I understand why it's useful to have reduced variables, especially as it is the basis for the theorem of corresponding states. However, I'm having a hard time understanding why the critical constants work best as the scale for reduced variables.

Unlike standard values for things like $\Delta G$, they clearly cannot be some arbitrary value since a substance is not necessarily a gas at any given state (hence the entire concept of a critical temperature). I'm more curious about why the critical temperature works so well with the law of corresponding states in the first place.

  • $\begingroup$ You got any better idea? Critical parameters do characterise p/T properties of substances in quite practical way. $\endgroup$
    – Mithoron
    Jan 30 at 21:57
  • 1
    $\begingroup$ What else to use? $\endgroup$ Jan 30 at 22:20

2 Answers 2


The reason to use the critical properties come from the two conditions that happen for pure fluids: \begin{align} \left(\frac{\partial p}{\partial V}\right)_T = 0 \hspace{1 cm} \left(\frac{\partial^2 p}{\partial V^2}\right)_T = 0 \tag{1,2} \end{align} i.e., the slope and the curvature of the pressure $p$ as a function of the molar volume $V$ (at constant temperature $T$) are zero. When you submit the vdW equation of state to these conditions, $a$ and $b$ are put in function of those critical properties. The math leads to \begin{equation} a = \frac{9RV_\mathrm{c}T_\mathrm{c}}{8} \hspace{2 cm} b = \frac{V_\mathrm{c}}{3} \tag{3,4} \end{equation}

Books do the proof in the natural or straightforward way. First they define the reduced variables and show that they are consistent with the theorem of the corresponding states.

But, we can do the opposite. We replace Eqs. (3) and (4) in the vdW equation of state and see what transformatitions we should choose in order to arrive at the theorem. We begin \begin{align} p &= \frac{RT}{V - b} - \frac{a}{V^2} \\ p &= \frac{RT}{V - V_\mathrm{c}/3} - \left(\frac{9RV_\mathrm{c}T_\mathrm{c}}{8}\right) \frac{1}{V^2} \\ p &= \left(\frac{RT}{V_\mathrm{c}}\right)\frac{1}{(V/V_\mathrm{c}) - 1/3} - \left(\frac{9RV_\mathrm{c}^2T_\mathrm{c}}{8V_\mathrm{c}}\right) \frac{1}{V^2} \\ p &= \left(\frac{RT}{V_\mathrm{c}}\right)\frac{1}{(V/V_\mathrm{c}) - 1/3} - \left(\frac{9RT_\mathrm{c}}{8V_\mathrm{c}}\right) \frac{1}{(V/V_\mathrm{c})^2} \\ p &= \left(\frac{RT_\mathrm{c}}{V_\mathrm{c}}\right) \left[ \frac{T/T_\mathrm{c}}{(V/V_\mathrm{c}) - 1/3} - \left(\frac{9}{8}\right) \frac{1}{(V/V_\mathrm{c})^2} \right] \\ \frac{pV_\mathrm{c}}{RT_\mathrm{c}} &= \frac{T/T_\mathrm{c}}{(V/V_\mathrm{c}) - 1/3} - \left(\frac{9}{8}\right) \frac{1}{(V/V_\mathrm{c})^2} \\ \frac{p_\mathrm{c}}{p_\mathrm{c}} \left(\frac{pV_\mathrm{c}}{RT_\mathrm{c}}\right) &= \frac{T/T_\mathrm{c}}{(V/V_\mathrm{c}) - 1/3} - \left(\frac{9}{8}\right) \frac{1}{(V/V_\mathrm{c})^2}\\ \frac{p}{p_\mathrm{c}} \underbrace{\left(\frac{p_\mathrm{c}V_\mathrm{c}}{RT_\mathrm{c}}\right)}_ {Z_\mathrm{c}} &= \frac{T/T_\mathrm{c}}{(V/V_\mathrm{c}) - 1/3} - \left(\frac{9}{8}\right) \frac{1}{(V/V_\mathrm{c})^2} \\ \left(\frac{p}{p_\mathrm{c}}\right)Z_\mathrm{c} &= \frac{T/T_\mathrm{c}}{(V/V_\mathrm{c}) - 1/3} - \left(\frac{9}{8}\right) \frac{1}{(V/V_\mathrm{c})^2} \\ \frac{p}{p_\mathrm{c}} &= \frac{1}{Z_\mathrm{c}} \left[ \frac{T/T_\mathrm{c}}{(V/V_\mathrm{c}) - 1/3} - \left(\frac{9}{8}\right) \frac{1}{(V/V_\mathrm{c})^2} \right] \tag{5} \end{align} If you remember, the vdW equation of state predicts a universal critical compressibility factor of $3/8$. Therefore, Eq. (5) yields \begin{align} \frac{p}{p_\mathrm{c}} &= \frac{8}{3} \left[ \frac{T/T_\mathrm{c}}{(V/V_\mathrm{c}) - 1/3} - \left(\frac{9}{8}\right) \frac{1}{(V/V_\mathrm{c})^2} \right] \\ \frac{p}{p_\mathrm{c}} &= \frac{8(T/T_\mathrm{c})}{3(V/V_\mathrm{c}) - 1} - \frac{3}{(V/V_\mathrm{c})^2} \tag{6} \\ \end{align} Eq. (6) suggests that the reduced variables must be in the form $x_\mathrm{r} \equiv x/x_\mathrm{c}$. This is the only way that we can arrive to a universal function, for which all fluids at the same reduced volume and temperature will lead to the same deviation from ideality.

