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Melting/freezing point depression of water vs. other liquids

  1. Jul 25, 2013 #1
    Hey

    I'm trying to derive the Gibbs-Thomson equation for the freezing/melting point depression of liquids inside a confined space, such as a cylindrical pore. This has been observed for many liquids, such as water, benzene, heptane, etc. Basically my question is, why does water, which expands in volume when frozen, also has its freezing point reduced in pores?

    Here's a more detailed account of my confusion. The Young-Laplace equation for the pressure drop [itex]ΔP[/itex] across a hemispherical interface with radius [itex]r[/itex] reads
    [itex]\Delta P = \frac{2 \gamma}{r}[/itex],
    with [itex]\gamma[/itex] as the surface tension of the interface. For a liquid meniscus, this is the surface tension of the liquid-vapour interface, but for an ice-water interface, this is the surface tension of the solid-liquid interface.

    So far so good. Now, to get the Gibbs-Thomson equation, the 'ol Clausius-Clapeyron relation is whipped out:
    [itex]\frac{\Delta P}{\Delta T} \approx \frac{dP}{dT} = \frac{L_T}{\Delta V}\frac{1}{T}[/itex],
    with [itex]L_T[/itex] as the total latent heat of transformation. For a phase transition from ice to water, [itex]\frac{\Delta P}{\Delta T}[/itex] is negative because [itex]\Delta V[/itex] is negative, unlike most liquids where [itex]\frac{\Delta P}{\Delta T}[/itex] is positive.

    Anyway, to get the Gibbs-Thomson equation, this is plugged into the Young-Lapace equation above for [itex]\Delta P[/itex], giving
    [itex]T(r=\infty) - T(r) = \Delta T = \frac{2 \gamma \Delta V T}{L_T r}[/itex].

    So since [itex]\Delta V[/itex] is negative for water, why isn't [itex]\Delta T[/itex] negative as well, causing confined water to melt at a temperature higher than 0 degrees? I'm guessing that [itex]\gamma[/itex] is negative, which in this case is the surface tension of the ice-water interface, where for other liquids in contact with their frozen phase, it's positive. This would cancel out the negative from the [itex]\Delta V[/itex].

    Thanks for any input.
     
  2. jcsd
  3. Jul 26, 2013 #2
    I think I know where I've gone wrong, just in case someone comes across this post in the future.

    The Clausius-Clapeyron equation only applies in the case of a flat boundary between two phases. In the macroscopic world, the discrepancy is so small it can be ignored. However, when deriving the Gibbs-Thomson equation with the corrected version of the CC equation, there is assumed to be no volume change, so [itex]\Delta V \rightarrow V[/itex]. This paper was pretty useful.
     
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