Van der Waals Critical Point and Constants

Recommended resources

The freely available resources this lesson will direct you to.

Hello! Welcome to our next lesson on the Van der Waals model.

In our previous session, we mapped out the liquid-gas phase diagram by defining the binodal (coexistence) and spinodal (stability limit) curves. We saw that these two curves meet at a single, unique point—the apex of the coexistence region. Today, we will focus entirely on this special point: the critical point.

Our goal for this lesson is to mathematically locate the critical point of the Van der Waals fluid and calculate the critical constants: the critical pressure $P_{c}$ , critical molar volume $V_{m, c}$ , and critical temperature $T_{c}$ , expressing them in terms of the Van der Waals parameters $a$ and $b$ . This is a cornerstone calculation that bridges the microscopic parameters of the model to the macroscopic, measurable properties of a fluid at criticality.

The Mathematical Conditions for the Critical Point

Let's begin with the Van der Waals equation of state for one mole of a fluid (using $V_{m}$ for molar volume):
$P = \frac{RT}{V _{m} - b} - \frac{a}{V _{m}^{2}}$
As we've discussed, for temperatures $T < T_{c}$ , the isotherms $P (V_{m})$ exhibit a local minimum and maximum. As $T$ approaches $T_{c}$ from below, these two extrema move closer together, finally merging at the critical point for the isotherm $T = T_{c}$ .

Van der Waals Equation and Critical Point — This diagram provides a concise summary. The plots show the isotherms, with the critical point C at the peak of the coexistence region. The conditions for this point are listed, along with the results of the derivation we are about to perform.

At this merger, the curve is momentarily flat and is also an inflection point. This provides the two mathematical conditions that uniquely define the critical point:

The slope of the isotherm is zero:
$(\frac{\partial P}{\partial V _{m}})_{T = T_{c}} = 0$
The curvature of the isotherm is also zero:
$(\frac{\partial ^{2} P}{\partial V _{m}^{2}})_{T = T_{c}} = 0$

These two conditions must be satisfied simultaneously at the critical point $(P_{c}, V_{m, c}, T_{c})$ .

To solidify this concept, let's watch a short video that introduces these conditions.

The van der Waals equation of state at the critical point

The following video segment clearly explains the physical reasoning behind the two mathematical conditions that define the critical point.

Watch the segment from 03:02 to 04:04. The speaker explains how the horizontal slope and the inflection point on the critical isotherm lead to the first and second derivative conditions.

Deriving the Critical Constants

Now we have a well-defined mathematical task: solve the two simultaneous equations above to find the critical constants. Let's proceed step-by-step.

Step 1: Calculate the derivatives

First, we differentiate the Van der Waals equation with respect to $V_{m}$ at constant $T$ :

(\frac{\partial P}{\partial V _{m}})_{T} = - \frac{RT}{( V _{m} - b ) ^{2}} + \frac{2 a}{V _{m}^{3}}

Next, we take the second derivative:

(\frac{\partial ^{2} P}{\partial V _{m}^{2}})_{T} = \frac{2 RT}{( V _{m} - b ) ^{3}} - \frac{6 a}{V _{m}^{4}}

Step 2: Set the derivatives to zero at the critical point

At the critical point $(V_{m, c}, T_{c})$ , both expressions are zero:

- \frac{R T _{c}}{( V _{m, c} - b ) ^{2}} + \frac{2 a}{V _{m, c}^{3}} = 0 ⟹ \frac{R T _{c}}{( V _{m, c} - b ) ^{2}} = \frac{2 a}{V _{m, c}^{3}} (Eq. 1)

\frac{2 R T _{c}}{( V _{m, c} - b ) ^{3}} - \frac{6 a}{V _{m, c}^{4}} = 0 ⟹ \frac{2 R T _{c}}{( V _{m, c} - b ) ^{3}} = \frac{6 a}{V _{m, c}^{4}} (Eq. 2)

Step 3: Solve the system of equations

We now have a system of two equations with two unknowns, $V_{m, c}$ and $T_{c}$ . A straightforward way to solve this is to divide Eq. 2 by Eq. 1. This will eliminate $R T_{c}$ and $a$ .

