Van der Waals Equation Calculator
Calculate the real-gas pressure from the Van der Waals equation. Enter the amount of gas, volume, temperature and the constants a and b to get the pressure, compressibility factor Z and the deviation from the ideal-gas law.
Estimate Real-Gas Pressure from the Van der Waals Equation
Van der Waals Equation Formula
Where:
- \(P\) — pressure of the real gas
- \(n\) — amount of gas (mol)
- \(V\) — volume of the container (L)
- \(T\) — absolute temperature (K)
- \(R\) — gas constant, \(0.082057\ \text{L·atm/(mol·K)}\)
- \(a\) — attraction constant, correcting for intermolecular attractions
- \(b\) — co-volume constant, correcting for the finite size of molecules
- \(Z\) — compressibility factor (\(Z = 1\) for an ideal gas)
The equation modifies the ideal gas law with two correction terms: the \(a\) term lowers the pressure because molecules attract each other, and the \(b\) term raises it because the molecules themselves take up space that is not available for movement.
Van der Waals Equation – Calculation Example
Example: 1 mol of nitrogen (N₂) in a 10 L container at 300 K. For nitrogen, \(a = 1.3708\ \text{L²·bar/mol²}\) and \(b = 0.0387\ \text{L/mol}\). Converting \(a\) to L²·atm/mol² gives \(a = 1.3708 \times 0.986923 = 1.3529\ \text{L²·atm/mol²}\).
\[ P = \frac{1 \times 0.082057 \times 300}{10 - 1 \times 0.0387} - 1.3529 \times \left(\frac{1}{10}\right)^2 = \frac{24.6171}{9.9613} - 0.01353 \\[/p]
\[ P = 2.4713 - 0.0135 = 2.4578\ \text{atm} \\[/p]
Result: the real-gas pressure is P = 2.4578 atm (about 2.4900 bar or 248.98 kPa). The ideal gas law would give 2.4617 atm, so the real pressure is 0.16% lower. The compressibility factor is Z = 0.9984 and the molar volume is 10 L/mol.
How to Use the Calculator
- Enter the Amount of Gas (n) and choose the unit (mol, mmol or kmol).
- Enter the Volume (V) of the container and select L, mL or m³.
- Enter the Temperature (T); you can type it in K, °C or °F and the calculator converts it to kelvin.
- Enter the Attraction Constant (a) for your gas. The default 1.3708 L²·bar/mol² is nitrogen; the unit can be L²·bar/mol², L²·atm/mol² or Pa·m⁶/mol².
- Enter the Co-Volume Constant (b) in L/mol, cm³/mol or m³/mol. For nitrogen it is 0.0387 L/mol.
- Pick the Pressure Unit for Results — atm, bar, kPa, MPa or psi — and read the pressure, the compressibility factor Z, the percentage deviation from the ideal-gas pressure, the molar volume and a step-by-step calculation trace.
What Is the Van der Waals Equation?
The ideal gas law, \(PV = nRT\), assumes that gas molecules are point particles that do not interact. Real molecules, however, occupy a finite volume and attract one another. In 1873 Johannes Diderik van der Waals proposed a modified equation of state that accounts for both effects, an achievement recognized with the 1910 Nobel Prize in Physics. The equation was the first to describe why gases condense into liquids and remains the classic textbook model for real-gas behaviour.
The correction \(a(n/V)^2\) represents the internal pressure: molecules about to hit the wall are pulled back by their neighbours, so the observed pressure is lower than the ideal value. The constant \(a\) grows with the strength of intermolecular attractions and is largest for polar, easily condensable gases such as water and ammonia. The term \(nb\) subtracts the volume occupied by the molecules themselves, effectively shrinking the free space in which they can move; \(b\) is roughly four times the actual molecular volume per mole.
Which correction dominates depends on the conditions. At moderate densities the attractions usually win and the real pressure is lower than the ideal-gas prediction. At very high densities the excluded volume takes over and the real pressure becomes higher. The compressibility factor \(Z = PV/nRT\) summarizes this: \(Z 1\) means the finite molecular volume dominates, and \(Z = 1\) corresponds to ideal behaviour.
Constants for Common Gases
| Gas | a (L²·bar/mol²) | b (L/mol) |
|---|---|---|
| Helium (He) | 0.0346 | 0.0238 |
| Nitrogen (N₂) | 1.3708 | 0.0387 |
| Oxygen (O₂) | 1.382 | 0.0319 |
| Methane (CH₄) | 2.300 | 0.0430 |
| Carbon dioxide (CO₂) | 3.657 | 0.0429 |
| Ammonia (NH₃) | 4.225 | 0.0371 |
| Water (H₂O) | 5.537 | 0.0305 |
Note that the table uses bar-based constants; switch the unit selector for a to L²·bar/mol² when using these values.
