Chem 170 1st Edition Lecture 10Outline of Last Lecture I. Last Lecture: Vapor Pressure, Clausius Clapeyron Equation, Boiling PointII. VocabularyIII. ExamplesOutline of Current Lecture IV. Last Lecture: VocabularyV. ExamplesCurrent LectureCh. 12 – Liquids and SolidsLast Lecture: Vocabulary:VaporizationH2O(l )→ H2O(g)ΔHvapCondensationH2O(g)→ H2O(l)FusionH2O(s)→ H2O(l)ΔHfusionFreezingH2O(l )→ H2O(s)Sublimation H2O(s)→ H2O(g )ΔHsup∆ E=q+w=q+P ∆ V → qp=∆ E+P ∆ V =∆ H∆ Hvap=Hvapor−Hliquid-A phase change is a physical change in the substance at a constant temperature.ex. If there is 10g of ice at -13°C, then how much heat is required to melt the ice to a liquid at 50°C?Ice at -13 C → Ice at 0 C q=ms ∆ T =(10 g)(2.11Jg × ℃)(0 ℃ −(−13 ℃))=316.5 J=0.3165 KJIce at 0 C →Liquid at 0 C (phase change: fusion)(10 g)(1 mol18 g)(6.01 KJ1 mol)=3.3 KJLiquid at 0 C → Liquid at 50 C(10 g)(4.18Jg × ℃)(50 ℃ −0 ℃)=2090 J =2.09 KJTOTAL: 0.3165 KJ +3.3 KJ +2.09 KJ =5.7 KJThese 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.ex. If there was a Styrofoam cup of 100g of water (100g water = 100ml water) at 22 C and 25g of ice at 0 C placed inside, then what is the final temperature?Thinking about it, there will only be two possible outcomes:1. The drink will have both liquid water and ice (0 C)2. The drink will only have liquid waterIce at 0 C →Liquid at 0 C (phase change: fusion)(25 g)(1mol18 g)(6.01 KJ1 mol)=8.34 KJ =8340 JLiquid at 22 C → Liquid at 0 C(100 g)(4.18Jg × ℃)(0 ℃ −22 ℃)=−9196 J =¿Therefore there is enough heat stored in the water to melt the ice [9196>8.34]qice=−qwater → qwater+qice=0(25 g)(1mol18 g)(6.01 KJ1 mol)+(25 g)(4.18Jg × ℃)(Tf−0 ℃)[(100 g)(4.18Jg × ℃)(Tf−22 ℃)]+¿¿(−9200 J)Tf+(8340 J)+(1043 J)TfTf=1.6 ℃Phase diagrams are graphs of the relationship between pressure (atm), temperature (Celsius) and the phase at which the substance is present
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