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§1 · Osmotic pressure from suspended particles

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A heuristic step, to an approximation

Why a suspended grain should press like a dissolved molecule

Should small bodies suspended in a liquid press on a wall that holds them back, as dissolved molecules do, and what would tell the two expectations apart?

The 3 printed paragraphs this passage explains

§1, paragraph 1: A dissolved substance held in part of a liquid by a wall that lets the solvent through presses on that wall with the osmotic pressure, which for a dilute solution obeys the gas law, pressure times volume equals R times T times the number of gram-molecules.
§1, paragraph 2: By classical thermodynamics, small suspended bodies held back by the same kind of wall should exert no force on it, because the free energy of the system seems not to depend on where the wall and the bodies are.
§1, paragraph 3: On the molecular-kinetic view a dissolved molecule differs from a suspended body only in size, so suspended bodies in slow irregular motion should press on the wall as dissolved molecules do, with osmotic pressure RT/N times their number per unit volume, N being the number of real molecules in a gram-molecule; §2 is to show that the theory leads there.

Take a liquid of total volume V. In part of it, a volume $V^*$ , dissolve z gram-molecules of a non-electrolyte, a substance that does not split into ions, and separate $V^*$ from the pure solvent by a wall that lets the solvent through but not the dissolved substance. The dissolved molecules push on that wall. The push on each unit of its area is the osmotic pressure p, and when $V^*/z$ is large enough, that is, when the solution is dilute, it obeys the law van 't Hoff found, which has the form of the gas law:

pV^*=RTz

p times V star equals R times T times z.

R is the gas constant and T the absolute temperature. Pressure is force per unit area: the force on the whole wall is p times the wall's area.

Now put small bodies suspended in the liquid into $V^*$ in place of the dissolved substance, bodies that also cannot pass through the wall. What does classical thermodynamics expect? Einstein states its answer together with its reason. At a fixed temperature the force on the wall follows from how the free energy of the system changes when the wall is moved. On the usual view the free energy depends on the total masses and kinds of the suspended substance, the liquid and the wall, and on pressure and temperature, but not on where the wall and the suspended bodies are. If moving the wall does not change the free energy, the wall feels no force. So, gravity aside, classical thermodynamics does not expect the suspended bodies to exert any force on the wall.

Open the foundation: Osmotic pressure and free energy

Einstein sets two effects aside so that the comparison is fair. Gravity, which would pull the bodies down, does not concern him here. The energy and entropy of the surfaces where the bodies meet the liquid (capillary forces) would also enter the free energy, but he assumes the moves considered do not change the size or nature of those surfaces, so they drop out.

The classical expectation is a coherent position, and for ordinary bodies it agrees with experience: a few pebbles held behind a sieve do not push on it measurably. The question is whether it still holds for bodies small enough to be jostled by the molecules of the liquid.

The molecular-kinetic theory of heat reaches a different view. On it, a dissolved molecule differs from a suspended body only in size (Einstein sets 'lediglich', only, in italics), and there is no reason why a number of suspended bodies should not give the same osmotic pressure as the same number of dissolved molecules. Jostled by the molecular motion of the liquid, the suspended bodies must perform an irregular motion in it, however slow; if the wall keeps them from leaving $V^*$ , they exert forces on it, just as dissolved molecules do.

With n suspended bodies in $V^*$ , so that there are $\nu = n/V^*$ of them in each unit of volume, and with neighbouring bodies far enough apart, the osmotic pressure should be

p=\frac{RT}{V^*}\frac{n}{N}=\frac{RT}{N}\,\nu

p equals R T over V star, times n over N, which equals R T over N, times nu.

where N is the number of real molecules in a gram-molecule. The first form is van 't Hoff's law with $z = n/N$ gram-molecules; the second says that the pressure depends on the number of bodies per unit volume and on the temperature, and not on their size or mass.

