Solve the Van der Waals equation (P + a·n²/V²)·(V − n·b) = n·R·T for pressure P, volume V, or temperature T – simply leave the desired variable blank. Choose a gas from the list or enter custom constants a and b. The calculator also shows the compressibility factor Z and a direct comparison to the ideal gas law.
The ideal gas law (P·V = n·R·T) assumes that gas particles have no volume of their own and do not attract each other. At high pressure or low temperature, this is no longer true – real gases then deviate noticeably from ideal behavior.
The Van der Waals equation corrects the ideal gas law for two effects: the particle volume (constant b) and mutual attraction between particles (constant a).
Van der Waals Equation
(P + a·n²/V²)·(V − n·b) = n·R·T
Attraction Constant (a)
Describes how strongly gas particles attract each other. The larger a is, the lower the actual pressure compared to an ideal gas.
Co-volume (b)
Corresponds to the intrinsic volume of one mole of gas particles. The actually available volume is smaller than the total volume by n·b.
Compressibility Factor (Z)
Z = P·V / (n·R·T) measures the deviation from ideal gas behavior. For an ideal gas, Z is always exactly 1.
Where do the constants a and b come from?
The stored values are taken from the CRC Handbook of Chemistry and Physics (tabulated there in L²·atm/mol², converted here to L²·bar/mol² via 1 atm = 1.01325 bar). Comparable values originating from the same source can also be found on the Wikipedia data page "Van der Waals constants (data page)" and can alternatively be derived directly from the critical data (T꜀, P꜀) in the NIST Chemistry WebBook (a = 27·R²·T꜀²/(64·P꜀), b = R·T꜀/(8·P꜀)). Because a and b are not natural constants but rather approximation parameters derived from experimental data, different reference sources typically vary by 1–3% in the final decimal place. The value for air is calculated as a molar-weighted 79%/21% average of N₂ and O₂, as air is a mixture and not uniformly tabulated in reference manuals.
1 mol of CO₂ is contained in a volume of 0.5 L at 300 K. What pressure is established, and how much does the result deviate from an ideal gas?
Given
n = 1 mol, T = 300 K, V = 0.5 L, a = 3.640 L²·bar/mol², b = 0.04267 L/mol
Solution
P = n·R·T/(V−n·b) − a·n²/V² ≈ 39.8 bar (real) compared to 49.9 bar (ideal) – a deviation of around 20%, because both intrinsic volume and attractive forces play a significant role at this small volume.
Most frequently, units of a and b are mixed up (e.g., bar instead of Pa) or temperature is entered in °C instead of Kelvin. Also make sure that n·b is never larger than V – the volume of the particles themselves cannot be exceeded.
Recommended Approach
First select the appropriate gas (or enter custom constants), convert all quantities to a uniform system of units (bar, L, K), and finally check the compressibility factor Z to estimate the magnitude of deviation.