Question

Increasing the temperature of a particular liquid from 298 K to 318 K causes its vapor pressure to double. What is the enthalpy of vaporization of this liquid

Answer 1

27.3 kJ/mol

Step-by-step explanation:

Step 1: Given data

• Temperature 1 (T₁): 298 K

• Vapor pressure 1 (P₁): P₁

• Temperature 2 (T₂): 318 K

• Vapor pressure 2 (P₂): 2 P₁

Step 2: Calculate the enthalpy of vaporization of this liquid

We will use the Clausius–Clapeyron equation.

ln (P₂/P₁) = -ΔHvap/R × (1/T₂ - 1/T₁)

ln 2 = -ΔHvap/(8.314 J/K.mol) × (1/318 K - 1/298 K)

ΔHvap = 2.73 × 10⁴ J/mol = 27.3 kJ/mol