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L²atm/mol²
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Intermolecular attraction correction
L/mol
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Molecular volume correction
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mol
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L
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Pressure (Van Der Waals)
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Sources & Methodology
Van Der Waals constants verified against NIST Chemistry WebBook and Atkins’ Physical Chemistry (10th edition).
NIST Chemistry WebBook — Thermophysical Properties
National Institute of Standards and Technology reference database for gas constants, thermodynamic properties, and equation of state parameters used to verify all Van Der Waals constant values.
Khan Academy — Real Gases: Van Der Waals Equation
Curriculum reference explaining the physical meaning of constants a and b, derivation of corrections from kinetic molecular theory, and comparison to ideal gas law behavior.
LibreTexts Physical Chemistry — Van Der Waals Equation
Peer-reviewed open textbook covering the Van Der Waals equation derivation, critical constants, and limitations of the equation near the critical point.
Equation: (P + a×n²/V²)(V − n×b) = n×R×T where R = 0.082057 L·atm/(mol·K). Solve for P: P = n×R×T/(V−n×b) − a×n²/V². Solve for T: T = (P + a×n²/V²)(V−n×b)/(n×R). Ideal gas comparison: P𝑖𝑑𝑒𝑎𝑙 = n×R×T/V. Compressibility factor Z = P×V/(n×R×T).
⏱ Last reviewed: March 2026
Van Der Waals Equation for Real Gases — Complete Guide
The Van Der Waals equation is the most widely taught correction to the ideal gas law. Proposed by Johannes Diderik van der Waals in 1873, it accounts for two physical realities the ideal gas law ignores: molecules have finite size and they exert attractive forces on each other.
The Van Der Waals Equation
(P + a×n²/V²) × (V − n×b) = n×R×T
P = pressure (atm) • V = volume (L) • T = temperature (K) • n = moles
R = 0.082057 L·atm/(mol·K) • a = attraction constant • b = volume constant
Pressure correction +a×n²/V²: Adds back the pressure lost due to attractive forces pulling molecules away from the wall.
Volume correction −n×b: Subtracts the physical space occupied by molecules from the container volume.
R = 0.082057 L·atm/(mol·K) • a = attraction constant • b = volume constant
Pressure correction +a×n²/V²: Adds back the pressure lost due to attractive forces pulling molecules away from the wall.
Volume correction −n×b: Subtracts the physical space occupied by molecules from the container volume.
Solving for Pressure
P = n×R×T / (V − n×b) − a×n² / V²
Example: 1 mol CO₂ (a=3.640, b=0.04267) at 300 K in 1.000 L:
P = (1×0.082057×300) / (1.000−0.04267) − 3.640×1² / 1.000²
P = 24.617 / 0.95733 − 3.640 = 25.715 − 3.640 = 22.08 atm
Ideal gas: P = nRT/V = 24.62 atm → difference = 2.54 atm (10.3%)
P = (1×0.082057×300) / (1.000−0.04267) − 3.640×1² / 1.000²
P = 24.617 / 0.95733 − 3.640 = 25.715 − 3.640 = 22.08 atm
Ideal gas: P = nRT/V = 24.62 atm → difference = 2.54 atm (10.3%)
Van Der Waals Constants for Common Gases
| Gas | a (L²·atm/mol²) | b (L/mol) | Boiling Point | Character |
|---|---|---|---|---|
| Helium (He) | 0.0341 | 0.02370 | −269°C | Nearly ideal |
| Hydrogen (H₂) | 0.2444 | 0.02661 | −253°C | Nearly ideal |
| Nitrogen (N₂) | 1.390 | 0.03913 | −196°C | Moderate |
| Oxygen (O₂) | 1.360 | 0.03183 | −183°C | Moderate |
| Carbon Dioxide (CO₂) | 3.640 | 0.04267 | −78°C (sub) | Significant |
| Water Vapour (H₂O) | 5.536 | 0.03049 | 100°C | Very large |
| Ammonia (NH₃) | 4.169 | 0.03707 | −33°C | Large |
| Methane (CH₄) | 2.253 | 0.04278 | −161°C | Moderate |
| Sulfur Dioxide (SO₂) | 6.803 | 0.05636 | −10°C | Very large |
| Benzene (C₆H₆) | 18.82 | 0.11930 | 80°C | Extreme |
💡 When to use Van Der Waals vs ideal gas law: Use the ideal gas law (PV=nRT) when pressure is below 5 atm and temperature is well above the boiling point. Switch to Van Der Waals when pressure exceeds 10 atm, temperature is within 100K of the boiling point, or when the gas has a large a value (polar molecules, hydrogen bonding). For extreme conditions near the critical point, consider the Redlich-Kwong or Peng-Robinson equations.
