In a physics class, I learned about Van Der Waals forces that allow geckos to stick.

  1. Do they have any relation to the Van Der Waals Equation relating gas pressure to temperature and volume?

3 Answers 3


The connection is simply this... Van der Waals postulated that there were attractive forces between gas molecules even when these weren't in contact. The $a/V^2$ term in his gas equation is a simple way to take account of such forces, without knowing how they vary with separation between molecules, beyond their being short-range. We now know that these forces really exist, and can calculate them for any separation, knowing electric dipole moments etc. for the molecules. The forces, then, are named after the man who famously made brilliant theoretical use of them, even though he didn't know how they arose. Indeed, in the 1870s, when he published his equation, even the existence of atoms was not acknowledged by all scientists; Ernst Mach (Mach number, Mach principle) and Friedrich Ostwald (dilution law) were well known sceptics.


No. Van der Waals forces are electromagnetic interactions due to microscopic charge fluctuations, wheres van der Waals equation is an empirical equation for non-ideal gas. Not only they refer to physically different phenomena, but the former are truly existing forces, whereas the latter is a mathematical ansatz.

It is necessary to add a clarification, since the answer by @PhilipWood may superficially appear as telling the opposite of what I have written. This answer gave a valuable historical perspective, and I upvoted it myself. Nevertheless, in the context of modern physics the places of VdW forces and VdW equation are very different: the forces are of great importance in many domains of physics, while the equation has mainly empirical value, largely divorced from the VdW's original motivations for writing it. In this respect I strongly recommend the answer by @user1271772.


The Van der Waals equation is a generalization of the ideal gas law $PV=nRT$:

$$ \tag{1} \left(P+a \frac{n^2}{V^2}\right)(V-n b)=n R T. $$

It is different from the Van der Waals forces, which are intermolecular forces such as the London dispersion interaction between two induced dipoles, with potential energy given by:

$$\tag{2} U(r) = -\frac{C_6}{r^6}. $$

Notice that not a single thing in Eq. 1 appears in Eq. 2.
$V$ is volume in Eq. 1, and $U$ is potential energy in Eq. 2.
$R$ is the gas constant in Eq. 1, and $r$ is the distance between the two objects experiencing the Van der Waals force in Eq. 2.

  • $\begingroup$ Well, you need a bit of statistical mechanics. I think on Wikipedia en.wikipedia.org/wiki/Van_der_Waals_equation there is a derivation of 'a' from C6, $\endgroup$
    – lalala
    Nov 30, 2020 at 8:32
  • $\begingroup$ @lalala vdW is an empirical equation. Whatever were the original motivations for wruting it, its value is mainly qualitative. If quantitative coefficients are necessary, they are measured rather than derived from the first principles. $\endgroup$ Nov 30, 2020 at 9:46

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