Open the foundation: Temperature and thermal energy

The two views disagree about something that can be looked for. If suspended bodies exert osmotic pressure, then wherever their number per unit volume varies, so does the pressure, and it pushes them toward thinner regions; §3 shows that this appears as diffusion, and §§4 and 5 predict how far a grain wanders in a given time. On the classical expectation there is no osmotic pressure to drive such a spread. Observing the predicted wandering, with the predicted size, would count for the molecular-kinetic view, and its absence against it. First, §2 shows that the molecular-kinetic theory really leads to the extended law.

Show every step here: Why a suspended grain should press like a dissolved molecule

Letters, as the paper prints them. V is the total volume of the liquid and $V^*$ the part of it behind the wall. z is the number of gram-molecules dissolved in $V^*$ ; a gram-molecule is what is now called a mole. p is the osmotic pressure, R the gas constant, T the absolute temperature. n is the number of suspended bodies, a count. ν is their number per unit volume, $\nu = n/V^*$ , and not a frequency. N is the number of real molecules in a gram-molecule, now called Avogadro's number. Three letters, three different numbers: n counts the bodies, ν counts them per unit volume, N counts the molecules in a gram-molecule.

The wall. It is semipermeable: the solvent, say water, passes through it; the dissolved substance does not. The solution is on one side, pure solvent on the other.
Osmotic pressure. The dissolved molecules push on the wall. The push on each unit of area is the osmotic pressure p. Pressure is force divided by area, so a wall of area A feels the force $pA$ .
Van 't Hoff's law. For a dilute solution, one in which $V^*/z$ is large, the pressure obeys the same law as an ideal gas of as many molecules in the same volume:

pV^*=RTz

p times V star equals R times T times z.

Reading it. Doubling the amount dissolved, z, doubles p. Doubling the volume $V^*$ at fixed z halves p. Raising the absolute temperature T raises p in proportion.
A worked case, with today's value of the gas constant, $R \approx 8.314$ joule per kelvin per gram-molecule: one gram-molecule in 22.4 litres at 273 K gives $p = RTz/V^* \approx 8.314 \times 273 / 0.0224 \approx 101\,300$ pascal, about one atmosphere, the pressure one gram-molecule of gas exerts in that volume at 0 °C.
Classical thermodynamics, for suspended bodies. Replace the dissolved substance by small bodies that float in the liquid and also cannot pass the wall. At a fixed temperature the force on the wall is found from the free energy F of the whole system: if moving the wall a little changes F, there is a force; if not, there is none.
On the usual view, F depends on the total masses and kinds of the suspended substance, the liquid and the wall, and on pressure and temperature. It does not depend on where the wall is, or where the bodies are. So moving the wall leaves F unchanged, and classical thermodynamics expects no force on the wall from the suspended bodies.
What is set aside. Gravity, which would make the bodies sink. And the energy and entropy of the surfaces between bodies and liquid, the capillary forces, removed by assuming that the moves considered do not change those surfaces. Without these set-asides the classical expectation would carry extra forces that have nothing to do with the question.

A tool for this step

Osmotic pressure and free energy

Why should grains floating in water push on a wall the way dissolved molecules do?

Put sugar water on one side of a membrane that lets water through but not sugar, and pure water on the other. Water flows toward the sugar until the sugar side stands higher: the sugar pushes outward on the membrane. That push on each square metre is the osmotic pressure.

For a dilute solution van 't Hoff found in 1887 that the pressure is the one an ideal gas of the same particles would exert in the same volume. With ν particles in each cubic metre at absolute temperature T, R the gas constant and N the number of molecules in a gram-molecule:

p = \frac{R\,T}{N}\,\nu

The osmotic pressure p equals R T over N, times the number of particles per unit volume, nu.

The paper's §1 asks why a grain floating in water, large enough to see under a microscope, should be any different. Classical thermodynamics expected no force at all from suspended grains: the free energy of the system did not seem to depend on where the wall or the grains were. On the kinetic theory of heat, the paper answers, a dissolved molecule and a suspended grain differ only in size. The grains are jostled into a slow, irregular motion, and a wall that keeps them in must be pushed as a wall that keeps molecules in is pushed. So n grains in a volume V*, ν = n/V* in each unit of volume, press on it with p = (RT/N)ν.

