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I am a mathematician and I want to simulate phase separation that occurs in a sodium silicate glass ($12.5\,\text{Na}_2\text{O}\cdot 87.5\,\text{SiO}_2$) at $T = 923\ \text{K}$ as it was done, for example, in Kim and Sanders (2020).

In particular, I want to do this simulation in two-dimensions. However, certain parameters that are implicitly required for the Cahn-Hilliard equation are only given in units that concern 3D simulations. For example, the molar volume for the mixture above can be computed to be \begin{equation} V_m = 25.13 \cdot 10^{-6}\ \frac{\text{m}^3}{\text{mol}}. \end{equation}

Question: How do I have to deal with that in a 2D simulation? From a geometrical point of view, I just would like to consider \begin{equation} (25.13 \cdot 10^{-6})^{\frac{2}{3}} \end{equation} as a corresponding quantity for a two-dimensional simulation (unit: $\frac{\text{m}^2}{\text{mol}^{(2/3)}}$ ?). The quantity mol refers basically to a 3D framework, isn't it?

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  • $\begingroup$ Define relation between molar volume and molar area. [m2/mol] $\endgroup$
    – Poutnik
    Commented Oct 30, 2022 at 15:58
  • $\begingroup$ My proposal simply comes from the geometric average but actually does not fit toe the notion of a mol. I still think that the notion of a mol is meat for 3D. Therefore, I have no idea how to deal with that notion in 2D. $\endgroup$
    – Henning
    Commented Oct 30, 2022 at 18:05
  • $\begingroup$ At least to me, "molar area" is not that common. And I have also no idea how to compute this quantity (without the corresponding mass density). $\endgroup$
    – Henning
    Commented Oct 30, 2022 at 18:07
  • $\begingroup$ Computing such things in 2D is not common either. You have to use 2D variants of 3D quantities. Similarly, density in kg/m3 would have 2D variant area density in kg/m2. $\endgroup$
    – Poutnik
    Commented Oct 30, 2022 at 18:17
  • $\begingroup$ Actually, in most cases simulations (especially simulating phase separations) start with 2D simulations. One paper is cited in the questions. Here is another one. And I was wondering where I can find the corresponding 2D parameter variants or how these 2D variants are computed. $\endgroup$
    – Henning
    Commented Oct 30, 2022 at 18:36

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Since one really wants to have the quantity (molar area) in the unit area per mole (and not per $\text{mol}^{\frac{2}{3}}$), the guessed solution $(25.13 \cdot 10^{-6})^{\frac{2}{3}}$ above is in fact not what we are looking for. Instead, it seems to be common to compute \begin{equation} A_m = \left(\frac{V_m}{N_A}\right)^{\frac{2}{3}} \cdot N_A = V_m^{\frac{2}{3}} \cdot N_A^{\frac{1}{3}}, \end{equation} where $N_A = 6.02 \cdot 10^{23}$ is the Avogadro constant.

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