On this page
In 1865 the size of an air molecule was, as Josef Loschmidt put it, simply assumed to be infinitesimally small, "whether millionths of a millimeter, or billionths, or even smaller", and had "never been discussed". Loschmidt, then teaching at a Vienna school, estimated it with two published numbers and three lines of algebra. This post covers who he was, how his argument works, his calculation redone with his own figures, and what a student can learn from it.
Who was Josef Loschmidt?
The Austrian Academy of Sciences' biographical dictionary (ÖBL) gives his birth as 15 March 1821 at Putschirn near Karlsbad in Bohemia, the son of a cottager and day labourer, and his death in Vienna on 8 July 1895. Only the support of the philosopher Franz Exner let him study mathematics and science. Seeing no hope of a university post, he went into industry: he and a colleague found a way to turn Chile saltpetre into the potassium nitrate used in gunpowder, and he later ran a paper factory. From 1856 he taught at a school in Vienna's Leopoldstadt district. Only after his molecule paper did his academic career begin: lecturer in physics in 1866, and full professor at Vienna from 1872 to 1891.
What did he need to know?
Loschmidt's paper, read to the Vienna Academy in 1865, is available in an English translation in the Journal of Chemical Education (1995). He started from the kinetic theory of Clausius and Maxwell, in which gas molecules fly in straight lines and bounce off each other like elastic spheres. Two quantities were by then within reach.
The mean free path L, the average distance between collisions. From gas viscosity, Maxwell had found 0.000062 mm for air, and O. E. Meyer's corrected work gave about 0.000140 mm. Loschmidt used 140 millionths of a millimetre.
The condensation coefficient ε, the fraction of a gas's volume that its molecules would fill if packed together, as in a liquid. For water vapour at 0 °C he listed 0.00081; for carbon dioxide, 0.00204.
Air could not then be liquefied, so he estimated the density of "liquid air" from similar compounds of nitrogen and oxygen, settling on 1.5 g/cm³. Since gaseous air is about 770 times lighter than water, he wrote ε = 1/(770 × 1.5) = 1/1155 = 0.000866.
How do two numbers give the size?
Let N be the number of molecules per unit volume and s the diameter. Loschmidt used two equations.
Clausius's mean free path formula. In his notation, 1 = (4/3)NπLs². Each molecule sweeps out a cylinder of cross-section πs²/4 and length L.
The definition of ε. Each sphere has volume πs³/6, so ε = Nπs³/6.
Divide the second by the first. N and π cancel, and you get ε = s/(8L), or
s = 8εL
The diameter is eight times the mean free path times the condensation coefficient. The unknown N has dropped out, which is the clever step.
Worked example: Loschmidt's own numbers
Diameter. s = 8 × 0.000866 × 0.000140 mm = 9.70 × 10⁻⁷ mm. Loschmidt wrote 0.000000969 mm, "in round numbers one millionth of a millimetre", that is, about 1 nm.
His error bar. He judged the value "neither too large nor too small by more than one power of 10."
Number per cubic metre. Rearrange ε = Nπs³/6 to N = 6ε/(πs³). With s = 9.70 × 10⁻¹⁰ m, s³ = 9.12 × 10⁻²⁸ m³, so N = (6 × 0.000866) / (3.1416 × 9.12 × 10⁻²⁸) ≈ 1.8 × 10²⁴ per m³.
Compare with today. NIST lists the Loschmidt constant, the number of molecules per cubic metre of an ideal gas at 273.15 K and 101.325 kPa, as exactly 2.686 780 111 × 10²⁵ m⁻³. The 1865 numbers give a value about 15 times too small, but in the right range for the first attempt. The ÖBL notes that his calculations led only to the order of 10²² for the related number of molecules in a mole.
The gap is large because N depends on 1/s³. If his ε were right, a diameter only about 2.5 times too large would explain the whole factor of 15. This is worth remembering whenever a cube appears in a formula.
What mistakes do students make here?
Mixing units. Keep s and L in the same unit, and convert s to metres before you cube it for N.
Using the modern formula with his factor. The mean free path post uses the JEE form l = 1/(√2 nπd²). Clausius's 1860 version had a factor 3/4 in place of 1/√2. Use the NCERT form in exams.
Confusing Loschmidt and Avogadro. Avogadro's hypothesis says equal volumes hold equal numbers of molecules; our post on Avogadro's hypothesis tells that story. Loschmidt's paper relied on it, then estimated how big that number per volume actually is.
If the idea of molecules sweeping out cylinders is hard to picture, the Fermi Simulation Lab has interactive simulations for ideas that are easier to see than read.
What can a student take from his way of working?
Combine two facts to remove an unknown. He could not measure N or s alone. Dividing two equations eliminated N.
State your uncertainty. He gave his answer to within a factor of ten and said so, rather than claiming more.
Check against the real world. He compared his size with the finest lines a microscope could resolve, about 4,000 per millimetre, and found those lines still 250 times wider than an air molecule.
Your job does not decide your science. He did this work as a school teacher, before any university post.
Next step: repeat the calculation with Maxwell's earlier L of 0.000062 mm and see how the diameter and N change.