§2 checks this inside the kinetic theory with free energy, F = E − TS: the energy minus the temperature times the entropy. At a fixed temperature a system settles where F can no longer decrease. Moving the wall that confines the grains by a small step changes F, and requiring that change to vanish gives the same pressure, (RT/N)ν.

The law holds for dilute suspensions and solutions, where the particles are far apart and do not act on one another. The paper states it for a large volume per gram-molecule; crowded particles depart from it.

Worked example: A sugar solution and a suspension of grains

Dissolve 0.1 gram-molecule of sugar in each litre: 100 in each cubic metre, so ν = 100 × 6.02 × 10²³ = 6.02 × 10²⁵ molecules per cubic metre.
At T = 293 K, RT/N = 8.314 × 293 ÷ 6.02 × 10²³ = 4.05 × 10⁻²¹ joules. Multiply by ν: p ≈ 2.4 × 10⁵ Pa, about 2.4 atmospheres, enough to hold up a column of water nearly 25 metres tall.
Now a suspension of grains, a million in each cubic millimetre: ν = 10¹⁵ per cubic metre. The same law gives p = 4.05 × 10⁻²¹ × 10¹⁵ ≈ 4 × 10⁻⁶ Pa.
Each grain counts exactly as much as one sugar molecule. The pressure is tiny only because there are fewer grains, by a factor of about 6 × 10¹⁰.

While the particles are few and far apart, the pressure depends on their number and the temperature, not on their size or mass. That is what lets §3 treat a grain like a molecule and weigh its jostling against a force.

Behind the partition: same count, same push

A wall that lets the liquid through but not the particles feels a pressure from them, p = νk_BT, while they are dilute. Only the number per cubic metre and the temperature enter, so a grain pushes exactly as much as a sugar molecule. The table is at 293 K.

What is behind the wall and how many per m³, the share of the volume they fill, and the pressure
Behind the wall	Share filled	Pressure, Pa
Sugar, 0.1 gram-molecule per litre: 6.02 × 10²⁵ per m³	dilute	2.4 × 10⁵
Grains of 0.5 μm radius, a million per mm³: 10¹⁵ per m³	5 × 10⁻⁴	4.0 × 10⁻⁶
Grains of 0.05 μm radius, a million per mm³: 10¹⁵ per m³	5 × 10⁻⁷	4.0 × 10⁻⁶
Grains of 0.5 μm radius, a billion per mm³: 10¹⁸ per m³	0.5	not given: too crowded for the dilute law

The two grain sizes at a million per cubic millimetre push equally, although one is a thousand times the volume of the other. The last row is refused rather than computed: grains filling half the volume press on one another, and p = νk_BT holds only when they do not. The osmotic partition laboratory lets you change the count, the volume and the radius and see which of them the pressure depends on.

What it shows, in words

At 293 kelvin, sugar dissolved at 0.1 gram-molecule per litre pushes on a partition with about 2.4 × 10⁵ pascals. A million grains per cubic millimetre push with about 4.0 × 10⁻⁶ pascals, whether their radius is 0.5 or 0.05 micrometres: the pressure counts particles and does not weigh or measure them. At a billion grains of 0.5 micrometres per cubic millimetre they would fill half the volume, and the dilute law no longer applies.

Where this lesson stops

This lesson stops at the pressure of a dilute suspension. How that pressure, balanced against a force on each grain, fixes how fast the grains spread is §3 of the Brownian paper.

Why the classical expectation is reasonable. For ordinary bodies it matches experience: a few pebbles held behind a sieve do not press on it measurably. It is a coherent position, and §1 treats it as one: the question it asks is whether the expectation still holds for bodies small enough to be jostled by molecules.
The molecular-kinetic view. A dissolved molecule and a suspended body differ only in size. Both are struck by the liquid's molecules and move irregularly, the body far more slowly. A body kept in by the wall strikes it now and then, and so pushes on it, as a dissolved molecule does.
So equally many bodies and molecules should give the same pressure. Take n bodies in $V^*$ . They amount to $z = n/N$ gram-molecules' worth of particles, because a gram-molecule holds N particles. Put that z into van 't Hoff's law and divide by $V^*$ ; since $\nu = n/V^*$ , the last step replaces $n/V^*$ by ν:

p=\frac{RT}{V^*}\frac{n}{N}=\frac{RT}{N}\,\nu

p equals R T over V star, times n over N, which equals R T over N, times nu.

