Van der Waals Equation
The Van der Waals equation corrects the ideal gas law for real gases by accounting for molecular size and intermolecular attractions.
The Formula
The Van der Waals equation, developed by Dutch physicist Johannes Diderik van der Waals in 1873, improves on the ideal gas law by accounting for two real-world effects that the simpler model ignores. Van der Waals received the Nobel Prize in Physics in 1910 for this work.
The first correction, the term a×n²/V², accounts for the attractive forces between gas molecules. Real molecules attract each other, which reduces the pressure below what an ideal gas would exert. Adding this term back into the pressure corrects for that reduction.
The second correction, n×b, accounts for the finite volume occupied by the gas molecules themselves. In an ideal gas, molecules are treated as point-sized particles with no volume. Real molecules take up space, so the available volume for movement is the total volume minus the space the molecules themselves occupy. The constants a and b are specific to each gas and must be looked up from a table.
Variables
| Symbol | Meaning | Unit |
|---|---|---|
| P | Pressure of the gas | Pa or atm |
| V | Volume of the container | m³ or L |
| n | Number of moles of gas | mol |
| R | Ideal gas constant = 8.314 | J/(mol·K) |
| T | Absolute temperature | K (Kelvin) |
| a | Van der Waals constant for intermolecular attraction (gas-specific) | L²·atm/mol² |
| b | Van der Waals constant for molecular volume (gas-specific) | L/mol |
Example 1
Calculate the pressure of 1 mole of CO₂ gas in a 1-liter container at 300 K. For CO₂: a = 3.640 L²·atm/mol², b = 0.04267 L/mol.
Rearrange: P = nRT / (V − nb) − an² / V²
First term: nRT / (V − nb) = (1 × 0.08206 × 300) / (1 − 0.04267) = 24.618 / 0.9573 ≈ 25.72 atm
Second term: an² / V² = 3.640 × 1² / 1² = 3.640 atm
P ≈ 25.72 − 3.64 ≈ 22.08 atm (ideal gas law would give 24.62 atm — a notable difference at high pressure)
Example 2
Estimate pressure of 2 moles of nitrogen (N₂) in a 5-liter container at 350 K. For N₂: a = 1.390 L²·atm/mol², b = 0.03913 L/mol.
First term: nRT / (V − nb) = (2 × 0.08206 × 350) / (5 − 2 × 0.03913) = 57.442 / 4.9218 ≈ 11.67 atm
Second term: an² / V² = 1.390 × 4 / 25 = 0.222 atm
P ≈ 11.67 − 0.22 ≈ 11.45 atm (ideal gas gives 11.49 atm — close at lower pressure)
When to Use It
Use the Van der Waals equation when:
- Working with gases at high pressures where the ideal gas law becomes inaccurate
- Working with gases near their condensation temperature (liquid phase transition)
- Modeling gases with strong intermolecular forces such as CO₂, NH₃, or water vapor
- Calculating conditions in industrial compressors, liquefied gas storage, and chemical reactors
- Studying real gas behavior in physical chemistry and thermodynamics courses