Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

In a didactic article, Victor Weisskopf estimated the size of molecules in a liquid from measurements of their surface tension and heat of vaporization. If atoms are exceedingly small, then only a small portion of them will be on the surface, so the surface tension will be small compared to heat of vaporization. If they are larger, surface tension will become more significant.

Specifically, he considers a molecule in the body of the liquid to have six bonds with neighbors while one on the surface has five. Boiling consists of breaking all bonds present while moving to the surface consists of breaking just one bond for each surface atom. With this, Weisskopf gets good estimates (within 10%) for the size of molecules/atoms of CCL4, Ne, and Ar. For water, his estimate too small by about a factor of two because water molecules are polar and can rearrange themselves on the surface to minimize energy, leading to a smaller surface tension and smaller estimate of molecular size.

However, he states that for Hg, Mg, Fe, and Cd the estimate for the atomic size comes out too big by a factor of three or so, meaning either the surface tension is anomalously high or the heat of vaporization is anomalously low. He says he doesn't know why this is the case, but I didn't find a response explaining it published in the journal.

Why does the estimate fail for these metals, and why does it give an atomic size too large, not too small? (It doesn't fail for all metals.)

link to article (behind paywall) http://ajp.aapt.org/resource/1/ajpias/v53/i1/p19_s1

share|improve this question

1 Answer 1

Looks like the following article might be relevant: Fluid Phase Equilibria 283 (2009) 89–92. The author offers some model that seems to be a good fit for a lot of elements. There is no data for Fe and Cd, the agreement looks good for Hg and fair for Mg.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.