Reading the result. The pressure is RT/N times ν, the number of bodies per unit volume. Their size and mass do not appear. A body a thousand times bigger than a molecule presses no harder, as long as there are equally many per unit volume.
The condition. Neighbouring bodies must be far enough apart, the same condition as a dilute solution. Crowded bodies, or bodies that act on one another, depart from the law.
For scale, with today's value of R/N, $1.38\times10^{-23}$ joule per kelvin, at T = 290 K, the 17 °C of the paper's §5 example, the factor RT/N is about $4.0\times10^{-21}$ joule. A suspension with $10^{16}$ grains per cubic metre, ten million per cubic millimetre, would press with about $4\times10^{-5}$ pascal: far too little to measure directly. The paper's test lies in the motion, in §§4 and 5.

A tool for this step

Temperature and thermal energy

What does a temperature say about the jostling of molecules and grains?

The air in a room at 20 °C is a crowd of molecules moving in every direction and colliding billions of times a second. Warm the room and, on average, they move faster. The kinetic theory of heat reads temperature this way: through the average energy of motion of the particles, taken over very many of them.

For any particle in equilibrium with its surroundings, the average energy of motion along each direction is the same:

\left\langle \frac{1}{2}\,m\,v_x^{2}\right\rangle = \frac{1}{2}\,k_B\,T

The average of one half m v x squared equals one half k B T.

Here m is the particle's mass, $v_x$ its speed along x, T the absolute temperature, and $k_B$ = 1.38 × 10⁻²³ joules per kelvin is Boltzmann's constant. It equals R/N, the gas constant shared out per molecule. The Brownian paper writes RT/N where a modern text writes $k_BT$ ; the paper's own letter k means the viscosity, not this constant.

The average is the point. One molecule's energy of motion changes at every collision, from almost nothing to several times the average. The temperature fixes how energy is shared out over many particles. It is not the energy of any one of them.

Nothing in the rule mentions size. A grain ten billion times heavier than a molecule has the same average energy of motion, so it moves more slowly, by the square root of the mass ratio. That is why a suspended grain belongs to the same thermal story as a dissolved molecule, as §1 of the paper insists.

Worked example: A nitrogen molecule and a half-micrometre grain at 20 °C

At T = 293 K, $k_BT$ = 1.381 × 10⁻²³ × 293 = 4.05 × 10⁻²¹ J, so the average energy of motion along one direction is half of that, 2.02 × 10⁻²¹ J.
A nitrogen molecule has a mass of 4.65 × 10⁻²⁶ kg. Setting $\frac{1}{2} m v_x^2$ equal to $\frac{1}{2} k_BT$ gives a typical speed along one direction of √(4.05 × 10⁻²¹ ÷ 4.65 × 10⁻²⁶), about 295 metres a second.
A grain of radius 0.5 μm and density 1200 kg/m³ has a mass of 6.3 × 10⁻¹⁶ kg, about 1.4 × 10¹⁰ times the molecule's. The same average energy gives it about 2.5 millimetres a second.
In water that motion is turned about within some 70 nanoseconds: the grain's mass divided by its drag coefficient 6πηa. That is far too brief to follow, so what a microscope sees is the net result of the zigzag.

This is why the Brownian paper asks for a displacement over a time rather than a speed: the thermal speed is real, but it turns about far faster than any observer can watch.

JavaScript is off, so the controls below cannot change anything. The construction shows its first setting, and its closing paragraph, “What it shows, in words”, describes the rest.

Try it: one spread of speeds, two scales

At any moment, the speeds of particles along one axis are spread in a bell curve. The temperature sets its width, and the width depends on the particle's mass. Label the same curve for a molecule or for a grain.

Label the axis for

At 293 K, a nitrogen molecule moves along one axis with a spread of about 295 m/s. The curve does not change; only the numbers on its axis do, by a factor of about 120 000.