Assumptions, Limitations and Tips
- Constants are not truly constant. The tabulated \(a\) and \(b\) are fitted to experimental data and are most reliable at moderate pressures (roughly up to a few hundred bar). At extreme densities the equation becomes qualitative rather than quantitative.
- Unreliable near condensation. Close to the critical point or in the two-phase region the equation can even predict a negative pressure or multiple volumes for one pressure. The calculator flags these cases with an error message instead of a number.
- Best for small, non-polar molecules. Noble gases, nitrogen and oxygen are described well; strongly associating gases such as water vapour deviate more, especially near saturation.
- The volume must exceed the co-volume. The term \(V - nb\) requires \(V > nb\); otherwise the model has no solution, which the calculator reports explicitly.
- When a = b = 0 the equation reduces exactly to the ideal gas law, which is a quick way to sanity-check your inputs.
If you are studying other gas-phase phenomena, you may also find the Van 't Hoff Equation Calculator for temperature dependence of equilibrium constants, the Graham's Law Calculator for diffusion rates, or the Clausius–Clapeyron Calculator for vapour pressure useful.
Frequently Asked Questions
How do I calculate the Van der Waals pressure step by step without a calculator?
First convert everything to consistent units (L, K, mol, L·atm for R = 0.082057). Then compute the two terms separately: the thermal term \( nRT/(V - nb) \) and the attraction term \( a(n/V)^2 \). Subtract the second from the first. Always check that \( V > nb \); if not, the equation has no physical solution. Comparing your result with \( nRT/V \) shows how far the gas is from ideal behaviour.
What is a normal value of the compressibility factor Z?
For an ideal gas \( Z = 1 \) exactly. Real gases at low pressure and high temperature typically give \( Z \) between about 0.95 and 1.05, which is why the ideal gas law works so well there. Values well below 1 (down to 0.2–0.7) indicate strong attractions, common near condensation; values above 1 appear at high density where the molecules' own volume dominates. Large deviations from 1 signal that the gas is far from ideal.
Why is the real-gas pressure sometimes lower and sometimes higher than the ideal-gas prediction?
Two corrections compete. The attraction term \( a(n/V)^2 \) reduces the pressure because molecules at the wall are pulled inward, while the excluded volume \( nb \) increases it by shrinking the free space. At moderate densities attractions usually win (Z 1). The temperature where the two effects exactly cancel (Z ≈ 1 over a wide pressure range) is called the Boyle temperature — about 323 K for nitrogen.
How do I convert the Van der Waals constant a between bar, atm and SI units?
The value of \( a \) depends on the pressure unit you intend to use: \( 1\ \text{L}^2\cdot\text{bar} = 0.986923\ \text{L}^2\cdot\text{atm} \) and \( 1\ \text{Pa}\cdot\text{m}^6 = 9.86923\ \text{L}^2\cdot\text{atm} \). For example, nitrogen has \( a = 1.3708\ \text{L}^2\cdot\text{bar/mol}^2 \), which equals \( 1.3529\ \text{L}^2\cdot\text{atm/mol}^2 \). Mixing these units is the most common source of wrong hand calculations, so always match \( a \) with the \( R \) you are using.
What are the most common mistakes when using the Van der Waals equation?
Three errors cause most wrong answers: entering temperature in °C instead of kelvin, pairing an \( a \) in L²·bar/mol² with \( R \) in L·atm/(mol·K), and forgetting that \( b \) must fit into the volume (\( V > nb \)). Also remember that \( a \) scales with the square of concentration, so doubling \( n \) in the same volume quadruples the attraction correction — it is not a linear term.
Can the Van der Waals equation be used to find volume or temperature instead of pressure?
Solving for pressure is direct, but solving for \( V \) or \( T \) is harder. For temperature, rearrange to \( T = (P + a(n/V)^2)(V - nb)/(nR) \), which is still explicit. For volume, the equation becomes a cubic in \( V \), which may have one or three real roots; below the critical temperature the smaller root approximates the liquid volume and the larger one the vapour volume. Iterative or cubic-formula methods are needed there.
How are the constants a and b related to the critical point of a gas?
From the Van der Waals equation, the critical constants follow directly: \( T_c = 8a/(27Rb) \), \( P_c = a/(27b^2) \) and \( V_c = 3nb \). This means you can estimate \( a \) and \( b \) from measured critical data even for gases missing from tables: \( a = 27R^2T_c^2/(64P_c) \) and \( b = RT_c/(8P_c) \). At the critical point the Van der Waals equation predicts \( Z_c = 3/8 = 0.375 \), whereas real gases mostly show 0.27–0.30.