# Osmotic pressure and free energy

In a dilute solution, dissolved molecules held behind a wall that lets water through push on it with the pressure an ideal gas of the same particles would exert: p = (RT/N)ν. §1 of the Brownian paper argues that suspended grains must push in the same way, and §2 checks it by the bookkeeping of free energy.

Written for this edition, not translated from Einstein. Editorial review pending.

## 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:

$$
p = \frac{RT}{N}\,\nu
$$

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

1. 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.
2. 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.
3. 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.
4. 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.

## 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.

## This lesson builds on

- [Entropy and temperature](/foundations/entropy-temperature/)
- [Quantities and units](/foundations/quantities-units/)
- [Powers of ten and physical units](/foundations/bridge-scientific-notation-units/)
- [Fractions and ratios](/foundations/bridge-fractions-ratios/)