I came across this with a lecturer while discussing the Gibbs-Adsorption Isotherm. The lecturer insisted that the unit for the concentration parameter in the the formula surface excess concentration should be $\mathrm{mol}$ as per literature. However, the formula itself would suggest otherwise.

The formula for calculating surface excess concentration is as following:

$$\Gamma = \frac{-1}{RT}\left(\frac{\delta \gamma}{\delta \ln C}\right) $$

where $\Gamma$ is the curface excess concentration in $\mathrm{mol/m^2}$, $R$ is the universal gas constant in $\mathrm{J/(mol\cdot K)}$, $T$ is the temperature in $\mathrm{K}$, $\gamma$ is the interfacial tension in $\mathrm{N/m}$, and C is the "molar concentration" as expressed in literature with the supposed unit of $\mathrm{mol/L}$.

When you work through the formula to calculate back the units for $C$ you will see that $C$ should be a dimensionless number to retain the $\mathrm{mol/m^2}$ units that are normally used for surface excess concentration. Does anyone know why the $C$ parameter in this equation is referred to as "molar concentration" in literature? Are these papers subsequently referring to surface excess concentration as having a unit of $\mathrm{L/m^2}$?

  • 2
    $\begingroup$ As a rule to remember: arguments of functions like $\exp$, $\sin$, $\log$, $\dots$ must always be dimensionless, since they can be written as a power series, which would add quantities of different dimensions together, and that would not make any physical sense. $\endgroup$
    – noah
    Aug 13, 2017 at 13:44

1 Answer 1


In your equation you use $\ln C$, $C$ have to be unitless as you pointed out for $\Gamma$ to have to proper unit. Futher we cannot consider $\ln$ to a unit in our unit analysis. Your equation should for clarity be written as:

$$\Gamma = \frac{-1}{RT}\left(\frac{\delta \gamma}{\delta \ln \left( \frac{C}{C^\circ} \right)}\right) $$

Here $C^\circ$ is a standrad concentration. Atleast in chemistry $C^\circ=1$ $\mathrm{mol\cdot dm^{-3}}$ [1]. In the equation the unit of $C$ have to match the unit of $C^\circ$ for the units to cancel out each other, and make the logarithm unitless as it should be.

[1] IUPAC Gold Book, standard concentration

  • $\begingroup$ Hi Erik. Thank you for your reply. I have edited the units of my question to match your answer. If I do use $\frac{C}{C^0}$ would both C values have the same base units and ultimately cancel each other out? $\endgroup$ Aug 13, 2017 at 18:09
  • $\begingroup$ @DeeCee yes, they should both have the same unit. The unit of $C^\circ$ is some standard value. Added this to my answer $\endgroup$ Aug 13, 2017 at 18:52

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