brownian motion

The equation explained

F = -\frac{R T}{N} \{ \lg J + n \lg V^* \}

In wordsThe free energy is minus the gas constant times the absolute temperature over the number of real molecules in a gram-molecule, times the logarithm of the factor added to the number of particles times the logarithm of the volume holding them.OverviewPut the value of the integral into the free energy. The logarithm turns the product into a sum, and the part that knows about the volume is separated from the part that does not: one term holds the unevaluated factor, the other the number of particles times the logarithm of the volume.Full explanationThe free energy was

F = -\frac{R T}{N} \lg B

, and

B

has just been found to be

J V^{*n}

. The logarithm of a product is the sum of the logarithms, and the logarithm of a power carries the exponent to the front, so

\lg B = \lg J + n \lg V^*

. The paper never evaluates

J

, and it does not have to. Whatever number it is, it does not change when the wall is moved, so it contributes nothing to a derivative with respect to

V^*

: the whole of the osmotic pressure will come from the second term. That the term is

n \lg V^*

, with

n

the number of particles and no other property of them, is why the result of the next display counts bodies and is indifferent to their size.Every step

F = - \frac{R T}{N} \lg B

The free energy from the first paragraph of the section.

F = - \frac{R T}{N} \lg ( J V^{*n} )

With the value of the integral just found.

\lg ( J V^{*n} ) = \lg J + \lg V^{*n}

The logarithm of a product is the sum of the logarithms.

\lg V^{*n} = n \lg V^*

A power inside a logarithm becomes a factor in front of it.

F = - \frac{R T}{N} \{ \lg J + n \lg V^* \}

The printed form. The first term inside the brace does not depend on

V^*

; the second holds the whole of that dependence.AlsoOverviewFull explanationEvery step

Read it in the paper