1. Copyright 1999, PRENTICE HALLChapter 111 Surface Tension

2. Copyright 1999, PRENTICE HALLChapter 112 Intermolecular Forces Bulk and Surface

3. Copyright 1999, PRENTICE HALLChapter 113 Phase Changes Energy Changes Accompanying Phase Changes All phase changes are possible under the right conditions (e.g. water sublimes when snow disappears without forming puddles). The sequence heat solid  melt  heat liquid  boil  heat gas is endothermic. The sequence cool gas  condense  cool liquid  freeze  cool solid is exothermic.

4. Copyright 1999, PRENTICE HALLChapter 114 Phase Changes Heating Curves

5. Copyright 1999, PRENTICE HALLChapter 115 Heating Curve Illustrated

6. Copyright 1999, PRENTICE HALLChapter 116 Vapor Pressure Explaining Vapor Pressure on the Molecular Level Dynamic Equilibrium: the point when as many molecules escape the surface as strike the surface. Vapor pressure is the pressure exerted when the liquid and vapor are in dynamic equilibrium.

7. Copyright 1999, PRENTICE HALLChapter 117 Vapor Pressure Volatility, Vapor Pressure, and Temperature If equilibrium is never established then the liquid evaporates. Volatile substances evaporate rapidly. The higher the temperature, the higher the average kinetic energy, the faster the liquid evaporates.

8. Copyright 1999, PRENTICE HALLChapter 118 Liquid Evaporates when no Equilibrium is Established

9. Copyright 1999, PRENTICE HALLChapter 119 Vapor Pressure Vapor Pressure and Boiling Point Liquids boil when the external pressure equals the vapor pressure. Temperature of boiling point increases as pressure increases. Two ways to get a liquid to boil: 1. 2. –Pressure cookers operate at high pressure. –The higher the pressure the higher the b.p. –reducing the cooking time required. Normal boiling point is the boiling point at 760 mmHg (1 atm).

10. The Ideal Gas Law PV=nRT P = Pressure V = Volume n = Number of Moles R = ideal gas constant T = temperature in Kelvin

12. R -The ideal gas constant Depends on unit of pressure 0.0821 L. Atm / K. mol 62.4 L. mmHg / K. mol (torr is mm Hg) 8.31 L. kPa / K. mol

13. Ideal Gas Law example problem Calculate the pressure of 1.65 g of helium gas at 16.0 o C and occupying a volume of 3.25 L? You will need g to moles and Celsius to Kelvin: 1.65 g He 1 mole He 4.0 g He = 0.413 mol He K = o C + 273 ; 16. 0 + 273 = 289 K For this problem you will need to pick an R value. For this problem I will choose to use the R value containing kPa. I picked it. You can’t do anything about it. So; just try and stop me. Plug and Chug baby, get ‘R done. Do it. Come on I dare ya. Get it - ‘R as in ideal gas constant

14. Ideal Gas Law example problem P x 3.25 L = 0.413 mol x 8.31 kPa. L x 289 K mol. K Do the algebra and solve; if you do it right, guess what? You get the answer right. Neat concept, huh? Maybe your mommy will give you a cookie. = 305 kPa Your turn How many moles of gas are present in a sample of Argon at 58 o C with a volume of 275 mL and a pressure of 0.987 atm.

15. Ideal Gas Law example problem Answer 0.987 atm x 0.275 L = n x 0.0821 L. Atm x 331K mol. K Do the dew; oops, I mean the algebra and presto; the answer with the correct number of sig figs is.. Do you know how to keep a so called chem student in suspense? Do ya? = 0.00999 mol Ar Congrats - you can plug and chug.

16. Gas Stoichiometry You can plug and chug. But, can you do…. For these problems you may need to do tracks first then use the ideal gas law or use the gas law first then the tracks.

17. Sample problem Ammonia (NH 3 ) gas can be synthesized from nitrogen gas + hydrogen gas. What volume of ammonia at 450 kPa and 80°C can be obtained from the complete reaction of 7.5 kg hydrogen? 7500 g H 2 1 mol H 2 2.02 g H 2 x = 2500 mol NH 3 2 mol NH 3 3 mol H 2 x PV = nRTP = 450 kPa, n = 2475 mol, T = 353 K (450 KPa) (2500 mol)(8.31)(353 K) = V = 16000 L NH 3 First we need a balanced equation: N 2 (g) + 3H 2 (g)  2NH 3 (g)

18. Your Turn: Hydrogen gas (and NaOH) is produced when sodium metal is added to water. What mass of Na is needed to produce 20.0 L of H 2 at STP? First we need a balanced equation: 2Na(s) + 2H 2 O(l)  H 2 (g) + 2NaOH(aq) 0.893 mol H 2 = 41.1 g Na 2 mol Na 1 mol H 2 x 22.99 g Na 1 mol Na x PV = nRT (8.31 kPaL/Kmol )(273 K) (101.3 kPa)(20.0 L) = n = 0.893 mol H 2 P= 101.3 kPa, V= 20.0 L, T= 273 K or # mol = 20.0 L x 1 mol / 22.4 L = 0.893 mol