Classifying regions of Van der Waal like gas

Question

Given the equation of state $$p+a\left(\frac{N}{V}\right)=\frac{Nk_BT}{V-bN} \tag 1$$ Taking into account of the fact that a realistic model requires $p \geq 0, V \geq Nb, N>0$ classify the solutions of the equation of state as a function of temperature.

My attempt is that

$$\left[p+a\left(\frac{N}{V}\right)\right]\left[\frac{V-bN}{V} \right]=\frac{Nk_BT}{V} $$ Let $n=\frac{N}{V}$,

$$\left[p+an\right]\left[1-bn \right]=nk_BT$$

$$p+an-pbn-abn^2=nk_BT$$

Let $\tilde{n}=\sqrt{ab}n$, $\tilde{n}^2=abn^2$,

$$\tilde{n}^2+\left(p\sqrt{\frac{b}{a}}-\sqrt{\frac{a}{b}}\right)\tilde{n}-p+\frac{k_BT}{\sqrt{ab}}=0$$

Let $\tilde{p}=\sqrt{\frac{b}{a}}p$,

$$\tilde{n}^2+\left(\tilde{p}-\sqrt{\frac{a}{b}}\right)\tilde{n}-\sqrt{\frac{a}{b}}p+\frac{k_BT}{\sqrt{ab}}=0$$

Let $\tilde{t}=\frac{k_BT}{\sqrt{ab}}$,

$$\tilde{n}^2+\left(\tilde{p}-\sqrt{\frac{a}{b}}\right)\tilde{n}-\sqrt{\frac{a}{b}}p+\tilde{t}=0$$

Now the discriminant is $b^2-4ac$ so

$$\left(\tilde{p}-\sqrt{\frac{a}{b}} \right)^2+4\sqrt{\frac{a}{b}}p-4\tilde{t}$$

$$\tilde{p}^2-2\sqrt{\frac{a}{b}}\tilde{p}+\frac{a}{b} +4\sqrt{\frac{a}{b}}p-4\tilde{t}$$

$$\left(\tilde{p}+\sqrt{\frac{a}{b}} \right)^2=4\tilde{t}$$

so ignoring the negative time solution we have

$$\tilde{p}=-\sqrt{\frac{a}{b}} +2\sqrt{\tilde{t}}$$

but I don't see what Im meant to be doing next.

After I that I need to show that there is a region of the $T-V^{-1}$ plane where this equation of state is not thermodynamically stable and determine the boundary of this region.

I know that for stability we need that

$$\left(\frac{\partial p}{\partial V} \right)_{T,N} \leq 0$$

but I dont see how this is usable in the current context.

Answer 1

First of all you can directly write $p$ as a function of $V$ without solving a quadratic equation:

$$ p = \frac{N k_B T}{V- N b } - a \frac{N}{V} $$

Then you take a look at the interesting points

$$\left(\frac{\partial p}{\partial V}\right)_{T,N} = 0 $$

The solutions to this equation tell you where the bulk modulus (or its inverse, the compressibility) might change the sign. That's how you can use this criterion to determine the boundaries of unstable regions.

EDIT: If I am not mistaken, the equation above has solutions

$$ \vartheta_\pm = b \frac{1}{1\pm \sqrt{\frac{k_B T}{a}}} = b \frac{1}{1\pm \sqrt{t}}$$,

with $\vartheta = V/N$ and the reduced temperature $t=T k_B /a $. In terms of this reduced variable and $v = \vartheta / b$ we can write

$$ \left(\frac{\partial p}{\partial \vartheta}\right)_{T,N} = \frac{a}{b^2} \left(\frac{1}{v^2}- \frac{t}{(v-1)^2} \right) \overset{!}{\leq} 0$$

So you know the critical points where the sign might change and you know the equation for checking which sign is present. The only thing you have left to do is check which sign is present in which region along the $\vartheta$ axis. This is only math left to do.. Since you know the sign along the $\vartheta$ axis you will automatically know the sign along the $\vartheta^{-1}$ axis.