CHEM 104 1st Edition Lecture 8Outline of Last Lecture I. Continued review of in class quiz on 02/02/2015A. Example questionsII. Heat vaporizationB. PracticeIII. Introduction to Vapor PressureC. Introduction to Clausius-Clapeyron equationOutline of Current Lecture IV. Practice on solving Clausius-Clapeyron equationsCurrent LectureI. Remember: everything has to be converted into the correct form before solving the equations, R always equals 8.314 J/K*mol, and the equation for Clausius-Clapeyron is:−∆ HvapR=ln (P2P1)1T2−1T1The formula for changing Celsius to kelvins is:℃ +273=KII. Practice on solving Clausius-Clapeyron equations:A. Example 1:Given: P2 = 780 mmHg; T2 = 30°C; T1 = 25°C; ∆Hvap = 42.0 kj/molFind: P1 = ?T2 = 30°C 30°C + 273 = 303 KT1 = 25°C 25°C + 273 = 298 K∆Hvap = 42.0 kj/mol 42.0*1000 = 42,000 J/mol−42,0008.314=ln (780P1)1303−1298First divide both Temperatures separately THEN SUBTRACT:−42,0008.314=ln (780P1)−0.0000553747Next, multiply -0.0000553747 by both sides:.2797=ln (780P1)Then, raise .2797 to e on both sides to cancel ln. Then you can rearrange the equation so you solve for P1:e.2797∗P1=780P1=780e.2797P1=
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