Van der Waals Equation Calculator [For Real Gases]

In the chemical industry, we constantly deal with real gases where the Ideal Gas Law won't work. That's because, in the real world, gases aren't "ideal." Their molecules have actual volume, and they attract each other, two key factors the ideal model ignores.

This is where the Van der Waals equation, which improves on the Ideal Gas Law by considering both molecular size and intermolecular forces

Van der Waals Equation Calculator

van der waals equation calculator

To make these more complex calculations fast and simple, Chemical Tweak's new Van der Waals Equation Calculator is here to help.

Van der Waals Calculator

Van der Waals Equation Calculator

Real Gas Equation of State
\left(P + \frac{an²}{V²}\right)(V - nb) = nRT
▼ Select Gas
▼ Van der Waals Constants a and b
Constant a: -
Constant b: -
▼ Critical Point (Reference)
▼ Gas Parameters (Enter values)
📘 About Van der Waals Equation:
The Van der Waals equation accounts for real gas behavior by including:
• a - correction for intermolecular attractive forces (bar·L²/mol²)
• b - correction for molecular volume exclusion (L/mol)

How to use:
1. Select a gas - critical point values are auto-filled for reference
2. Enter values in "Gas Parameters" section
3. Leave exactly one field empty to calculate it
4. Click Calculate to see results

How to Use the Calculator

  1. Select Your Gas: Start by choosing your gas from the dropdown in Section 1.
  2. Review Constants: The calculator will automatically load the critical a and b constants (Section 2) and the critical point data (Section 3) for that gas.
  3. Enter Your Parameters: Go to Section 4 and fill in the values for your specific problem.
  4. Calculate: Leave exactly ONE of the four fields (Pressure, Volume, Temperature, or Amount) empty. Click on"Calculate," and the tool will solve for the missing variable.

For ideal conditions where gases follow PV=nRT, you can try our Ideal Gas Law Calculator

What is Van der Waals Equation?

The Ideal Gas Law is simple and useful, but it's built on two assumptions that aren't always true:

  1. It assumes gas particles have no volume (they are just points).
  2. It assumes there are no attractive or repulsive forces between particles.

These assumptions work well at low pressures and high temperatures, where particles are far apart and moving fast. But when you increase the pressure or lower the temperature, these factors become impossible to ignore.

    \[\left( P + \frac{a n^2}{V^2} \right)(V - nb) = nRT\]

P = Pressure of the gas (bar or atm)
V = Volume of the gas (L or m³)
n = Number of moles of the gas (mol)
T = Absolute temperature (K)
R = Universal gas constant = 0.08314 L·bar·mol⁻¹·K⁻¹ (or 8.314 J·mol⁻¹·K⁻¹)
a = Van der Waals constant for attraction between molecules (bar·L²·mol⁻²)
b = Van der Waals constant for volume excluded by a mole of gas molecules (L·mol⁻¹)

Example: Finding the Real Pressure of a CO₂ Tank

You have a 10-liter reactor and need to charge it with 5 moles of Carbon Dioxide (CO₂). The process operates at 350 K (76.85°C). What is the actual pressure you should expect inside the reactor?

Values & Constants:

  • Gas: CO₂
  • n: 5 mol
  • V: 10 L
  • T: 350 K
  • R: 0.08314 L·bar/(mol·K)
  • a: 3.658 L²·bar/mol²
  • b: 0.04267 L/mol

Step 1: Calculate the Terms

Volume Correction ((V - nb)):

    \[\begin{aligned}\frac{a n^2}{V^2}&= \frac{(3.658\ \mathrm{L^2\cdot bar\cdot mol^{-2}})\times(5\ \mathrm{mol})^2}{(10\ \mathrm{L})^2} \&= \frac{91.45\ \mathrm{L^2\cdot bar}}{100\ \mathrm{L^2}} \&= 0.9145\ \mathrm{bar}\end{aligned}\]

Pressure Correction (\frac{a n^2}{V^2}):

    \[\begin{aligned}\frac{a n^2}{V^2}&= \frac{(3.658\ \mathrm{L^2\cdot bar\cdot mol^{-2}})\times(5\ \mathrm{mol})^2}{(10\ \mathrm{L})^2} \&= \frac{91.45\ \mathrm{L^2\cdot bar}}{100\ \mathrm{L^2}} \&= 0.9145\ \mathrm{bar}\end{aligned}\]

Numerator ((nRT)):

    \[\begin{aligned}nRT &= (5\ \mathrm{mol})\times(0.08314\ \mathrm{L\cdot bar\cdot mol^{-1}\cdot K^{-1}})\times(350\ \mathrm{K}) \&= 145.4975\ \mathrm{L\cdot bar}\end{aligned}\]

Step 2: Substitute and Solve for Pressure ((P)):

    \[P = \frac{145.4975 \text{ L·bar}}{9.78665 \text{ L}} - 0.9145 \text{ bar}\]

    \[P = 14.8671 \text{ bar} - 0.9145 \text{ bar}\]

    \[\boxed{P = 13.9526 \text{ bar}}\]

1. What do A and B represent in van der Waals?

Ans - a represents the attractive forces between gas molecules and b represents the finite volume occupied by gas molecules

2. What is the importance of van der Waals forces?

Ans - Van der Waals forces are weak attractions between molecules that affect how real gases and liquids behave. They help explain why gases deviate from ideal behavior and why substances condense or liquefy under certain conditions.

3. What is the unit of constant A and B?

Ans - The unit of ‘a’ is L²·bar·mol⁻² (or atm·L²·mol⁻²), and the unit of ‘b’ is L·mol⁻¹.

Wrapping Up

This was the Van der Waals Equation Calculator, which will help the students and process engineers for faster calculation and to understand the importance. If you have any doubts, feel free to use the comment section.