physical chemistry - Relation between the virial coefficients and van der Waals constants

I don't know where to begin with this question.

A function $f(x)$ can be expressed as a special power series called the Taylor expansion. The common expansion is $(1-x)^{-1}=1+x+x^2$ Note that the higher terms become neglibible small and usually are not included in expansion. Find the correlation between the virial coefficients and $a$ and $b$ in the corresponding van der Waals equation of state. Start with the equation shown below and compare it with the virial equation of state.

$$ p=\frac{RT\left(1-\frac{b}{V_{\mathrm m}}\right)^{-1}}{V_{\mathrm m}}- \frac{a}{V_{\mathrm m}^2}$$

The virial equation of state is

$$ p=\frac{nRT}{V}\left(1+\frac{nB}{V}+\frac{n^2C}{V^2}+\ldots\right)$$

I get that there are $A$, $B$, $C$… coefficients, and that the $A$ coefficient should be 1.

The best I have is that:

$$\begin{align} A &= 1 \\ B &= -\left(\frac{b}{V_{\mathrm m}}\right)^{-1} \\ C &= -\frac{a}{V_{\mathrm m}^2} \end{align}$$

If I were to relate that to the terms in the Taylor expansion, then $p=(1-x)^{-1}$ and then $A$, $B$, $C$… are the terms on the right side of the equation.

Is this the correct approach? Can the virial coefficients have a negative value?

Answer

You have to bring both equations into the same form and then compare the coefficients. You seem not to have used the tip with the Taylor expansion and I guess that's why you got an incorrect result.

As you wrote: The virial equation of state is

\begin{align} p &=\frac{nRT}{V} \left(1+\frac{nB}{V}+\frac{n^2C}{V^2}+ \ldots \right) \\ &= \frac{RT}{V_{\mathrm{m}}} \left(1+ B\frac{1}{V_{\mathrm{m}}}+ C\frac{1}{V_{\mathrm{m}}^2}+ \ldots \right) \end{align}

and the van der Waals equation is

\begin{align} p &=\frac{RT(1-\frac{b}{V_{\mathrm m}})^{-1}}{V_{\mathrm m}}- \frac{a}{V_{\mathrm m}^2} \ . \end{align}

Now, you have to bring the van der Waals equation into the same form as the virial equation by factoring out $\frac{RT}{V_{\mathrm{m}}}$

\begin{align} p &= \frac{RT}{V_{\mathrm{m}}}\left( \left(1 - \frac{b}{V_{\mathrm{m}}} \right)^{-1} - \frac{a}{R T V_{\mathrm{m}}} \right) \ , \end{align}

using the Taylor expansion $\left(1 - \frac{b}{V_{\mathrm{m}}} \right)^{-1} = 1 + \frac{b}{V_{\mathrm{m}}} + \frac{b^2}{V_{\mathrm{m}}^2} + \ldots$ and ordering the terms according to their power in $\frac{1}{V_{\mathrm{m}}}$

\begin{align} p &= \frac{RT}{V_{\mathrm{m}}}\left( 1 + \left(b - \frac{a}{R T} \right) \frac{1}{V_{\mathrm{m}}} + b^{2} \frac{1}{V_{\mathrm{m}}^{2}} + \ldots \right) \ , \end{align}

so that you can compare the coefficients in both equations to get the result

\begin{align} B &= b - \frac{a}{R T} \\ C &= b^2 \ . \end{align}