Van der Waals Equation Calculator
Solve the van der Waals equation for the pressure of a real gas (CO₂, N₂, O₂, H₂O, NH₃…) and compare it with the ideal gas law — compressibility factor Z, attraction and repulsion terms and critical constants.
How it works
(P + a n²/V²)(V − n b) = n R T → P = n R T ÷ (V − n b) − a n² ÷ V² · R = 0.083145 L·bar/(mol·K) · Z = P V ÷ (n R T) · T_c = 8a ÷ (27 R b), P_c = a ÷ (27 b²), V_c = 3 b
The van der Waals equation (1873) corrects the ideal gas law for two facts: molecules take up space (the excluded volume b per mole) and they attract each other (the a term, which lowers the pressure on the walls). It describes real gases well at moderate pressures and predicts a critical point. The constants a and b are the standard tabulated values (CRC Handbook) in L²·bar/mol² and L/mol. At high densities or near condensation, more accurate equations of state such as Peng–Robinson are preferred.
Worked example
One mole of CO₂ in a 1 L vessel at 300 K (a = 3.640 L²·bar/mol², b = 0.04267 L/mol): the repulsion term is 0.083145 × 300 ÷ (1 − 0.04267) = 26.055 bar and the attraction term is 3.640 bar, so P = 22.415 bar. The ideal gas law would predict 24.943 bar, so Z = 0.8986 — the attraction between CO₂ molecules dominates.
Sources: LibreTexts Chemistry — The van der Waals equation · Wikipedia — Van der Waals equation
Frequently asked questions
What do a and b mean in the van der Waals equation?
a measures the attraction between molecules (larger for polar or big molecules, e.g. 5.5 for water vs 0.035 for helium), and b is the volume excluded by one mole of molecules themselves.
When does a real gas differ most from an ideal gas?
At high pressure (molecules are close together, so their size matters) and at low temperature (attractions matter, especially near condensation). At room conditions most gases are within 1% of ideal.
What is the compressibility factor Z?
Z = PV/(nRT). Z = 1 for an ideal gas; Z < 1 means attractions dominate and Z > 1 means molecular size dominates.
How do I convert a from L²·atm/mol² or Pa·m⁶/mol²?
1 L²·bar/mol² = 0.98692 L²·atm/mol² = 0.1 Pa·m⁶/mol². b in L/mol equals 0.001 m³/mol.
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This calculator is for educational purposes. Double-check important results before acting on them.