Deriving an alternative expression for the Van der Waals equation using given parameters


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> The question is to find out an alternative expression for the Van der Waals equation of corresponding states for 1 mole of gas using the parameters given as $$A=\frac{P}{P\_c},\quad B=\frac{V}{V\_c},\quad C=\frac{T}{T\_c} ,$$ where $P\_c,T\_c,V\_c$ represent the critical pressure, temperature and volume respectively.
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I am aware of the simple form of the Van der Waals equation which is $$\left(P+\frac{a}{V^2}\right)\left(V-b\right)=RT .$$ The values of $P\_c , T\_c \text{ and } V\_c$ can also be written down in terms of the Van der Waals constants but I have no idea on how to proceed after this. I tried to substitute the known values but it made things complicated. Is there any other approach to this problem? 




$\def\p{\partial} \require{enclose}$Start with the van der Waals equation $(1)$.


$$\left(P + a\frac{n^2}{V^2}\right)\left(V - bn\right) = nRT\tag1$$


Note that you have a slightly different form of the equation, namely you have used molar volume $V\_m$. Divide both sides of $(1)$ with $n$, use $V\_m = V/n$, and $n^2/V^2 = 1/V\_m^2$ to reach your variant.


Values for critical parameters in terms of $a, b, n, R$
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The critical point $(P\_c, T\_c, V\_c)$ is a critical point and an inflection point of the isotherm by definition.$^{[a]}$ Therefore we arrive at two equations;


$$\left(\frac{\p P}{\p V}\right)\_{T\ =\ T\_c} = 0,\tag2$$
$$\left(\frac{\p^2 P}{\p V^2}\right)\_{T\ =\ T\_c} = 0.\tag{2'}$$


Equation $(2)$ is a sufficient condition for a critical point (mathematically, a derivative's non-existance works too). Equation $(2')$ gives a necessary condition for an inflection point.


So our first step should be to find the equation $P = P(T, V)$. Just use equation $(1)$.


$$P = \frac{nRT}{V-nb} - a\frac{n^2}{V^2}\tag{3}.$$


Apply conditions $(2)$ and $(2')$ on $(3)$.


$$\left(\frac{\p P}{\p V}\right)\_T = -\frac{nRT}{(V - bn)^2} + 2a\frac{n^2}{V^3} \overset{(2)}{=} 0\tag4$$


$$\left(\frac{\p^2 P}{\p V^2}\right)\_T = \frac{2nRT}{(V-bn)^3} - 6a\frac{n^2}{V^4} \overset{(2')}{=} 0\tag{4'}$$


Both $(4)$ and $(4')$ are equal to zero. Thus we may multiply either equation by non-matching real numbers, and still maintain equality: $(4) = (4')$. We will use this to our advantage by multiplying $(4)$ with $2/(V-bn)$ and $(4')$ by $-1$.


$$-\frac{2nRT}{(V - nb)^3} + 4a\frac{n^2}{V^3(V- nb)} \overbrace{=}^{(4)\ =\ (4')} -\frac{2nRT}{(V - nb)^3} + 6a\frac{n^2}{V^4}\tag5$$


$$\frac{2}{V - nb} = \frac{3}{V} \implies 3V - 3nb = 2V \implies \enclose{box}[mathcolor="green"]{V\_c = 3nb}\tag{5'}$$


Plug result $(5')$ into equation $(4')$.


$$-\frac{nRT}{(3nb - nb)^2)} + 2a\frac{n^2}{27n^3b^3} = 0 \implies -\frac{RT}{4} + \frac{2a}{27b} = 0 \implies \enclose{box}[mathcolor="red"]{T\_c = \frac{8a}{27bR}}\tag6$$


The outcomes $(5')$ and $(6)$ are to be placed into equation $(3)$.


$$P\_c = nR\frac{8a}{27bR} : 2nb - a\frac{n^2}{9n^2b^2} \implies \enclose{box}[mathcolor="blue"]{P\_c = \frac{a}{27b^2}}\tag7$$


Values for $a, b, R$ in terms of critical parameters
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$$b \overset{(5')}{=} \frac{1}{3n}V\_c\tag{8a}$$
$$a \overset{(7)}{=} 27b^2P\_c \overset{(8a)}{=} \frac{3P\_cV\_c^2}{n^2}\tag{8b}$$
$$R \overset{(6)}{=} \frac{8a}{27bT\_c} \overset{(8a, 8b)}{=} \frac{8P\_cV\_c}{3nT\_c}\tag{8c}$$


Reduced van der Waals equation
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Replace the values of $a, b, R$ from equations $(8a)$$-$$(8c)$ into the original equation $(1)$. Immediately after this substitution multiply both sides by $3/(P\_cV\_c)$. You should have the result


$$\left[\frac{P}{P\_c} + 3\left(\frac{V\_c}{V}\right)^2\right]\left(3\frac{V}{V\_c} - 1\right) = 8\frac{T}{T\_c}.\tag9$$


Define $P/P\_c = \pi, V/V\_c = \varphi, T/T\_c = \tau$. Finally, enjoy the reduced van der Waals equation


$$\enclose{box}[mathcolor="orange"]{\left(\pi + \frac{3}{\varphi^2}\right)(3\varphi - 1) = 8\tau}.\tag{10}$$


* What does equation $(10)$ imply by there being only *reduced* quantities $\pi, \varphi, \tau$?



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>  **Hint:** [theorem of corresponding states](https://en.wikipedia.org/wiki/Theorem_of_corresponding_states).
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$[a]$ While I believe this to be technically true, I also think it deserves a question of its own. The fact that a phase critical point physically *always has to* (?) be a inflection point confuses me a bit, so is best left to another answerer.


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The *a,b,c* in your question are the reduced values not the *a, b* of the vdw eqn and so its best to write them as $T\_R=T/T\_c,~ V\_R=V/V\_c, ~ P\_R=P/P\_c$. 


The critical constants $T\_c, P\_c, V\_c$ can be found from the vdw eqn. by writing it as a cubic in volume. The roots are real for temperatures below the critical temperature $T<T\_C$, ($T\_c$ itself is a point of inflexion)$^\*$. 


The result after some messing around with the algebra is 
$$V\_c=3b, ~ P\_c=\frac{a}{27b^2}, ~T\_c=\frac{8a}{27bR}$$


You can then substitute into the vdw equation and you should get an equation that depends only on reduced values. Have a go at this for yourself, if you get stuck add a comment :)


$\*$ The solution to a cubic is a well known problem and several web pages describe this, but its very messy. It is easier in this particular case to use a 'trick' based on the physics which is that at the critical volume $(V-V\_c)^3=0$ , expand this out as a cubic and compare it term for term with the expansion of the vdw equation. The critical values given above are then apparent.


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