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Osmotic pressure, question on derivation (TD)

  1. May 17, 2014 #1
    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
     
  2. jcsd
  3. May 17, 2014 #2
    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.
     
  4. May 17, 2014 #3
    Would you mind elaborating on the taylor expansion?
     
  5. May 17, 2014 #4
    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.
     
  6. May 17, 2014 #5
    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 [itex]1-v/V_0[/itex]. 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?

    $$
    V_0/V - v/V
    $$
     
    Last edited: May 17, 2014
  7. May 17, 2014 #6
    As I said, you can replace 1/(V0 - nv) = (1/V0)1/(1 - nv/V0) with (1/V0)(1 + nv/V0).
     
  8. May 18, 2014 #7
    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.
     
  9. May 19, 2014 #8
    bump

    also meant to say (V/V_0)^n
     
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