Van der Waals Isotherms: Unphysical Regions and Phase Separation

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Hello! Welcome back to our study of the Van der Waals model.

In our last lesson, we derived the Van der Waals (vdW) equation of state by incorporating two physically motivated corrections—excluded volume and intermolecular attraction—into the ideal gas law. We saw how this equation can be justified both heuristically and through a more formal statistical mechanics derivation.

Today, we will analyze the predictions of the vdW equation by plotting its isotherms—graphs of pressure versus volume at constant temperature. This will lead us to discover a peculiar, non-monotonic behavior below a certain critical temperature. Our main goal is to identify this seemingly "unphysical" region and understand its profound connection to the phenomenon of phase separation, where the system splits into distinct liquid and gas phases.

From Equation to Graph: The Van der Waals Isotherms

Let's start with the vdW equation we derived:
$(P + a \frac{N ^{2}}{V ^{2}}) (V - N b) = N k_{B} T$
To plot pressure as a function of volume, we can rearrange it as:
$P (V, T) = \frac{N k _{B} T}{V - N b} - a \frac{N ^{2}}{V ^{2}}$
This equation describes a surface in P-V-T space. An isotherm is a slice through this surface at a constant temperature $T$ . Let's examine the shape of these curves.

As you can see in the diagram, the isotherms show three distinct behaviors depending on the temperature:

High Temperature ( $T > T_{c}$ ): The first term, representing kinetic pressure, dominates. The curve is a monotonically decreasing function, qualitatively similar to an ideal gas. As you compress the fluid, the pressure smoothly increases.
Low Temperature ( $T < T_{c}$ ): The second term, representing attractive forces, becomes significant. Its competition with the first term creates a non-monotonic "S-shaped" curve, often called a "van der Waals loop."
Critical Temperature ( $T = T_{c}$ ): This is the boundary case where the "S-shape" flattens out into a single point of inflection with a horizontal tangent.

The appearance of the loop at low temperatures is the most interesting feature. It suggests that for a given pressure in a certain range, there can be three possible volumes the system could occupy. Let's explore this.

van der Waals Phase Behavior

To visualize how these isotherms emerge from the equation, the following video provides a helpful explanation. It highlights that the vdW equation is cubic in volume, which is the mathematical origin of the one- and three-solution regimes.

Please watch from 1:25 to 5:34. The first part (until 3:02) explains why the equation can have one or three solutions for the volume. The second part (from 3:02) illustrates how this translates into the different shapes of the P-V isotherms at high and low temperatures.

The Unphysical Region and Thermodynamic Stability

As the video and the isotherm plot show, for $T < T_{c}$ , there is a segment of the curve where pressure increases with volume, meaning $(\partial P / \partial V)_{T} > 0$ . This is physically bizarre. Imagine compressing a piston in a cylinder containing such a substance; instead of pushing back harder, the pressure would drop, causing the piston to slam inwards. This suggests a profound instability.

Let's formalize this intuition. The isothermal compressibility, $κ_{T}$ , measures the relative change in volume in response to a change in pressure:
$κ_{T} = - \frac{1}{V} (\frac{\partial V}{\partial P})_{T}$
For any stable, real-world substance, increasing the pressure must decrease the volume, so $κ_{T}$ must be positive. A region where $(\partial P / \partial V)_{T} > 0$ implies that $κ_{T} < 0$ . This cannot correspond to a stable, uniform equilibrium state.

5.1 Liquid-Gas Transition

The following reading from David Tong's lecture notes provides a crisp explanation of why this region of the vdW isotherm is unphysical and must be replaced by something else.

Please read the section '5.1 Liquid-Gas Transition', up to but not including the subsection '5.1.1 Phase Equilibrium'. Focus on the interpretation of the three solutions and the argument based on the positive gradient (( abla P > 0)) for why the middle solution is unstable.

A Deeper Look: The Helmholtz Free Energy

The instability argument based on compressibility is powerful. We can gain deeper insight by examining the system from the perspective of its free energy. For a system at constant temperature $T$ and volume $V$ , the condition for stable equilibrium is that the Helmholtz free energy $A (T, V, N)$ is at a global minimum.