Shaded in the middle, under half a width either way: the particles that have, at that moment, less than a quarter of the average energy of motion along the axis, about 38 per cent of them. Outlined at the two ends, beyond two widths: those with more than four times the average, about 5 per cent. Collisions move each particle about the whole curve.

What it shows, in words

At 293 kelvin a nitrogen molecule's speed along one axis is spread with a width of about 295 metres a second, and a half-micrometre grain's with a width of about 2.5 millimetres a second. Measured in its own width the spread is the same bell curve for both, with the same average energy of motion along the axis, 2.02 × 10⁻²¹ joules. At any moment about 38 per cent of the particles have less than a quarter of that average and about 5 per cent more than four times it. The temperature fixes the spread, not the energy of any one particle.

Where this lesson stops

This lesson stops at the average energy of motion. How the jostling of many molecules makes a grain wander, and how far, is §4 of the Brownian paper.

What tells the views apart. On the molecular-kinetic view, wherever the number per unit volume varies, the pressure varies, and it pushes the bodies toward thinner regions: §3 shows that this is diffusion, and §§4 and 5 predict how far a grain wanders in a given time. On the classical expectation there is no osmotic pressure to drive such a spread.
So observing the predicted wandering, with its predicted size, would count for the molecular-kinetic view, and its absence against it. §1 only states the expectation; §2 shows that the molecular-kinetic theory really leads to it.

Modern qualifications

Van 't Hoff stated the law for dilute solutions in 1887 (Zeitschrift für physikalische Chemie 1, p. 481), drawing on Wilhelm Pfeffer's measurements of 1877 with membranes that pass water but not sugar (Osmotische Untersuchungen). The paper takes the law as known and cites neither.

The expectation Einstein attributes to classical thermodynamics is his own characterization of 'the usual view' (der üblichen Auffassung); the paper names no one who held it. The paper does not call that view mistaken. It says the molecular-kinetic theory leads to a different one, and §2 shows how.

A modern lens: R/N is what is now written $k_B$ , Boltzmann's constant, so the extended law reads $p = \nu k_B T$ , the ideal-gas law for particles of any size. The paper's N is what is now called Avogadro's number; since 2019 it has an exact defined value, and in 1905 it was known only roughly. §5 of this paper proposes a way to measure it.

Later evidence: Jean Perrin's measurements of 1908 and 1909 found that the grains of a suspension at rest thin out with height as a gas of very heavy molecules would, which is what the extended law predicts once gravity is put back. That is later evidence, dated after this paper, and not a premise of it.

Assumptions and limits: Why a suspended grain should press like a dissolved molecule

Assumed here

Van 't Hoff's law for a dilute solution of a non-electrolyte held behind a wall that passes the solvent only: pV* = RTz, for large enough V*/z.
On the molecular-kinetic theory of heat, a dissolved molecule differs from a suspended body only in size.
The suspended bodies are few enough that neighbouring bodies are far apart.

What this does not establish

Gravity is set aside, and so are the energy and entropy of the surfaces between bodies and liquid (capillary forces), on the assumption that the moves considered do not change those surfaces.
§1 states the molecular-kinetic expectation; §2 derives it from the theory. Neither section says which view nature follows; that is for observation.
The law is for great dilution. Crowded bodies, or bodies that act on one another, depart from it.

Source context: Annalen der Physik (4), 17, 549–560 (1905), §§1–5. Bibliographic pointer; this preview is not a source transcription or translation.

Explanation

Einstein's displacement argument: the typical distance grows with the square root of time, not with time itself. Every explanation stays on the page when reading-only is on.

Static worked case

For radius 0.5 μm, viscosity 1.35×10⁻³ Pa·s, and T = 290.15 K, the RMS displacement is about 0.8 μm in one second. This static worked case stays in the markup; loading the live ensemble does not replace it.

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Notes and laboratory

Page 549 of Annalen der Physik, volume 17, where §1 begins. — §1 begins on page 549 of Annalen der Physik, volume 17. Read it in Einstein’s German

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