Interactive Lecture
Van der Waals, Redlich-Kwong Models, and Compressibility Factor
by Dr Arnaldo Delli Carri
The ideal gas law \(PV = nRT\) works well at low pressures and high temperatures, but real gases deviate significantly at high pressures or near condensation points.
Real gases experience intermolecular forces (attractive) and molecular volume (repulsive) that the ideal gas law ignores. Equations of state correct for these effects.
\[PV = nRT\]
Assumptions: No intermolecular forces, negligible molecular volume
The Van der Waals equation accounts for molecular size and attractions:
\[\left(P + \frac{an^2}{V^2}\right)(V - nb) = nRT\]
Or per mole: \(\left(P + \frac{a}{v^2}\right)(v - b) = RT\)
Where:
Physical meaning:
An improvement over Van der Waals with better accuracy:
\[P = \frac{RT}{v-b} - \frac{a}{\sqrt{T}\,v(v+b)}\]
Or: \(\left(P + \frac{a}{\sqrt{T}\,v(v+b)}\right)(v - b) = RT\)
More accurate for high pressures and wider temperature ranges
\[Z = \frac{PV}{nRT} = \frac{Pv}{RT}\]
For ideal gas: \(Z = 1\)
For real gas: \(Z \neq 1\)
Critical Properties (Van der Waals):
\[P_c = \frac{a}{27b^2}, \quad v_c = 3b, \quad T_c = \frac{8a}{27Rb}\]
Reduced Properties:
\[P_r = \frac{P}{P_c}, \quad v_r = \frac{v}{v_c}, \quad T_r = \frac{T}{T_c}\]
Law of Corresponding States: Gases at same \((P_r, T_r)\) have same \(Z\)
Explore how real gas molecules behave differently from ideal gas predictions. Adjust temperature and volume to see molecular interactions and their effect on the P-V diagram. Drag the container wall or click on the P-V diagram to change state.
| Model | Equation | Accuracy |
|---|---|---|
| Ideal Gas | \(PV = nRT\) | Low P, high T |
| Van der Waals | \((P + \frac{a}{v^2})(v-b) = RT\) | Moderate P, T |
| Redlich-Kwong | \(P = \frac{RT}{v-b} - \frac{a}{\sqrt{T}\,v(v+b)}\) | High P, wide T |
| Compressibility | \(Z = \frac{Pv}{RT}\) | All conditions |
For CO₂ (a = 0.3658 Pa·m⁶/mol², b = 4.286×10⁻⁵ m³/mol) at T = 300 K, v = 0.001 m³/mol, find pressure using the Van der Waals equation.
Show Solution
\(P = \frac{RT}{v-b} - \frac{a}{v^2}\)
\(P = \frac{8.314 \times 300}{0.001 - 4.286 \times 10^{-5}} - \frac{0.3658}{0.001^2}\)
\(P = \frac{2494.2}{9.571 \times 10^{-4}} - 365{,}800 = 2.606 \times 10^6 - 3.658 \times 10^5 = 2.24 \text{ MPa}\)
Ideal gas would give: \(P = RT/v = 2.494 \text{ MPa}\) (11% error!)
Answer: \(P \approx 2.24\,\text{MPa}\), about 11% below the ideal-gas value of 2.494 MPa.
Calculate Z for nitrogen at P = 10 MPa, T = 300 K. N₂ Van der Waals constants: a = 0.137 Pa·m⁶/mol², b = 3.87×10⁻⁵ m³/mol.
Show Solution
Starting estimate — ideal gas:
\(v_0 = \frac{RT}{P} = \frac{8.314 \times 300}{10 \times 10^6} = 2.494 \times 10^{-4} \text{ m}^3/\text{mol}\)
Fixed-point iteration — rearrange the Van der Waals equation \(P = \frac{RT}{v-b} - \frac{a}{v^2}\) as \(v = \frac{RT}{P + a/v_0^2} + b\):
\(\dfrac{a}{v_0^2} = \dfrac{0.137}{(2.494 \times 10^{-4})^2} = 2.202 \times 10^6\text{ Pa}\)
\[v_1 = \frac{2494.2}{10{,}000{,}000 + 2{,}202{,}000} + 3.87\times10^{-5} = 2.044 \times 10^{-4} + 3.87 \times 10^{-5} = 2.431 \times 10^{-4} \text{ m}^3/\text{mol}\]
Repeating the iteration converges to \(v \approx 2.403 \times 10^{-4}\text{ m}^3/\text{mol}\) (at which \(P = 10\text{ MPa}\)).
Compressibility factor:
\(Z = \frac{Pv}{RT} = \frac{10 \times 10^6 \times 2.403 \times 10^{-4}}{8.314 \times 300} = \frac{2403}{2494.2} = 0.963\)
Z < 1 indicates that attractive intermolecular forces are dominant at this pressure and temperature, compressing the gas below the ideal-gas volume.
Answer: \(Z \approx 0.963\).
For a gas with a = 0.5 Pa·m⁶/mol², b = 5×10⁻⁵ m³/mol, find critical temperature and pressure.
Show Solution
\(T_c = \frac{8a}{27Rb} = \frac{8 \times 0.5}{27 \times 8.314 \times 5 \times 10^{-5}} = 356.4 \text{ K}\)
\(P_c = \frac{a}{27b^2} = \frac{0.5}{27 \times (5 \times 10^{-5})^2} = 7.41 \text{ MPa}\)
Answer: \(T_c \approx 356.4\,\text{K}\), \(P_c \approx 7.41\,\text{MPa}\).