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

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.

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 $a, b, n, R$

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

$$b \overset{(5')}{=} \frac{1}{3n}V_c\tag{8a}$$ $$a \overset{(7)}{=} 27b^2P_c \overset{(8a)}{=} \frac{3P_cV_c^2}{n^2}\tag{8b}$$ $$R \overset{(6)}{=} \frac{8a}{27bT_c} \overset{(8a, 8b)}{=} \frac{8P_cV_c}{3nT_c}\tag{8c}$$

Reduced van der Waals equation

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 $\pi, \varphi, \tau$?

$[a]$ 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.

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.

• Thanks for your answer.I tried to go in for direct substitution it made it cumbersome .Meanwhile I just got hold of the reduced form as $[A-\frac{3}{B}][3B-1]=8RC$.Is there any method to arrive at this which requires ingenuity and not just plain substitution. – Pink Feb 21 '17 at 13:10
• Its almost correct. Use the method given at the bottom of my text is by far the easiest. – porphyrin Feb 21 '17 at 13:16
• I just noticed you changed A, B C which confused me. You should get in your notation $(A+3/B^2)(3B-1)=8C$ – porphyrin Feb 21 '17 at 13:19
• yes sorry for that .I just read your first line and edited the question to avoid confusion with van der wall constants – Pink Feb 21 '17 at 13:21
• Can you please explain a bit more on your last para.I am aware of the 'trick' you mentioned and couldnot grasp it properly. – Pink Feb 21 '17 at 13:22