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?


2 Answers 2


$\def\p{\partial} \require{enclose}$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 $\pi, \varphi, \tau$?

Hint: theorem of corresponding states.

$[a]$ 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.

  • $\begingroup$ 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. $\endgroup$
    – Pink
    Feb 21, 2017 at 13:10
  • $\begingroup$ Its almost correct. Use the method given at the bottom of my text is by far the easiest. $\endgroup$
    – porphyrin
    Feb 21, 2017 at 13:16
  • $\begingroup$ I just noticed you changed A, B C which confused me. You should get in your notation $(A+3/B^2)(3B-1)=8C$ $\endgroup$
    – porphyrin
    Feb 21, 2017 at 13:19
  • $\begingroup$ yes sorry for that .I just read your first line and edited the question to avoid confusion with van der wall constants $\endgroup$
    – Pink
    Feb 21, 2017 at 13:21
  • $\begingroup$ Can you please explain a bit more on your last para.I am aware of the 'trick' you mentioned and couldnot grasp it properly. $\endgroup$
    – Pink
    Feb 21, 2017 at 13:22

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