in thermal equilibrium because some of its fastest molecules have escaped.
If the force field is turned back on after they escape, so the liquid is an isolated state again, why is it no longer in thermal equilibrium?
(Or, depending on how you want to define things - was the system ever in thermal equilibrium, if molecules escaped when you lowered the force field?)
Sorry, I misused the terminology. I was talking about thermalization, in which (I assume) particle speed tends toward a Boltzmann distribution. Once you chop off the tail of the Boltzmann distribution, the system can no longer be called "thermalized". What I meant by thermal equilibrium was that the particles in the liquid have reached whatever distribution of energy that corresponds with equipartition.

#### mjc123

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##### Re: Describing evaporation without entropy — what about Helmholtz?
« Reply #3 on: November 09, 2015, 08:41:15 AM »
Quote
What I meant by thermal equilibrium was that the particles in the liquid have reached whatever distribution of energy that corresponds with equipartition.
In other words, the entropy of the liquid is maximised. The system evolves, while maintaining its total internal energy constant (because it is isolated), to its most probable (highest entropy) distribution of energy. This happens as the result of billions of molecular collisions, each of which can be described in terms of the energy of the individual molecules, but the overall outcome for the ensemble of billions of molecules is the most probable distribution of energies, i.e. the state of maximum entropy. Thus the state of the liquid cannot be defined by its energy alone, you need the entropy too.
The same applies to the vapour, and to the distribution of molecules between the liquid and vapour, with the added complication that it takes energy to transfer a molecule from the liquid to the vapour. The energy consumed/released by evaporation/condensation goes to lower/raise the temperature of the isolated system, and hence lower/raise its entropy, in addition to the entropy change due to going from liquid to gas or vice versa. Equilibrium will be achieved by maximising the entropy of the system. (If the system was not thermally insulated, and was maintained isothermal by exchange of heat with the surroundings, the entropy of the universe would be maximised, which would be equivalent to minimising the Helmholtz energy for the system. But free energy seems not to be a useful tool when talking about non-isothermal processes - see http://www.chemicalforums.com/index.php?topic=81854.0)

#### Enthalpy

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##### Re: Describing evaporation without entropy — what about Helmholtz?
« Reply #4 on: November 09, 2015, 09:56:48 AM »
Step 2: Now the liquid is no longer in thermal equilibrium because some of its fastest molecules have escaped. The same can be said for the vapor, except it lacks slow ones.

No. The fastest molecules in the liquid evaporate, but when leaving the liquid, they lose this speed, because the other molecules in the liquid attract them.

In an extreme case, when starting with a distribution like exp(-E/kT), if you subtract the same (vaporization) energy H to every molecule, you get again the same distribution exp(-E/kT), just with a pre-exponential factor.

#### MackTuesday

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##### Re: Describing evaporation without entropy — what about Helmholtz?
« Reply #5 on: November 09, 2015, 10:30:56 PM »
Quote
What I meant by thermal equilibrium was that the particles in the liquid have reached whatever distribution of energy that corresponds with equipartition.
In other words, the entropy of the liquid is maximised. The system evolves, while maintaining its total internal energy constant (because it is isolated), to its most probable (highest entropy) distribution of energy. This happens as the result of billions of molecular collisions, each of which can be described in terms of the energy of the individual molecules, but the overall outcome for the ensemble of billions of molecules is the most probable distribution of energies, i.e. the state of maximum entropy. Thus the state of the liquid cannot be defined by its energy alone, you need the entropy too.

