# Computing the molar volume for a 2D-simulation of phase separation

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 $$$$V_m = 25.13 \cdot 10^{-6}\ \frac{\text{m}^3}{\text{mol}}.$$$$

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 $$$$(25.13 \cdot 10^{-6})^{\frac{2}{3}}$$$$ as a corresponding quantity for a two-dimensional simulation (unit: $$\frac{\text{m}^2}{\text{mol}^{(2/3)}}$$ ?). The quantity mol refers basically to a 3D framework, isn't it?

• Define relation between molar volume and molar area. [m2/mol] Commented Oct 30, 2022 at 15:58
• 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. Commented Oct 30, 2022 at 18:05
• 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). Commented Oct 30, 2022 at 18:07
• 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. Commented Oct 30, 2022 at 18:17
• 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. Commented Oct 30, 2022 at 18:36

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 $$$$A_m = \left(\frac{V_m}{N_A}\right)^{\frac{2}{3}} \cdot N_A = V_m^{\frac{2}{3}} \cdot N_A^{\frac{1}{3}},$$$$ where $$N_A = 6.02 \cdot 10^{23}$$ is the Avogadro constant.