# of particles Chem 104 1st Edition Lecture 7 Outline of Last Lecture II. Review of in class quiz on 02/02/2015A. Example questionsIII. Introduction to heat of vaporizationOutline of Current Lecture IV. Continued review of in class quiz on 02/02/2015B. Example questionsV. Heat vaporizationC. PracticeVI. Introduction to Vapor PressureD. Introduction to Clausius-Clapeyron equationCurrent LectureI. On the quiz Monday 02/02/2015, there will be calculations.A. Example 1:a. Calculate the mass of water that can be vaporized with 155 kJ od heat at 100° C.Given: 155 kJFind: g H2OProcess: kJ » mol H2O » g H2O155 kJ × 1 mol H2O40. 7 kJ+18.02 g1 mol=68.6 g H2OA way to remember how to switch between mass and moles is below. You will need to remember this for future classes:# particles ÷ 6.02*1023Mole × 6.02*1023Mol × molar massMass ÷ molar massmolmassB. Example 2:a. Calculate the amount of heat needed to vaporize 90.0g of C3H7OH at its boiling point.Given: 90.0gFind: kJProcess: g » mol » kJ90.0 g ×1 mol60.09 g×39.9 kJ1 mol=59.8 kJII. Vapor pressure: pressure exerted by vapor when it is in dynamic equilibrium with its liquid. If the bonds are weak then more vapor will appear. The weaker the attractive forces, then the higher the vapor pressure. The higher the vapor pressure, the more volatile the liquid. Also, increasing the temperature increases the number of molecules that will escape and vaporize. This also increases vapor pressure.A. Clausius-Clapeyron Equation:ln P=−∆ HvapR×1T+CIn this equation:P = vapor pressureR = gas constantT = temperatureC = constantR will always be given to you. However, for reference R = 8.314 J/mol. Below is the same equation put into an easier form of practice. This equation will most likely be used more:lnP2P1=−∆ HvapR(1T2−1T1) A brief review of ln laws:ln 1=0Ln x > 0 (there are no negative ln)L n e=1Ln x * y = ln x + ln yL nxy= ln x−ln yL n xn=n*ln
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