Osmotic pressure, question on derivation (TD)

[nossren]

I'm currently reading about thermodynamics and osmosis and I happened to stumble across this paper. There is one thing I don't really understand, though.. In chapter 8 the author wishes to give a thermodynamic explanation of the osmotic pressure so I've been reading through the derivation. When calculating the thermodynamic probability W (page 20), how do I go from (11) to (12)?

http://arxiv.org/ftp/physics/papers/0305/0305011.pdf

 

[dauto]

since v/V is small, the author expanded the expression in a Taylor series expansion and dropped any terms that are quadratic (or higher powers) in v/V, keeping only the independent and linear terms. This is called linearizing the equation - a very common device.

 

[nossren]

Would you mind elaborating on the taylor expansion?

 

[dauto]

a Taylor expansion is an expansion of a function in a power series. It is taught in calculus class. All you need here is the fact that 1/(1 + v/V) can be replaced with 1 - v/V when v << V. Do that and collect all the linear terms. For instance

(1+v/V)(1+2v/V)(1+3v/V) = 1 + (1+2+3) v/V + ...

where the dots represent quadratic or higher terms that were dropped from the equation.

 

[nossren]

I'm still not following. I solved it for the numerator, which yielded 1-(1+2+3+...)v/V. By reverse engineering (12), the denominator must be 1-v/V_0. But since the numerator was divided by V to obtain 1-v/V the denominator must too. How do I solve it for the denominator?

$$
V_0/V - v/V
$$

 

[dauto]

As I said, you can replace 1/(V₀ - nv) = (1/V₀)1/(1 - nv/V₀) with (1/V₀)(1 + nv/V₀).

 

[nossren]

I did that, and now I get (after dropping higher power terms)
$$
(V/V_0)^{n-1} \cdot (1 - v/V + v/V_0 -2v/V + 2v/V_0 - ...) = (V/V_0)^{n-1} \cdot (1 - (1+2+3+...)v/V + (1+2+3+...)v/V_0)
$$
which leaves me with a (V/V_0)^(n-1) in front of the wanted expression. I don't see how to get around this.

 

[nossren]

bump

also meant to say (V/V_0)^n