Chem 104 1st Edition Lecture 9 Outline of Last Lecture I. Practice on solving Clausius-Clapeyron equationsOutline of Current Lecture I. Vapor Pressure vs. temperatureII. Boiling PointIII. Supercritical FluidA. The critical temperature and the critical pressureIV. Stages of solid, gas, and liquidA. Sublimation and depositionB. The different stages and their importanceV. HeatA. Heat of fusion and its process to solve for itB. The heat curve of water and its segment processCurrent LectureI. Weaker attractive forces results in a higher vapor pressure.II. The boiling point is the external pressure. The normal boiling point is the temperature at which vapor pressure of a liquid 1 atm. As the external pressure decreases, the boiling point of the liquid decreases. Equations include: q=mliquid∗Cs∗∆ T. Temperature is in Kelvins and Cs is significant to the liquid. III. At some temperature, the meniscus between the liquid and vapor disappears and the states commingle to form supercritical fluid. These fluids have the properties of both gases and liquids.IV. The critical point describes the critical temperature and critical pressure. The critical temperature is the temperature required to produce superficial fluid. Critical pressure is thepressure at the critical temperature.V. The stages of a gas, liquid, and solid are as follows:I. Remember that thermal energy is heat and allows the molecules to vibrate in a solid. The solid and liquid phases exist in dynamic equilibrium in a closed container. The energeticsconcerning melting are endothermic because energy is lost.II. the heat of fusion is the amount of energy required to melt one mole of a solid. The equations to remember:∆ Hsublimation=∆ Hfusion+∆ Hvaporization∆ Hcrystallization=−∆ Hfusionq=mass∗Cs∗∆TIII. The heat curve of water is measured in segments. If you wanted to measure the heatingof water from -25°C to 100°C (its boiling point) and then 125°C. You would calculate the heatthrough about 5 segments.Segment 1:Heating 1.00 mole of ice at -25°C up to its melting point at 0.0°CGiven: mass of 1.00 mole ice = 18.0g; Cs = 2.09 J/mol °Cq=mliquid∗Cs∗∆ TInput the variables into the equation:depositionsublimationvaporizationcondensationmeltingfreezingSolidliquidgasq=(18.0 g)∗(2.09Jg °C)∗(0.0 ° C−(−25.0 °))Solve for q then convert J into kj:q=941 J =.941 kjSegment 2:Melting 1.00 mole of ice at the melting point 0.0°CGiven: n = 1.00 mole of ice; ∆Hfus = 6.02 kj/molEquation:q=n∗∆ HfusInput the variables into the equation and solve for q:q=(1.00 mol)∗(6.02kjmol)q=6.02 kjSegment 3:Heating 1.00 mole of water at 0.0°C up to the boiling point of 100.0°CGiven: mass of 1.00 mole of water = 18.0g; CL = 4.18 J/mol*°CEquation:q=mliquid∗CL∗∆TInput variables into equation and solve for q. Be sure to convert from J to kj:q=(18.0 g)∗(4.18Jg∗° C)∗(100.0° C−(0.0 °C))q=7.52∗103J =7.52 kjSegment 4:Boiling 1.00 mole of water at the boiling point 100.0°CGiven: n = 1.00 mole of ice; ∆Hfus = 40.7 kJ/molEquation:q=n∗∆ HvapInput variables into equation and solve for q:q=(1.00 mol)∗(40.7kJmol)q=40.7 kjSegment 5:Heating 1.00 mole of steam at 100.0°C up to 125.0°C.Given: mass of 1.00 mole of water = 18.0g; CV = 2.01 J/mol*°CEquation:q=mass∗CV∗∆TInput the variables into the equation and solve for q. Be sure to convert J to kj:q=(18.0 g)∗(2.01Jg∗° C)∗(125.0 °C−(100.0 °C))q=904 J =.904 kj.904 kj is the final answer to how much energy is needed to measure the heating curve of water from -25°C to
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