Physical Meaning of a and b
The constant a measures intermolecular attraction strength. Polar molecules and molecules capable of hydrogen bonding have large a values. Water (a=5.536) forms extensive hydrogen bond networks. SO₂ (a=6.803) has strong dipole interactions. Noble gases like He (a=0.0341) have only weak London dispersion forces. A large a means the pressure is significantly lower than ideal because molecules are being pulled back from the walls.
The constant b measures effective molecular volume — the space unavailable to other molecules. Larger, heavier molecules have larger b values. Benzene (b=0.119) is nearly 5 times larger than helium (b=0.0237). The correction −nb reduces the effective free volume, which increases pressure above what ideal gas law predicts alone.
Compressibility Factor Z
The compressibility factor Z = PV/(nRT) measures how much a gas deviates from ideal behaviour. Z = 1 is ideal. For most gases at moderate conditions, Z ranges from 0.9 to 1.1. At high pressure and low temperature, attractive forces dominate and Z < 1 (gas is more compressed than ideal). At very high pressure, molecular volume dominates and Z > 1 (gas takes up more space than ideal). Van Der Waals predicts Z correctly for moderate deviations.
Frequently Asked Questions
The Van Der Waals equation is (P + a×n²/V²)(V − n×b) = nRT. It corrects the ideal gas law for two real-gas effects: intermolecular attractions (constant a) that reduce actual pressure, and finite molecular volume (constant b) that reduces available free volume. Proposed in 1873 by Johannes Van Der Waals, it earned him the 1910 Nobel Prize in Physics.
Solve for P: P = nRT/(V−nb) − an²/V². For 1 mol CO₂ (a=3.640, b=0.04267) at 300K in 1L: P = (1×0.08206×300)/(1−0.04267) − 3.640/1 = 24.62/0.957 − 3.640 = 25.72−3.64 = 22.08 atm. Compare to ideal: nRT/V = 24.62 atm. The difference of 2.54 atm shows CO₂’s significant intermolecular attractions.
Constant a (L²·atm/mol²) represents the strength of intermolecular attractive forces. Larger a = stronger attractions = gas deviates more from ideal at low temperatures. Constant b (L/mol) represents the effective volume of one mole of gas molecules — the volume unavailable to other molecules. Larger b = larger molecules. He has a=0.034 (tiny, weak forces) while water has a=5.536 (hydrogen bonding, strong forces).
Ideal behaviour occurs at low pressure (below 5 atm) and high temperature (well above boiling point). Real (non-ideal) behaviour is significant at high pressure (above 10 atm), low temperature (near boiling point), and for gases with large a values (polar, hydrogen-bonding molecules). At the limit P→0 or T→∞, all real gases approach ideal behaviour because molecules are far apart and interactions become negligible.
Ideal gas law PV = nRT assumes: (1) point-sized molecules with zero volume, and (2) no intermolecular forces. Van Der Waals adds: (1) pressure correction +an²/V² to account for reduced pressure due to attractive forces, and (2) volume correction −nb to account for the space molecules themselves occupy. When a=0 and b=0, Van Der Waals reduces exactly to PV = nRT.
Among common gases, benzene (C₆H₆) has a=18.82 due to its large size and polarisability. Among gases you are likely to encounter in chemistry: SO₂ (a=6.803), H₂O (a=5.536), NH₃ (a=4.169), CO₂ (a=3.640). At the extreme end, large organic molecules can have a values exceeding 20 L²·atm/mol². Noble gases have the smallest a values: He=0.034, Ne=0.211, Ar=1.345.
Yes. For temperature: T = (P + an²/V²)(V−nb)/(nR). Solving for volume requires solving a cubic equation in V, which has up to three real roots — two physically meaningful ones (gas and liquid phases) below the critical temperature. This calculator solves for pressure and temperature directly. Volume requires numerical iteration and is not currently included.
Z = PV/(nRT). Ideal gas: Z = 1. Real gas: Z ≠ 1. Van Der Waals predicts Z from its equation: once P is calculated, Z = P×V/(n×R×T). Z < 1 means attractive forces are dominant (gas more compressed than ideal). Z > 1 means molecular volume dominates (gas less compressed than ideal). For most gases at low to moderate pressure, Z is between 0.9 and 1.1.
Van Der Waals works well for moderate deviations from ideal behaviour but has several limitations: (1) It predicts a critical compressibility Zc = 3/8 = 0.375 for all gases, while real gases have Zc between 0.23 and 0.31. (2) It is inaccurate near the critical point where fluctuations are important. (3) Constants a and b are temperature-dependent in reality. For engineering applications at high pressure, the Peng-Robinson or Redlich-Kwong equations are more accurate.
Van Der Waals constants use atm as the pressure unit when listed in L²·atm/mol². Conversions: 1 atm = 101.325 kPa = 1.01325 bar = 760 mmHg = 14.696 psi. To use SI units (Pa, m³), multiply a by 0.101325 to convert from L²·atm/mol² to m²·Pa/mol² × 10⁻³, and use R = 8.3145 J/(mol·K). Most textbook problems use atm and litres, which is why this calculator uses those units.
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