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  1. Home
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  4. /Van der Waals Equation Calculator

Van der Waals Equation Calculator

Last updated: March 17, 2026

Calculator

Results

Van der Waals Pressure

—

atm

Ideal Gas Pressure

—

atm

Van der Waals Volume

—

L

Van der Waals Temperature

—

K

Pressure Correction (a·n²/V²)

0.00715

atm

Volume Correction (n·b)

0.04267

L

Deviation from Ideal

—

Results

Van der Waals Pressure

—

atm

Ideal Gas Pressure

—

atm

Van der Waals Volume

—

L

Van der Waals Temperature

—

K

Pressure Correction (a·n²/V²)

0.00715

atm

Volume Correction (n·b)

0.04267

L

Deviation from Ideal

—

The Van der Waals Equation Calculator applies the modified gas equation that accounts for intermolecular forces and finite molecular volume. While the ideal gas law ($$PV = nRT$$) assumes point-like, non-interacting particles, real gases deviate significantly at high pressures and low temperatures. The Van der Waals equation introduces two correction parameters to model this behavior more accurately.

The Van der Waals equation is written as:

$$\left(P + \frac{an^2}{V^2}\right)(V - nb) = nRT$$

where a accounts for attractive intermolecular forces (reducing effective pressure) and b accounts for the finite volume occupied by gas molecules themselves. Each gas has unique a and b constants determined experimentally. For example, CO₂ has a = 3.592 L²·atm/mol² and b = 0.04267 L/mol.

This equation was proposed by Johannes Diderik van der Waals in 1873 and earned him the Nobel Prize in Physics in 1910. It represents one of the earliest and most important equations of state for real gases, bridging the gap between ideal behavior and experimental observations.

Visual Analysis

How It Works

The calculator solves the Van der Waals equation for pressure, volume, or temperature. When solving for pressure:

$$P = \frac{nRT}{V - nb} - \frac{an^2}{V^2}$$

The first term represents the kinetic pressure (corrected for molecular volume), while the second term subtracts the pressure reduction due to intermolecular attractions. The calculator also computes the ideal gas pressure $$P_{ideal} = \frac{nRT}{V}$$ for direct comparison.

When solving for temperature:

$$T = \frac{\left(P + \frac{an^2}{V^2}\right)(V - nb)}{nR}$$

The pressure correction $$\frac{an^2}{V^2}$$ quantifies how much intermolecular attraction reduces pressure below the ideal value. The volume correction $$nb$$ quantifies the excluded volume occupied by molecules. The percent deviation shows how far the real gas behavior is from ideal predictions, which grows larger at high pressures, low temperatures, and near the critical point of the gas.

Understanding Your Results

A positive percent deviation indicates the Van der Waals pressure differs from ideal. Large deviations (>5%) suggest the gas is far from ideal conditions. Gases with large a values (like water vapor, a = 5.536) have strong intermolecular forces and deviate more. Gases with small a (like helium, a = 0.0342) behave nearly ideally. At very high temperatures and low pressures, both models converge because thermal energy overwhelms intermolecular forces and the gas occupies a large volume relative to molecular size.

Worked Examples

CO₂ at 300 K in a 10 L container

Inputs

solve forpressure
n1
V10
T300
P1
a3.592
b0.04267

Results

vdw pressure2.4272
ideal pressure2.4618
pressure correction0.03592
volume correction0.04267
percent deviation1.4

At moderate conditions, CO₂ deviates only ~1.4% from ideal behavior.

NH₃ at 400 K in a 5 L container (2 mol)

Inputs

solve forpressure
n2
V5
T400
P1
a4.17
b0.03707

Results

vdw pressure12.6632
ideal pressure13.1296
pressure correction0.6672
volume correction0.07414
percent deviation3.55

NH₃ has strong hydrogen bonding (high a value), so it deviates ~3.6% from ideal at these conditions.

Frequently Asked Questions

The constant a (in L²·atm/mol²) measures the strength of intermolecular attractive forces. Larger a means stronger attractions. The constant b (in L/mol) represents the effective volume of one mole of gas molecules. These are empirically determined for each gas.

The ideal gas law fails at high pressures (molecules close together, volume correction matters), low temperatures (slow molecules feel attractions more), and near the critical point where gases liquefy. Under these conditions, the Van der Waals equation provides much better predictions.

Constants are tabulated in chemistry references. Common values: He (a=0.0342, b=0.02370), H₂ (a=0.2444, b=0.02661), N₂ (a=1.390, b=0.03913), O₂ (a=1.360, b=0.03183), CO₂ (a=3.592, b=0.04267), H₂O (a=5.536, b=0.03049), NH₃ (a=4.170, b=0.03707).

Yes, qualitatively. Below the critical temperature, the equation produces an S-shaped PV curve with three roots, corresponding to liquid, unstable, and vapor phases. The critical point can be derived: $$T_c = \frac{8a}{27Rb}$$, $$P_c = \frac{a}{27b^2}$$, $$V_c = 3nb$$.

Solving the Van der Waals equation for volume requires solving a cubic equation $$V^3 - (nb + nRT/P)V^2 + (an^2/P)V - an^3b/P = 0$$. The calculator uses an approximation $$V \approx \frac{nRT}{P} + nb$$ that works well at moderate conditions.

No. More accurate equations include the Redlich-Kwong, Peng-Robinson, and virial equations. However, Van der Waals remains widely used in education because it clearly shows the physical meaning of each correction term.

Sources & Methodology

Van der Waals, J.D. (1873). "Over de Continuïteit van den Gas- en Vloeistoftoestand"; Atkins, P. & de Paula, J. "Physical Chemistry" (11th ed., Oxford); NIST Chemistry WebBook for Van der Waals constants.
R

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