X-Message-Number: 5130
Date: 07 Nov 95 16:41:52 EST
From: "Steven B. Harris" <>
Subject: SCI.CRYONICS:Okay, you asked for Thermo
Perhaps I can add something to the discussion of pressure and
freezing points, since there seems to be a lot of confusion.
Hopefully the following won't come across to all as "spur-of-the-
-moment, often late at night, stream-of-consciousness stuff,
ramblings mainly to help one's own thoughts along, unedited." I
agree there is nothing more sad than the depressed guy who has to
admit "I've written two children's books; but not on purpose..."
Ettinger: "On previous occasions Brian has said that, when
pressure is released on a water solution at low temperature, it
may partly freeze, but then the released heat of fusion will melt
it again, at least partly. Similarly, now he says that adding a
little pressure to ice at the freezing point will cause
only a little melting, the energy required coming from the
internal energy of the system and lowering the temperature,
tending to cause refreezing. This certainly sounds sensible.<<
Comment: Yes, and I also agree. I'll discuss this more below.
Brian is certainly right that when ice is pressurized, the energy
to melt it is mostly not PV work. Only a bit of pressure is
enough to melt all the ice you like at 0 C., but the heat for
this must come out of the environment (from a skater, say), and
this is thus NOT an adiabatic process by definition. If the
system was kept insulated, only enough ice would melt to remove
enough heat to lower the temp to the new equilibrium temp for
that pressure (not much), and then (as Brian says) the process
would halt.
Ettinger "but I was troubled by trying to picture the
mechanism.. How does added pressure lower the freezing point? One
way to think of it is that pressure tends to compress, and water
is denser than ice. But then we remember that the heat of fusion
is sometimes thought of as the binding energy of molecules in
crystal formation, similar to the binding energy of exothermic
chemical reactions. If we push on the crystals, tending to break
the bonds between molecules, why should those or neighboring
molecules obligingly give up their thermal energy?<<
Comment: they don't-- in melting, molecules of a crystal
ABSORB thermal energy. They do so because at a particular
temperature, increasing entropy of the system (solid to liquid)
balances the decreasing entropy of the universe (or the rest of
the closed system, if this is adiabatic) when it supplies the
heat to do the melting (heat of fusion). The point of zero
change in free energy with phase change, which tells us where
phase equilibrium is, is just another way of keeping track of
this same entropy balance.
The reason melting points of pure substances change with
pressure (which seems to be the question du jour here) is MAINLY
that the different amounts of pressure-volume work due to the
phase change at different pressures, either adds, or subtracts,
differing amounts of entropy and free energy to the process.
This gets lumped in with the effects of the thermal energy
necessary to break the molecular bonds of the crystal.
lower-order effects (*usually* they are lower-order) come from
the fact that the compressibility of the different phases may not
be quite the same (and thus the volume change with phase change
is not quite independent of pressure), and also because the
enthalpy of phase change (which automatically takes into account
the PV work during the change) is *still* not quite independent
of temperature, due to similar differences in the way heat
capacities of the different phases vary with temperature. With
water there are gentle curves at high temperatures due to this
latter effect, but still the first order PV work effects are
predominant over small ranges and at low temps, and thus the
pressure-temperature (P-T) diagram phase boundaries are often
reasonably linear. The relevant thermodynamic relationship is the
Clapeyron equation, which is exact:
dP/dT = delta H/(T*delta V)
Where P is pressure and T is the temperature at equilibrium for
the phase change, and delta H is the specific change in enthalpy
(heat produced or absorbed) and delta V the specific change in
volume, during the phase change.
For example, let us do the calculation for compression of
water over the range Brian Wowk gave (1000 atm). We will assume
that delta H and delta V are constant and that T is 273 K. Delta
H for water is about -330 J/g (the amounts of substance cancel
since these are specific quantities, so I may as well use grams
here as moles), and the delta V is about 10% of a cc/gram, or
10^-7 m^3/g. That gives us a dP/dT of - 1.2 x 10^7 N/m^2/degree
K = -120 Atm/degree K = 120 Atm/ -degree K. For 1000 atm that's
a freezing point depression of -8.3 K, as opposed to the figure
Brian gave of -9 K. Close enough.
Please note that when we try to integrate the Clapeyron
equation (even making the simple assumptions above) we get a
constant of integration which we can only put in by knowing
(measuring) the actual equilibrium point SOMEWHERE on the P-T
phase diagram (as in the above example case, when we knew a
priori that it was 0 degrees and 1 atm for water/ice). Thermo-
dynamics tells us how phase change points CHANGE with P and T
(the slope), but not where to start. Nor does thermo tell us
anything about what the different phases will be. These are all
bulk properties of matter which are in theory calculable from
first principles (quantum mechanics), but which nobody has the
computational power to do at present for any system (so far as I
know). If mother nature decides that another phase of crystal is
"coming up," with a different P or T, our thermo calculations
past that point are no good, although we can predict in general
the *directions* of some phase changes from Le Chatelier's
principle, if we know the sign of the enthalpy and volume changes
of those transitions. Does the substance expand when changing
phase, or does it contract? If it contracts, then pressure
increases will favor than phase change. Does it absorb heat?
Then higher temperatures will favor that phase change.
