van der Waals equation
Gaseous State · Chemistry
Study notes
For CO₂, a = 3.59 L²·atm/mol², b = 0.0427 L/mol. Find P for 1 mol in 1 L at 300 K. Ideal: P = nRT/V = 1 × 0.0821 × 300/1 = 24.63 atm. van der Waals: P = nRT/(V − nb) − a(n/V)². Step 1: V − nb = 1 − 0.0427 = 0.9573 L. Step 2: 0.0821 × 300/0.9573 = 25.73 atm. Step 3: Minus a(1)² = 3.59 → P = 22.14 atm. Real P < ideal: attractions win at this density.