Answer:
27.3 kJ/mol
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
