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)

L²·atm/mol²

L/mol

atm

L

mol

K

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)

Preal = Pideal - a(n/V)²
Attractions reduce pressure below ideal

Volume Correction (b)

Vavailable = 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: Hâ‚‚O, NH₃ with strong IMFs
  • •Large Molecules: COâ‚‚, 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

Gasab
He0.03460.0238
Hâ‚‚0.24520.0265
Nâ‚‚1.3700.0387
Oâ‚‚1.3820.0319
COâ‚‚3.6580.0429
Hâ‚‚O5.5370.0305
NH₃4.2250.0371
CHâ‚„2.3030.0431

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)
  • →Hâ‚‚O (5.537) > COâ‚‚ (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
  • →CHâ‚„ (0.0431) > Nâ‚‚ (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:

  • âš High pressure (> 10 atm)
  • âš Low temperature (near boiling point)
  • âš Polar gases (Hâ‚‚O, NH₃, HCl)
  • âš Large molecules (COâ‚‚, hydrocarbons)
  • âš 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:

ConditionZ
Ideal gas1.00
Hâ‚‚ at 200 atm1.07
Nâ‚‚ at 50 atm0.98
COâ‚‚ at 200 atm0.75

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