Critical Constants and the Critical Point
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You should be comfortable manipulating the van der Waals equation and understanding how intermolecular forces and molecular volume deviate from ideal gas behaviour.
Why this matters
The critical point defines the boundary beyond which distinct liquid and gas phases cease to exist.
In the previous lecture, we introduced the van der Waals equation to account for the finite size of molecules and the attractive forces between them. While this equation improves our description of real gases, it also reveals a profound physical limit: the critical point. At this specific temperature and pressure, the distinction between liquid and gas disappears entirely. Understanding this concept is crucial not only for thermodynamics but also for practical applications such as supercritical fluid extraction, where substances like carbon dioxide are used as solvents in a state that possesses the density of a liquid but the viscosity of a gas. This lecture explores how the critical constants are derived from the van der Waals equation and what they signify physically.
The Disappearance of the Meniscus
Imagine heating a sealed container of liquid water. As the temperature rises, the liquid expands and the vapour above it becomes denser. Eventually, at a specific temperature known as the critical temperature, the densities of the liquid and the vapour become identical. At this moment, the boundary surface, or meniscus, between the two phases vanishes. You can no longer distinguish which part of the container is liquid and which is gas. This state is called the supercritical fluid. If you continue to heat the substance, it remains a single, homogeneous phase regardless of pressure. The critical point is therefore the highest temperature at which a substance can exist as a liquid. Above this temperature, no amount of pressure can liquefy the gas. This is a fundamental departure from the ideal gas law, which assumes no intermolecular attractions and thus no phase transitions. The van der Waals equation, by incorporating attractive forces, naturally predicts this phenomenon. It shows that the isotherms (curves of constant temperature) for a real gas exhibit a region of instability where the pressure decreases as volume decreases, a mathematical artefact that corresponds to the physical phase transition. The critical point is the inflection point on the critical isotherm, where the slope and the curvature of the pressure-volume curve are both zero. This mathematical condition allows us to calculate the critical constants directly from the van der Waals parameters \(a\) and \(b\), linking microscopic molecular properties to macroscopic observable limits.
Mathematical Conditions for the Critical Point
To locate the critical point mathematically, we examine the van der Waals equation for one mole of gas, where \(n=1\). The equation is \(\left(P+a\dfrac{1}{V^2}\right)(V-b)=RT\). We can rearrange this to express pressure \(P\) as a function of volume \(V\) and temperature \(T\): \[ P = \frac{RT}{V-b} - \frac{a}{V^2} \] The critical point is defined as the point on the critical isotherm where the curve has an inflection point. At an inflection point, the first derivative of pressure with respect to volume is zero (the slope is horizontal), and the second derivative is also zero (the curvature changes sign). These conditions are: \[ \left(\frac{\partial P}{\partial V}\right)_T = 0 \quad \text{and} \quad \left(\frac{\partial^2 P}{\partial V^2}\right)_T = 0 \] Let us compute the first derivative with respect to \(V\), treating \(T\) as constant: \[ \frac{\partial P}{\partial V} = -\frac{RT}{(V-b)^2} + \frac{2a}{V^3} \] Setting this to zero gives: \[ \frac{RT}{(V-b)^2} = \frac{2a}{V^3} \quad \dots (1) \] Now, we compute the second derivative: \[ \frac{\partial^2 P}{\partial V^2} = \frac{2RT}{(V-b)^3} - \frac{6a}{V^4} \] Setting this to zero gives: \[ \frac{2RT}{(V-b)^3} = \frac{6a}{V^4} \quad \dots (2) \] By dividing equation (2) by equation (1), we can eliminate \(RT\) and \(a\). \[ \frac{\frac{2RT}{(V-b)^3}}{\frac{RT}{(V-b)^2}} = \frac{\frac{6a}{V^4}}{\frac{2a}{V^3}} \] \[ \frac{2}{V-b} = \frac{3}{V} \] Solving for \(V\), we find the critical volume \(V_c\): \[ 2V = 3(V-b) \implies 2V = 3V - 3b \implies V_c = 3b \] This result indicates that the critical molar volume is three times the excluded volume parameter \(b\). This is a direct consequence of the specific form of the van der Waals equation. At the critical point, then, the free volume \(V_c - b\) is twice the excluded volume \(b\).
Determining Critical Temperature and Pressure
Having found \(V_c = 3b\), we can substitute this back into equation (1) to find the critical temperature \(T_c\). \[ \frac{RT_c}{(3b-b)^2} = \frac{2a}{(3b)^3} \] \[ \frac{RT_c}{4b^2} = \frac{2a}{27b^3} \] Solving for \(T_c\): \[ T_c = \frac{2a}{27b^3} \cdot \frac{4b^2}{R} = \frac{8a}{27bR} \] This equation shows that the critical temperature is directly proportional to the strength of intermolecular attractions (\(a\)) and inversely proportional to the molecular size (\(b\)). Substances with strong attractive forces, such as water or ammonia, have higher critical temperatures than substances with weak forces, such as helium or hydrogen. Next, we substitute \(V_c\) and \(T_c\) into the original pressure equation to find the critical pressure \(P_c\). \[ P_c = \frac{R T_c}{V_c - b} - \frac{a}{V_c^2} \] Substituting \(T_c = \frac{8a}{27bR}\) and \(V_c = 3b\): \[ P_c = \frac{R \left(\frac{8a}{27bR}\right)}{3b - b} - \frac{a}{(3b)^2} \] \[ P_c = \frac{\frac{8a}{27b}}{2b} - \frac{a}{9b^2} \] \[ P_c = \frac{8a}{54b^2} - \frac{a}{9b^2} = \frac{4a}{27b^2} - \frac{3a}{27b^2} \] \[ P_c = \frac{a}{27b^2} \] Thus, the three critical constants are expressed solely in terms of the van der Waals parameters \(a\) and \(b\): \[ V_c = 3b, \quad T_c = \frac{8a}{27bR}, \quad P_c = \frac{a}{27b^2} \] These relationships allow us to estimate the critical constants of a gas if we know its \(a\) and \(b\) values, or conversely, to determine \(a\) and \(b\) if the critical constants are known experimentally. This is particularly useful because \(a\) and \(b\) are often determined by fitting experimental data, and the critical point provides a robust anchor for these parameters.
