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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

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  • $\begingroup$ I have the same question and I still don't get it. Z = PV/RT (N=1) And I've read that the expansion has to be when P→0, or when V→∞, so (1/V)→0. Ok, fine, but I can't expand in a series of Taylor in terms of P or V because in this form they don't have a second (and third, fourth and so on) derivative: Z'(P) = V/RT Z''(P) = 0 Z'(V) = P/RT Z''(V) = 0 From what I know, its a expansion (Maclaurin) of: 1/(1-x) which gives (1 + x + x² + x³ +...) And I don't see a way to have PV/RT to be written in this form. Tomorrow I'll ask my professor and I'll come back if I get a clear answer. $\endgroup$
    – Isabela
    Oct 24, 2022 at 4:23

2 Answers 2

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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.

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  • $\begingroup$ 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, 2015 at 16:03
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    $\begingroup$ Are you asking about the mechanics of the Taylor series expansion specifically? $\endgroup$ Jun 24, 2015 at 16:19
  • $\begingroup$ Exactly, this is where my confusion is. $\endgroup$
    – wuschi
    Jun 24, 2015 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 $B_{2V}$). $\endgroup$ Jun 24, 2015 at 17:51
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The expansion in a Taylor series of is not made starting from an ideal gas equation, it's actually made from the equation for the pressure obtained from statistical mechanics, $pV = kT\ln Z$, in the grand canonical ensemble.

The derivation of the virial expansion is shown in McQuarrie (ref 1). I had misunderstood (probably you too) based on what I read in some site or videos. The ideal gas equation is obtained from the virial expansion and not the opposite.

References

  1. McQuarrie, Donald A.. The Virial Equation of State from the Grand Partition Function, Section 12-1, p. 224, in Statistical Mechanics, ISBN 06-044366-9.
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