Zan Divine
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The special case of distilling metals in vacuo
Don't you hate it when something you took for granted is shattered by special cases? Well, here's another one....
I've been interested in purifying the more volatile group 1 elements. A question that baffled me was why the less volatile and much higher boiling
impurity metals like sodium (present as an impurity in my s.m.) should come over with cesium or rubidium in the first place. Eventually I started
digging through the available literature to understand the mechanism.
With most liquids, if heat is supplied, the liquid's temperature rises to the boiling point, then remains constant while bubbles of vapor form beneath
the surface and boiling starts. If the rate of heating of the liquid is increased, the temperature remains constant and the rate of boiling increases.
The rate of boiling is determined only by the rate of heat transfer to the liquid and the actual temperature at which the boiling occurs is determined
by the partial pressures of the volatile constituents in the liquid and the total pressure above the surface.
In contrast, when boiling metals under a vacuum, the partial pressures are generally so small that boiling does not occur because at even a fraction
of a millimeter below the surface of the metal because the hydrostatic pressure is usually too great to permit the formation of a pocket of metal
vapor. The high thermal conductivity of metals also tends to prevent the local superheating necessary for bubbles to form. Thus, evaporation of
volatile constituents can only happen from the surface of the molten metal.
In this situation, the rate of evaporation depends upon;
1) the surface area of the metal
2) the surface concentration of volatile constituents
3) the surface temperature
4) the partial pressures of volatile constituents immediately above the surface
Increasing the heating does not increase the evaporation rate at constant temperature, but instead raises the temperature of the liquid until an
increased rate of evaporation (boiling) can balance the rate of heat input. When attempting to separate metals by fractional distillation this
directly translates to less clean separations.
This also suggests to me that the two factors that almost always work in opposition to one another, yield and purity, are even more interdependent for
boiling metal based syntheses.
[Edited on 7/29/2012 by Zan Divine]
He who makes a beast of himself gets rid of some of the pain of being a man. --HST
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blogfast25
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Quote: Originally posted by Zan Divine  |
Increasing the heating does not increase the evaporation rate at constant temperature, but instead raises the temperature of the liquid until an
increased rate of evaporation (boiling) can balance the rate of heat input. When attempting to separate metals by fractional distillation this
directly translates to less clean separations.
This also suggests to me that the two factors that almost always work in opposition to one another, yield and purity, are even more interdependent for
boiling metal based syntheses.
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Hmmm. You have literature or empirical evidence for this, Zan? Seems pretty counter-intuitive to me. You’re essentially saying alkali metals in near
vacuum superheat until the (total) enthalpy of vaporisation per unit of time matches the power input (assuming no heat losses). But that should be
true also at true boiling point (at that applicable pressure).
Superheating would favour less volatile components to come over, granted.
Are more prosaic explanations like mechanical entrainent considered (you do need bubbles for that, of course)?
[Edited on 29-7-2012 by blogfast25]
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Zan Divine
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The best literature reference I can give is:
AIME "Extractive Metallurgy Division - Vacuum Dezincing of Desilverized Lead Bullion"
Gokcen, T. R. A.
Which should be available in whole or in part simply by a Google search.
Although the metal doesn't boil as such, the evolution of the dissolved gasses in the metal may cause mechanical carry-over. That's why the reaction
mixtures are always topped off with a wad of SS wool.
It's hard (for me, anyway) to say what part entrainment might play. In the absence of the mechanical churning of boiling, this may be a relatively
less important effect.
[Edited on 7/29/2012 by Zan Divine]
He who makes a beast of himself gets rid of some of the pain of being a man. --HST
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watson.fawkes
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I can comment on this as a physics problem.
The first point is that vapor pressure is an equilibrium value. The equilibrium is between evaporation and condensation. More precisely, the
equilibrium vapor pressure is when the rate of evaporation is the same as that of condensation. These two rates, however, are not governed by the same
kinds of energies. Evaporation rate is about overcoming the surface tension so that an atom will evaporate (or desorb) from its surface. Condensation
rate, on the other hand, is simply about colliding with the surface inelastically.
The second point is that vacuum purification is not an equilibrium process. At high vacuum, you have close to the same evaporation rates, but
you do not have the same condensation process. At high vacuum, you don't have the same energy exchanges between particles as a result of collisions.
There's still a definable pressure, but only in a moving frame of reference with a velocity that's pointing away from the surface. The average motion
in this non-equilibrium situation is, thus, mostly not directed back at the evaporation surface, so the effective condensation rates drops toward
zero.
The third point is that lower-mass atoms have a higher mean velocity than higher-mass ones. The average velocity is proportional to m-1/2
for an ideal gas, as given by the Maxwell-Boltzmann distribution. Given two element vapors at the same pressure, the lighter one will have both a higher rate of condensation and
thus also a higher rate of evaporation.
The combination of all this is that when you drop the pressure in a vacuum chamber, the rate of lighter elements coming off the liquid will be
significantly higher than their vapor pressures at equilibrium would indicate. I've simplified a lot, but an ideal gas model gets the gist of the gas
phase across. At this juncture, I'll address the specific claim: | Quote: Originally posted by Zan Divine | In this situation, the rate of evaporation depends upon;
1) the surface area of the metal
2) the surface concentration of volatile constituents
3) the surface temperature
4) the partial pressures of volatile constituents immediately above the surface |
Item 1 affects the total
rate, but not the relative rate. Item 2 gives the baseline concentration that the relative rate of evaporation changes. Item 3, through experimental
data, determines partial pressures for the individual species that are related as above. Item 4, though, is right out; it's the wrong concept to use
directly.
For Rb vs. Na, the square root of the mass ratio is about 1.9; that for Cs is about 2.4. These represent effective multipliers on partial pressures
that convert them into more realistic evaporation rates. Because these multipliers are greater than one, vapor evaporation will tend to concentrate Na
in the vapor over that in the melt. Taking a cue from other distillation processes, that means that the concentration of Na in the melt is getting
smaller, which gives you an alternate purification means.
These multipliers are only guesses on what the realistic evaporation rates are. While I talked about the vapor phase explicitly, I was implicitly
using Raoult's law that the evaporation rates were proportional. I don't know how true this is in the present case, because we're dealing with liquid
metals.
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