The Critical Compressibility Factor
A key dimensionless quantity in thermodynamics is the compressibility factor \(Z\), defined as \(Z = \frac{PV}{nRT}\). For an ideal gas, \(Z\) is always 1. For real gases, \(Z\) varies with pressure and temperature. At the critical point, we can calculate the critical compressibility factor \(Z_c\) using the van der Waals results derived above. \[ Z_c = \frac{P_c V_c}{R T_c} \] Substituting the expressions for \(P_c\), \(V_c\), and \(T_c\): \[ Z_c = \frac{\left(\frac{a}{27b^2}\right)(3b)}{R \left(\frac{8a}{27bR}\right)} \] Simplifying the numerator: \[ \text{Numerator} = \frac{3a}{27b} = \frac{a}{9b} \] Simplifying the denominator: \[ \text{Denominator} = \frac{8a}{27b} \] Therefore: \[ Z_c = \frac{\frac{a}{9b}}{\frac{8a}{27b}} = \frac{a}{9b} \cdot \frac{27b}{8a} = \frac{27}{72} = \frac{3}{8} \] \[ Z_c = 0.375 \] This is a universal value predicted by the van der Waals equation. It means that for any gas described by this model, the product of critical pressure and critical volume is exactly \(3/8\) of \(R T_c\). However, experimental values for real gases vary. For example, for nitrogen, \(Z_c\) is approximately 0.29, and for carbon dioxide, it is approximately 0.27. The fact that the van der Waals prediction (0.375) is higher than most experimental values indicates that the simple van der Waals model overestimates the compressibility at the critical point. This discrepancy arises because the van der Waals equation treats attraction as a crude average (mean-field) correction and repulsion as hard spheres, whereas real intermolecular potentials are more complex. Nevertheless, the derivation of \(Z_c\) is a powerful demonstration of how microscopic parameters constrain macroscopic behaviour.
The result
The van der Waals equation
Real gas molecules have finite size and attract one another — both directly contradicting the idealisations kinetic-theory-pressure used to derive the ideal gas law. Correcting for finite molecular volume (subtracting an excluded volume \(nb\) from the container volume) and for intermolecular attraction (adding a pressure correction \(a(n/V)^2\) to the measured pressure) gives the van der Waals equation, \(\left(P+a\dfrac{n^2}{V^2}\right)(V-nb)=nRT\), with \(a\) and \(b\) empirically determined constants specific to each gas, unlike the universal \(R\).
A common misconception is that the critical point is a phase transition in the same sense as boiling or melting. In reality, at the critical point, the latent heat of vaporisation becomes zero, and the distinction between liquid and gas disappears continuously. There is no discontinuous jump in density or enthalpy. This is a second-order phase transition, unlike the first-order transitions observed at lower temperatures. Confusing these leads to incorrect assumptions about energy changes during supercritical processes. C
Worked example
Problem. Given the van der Waals constants for carbon dioxide as \(a = 3.59\ \mathrm{L^2\,atm\,mol^{-2}}\) and \(b = 0.0427\ \mathrm{L\,mol^{-1}}\), calculate the critical temperature \(T_c\) and critical pressure \(P_c\). Use \(R = 0.08206\ \mathrm{L\,atm\,K^{-1}\,mol^{-1}}\).
- Identify the formula for critical temperature: \(T_c = \frac{8a}{27bR}\).
- Substitute the given values into the equation: \(T_c = \frac{8 \times 3.59}{27 \times 0.0427 \times 0.08206}\).
- Calculate the numerator: \(8 \times 3.59 = 28.72\).
- Calculate the denominator: \(27 \times 0.0427 \times 0.08206 \approx 0.09461\).
- Compute \(T_c\): \(T_c = \frac{28.72}{0.09461} \approx 303.6\ \mathrm{K}\).
- Identify the formula for critical pressure: \(P_c = \frac{a}{27b^2}\).
- Substitute the values: \(P_c = \frac{3.59}{27 \times (0.0427)^2}\).
- Calculate the denominator: \(27 \times 0.001823 \approx 0.04923\).
- Compute \(P_c\): \(P_c = \frac{3.59}{0.04923} \approx 72.9\ \mathrm{atm}\).
Units check. The units for \(T_c\) are \(\mathrm{K}\) and for \(P_c\) are \(\mathrm{atm}\), consistent with the input constants.
These estimates are close to the measured critical constants of carbon dioxide, 304.1 K and 72.8 atm, which is no surprise: tabulated van der Waals constants are usually fitted to the critical point.
Three quick tests
- What is the value of the critical compressibility factor \(Z_c\) predicted by the van der Waals equation?
Answer
0.375 or 3/8. - If the van der Waals constant \(a\) increases, what happens to the critical temperature \(T_c\)?
Answer
It increases, as \(T_c\) is directly proportional to \(a\). - Why is the critical point considered a second-order phase transition?
Answer
Because there is no latent heat and no discontinuity in density; the transition is continuous.
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Every lecture and every result in this unit, with prerequisites and threads.
Lecture notes checked line by line by the editor; every number in the worked example was recomputed.