Science, Tech, Math › Science Clausius-Clapeyron Equation Example Problem Predicting Vapor Pressure Share Flipboard Email Print imagenavi / Getty Images Science Chemistry Basics Chemical Laws Molecules Periodic Table Projects & Experiments Scientific Method Biochemistry Physical Chemistry Medical Chemistry Chemistry In Everyday Life Famous Chemists Activities for Kids Abbreviations & Acronyms Biology Physics Geology Astronomy Weather & Climate By Todd Helmenstine Todd Helmenstine is a science writer and illustrator who has taught physics and math at the college level. He holds bachelor's degrees in both physics and mathematics. our editorial process Todd Helmenstine Updated December 09, 2019 The Clausius-Clapeyron equation is a relation named for Rudolf Clausius and Benoit Emile Clapeyron. The equation describes the phase transition between two phases of matter that have the same composition. Thus, the Clausius-Clapeyron equation can be used to estimate vapor pressure as a function of temperature or to find the heat of the phase transition from the vapor pressures at two temperatures. When graphed, the relationship between temperature and pressure of a liquid is a curve rather than a straight line. In the case of water, for example, vapor pressure increases much faster than temperature. The Clausius-Clapeyron equation gives the slope of the tangents to the curve. This example problem demonstrates using the Clausius-Clapeyron equation to predict the vapor pressure of a solution. Problem The vapor pressure of 1-propanol is 10.0 torr at 14.7 °C. Calculate the vapor pressure at 52.8 °C.Given:Heat of vaporization of 1-propanol = 47.2 kJ/mol Solution The Clausius-Clapeyron equation relates a solution's vapor pressures at different temperatures to the heat of vaporization. The Clausius-Clapeyron equation is expressed byln[PT1,vap/PT2,vap] = (ΔHvap/R)[1/T2 - 1/T1]Where:ΔHvap is the enthalpy of vaporization of the solutionR is the ideal gas constant = 0.008314 kJ/K·molT1 and T2 are the absolute temperatures of the solution in KelvinPT1,vap and PT2,vap is the vapor pressure of the solution at temperature T1 and T2 Step 1: Convert °C to K TK = °C + 273.15T1 = 14.7 °C + 273.15T1 = 287.85 KT2 = 52.8 °C + 273.15T2 = 325.95 K Step 2: Find PT2,vap ln[10 torr/PT2,vap] = (47.2 kJ/mol/0.008314 kJ/K·mol)[1/325.95 K - 1/287.85 K]ln[10 torr/PT2,vap] = 5677(-4.06 x 10-4)ln[10 torr/PT2,vap] = -2.305take the antilog of both sides 10 torr/PT2,vap = 0.997PT2,vap/10 torr = 10.02PT2,vap = 100.2 torr Answer The vapor pressure of 1-propanol at 52.8 °C is 100.2 torr.