If you want to be creative, such as $T_\mathrm{r} \equiv T / \sqrt{T_\mathrm{c}T_\mathrm{t}}$, where $T_\mathrm{t}$ is the triple temperature, this won't work. $T_\mathrm{t}$ will be in some part of the equation, and we will have to say that "all fluids at the same reduced temperature, reduced volume, and triple temperature will lead to the same deviation from ideality". This is not the statement of the theorem, because it adds one more thing. You are left with no other choice.

I'm more curious about why the critical temperature works so well with the law of corresponding states in the first place.

Unfortunately, this principle works well with very simple fluids. I have used much more complex equations of states and is unreliable for most organic compounds that are moderately-highly polar such as ammonia, all alcohols, aromatic compounds with some groups, amines, etc. When in the last paragraph I sad 'add one more thing', this is exactly what some other people did. You can add another parameter to this theorem, Kenneth Pitzer did it in the 1955 with the accentric factor and created a 'three-parameter theorem of corresponding states'. Still, for polar fluids, it sometimes fails pretty bad.

  • $\begingroup$ Do not you have the derivatives (3,4) reciprocal ? there should be dp/dV and d2p/dV^2 and not dV/dp and d2V/dp^2 $\endgroup$
    – Poutnik
    Jan 31 at 16:16
  • 1
    $\begingroup$ @Poutnik Yeah, you are right.... How come I didn't notice that? Thank you very much for the observation... $\endgroup$ Jan 31 at 16:22
  • $\begingroup$ Perhaps we are inclined to consider p as independent and V as dependent variable, so volume derivative per pressure seems more natural. $\endgroup$
    – Poutnik
    Jan 31 at 16:32
  • $\begingroup$ Yes, that's how it goes. Equations of states are explicit in $V$ and $T$, and when you see the $p-V$ graph, you see the maximum righ there. Actually, we can think of $V$ (independent variable) as a function of $p$ and $T$, but you may need like 5 papers to write those three roots. No one does that. $\endgroup$ Jan 31 at 16:37

In context of the van der Waals equation, it has the clear mathematical analysis reasoning.

The function $p=\text{f}(V)_{T=\text{const}}$ must have at $T_\mathrm{c}$ and $p_\mathrm{c}$ an inflection point and zero slope, so the first and second derivative of pressure per volume must be zero. It also means the infinite derivative of volume per pressure.

It allows splitting the unique value of volume for given temperature and pressure to two values ( the 3rd without physical meaning) below the critical temperature.

It then comes naturally as advantageous to express the equation in terms of the reduced parameters wrt the critical ones.


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