From the fundamental thermodynamic relation $d A = - S d T - P d V + μ d N$ , we know that at constant $T$ and $N$ :
$P = - (\frac{\partial A}{\partial V})_{T}$
Differentiating with respect to $V$ once more gives:
$(\frac{\partial P}{\partial V})_{T} = - (\frac{\partial ^{2} A}{\partial V ^{2}})_{T}$
The stability condition of negative compressibility ( $κ_{T} > 0$ ) is therefore equivalent to:
$(\frac{\partial ^{2} A}{\partial V ^{2}})_{T} > 0$
This is the mathematical condition for the function $A (V)$ to be convex (curving upwards). A region where the curve is concave (curving downwards) is locally unstable. The system can lower its free energy by moving away from that state. The unphysical region of the vdW isotherm corresponds precisely to a concave region in the Helmholtz free energy.

Thermodynamics of Real Gases

This reading provides a more formal analysis of the instability using the Helmholtz free energy. It demonstrates how the S-shaped isotherm directly implies a non-convex free energy function.

Please read Section 15.3, 'Helmholtz free energy' (pages 88-90). Focus on the argument connecting the stability condition (\partial P/\partial V)_T < 0 to the convexity of the free energy, (\partial^2 F/\partial V^2)_T > 0. Note that this text uses 'F' for Helmholtz free energy, which we are calling 'A'.

The Resolution: Phase Separation

So, if the system cannot exist in the uniform states described by the unstable part of the isotherm, what happens instead?

The system finds a way to achieve a lower free energy. It does this by separating into two distinct, stable phases: a dense liquid phase (small volume) and a low-density gas phase (large volume). Both phases coexist at the same temperature $T$ and the same pressure $P$ .

Phase Separation in the Van der Waals Model — This diagram illustrates the connection between the P-V isotherm (top), the Helmholtz free energy A(V) (middle), and the physical state of the fluid (bottom) for T < T_c. The unphysical region in the P-V curve corresponds to the concave part of the A(V) curve. The system avoids this high-energy state by splitting into two phases (liquid at volume V₁ and gas at volume V₂). This coexistence state corresponds to the straight 'common tangent' line on the A(V) plot, which lies below the unstable curve, representing a state of lower free energy.

As the volume of the container is changed from $V_{1}$ to $V_{2}$ , the system doesn't follow the vdW loop. Instead, it moves along the horizontal line of constant pressure. What changes is the proportion of liquid and gas. At volume $V_{1}$ , the system is 100% liquid. As volume increases, more liquid turns into gas, until at volume $V_{2}$ , the system is 100% gas.

The unstable part of the vdW curve is thus replaced by a horizontal tie-line representing a region of liquid-gas coexistence. This is the vdW model's prediction of a first-order phase transition.

Test your understanding!

Imagine you have a vdW fluid in a cylinder with a piston, held at a temperature $T < T_{c}$ . You start at a large volume, where the fluid is a gas, and slowly compress it. According to the analysis above, what would you observe happening to the pressure as the volume decreases?

Show answer

Gas Phase: Initially, as you decrease the volume, the pressure will rise, following the upper part of the vdW isotherm.
Phase Coexistence: At a specific volume ( $V_{g a s}$ ), droplets of liquid will start to form. As you continue to compress the system, the pressure will remain constant. The decreasing volume is accommodated by more of the gas condensing into the higher-density liquid phase. You would be moving left along the horizontal tie-line in the P-V diagram.
Liquid Phase: Once all the gas has condensed into liquid (at volume $V_{l i q u i d}$ ), the system is again a single phase. Further compression will cause a very sharp rise in pressure, as you are now moving up the steep, liquid part of the isotherm.

Conclusion

In this lesson, we have explored the rich behavior contained within the Van der Waals equation of state. By analyzing the isotherms, we have made a crucial connection between a mathematical feature of the model and a profound physical phenomenon.

Key Takeaways:

Below a critical temperature $T_{c}$ , the vdW isotherms exhibit a non-monotonic "S-shape" or loop.
The region of this loop with a positive slope, $(\partial P / \partial V)_{T} > 0$ , is unphysical because it corresponds to a negative isothermal compressibility ( $κ_{T} < 0$ ), representing a mechanically unstable state.
From a thermodynamic standpoint, this unstable region corresponds to a concave portion of the Helmholtz free energy curve ( $(\partial^{2} A / \partial V^{2})_{T} < 0$ ), which is not a stable equilibrium state.
The physical system resolves this instability by undergoing phase separation. It splits into coexisting liquid and gas phases, which is represented on the P-V diagram by a horizontal line of constant pressure that replaces the unphysical loop.

Preview of the Next Lesson:

We have established that a phase transition occurs and replaces the unstable loop with a flat line at a constant pressure. But how do we determine the exact pressure at which this happens? In the next lesson, we will introduce the Maxwell equal-area construction, a clever and elegant rule that allows us to pinpoint the exact coexistence pressure for any given temperature below $T_{c}$ .