The same applies to the vapour, and to the distribution of molecules between the liquid and vapour, with the added complication that it takes energy to transfer a molecule from the liquid to the vapour. The energy consumed/released by evaporation/condensation goes to lower/raise the temperature of the isolated system, and hence lower/raise its entropy, in addition to the entropy change due to going from liquid to gas or vice versa. Equilibrium will be achieved by maximising the entropy of the system. (If the system was not thermally insulated, and was maintained isothermal by exchange of heat with the surroundings, the entropy of the universe would be maximised, which would be equivalent to minimising the Helmholtz energy for the system.
Yes, entropy changes, but with the model I present, you don't need to quantify the change in order to know which direction the process follows. All you need to know is the energy distributions of the liquid and vapor. You chop off the tail of one and the head of the other (and you can know exactly where to chop if you've previously measured the ΔU of vaporization), take some proportion based on an estimate of the volume of the interface between the different phases, and compare the number of particles between the two sets.
Quote
But free energy seems not to be a useful tool when talking about non-isothermal processes...
I don't see the truth of that, because for example, ΔG = ΔH - Δ(TS). So there exists a formulation that doesn't require a constant T.

#### MackTuesday

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##### Re: Describing evaporation without entropy — what about Helmholtz?
« Reply #6 on: November 09, 2015, 10:36:37 PM »
Step 2: Now the liquid is no longer in thermal equilibrium because some of its fastest molecules have escaped. The same can be said for the vapor, except it lacks slow ones.

No. The fastest molecules in the liquid evaporate, but when leaving the liquid, they lose this speed, because the other molecules in the liquid attract them.

In an extreme case, when starting with a distribution like exp(-E/kT), if you subtract the same (vaporization) energy H to every molecule, you get again the same distribution exp(-E/kT), just with a pre-exponential factor.
Ah, OK. So you're saying slow particles are likely more common in the vapor than in the Boltzmann distribution, at least at first, right?

#### mjc123

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##### Re: Describing evaporation without entropy — what about Helmholtz?
« Reply #7 on: November 10, 2015, 07:22:05 AM »
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All you need to know is the energy distributions of the liquid and vapor.
If you know that, you know the entropy, don't you? I think the two models contain the same information, but described in different ways. On your model, you need not only the total energy (i.e. U) but also the distribution of energies (which is related to S), so you effectively bring in the Second Law as well as the First. As I said, on the molecular level the behaviour can be explained in terms of energy, but to describe the behaviour of large collections of molecules you need the statistics, which brings in the entropy or the energy distribution. Entropy is a useful concept to describe this, and the "entropy of the universe increases" criterion is a simple concept to apply (if not always easy to calculate).
Quote
You chop off the tail of one and the head of the other (and you can know exactly where to chop if you've previously measured the ΔU of vaporization), take some proportion based on an estimate of the volume of the interface between the different phases, and compare the number of particles between the two sets.
I'm not sure that this is any simpler than calculating ΔU and ΔS, if not harder. If you know ΔU and ΔS (assuming to a first approximation that they are independent of temperature) you can calculate equilibrium vapour pressure vs temp, so in any state of the system you know which way it will go.
Quote
I don't see the truth of that, because for example, ΔG = ΔH - Δ(TS). So there exists a formulation that doesn't require a constant T.
Yes, ΔG can be defined, but ΔG < 0 is not useful as a criterion - the thread I quoted contains an example where ΔG > 0 although ΔSuniverse > 0.

#### Enthalpy

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##### Re: Describing evaporation without entropy — what about Helmholtz?
« Reply #8 on: November 10, 2015, 11:29:40 AM »
Step 2: Now the liquid is no longer in thermal equilibrium because some of its fastest molecules have escaped. The same can be said for the vapor, except it lacks slow ones.

No. The fastest molecules in the liquid evaporate, but when leaving the liquid, they lose this speed, because the other molecules in the liquid attract them.

In an extreme case, when starting with a distribution like exp(-E/kT), if you subtract the same (vaporization) energy H to every molecule, you get again the same distribution exp(-E/kT), just with a pre-exponential factor.
Ah, OK. So you're saying slow particles are likely more common in the vapor than in the Boltzmann distribution, at least at first, right?

I say that, in this simplified case, and for a limited amount of evaporated compound, the vapour follows the same Boltzmann distribution at the same temperature.