# 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?

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}

• Thank you! This makes a lot more sense. However, Why is there no $a$ in the C term? That is, why is it not something like $b^2-\frac{a^2}{(RT)^2}$? – John Snow Sep 11 '14 at 0:39
• @JohnSnow That's the result of the calculation. The constants $B$ and $C$ don't have to show any similarity. – Philipp Sep 11 '14 at 1:07