\frac{2 R T _{c} / ( V _{m, c} - b ) ^{3}}{R T _{c} / ( V _{m, c} - b ) ^{2}} = \frac{6 a / V _{m, c}^{4}}{2 a / V _{m, c}^{3}}

Simplifying both sides gives:

\frac{2}{V _{m, c} - b} = \frac{3}{V _{m, c}}

Solving for $V_{m, c}$ yields our first critical constant:

2 V_{m, c} = 3 (V_{m, c} - b) ⟹ 2 V_{m, c} = 3 V_{m, c} - 3 b ⟹ V_{m, c} = 3 b

The critical molar volume is simply three times the excluded volume parameter $b$ .

Now, substitute this result back into Eq. 1 to find $T_{c}$ :

\frac{R T _{c}}{( 3 b - b ) ^{2}} = \frac{2 a}{( 3 b ) ^{3}} ⟹ \frac{R T _{c}}{( 2 b ) ^{2}} = \frac{2 a}{27 b ^{3}} ⟹ \frac{R T _{c}}{4 b ^{2}} = \frac{2 a}{27 b ^{3}}

Solving for $T_{c}$ :

T_{c} = \frac{4 b ^{2}}{R} \cdot \frac{2 a}{27 b ^{3}} ⟹ T_{c} = \frac{8 a}{27 R b}

Finally, we substitute both $V_{m, c}$ and $T_{c}$ back into the original Van der Waals equation to find $P_{c}$ :

P_{c} = \frac{R T _{c}}{V _{m, c} - b} - \frac{a}{V _{m, c}^{2}} = \frac{R}{3 b - b} (\frac{8 a}{27 R b}) - \frac{a}{( 3 b ) ^{2}}

P_{c} = \frac{1}{2 b} \frac{8 a}{27 b} - \frac{a}{9 b ^{2}} = \frac{8 a}{54 b ^{2}} - \frac{a}{9 b ^{2}} = \frac{4 a}{27 b ^{2}} - \frac{3 a}{27 b ^{2}} ⟹ P_{c} = \frac{a}{27 b ^{2}}

The following video walks through this entire derivation, which will help reinforce the algebraic steps.

The van der Waals equation of state at the critical point

Let's watch this derivation being performed. Seeing the steps laid out in a different format can help solidify your understanding of the process.

Watch from 04:04 to 12:47. The video systematically derives the expressions for the critical molar volume, critical temperature, and critical pressure, just as we did above.

An Alternative Approach

Given your mathematical background, you may appreciate an alternative, more algebraic method for finding $V_{m, c}$ . It avoids calculus by considering the roots of the Van der Waals equation when written as a cubic polynomial in $V_{m}$ . At the critical point, the three real roots of the cubic equation must all coalesce into a single value, $V_{m, c}$ .

Properties of Gases

The following resource presents this elegant alternative derivation.

Please read the section from just before equation 6.3.7 to equation 6.3.12. Pay special attention to the boxed-off text (equations 6.3.10a and 6.3.10b) that explains how to find V_{m,c} by comparing polynomial coefficients. It's a neat trick that highlights the structure of the equation.

The Critical Compression Factor, $Z_{c}$

We have now expressed $P_{c}, V_{m, c}, T_{c}$ in terms of $a$ and $b$ . A very interesting consequence is that we can form a dimensionless combination of these constants, known as the critical compression factor, $Z_{c}$ .

