Van der Waals Calculator

Calculate real gas behavior accounting for molecular size and intermolecular attractions

Van der Waals Calculator

Van der Waals Equation: [P + a(n/V)²](V - nb) = nRT
Accounts for molecular size (b) and intermolecular attractions (a)

Solve For

Select Gas (Quick Fill Constants)

Constant a

L²·atm/mol²

Constant b

L/mol

Pressure (P) ← Solving for this

atm

Volume (V)

Moles (n)

mol

Temperature (T)

Van der Waals Constants:

• He: a=0.0346, b=0.0238

• H₂: a=0.2452, b=0.0265

• N₂: a=1.370, b=0.0387

• O₂: a=1.382, b=0.0319

• CO₂: a=3.658, b=0.0429

• H₂O: a=5.537, b=0.0305

Understanding Van der Waals Equation

The Van der Waals equation corrects the ideal gas law for real gas behavior by accounting for molecular volume and intermolecular attractions.

[P + a(n/V)²](V - nb) = nRT

a = attraction parameter (L²·atm/mol²)
b = volume parameter (L/mol)
Reduces to PV = nRT when a,b = 0

The Corrections

Pressure Correction (a)

P_real = P_ideal - a(n/V)²
Attractions reduce pressure below ideal

Volume Correction (b)

V_available = V - nb
Molecules occupy space, reducing free volume

Where It's Used

•High Pressure: Industrial gas storage and processing
•Low Temperature: Liquefaction and cryogenics
•Polar Gases: H₂O, NH₃ with strong IMFs
•Large Molecules: CO₂, hydrocarbons

Example Calculation

Problem:

Calculate the pressure of 1.00 mol CO₂ in a 1.00 L container at 300 K using both ideal and Van der Waals equations. For CO₂: a = 3.658 L²·atm/mol², b = 0.0429 L/mol

Ideal Gas Law:

P = nRT/V
P = (1.00)(0.08206)(300)/1.00
P = 24.62 atm

Van der Waals Equation:

P = [nRT/(V-nb)] - a(n/V)²
P = [(1.00)(0.08206)(300)/(1.00-0.0429)] - 3.658(1.00)²
P = 24.62/0.9571 - 3.658
P = 25.72 - 3.66 = 22.06 atm

Result:

Van der Waals pressure is 2.56 atm lower (10.4% difference) due to intermolecular attractions dominating over volume effects.

Van der Waals Constants

Gas	a	b
He	0.0346	0.0238
H₂	0.2452	0.0265
N₂	1.370	0.0387
O₂	1.382	0.0319
CO₂	3.658	0.0429
H₂O	5.537	0.0305
NH₃	4.225	0.0371
CH₄	2.303	0.0431

Units: a in L²·atm/mol², b in L/mol

Interpreting Constants

Constant 'a' (Attraction):

→Measures strength of intermolecular forces
→Larger a = stronger attractions (polar molecules, H-bonding)
→H₂O (5.537) > CO₂ (3.658) > He (0.0346)
→Reduces pressure below ideal value

High 'a' gases are easier to liquefy because molecules attract each other strongly.

Constant 'b' (Volume):

→Represents volume excluded by molecules
→Larger b = larger molecular size
→CH₄ (0.0431) > N₂ (0.0387) > He (0.0238)
→Increases pressure above ideal value

'b' is approximately 4× the actual molecular volume (accounts for excluded volume in collisions).

When to Use Van der Waals vs Ideal Gas Law

Ideal Gas Law Works Well:

✓Low pressure (< 10 atm)
✓High temperature (> 0°C)
✓Noble gases (He, Ne, Ar)
✓Small, nonpolar molecules
✓Error < 5% acceptable

Use Van der Waals When:

⚠High pressure (> 10 atm)
⚠Low temperature (near boiling point)
⚠Polar gases (H₂O, NH₃, HCl)
⚠Large molecules (CO₂, hydrocarbons)
⚠Precision required

Compressibility Factor

Z = PV/(nRT)

Z = compressibility factor

• Z = 1: Ideal gas behavior
• Z > 1: Repulsions dominate (high P)
• Z < 1: Attractions dominate (low T)
• Van der Waals helps predict Z

Typical Z Values:

Condition	Z
Ideal gas	1.00
H₂ at 200 atm	1.07
N₂ at 50 atm	0.98
CO₂ at 200 atm	0.75

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