CHEM 122 1st Edition Lecture 6Outline of Last LectureI. Molar Heat CapacityII. Evaporation of a LiquidIII. Vapor PressureIV. Boiling of a Liquida. Effect of Pressure on Boiling Point.Outline of Current Lecture I. Clausius-Clapeyron Equationa. Relate to a straight lineII. Phase Diagramsa. Phase Diagram Components Current LectureClausius-Clapeyron Equation:How is the heat of vaporization measured?By measuring the vapor pressure of the liquid at various temperatures.lnP=−∆ HvapRT+CP: Vapor Pressure of liquidChange in Hvap: Heat of vaporizationR: Universal Gas Constant (8.314 J/mol K)T: Kelvin temperatureThese notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best Used as a supplement to your own notes, not as a substitute.C: Constant for that particular liquidCompare this equation to the equation of a straight line: y=mx+bP=y 1/T=x -change in Hvap=M C=bNew equation in slope form:lnP=(∆ HvapR)1T+CSince the slope is equal to the change in the heat evaporation divided by the universal gas constant you can determine the heat evaporation if you know the slope of the line.Slope=−∆ HvapR∆ Hvap=−slope(R)When determining the heat evaporation when given 2 pressures and 2 temperatures simply make 2 equations and subtract them.lnP1−lnP 2=∆ HvapR(1T 1)−(∆ HvapR(1T 2))Collect the terms, simply the equation and your final equations for finding the heat vaporizationbetween 2 pressures and temperatures becomes:lnP 1P 2=∆ HvapR(1T 1−1T 2)Phase Diagrams:Illustrates which phase is present at a given pressure and temperature.- Triple Point: the temperature and pressure where all three phases are at equilibrium.- Critical temperature: the highest temperature, which a substance can exist as a liquid.- Critical Pressure: the pressure required to bring about liquefaction at this critical temperature.- Critical Point: the point with critical pressure and critical
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