# Expansion of Van der waals for small pressure

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## Homework Statement

In the VDW eqn of state, ##(P + a/v^2)(v-b)=RT## write Pv as a function of {P,T} and by expanding the result in powers of P or otherwise show that the first terms of a Virial expansion in powers of P for a VDW gas are given by $$Pv = RT + \left(b - \frac{a}{RT}\right)P + O(P^2)$$

Taylor expansion

## The Attempt at a Solution

I have already solved this (I think) via a method belonging in the 'otherwise' category. However, I wish to also solve it via the method outlined. The hint is that we can express y =Pv, x=P and write ##y = y(0,T) + y'(0,T)x + O(x^2)##
This can be rewritten as $$Pv = Pv(0,T) + \frac{d}{dP} (Pv)|_{P=0}P + O(P^2)$$, so essentially we are expanding the VDW eqn about small pressures. The first term on the RHS I think should be RT, which makes sense, yet I don't see how it comes about from subbing P=0 into VDW. The differentiation of the second term gives $$b - a\frac{d}{dP} \frac{1}{v} + ab\frac{d}{dP} \frac{1}{v^2} = b -a \frac{d}{dv}\frac{1}{v}\frac{dv}{dP} + ab\frac{d}{dv} \frac{1}{v^2} \frac{dv}{dP}$$ but I am not sure how to continue from here.
Many thanks.

Mentor
Try rewriting the equation as:

$$Pv+\frac{a}{v}-Pb-\frac{ab}{v^2}=RT$$

Next, reexpress this as:
$$Pv+\frac{aP}{Pv}-Pb-\frac{abP^2}{(Pv)^2}=RT$$

Now, for convenience, substitute Pv = x:

$$x+\frac{aP}{x}-Pb-\frac{abP^2}{x^2}=RT$$

You know that, at P = 0, x = RT

So do your Taylor series expansion about this point. Take the derivative of the equation with respect to P, and then solve for the derivative of x with respect to P. Evaluate this derivative at P=0, x = RT.

Chet

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