Estimating Enthalpy of Vaporization from Two Data Points

Estimating a liquid's enthalpy of vaporization from two vapor-pressure readings means converting both temperatures to kelvin and substituting both temperature-pressure pairs into the two-point form of the Clausius-Clapeyron equation, then solving for the enthalpy term.

Isooctane's vapor pressure is P1 = 10.0 kPa at T1 = 34.0 °C (307.2 K) and P2 = 100.0 kPa at T2 = 98.8 °C (372.0 K), with the gas constant R = 8.314 J/(mol·K). Substituting these named values into ln(P2/P1) = -(ΔHvap/R)(1/T2 - 1/T1) gives ln(10.0) = -(ΔHvap/8.314 J/(mol·K))(1/372.0 K - 1/307.2 K); carrying the units through, the kelvin terms cancel inside the parentheses and the joules cancel against R, leaving ΔHvap in kJ/mol. Solving isolates an enthalpy of vaporization near 33.8 kJ/mol — a positive value of the size expected for a nonpolar liquid rather than an implausibly large or negative one, consistent with the anchor's own worked value for this compound.