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What is the Van der Waals Equation?

Real gases deviate from ideal gas behavior due to intermolecular forces and finite molecular volume. The Van der Waals equation corrects the ideal gas law with two constants: a (intermolecular attractions) and b (molecular volume), providing accurate predictions for real gases.

Short answer

The Van der Waals equation is (P + a/V_m²)(V_m − b) = RT, where a accounts for intermolecular forces and b for molecular volume. It predicts real gas behavior more accurately than the ideal gas law.

Pressure vs Molar Volume: Ideal vs Real Gas

x: Molar Volume V_m (L/mol) · y: Pressure P (atm)Ideal GasVan der Waals (CO₂)

Try it: interactive calculator

Pressure P

16.159atm

= (24.5 / (1.5 - 0.04)) - (1.4 / (1.5 * 1.5))

RT (molar gas energy)24.5 L·atm/molMolar volume V_m1.5 L/molConstant a (attraction)1.4 atm·L²/mol²Constant b (volume)0.04 L/mol

Step-by-step worked examples

Calculate pressure for CO₂ at 25°C with V_m = 1 L/mol. Use a = 3.64 atm·L²/mol² and b = 0.04267 L/mol.

T = 25 + 273 = 298 K
RT = 0.08206 × 298 = 24.45 L·atm/mol
(P + 3.64/1²)(1 − 0.04267) = 24.45
(P + 3.64)(0.95733) = 24.45
P + 3.64 = 25.55
P = 21.91 atm

N₂ gas at 20°C, V_m = 2 L/mol. Constants: a = 1.408, b = 0.0391 L/mol. Find P.

T = 20 + 273 = 293 K
RT = 0.08206 × 293 = 24.04 L·atm/mol
(P + 1.408/4)(2 − 0.0391) = 24.04
(P + 0.352)(1.9609) = 24.04
P + 0.352 = 12.25
P ≈ 11.9 atm

Compare ideal vs real for O₂: at 1 L/mol, 300 K, a = 1.378, b = 0.03183 L/mol.

Ideal: P = RT/V_m = 24.618/1 = 24.62 atm
Real: (P + 1.378/1)(1 − 0.03183) = 24.618
(P + 1.378)(0.96817) = 24.618
P + 1.378 = 25.43
P = 24.05 atm (lower due to attractions)

Flashcards

Quick quiz

Q1.Which factor makes pressure lower in real vs ideal gas?

Correct answer: B. Attractions between molecules reduce observed pressure compared to ideal prediction.

Q2.In the Van der Waals term (P + a/V_m²), why add a/V_m² to P?

Correct answer: B. Attractions reduce pressure; we add this correction to find the ideal pressure that would exist without attractions.

Q3.Which condition makes ideal gas law most accurate?

Correct answer: B. At low pressure and high temperature, gases behave ideally (b negligible, attractions weak).

Q4.If a = 0 and b = 0 in Van der Waals, what remains?

Correct answer: A. Setting both constants to zero recovers PV = nRT.

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Common mistakes

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Thinking b corrects pressure directly. — Correct: b corrects volume; (V − b) is the effective volume.

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Using the same units for P, V, R without checking. — Correct: Units must match: if a in atm·L²/mol², use L and atm.

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Ignoring both a and b because they're small. — Correct: At high pressures or low T, these corrections become significant.

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Confusing which constant causes which deviation. — Correct: a = attraction (lowers P); b = volume (raises P).

FAQ

What is the Van der Waals equation formula?

(P + a/V_m²)(V_m − b) = RT. The a term corrects for intermolecular forces; b corrects for molecular volume.

Why do real gases deviate from ideal behavior?

Real gas molecules attract each other (a term) and occupy finite space (b term), unlike ideal gas assumptions.

When is Van der Waals equation most useful?

At high pressures or low temperatures where intermolecular forces and molecular volume become significant.

How do I find constants a and b for a gas?

They're provided in tables. Alternatively, they can be calculated from critical point data: a = 27R²T_c²/(64P_c) and b = RT_c/(8P_c).