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PSU CHEM 110 - Phase changes & heat capacity

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Sheets Page 1 Lecture 26 Lecture 26: Liquids 1: phase changes & heat capacity Read: BLB 5.5; 11.4 HW: BLB 5:48,49,51; 11:33,37,39 Know: • viscosity, surface tension • cohesive & adhesive forces • phase changes • heat capacity • calorimetry Need help?? Get help!! TAs in CRC (211 Whitmore) and SI—hours on Chem 110 website; my office hours (Mon 12:30-2 & Tues 10:30-12 in 324 Chem Bldg [or 326 Chem]) Bonus deadline for BST #8: Intermolecular forces, March 26 Bonus deadline for BST #9: Solutions & dilutions, April 2 Exam 3: Monday, April 6 @ 6:30 Check out the grade-u-lator @ http://courses.chem.psu.edu/chem110/spring/grade.htmSheets Page 2 Lecture 26 Manifestations of IM forces • cohesive forces: forces between molecules for example: viscosity: resistance to flow surface tension: E needed to increase vapor pressure (more next time) BP (MP) ΔHvap (ΔHsub) • adhesive forces: forces between a & the surface for example: water–glass interactions water–oil interactions meniscus formation capillary action glue non-stick surfacesSheets Page 3 Lecture 26 Properties of liquids • viscosity: resistance to flow; IM forces ↑, viscosity • surface tension: energy needed to increase surface area; IM forces ↑, surface tension • IM interactions are energetically favorable; that is, heat is required to break them—the more interactions, the better • BUT surface molecules have fewer interactions; minimize energy by minimizing surface areaSheets Page 4 Lecture 26Sheets Page 5 Lecture 26 Kinetic energy & intermolecular forces • kinetic energy (KE) prevents molecules from interacting with each other, i.e., KE attractive intermolecular forces (IMF) gas: KE >> IMF liquid: KE ≈ IMF solid: KE << IMF • KE ∝ T • energy IN (heating): solid → liquid → gasSheets Page 6 Lecture 26 Phase changes & temperature • energy change: ΔHsub, ΔHfus, ΔHvapSheets Page 7 Lecture 26 Phase diagrams Go to: http://treefrog.fullerton.edu/chem/LS/phased.html • phase diagram: graphically summarizes conditions where equilibria exist between phases; shows the relationship between the phases as functions of T & P (for example) normal melting point (P = 1 atm) pressure dependence of melting point normal boiling point (P = 1 atm) pressure dependence of boiling point critical point triple point supercritical fluid coexistence curvesSheets Page 8 Lecture 26 Calorimetry See: http://treefrog.fulle rton.edu/chem/LS/coolheat.htm l & check out heating curve & phase diagram • experimental measure of heat flow; used to determine ΔHrxn • molar heat capacity: amount of heat required to raise 1 mole of substance 1°C (or 1 K); an intensive variable • specific heat (capacity): amount of heat required to raise 1 g of substance 1°C (or 1 K); an intensive variable; for H2O C = 4.184 J/(g °C) q = C m ΔT q = quantity of heat (measure of energy) C = specific heat (heat capacity per gram) [energy/(mass•°C)] [see BLB Table 5.2] m = mass ΔT = Tfinal – Tinitial m•C = heat capacity (an extensive property–that is, it depends on how much stuff you have)Sheets Page 9 Lecture 26 Heating curve when heat is added to a system: 2 types of changes occur 1. within single phase (in blue), changes are ; q = n CpΔT • since T ↑, kinetic energy ↑, enthalpy (ΔH) ↑, molecular motion ↑; separation between molecules ↑, molecular attractions ↓, & molecular order ↓ 2. between phases, ( in red) abrupt changes from one physical state to another; q = n ΔHx (x = melting (⇐fusion), vaporization) • since T is constant, kinetic energy is constant, BUT enthalpy (ΔH) ↑, molecular separation ↑, molecular attractions ↓ & molecular order ↓Sheets Page 10 Lecture 26 Example: use your units!!, very important!!! 2 moles of ice at −25°C are heated to 125°C. How much energy is needed? Cp(ice) = 37.6 J/mol·K Cp(water) = 75.3 J/mol·K Cp(steam) = 33.1 J/mol·K ΔHfusion = 6.02 kJ/mol ΔHvapor = 40.67 kJ/mol 1. ice –25°C ! 0°C: "T = q1=2. ice 0°C ! water 0°C: "T = q2=3. water 0°C ! 100°C: "T = q3=4. water 100°C ! steam 100°C: "T = q4=5. steam 100°C ! 125°C: "T = q5=q =#Sheets Page 11 Lecture 26 Example: A swimmer emerging from a pool is covered with a film containing 75 g of water. (a) How much heat must be supplied to evaporate this water? (b) Why does the swimmer feel a chill? ΔHvapor = 40.67 kJ/mol water film = system; swimmer = surroundingsSheets Page 12 Lecture 26 Coffee cup calorimeter (constant pressure calorimetry) [• heat lost by the reaction (qrxn) is equal in magnitude to but opposite in sign to heat gained by the solution (qsoln) (and vice versa) (more also later)—weʼll hit this again when we cover thermochemistry!!] • we can measure temperature and calculate enthalpy (heat lost or gained)Sheets Page 13 Lecture 26 Before next class: Read: BLB 11.3; 11.5–11.6 HW: BLB 11:4,6,31,43,45,47,50,53,56 Sup 11:5–13 Know: • phase changes • vapor pressure & boiling points • concentrations & dilutions Answers: p. 10: qtot = 112 kJ p. 11: qtot = 169.5


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PSU CHEM 110 - Phase changes & heat capacity

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