Question

What is the Taylor series expansion of z(V) = V/(V-b) in the limit V >> b?

Answer 1

Taylor? Are you sure you don't want Laurent? 'z' just sets off alarms.

\(z(V) = \dfrac{V}{V-b} = \dfrac{1}{\dfrac{V-b}{V}} = \dfrac{1}{1 - \dfrac{b}{V}}\)

You're almost done.

Answer 2

Ok, let me clarity. The compression factor z for a van der Waals gas is given as:

z=V/(V-b)-a/(VRT)

I need to expand the first term of this expression using Taylor series expansion and use the first two terms of the polynomial to get

z = b-a/RT.

The big simplifying assumption here is that I'm allowed to treat the quantity V-b = V because V >> b.

I'm not getting the answer I'm suppose to.

Answer 3

So basically I need to turn V/(V-b) into b using Taylor series expansion

Answer 4

z'(V)=1/(V-b)-V/(V-b)^2

So, using the definition of Taylor series

z=V/(V-b)+[1/(V-b)-V/(V-b)^2](V-????)

I'm stuck because I don't know what value I'm making the expansion around. Or something (maybe I don't need a value). I'm supposed to use this simplification I mentioned too, but...I'm lost.

Answer 5

Well, it seems to me that V >>b so you can just use "V" is not very interesting. Let's just remember that the asymptote at V = b is nowhere near where we are creating the Taylor Series.

\(z'(V) = -\dfrac{b}{(V-b)^{2}}\)

\(z"(V) = +\dfrac{2b}{(V-b)^{3}}\)

\(z^{[3]}(V) = -\dfrac{6b}{(V-b)^{4}}\)

\(z^{[4]}(V) = +\dfrac{24b}{(V-b)^{5}}\)

etc...

You WILL have to pick a point, maybe V = a, just to emphasize that it still isn't 'b'.