Annus MirabilisEinstein in 1905
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brownian motion

The equation explained

p=−∂F∂V∗=RTV∗nN=RTNν.p = -\frac{\partial F}{\partial V^*} = \frac{R T}{V^*} \frac{n}{N} = \frac{R T}{N} \nu.
In wordsThe osmotic pressure is how much the free energy falls when the volume holding the particles grows, which comes to the gas constant times the absolute temperature times the number of particles over the number of real molecules in a gram-molecule and over that volume; that is, R T over N times the number of particles in unit volume.OverviewDifferentiating the free energy with respect to the volume gives the pressure on the wall, and it is exactly the expression section 1 proposed for suspended bodies. The unevaluated factor drops out, because it does not change when the wall moves.Full explanationThe pressure a wall feels is the rate at which the free energy falls as the volume behind it grows, p=−∂F/∂V∗p = -\partial F / \partial V^*. Only the term nlg⁡V∗n \lg V^* depends on V∗V^*, and the derivative of lg⁡V∗\lg V^* is 1/V∗1/V^*, so the derivative of the free energy is −RTN⋅nV∗-\frac{R T}{N} \cdot \frac{n}{V^*} and the pressure is its negative. Writing ν=n/V∗\nu = n / V^* for the number of particles in unit volume gives the second form, p=RTNνp = \frac{R T}{N} \nu, which is the expression section 1 wrote down for suspended bodies and promised to justify here. The paragraph after the display draws the conclusion: the existence of the osmotic pressure is a consequence of the molecular-kinetic theory of heat, and on that theory dissolved molecules and suspended bodies of equal number behave, in respect of osmotic pressure, completely alike at great dilution. Nothing in the derivation asked how large the particles are. What entered was their number, the homogeneity of the liquid, the absence of forces, and the independence of their motions.Every stepp=−∂F∂V∗p = - \frac{\partial F}{\partial V^*}The pressure on the wall as the fall of the free energy with the volume behind it. The paper prints this as the first member of the display.F=−RTN{lg⁡J+nlg⁡V∗}F = - \frac{R T}{N} \{ \lg J + n \lg V^* \}The free energy from the display before.∂∂V∗lg⁡J=0\frac{\partial}{\partial V^*} \lg J = 0JJ was shown not to depend on V∗V^*, so its term contributes nothing to the derivative. This is where the work of the middle of the section is spent.∂∂V∗lg⁡V∗=1V∗\frac{\partial}{\partial V^*} \lg V^* = \frac{1}{V^*}The derivative of the logarithm.∂F∂V∗=−RTN⋅nV∗\frac{\partial F}{\partial V^*} = - \frac{R T}{N} \cdot \frac{n}{V^*}The constant factor and the nn carried through the derivative of the second term.p=RTV∗nNp = \frac{R T}{V^*} \frac{n}{N}Its negative, which is the pressure: the first printed value.ν=nV∗\nu = \frac{n}{V^*}The number of particles in unit volume, as section 1 defined it.p=RTNνp = \frac{R T}{N} \nuThe second printed value. It is the expression of section 1, now derived rather than proposed.Also

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