Van der Waals Gas Pressure Calculator: Real Gas Pressure

This calculator finds the pressure of a real gas with the van der Waals equation, which corrects the ideal gas law for molecular volume and attraction. Enter the amount, temperature, volume and the gas constants a and b; you get the van der Waals pressure, the ideal gas pressure and the difference between them.

Formula

P = n × R × T ÷ (V − n × b) − a × n² ÷ V², with T = t + 273.15.

• P: pressure in bar
• n: amount of gas in moles; V: volume in litres
• R = 0.0831446261815324 L·bar/(mol·K), the molar gas constant 8.31446261815324 J/(mol·K) divided by 100 (1 bar = 100 kPa, 1 kPa·L = 1 J)
• T: absolute temperature in kelvin; t: temperature in °C
• a (L²·bar/mol²): attraction term; b (L/mol): volume excluded per mole of molecules
• Defaults are the tabulated values for carbon dioxide, a = 3.640 L²·bar/mol² and b = 0.04267 L/mol

Worked Example

1 mol of carbon dioxide at 25 °C in a 2 L container.

1. T = 25 + 273.15 = 298.15 K; n × R × T = 1 × 0.0831446 × 298.15 = 24.7896 L·bar
2. V − n × b = 2 − 1 × 0.04267 = 1.95733 L
3. Repulsion term: 24.7896 ÷ 1.95733 = 12.6650 bar; attraction term: 3.640 × 1² ÷ 2² = 0.91 bar
4. P = 12.6650 − 0.91 = 11.7550 bar
5. Ideal gas: 24.7896 ÷ 2 = 12.3948 bar; difference: (11.7550 − 12.3948) ÷ 12.3948 × 100 = -5.16 %

Assumptions

• The equation is an approximation: good for moderate deviations from ideal behaviour, not accurate near the critical point or for liquids.
• The constants a and b are for the gas in question; values differ slightly between data tables, so use the ones your course or method gives.
• The volume is larger than n × b; if it is not, the equation has no physical meaning and the calculator shows “Cannot compute”.

How to read the result

A negative difference means the attractions between molecules dominate and the real gas pushes less on the walls than an ideal gas would, which is typical for CO₂ at room temperature. For an ideal-gas calculation in kPa and atm, use the ideal gas law calculator.

Limits and common mistakes

• Using a in L²·atm/mol² with R in L·bar/(mol·K) mixes units; for CO₂ a is about 3.59 L²·atm/mol², which equals 3.640 L²·bar/mol².
• Entering the temperature in kelvin in the °C field adds 273.15 twice.

Frequently Asked Questions

What is the pressure of 1 mol CO2 at 25 °C in 2 L?

About 11.755 bar by van der Waals, against 12.395 bar for an ideal gas.

What do a and b mean?

a measures the attraction between molecules; b is the volume taken up by one mole of the molecules themselves.

Why does it say Cannot compute?

The volume is not larger than n × b, the space the molecules themselves occupy, so the equation has no physical solution.

in n | Amount of gas | mol | 1 | 0
in t | Temperature | °C | 25 | -273.15
in v | Volume | L | 2 | 0.000001
in a | Constant a | L²·bar/mol² | 3.640 | 0
in b | Constant b | L/mol | 0.04267 | 0
out p | Van der Waals pressure | bar | n * 0.0831446261815324 * (t + 273.15) / (v - n * b) - a * n ^ 2 / v ^ 2 | 4
out pid | Ideal gas pressure | bar | n * 0.0831446261815324 * (t + 273.15) / v | 4
out dev | Difference from ideal | % | (p - pid) / pid * 100 | 2