I don't get how Taylor expansion over $ \frac{1}{\overline{V}} $ of Virial equation of state is caried out: $Z = \frac{P\overline{V}}{RT} $ which yields $$ Z = 1 + \frac{B_{2V}(T)}{\overline{V}} + \frac{B_{3V}(T)}{\overline{V^2}} + ... $$ I thought I understood calc2 Taylor expansion of form $$f(x) = \sum\limits_{k=0}^\infty \frac{f^{(k)}(0)}{k!} x^k $$ so maybe I was in illusion. What's is $Z$ function of? At what expansion point do we expa


Here, Z is explicitly a function of temperature.

At what expansion point do we expand?

Can you clarify what you mean? I can state that it's common to truncate this particular expansion at the second term, so we'd have:

$$\ce Z = 1 + {B_{2V}(T)\over \overline{V}}$$

The virial coefficient in this case accounts for deviation from the ideal case by accounting for attractive and repulsive forces between the particles in the system.

  • $\begingroup$ I'm sorry I have badly formulated my question, but thanks for the answer. I mean if we expand around some particular point (like 0 or any other fixed value). I have reformulated my question. So Z is function of T, how do we carry out expansion over $ \frac{1}{\overline{V}}? $ $\endgroup$ – wuschi Jun 24 '15 at 16:03
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    $\begingroup$ Are you asking about the mechanics of the Taylor series expansion specifically? $\endgroup$ – Todd Minehardt Jun 24 '15 at 16:19
  • $\begingroup$ Exactly, this is where my confusion is. $\endgroup$ – wuschi Jun 24 '15 at 16:35
  • $\begingroup$ The unity term corresponds to the $k = 0$ case, and the next term (where $k = 1$) is the first derivative term (the negative sign from differentiation is incorporated in $B_{2V}$). $\endgroup$ – Todd Minehardt Jun 24 '15 at 17:51

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