Water here again provides a nice example. Our familiar low
pressure water-ice (the polymorph called "Ice-I") is less dense
than water, so pressure increases favor melting. If you take
ordinary water ice at -9 C and compress it to 1000 atm or so
(corresponding to the deepest ocean pressures), it will melt
completely (taking up heat, so long as you keep the temp at -9),
until it is liquid again. If you continue to compress at the
same temperature, however (removing heat as necessary), you will
eventually, at about 4,000 atmospheres, find that your liquid
water has frozen once again at -9 C to a new ice polymorph with a
structure DENSER than water, called Ice-V (perhaps going through
some metastable Ice-IV first). With even more compression you
get Ice-VI, then (somewhere above 20,000 atm) Ice-VII-- each
solid phase denser than the last [Vonnegut fans will know that if
you manage to make Ice-IX at normal conditions, you may be
letting yourself in for some major zah-mah-ki-bo].
But you never know what you're going to get in a phase
diagram until you explore it. For instance, at lower temps if
you compress Ice-I, you will make Ice-II or Ice-III (different
crystal structures). At higher temps-- say if you compress a
live human at 37 C.-- nothing happens (besides death) until you
get to about 10,000 atm, at which time you freeze directly from
liquid to Ice-VI, still at 37 C, without going through other
phases.
Now, if we actually DID this to a person, there is no reason
to think that we'd do much less damage than if we froze him at
-196 C. When water freezes, the major damage is NOT done by
expansion, so we don't save anything by freezing at pressure to a
denser ice polymorph. Perhaps a bit less damage would be done by
differential thermal effects (if you left the person frozen solid
at high temps!), but you'd be left with the baro-trauma and the
osmotic trauma.
Liquid water has a lot of new strong hydrogen bonds which
form upon any kind of freezing, and therefore you always get a
lot of heat output on freezing from liquid water to crystal
solid, no matter what ice-polymorph you freeze to. If you quick-
freeze at some point far from equilibrium, you still have all
this heat of fusion to deal with (as Brian notes), as well as all
the osmotic effects (unless the freezing is an incredibly fast
flash freeze-- only possible with thin layers of tissue). The
osmotic effects are severe with fast (verses flash) freezing, due
to lack of time for equilibration between cell insides and
outsides, but still enough time for residual liquid water with
high osmolality to do damage. Pressure doesn't fix this.
Specifically, as to the question of what would happen if you
pressurized liquid water to 1000 atm, extracted amount of heat Q
to take it down to -9 C, and then took 999 atm of pressure off,
the answer is that (save for small second-order compression
effects) you'd get the same thing happening as if you supercooled
ordinary water to -9 and then seeded it with a crystal to allow
some of it to quick freeze. The ordinary heat of fusion would be
released, MINUS the heat Q you already had to extract to cool
liquid water from 0 to -9 C, which is only about 1/9th as much as
you have to extract to freeze it completely at 0 C. Thus, for
the ice you formed, you'd still get 8/9ths of the ordinary 0 C
heat of fusion out, and all at once. This would heat the rest of
your sample. If you did this adiabatically (in a thermos bottle,
say, where heat could neither enter or leave) this would freeze
1/9th of your sample very rapidly, expanding that part and
heating the whole mix to 0 C, and then the process would stop, at
a mix of 1/9ths ice and 8/9ths liquid water at 0 C (this equilib-
rium temp set uniquely by the ambient pressure, the substance,
and the amount of energy in the system-- see the Gibbs phase
rule). This would leave you in the same place as if you had
extracted the same initial amount of heat Q from water at 0 C and
1 atm, allowing some to freeze as you went. You can't fool
Mother Nature. You can stay a long time at non-equilibrium
conditions, but once you get to equilibrium, you're always at the
same place, no matter how you did it. I'm happy that Ettinger
and Wowk now seem to agree on this.
Ettinger: "Three main practical questions remain, as well as
several subsidiary ones. One is the temperature to which the
body would have to be reduced, and the pressure required, if it
is suddenly to freeze completely upon release of pressure. (In
the neighborhood of - 100 C?)"
Comment: this is a good guess, and the laws of thermodynamics
don't forbid it (I know Bob must remember more thermo than he
lets on-- I have a picture of him with the constant pressure heat
capacity integral written on the blackboard behind him).
However, unfortunately the nature of water and its phase diagram
(those funny bulk properties due to quantum properties) DO forbid
this for H2O. Without supercooling (i.e., if you keep only to
equilibrium conditions), you cannot cool pure liquid water at ANY
pressure below about -20 C, which corresponds to a pressure of
about 2,000 atmospheres. Applying either higher pressure or
lower temperature at that point gives you Ice-III. There is no
way to predict this a priori, but that's what happens. Mother
Nature hath spoken.
Ettinger: "The second is whether, with this procedure,
conservation of information is clearly improved over other
procedures (even though damage by some criteria might be worse)."
I've already answered this as well as I can. Pressure
chambers MAY be useful perhaps one day to *avoid* freezing during
rewarming from vitrified states, or in producing pure vitrified
states, without freezing, at lower concentrations of
cryoprotectant. I cannot at present see any way that they will
save us any trauma from freezing itself.
Steve Harris
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