Z_{c} = \frac{P _{c} V _{m, c}}{R T _{c}}

Let's substitute our derived expressions:

Z_{c} = \frac{( \frac{a}{27 b ^{2}} ) ( 3 b )}{R ( \frac{8 a}{27 R b} )} = \frac{\frac{3 ab}{27 b ^{2}}}{\frac{8 a R}{27 R b}} = \frac{\frac{a}{9 b}}{\frac{8 a}{27 b}}

Z_{c} = \frac{a}{9 b} \cdot \frac{27 b}{8 a} = \frac{27}{72} = \frac{3 \times 9}{8 \times 9} = \frac{3}{8}

Thus, the Van der Waals model predicts a universal value for the critical compression factor:

Z_{c} = \frac{3}{8} = 0.375

This prediction is independent of the specific gas parameters $a$ and $b$ . It suggests that $P_{c} V_{m, c} / (R T_{c})$ should be the same for all substances that follow the Van der Waals equation. Experimentally, most real gases have $Z_{c}$ values in the range 0.25–0.31. For example, for water, $Z_{c} \approx 0.23$ , and for CO₂, $Z_{c} \approx 0.27$ .

The vdW prediction of 0.375 is in the right ballpark but is consistently high. This is one of the key quantitative failings of the model, but the fact that it predicts a universal constant at all is a remarkable success.

The van der Waals equation of state at the critical point

The video we watched earlier concludes by calculating Z_c and discussing its comparison with experimental values.

Watch the final segment from 13:22 to 16:11. This part covers the calculation of Z_c=3/8 and puts this value into the context of experimental results for real gases.

Test your understanding!

You can invert the relationships we found for $(P_{c}, V_{m, c}, T_{c})$ to express the Van der Waals parameters $(a, b)$ in terms of the critical constants. This is practically useful, as the critical constants are experimentally measurable. Try to derive the expressions for $a$ and $b$ in terms of $T_{c}$ and $P_{c}$ .

Show answer

We start with our results:

$T_{c} = \frac{8 a}{27 R b}$
$P_{c} = \frac{a}{27 b ^{2}}$

From Eq. 2, we can write $a = 27 b^{2} P_{c}$ . Substitute this into Eq. 1:

T_{c} = \frac{8 ( 27 b ^{2} P _{c} )}{27 R b} = \frac{8 b P _{c}}{R}

Now, we can solve this for $b$ :

b = \frac{R T _{c}}{8 P _{c}}

With $b$ determined, we can substitute it back into the expression for $a$ :

a = 27 P_{c} (\frac{R T _{c}}{8 P _{c}})^{2} = 27 P_{c} \frac{R ^{2} T _{c}^{2}}{64 P _{c}^{2}}

a = \frac{27 R ^{2} T _{c}^{2}}{64 P _{c}}

So, if you measure the critical temperature and pressure of a gas, you can estimate its Van der Waals parameters $a$ and $b$ .

Conclusion

In this lesson, we have analytically pinpointed the critical point of the Van der Waals fluid. By applying the mathematical conditions for a horizontal inflection point, we have connected the model's microscopic parameters to the macroscopic critical constants.

Key Takeaways:

The critical point is defined by the simultaneous conditions $(\partial P / \partial V_{m})_{T} = 0$ and $(\partial^{2} P / \partial V_{m}^{2})_{T} = 0$ .
Solving this system for the Van der Waals equation yields the critical constants:
- Critical Volume: $V_{m, c} = 3 b$
- Critical Temperature: $T_{c} = \frac{8 a}{27 R b}$
- Critical Pressure: $P_{c} = \frac{a}{27 b ^{2}}$
These constants combine to form a universal dimensionless quantity, the critical compression factor $Z_{c} = \frac{P _{c} V _{m, c}}{R T _{c}} = \frac{3}{8}$ , which is a key prediction of the model.

Preview of the Next Lesson:
The existence of these critical constants allows us to define a set of "reduced" variables for pressure, volume, and temperature ( $p_{r} = P / P_{c}$ , etc.). In our next lesson, we will substitute these reduced variables into the Van der Waals equation. We will discover that the parameters $a$ and $b$ completely disappear, leading to a universal equation of state known as the Law of Corresponding States. This is our first encounter with the powerful concept of universality in critical phenomena.

Van der Waals Critical Point and Constants

Recommended resources

The Mathematical Conditions for the Critical Point

Deriving the Critical Constants

An Alternative Approach

The Critical Compression Factor, Zc​

Conclusion

The Critical Compression Factor, $Z_{c}$