1. main difference between phases is… the distance between particles… Phases (s, l, g) …which depends on two competing quantities: IMAFs (intermolecular attractive forces between particles) KE (kinetic energy of the particles) vs. GASLIQUIDSOLID compressible

2. What happens to added heat? Heating Curves added heat breaks IMAFs & increases distance (↑PE) same temp during phase change (=KE avg ) WS #1-3 (2 min)

3. heat of vaporization 100 o C 0oC0oC Heating Curve of Water Time (or Heat Added) gas liquid solid ( ∆ H fus ) heat required to melt (or freeze) heat required to vaporize (or condense) heat of fusion ( ∆ H vap ) (2260 J) (418 J) (334 J) 4.18 J/g o C Why is ∆ H vap >>> ∆ H fus ? l  g takes more energy to break IMAFs than s  l Temp ( o C) WS #4 1 gram of water

4. 100 o C 0oC0oC Cooling Curve of Water Time (or Heat Removed) gas liquid solid Temp ( o C) The phase change sequence is reversible. condensation boiling freezing melting WS #5-7 (3 min)

5. Atmospheric Pressure (weight of air) (Force) empty space (a vacuum) h 760 mm 1 atm = 760 mmHg = 760 torr = 101.3 kPa 1 N 1 m 2 Units (at sea level) Pressure STP (standard T & P) 273 K 1 atm (0 o C) P = FAFA

6. 1)constant particle motion 2)Pressure caused by collisions 3)IMAFs are negligible 4)Volume of particles is negligible compared to total volume (V gas = V container – V particles ) 5)Average KE directly proportional to Kelvin Temp (  T,  KE) Kinetic-Molecular Theory (KMT) motion of particles (energy, collisions)

7. R = 0.08206 L∙atm / mol ∙ K (changes in P,V,T,n) P ↑, V ↓ T ↑, V ↑ T ↑, P ↑ n ↑, V ↑ inverse NOT given PV = nRT P2V2n2T2P2V2n2T2 P1V1n1T1P1V1n1T1 = direct P A = P total x X A X A = moles of A total moles P total = P A + P B + P C + … X A : mol fraction Molar Volume: 22.4 L/mol @STP given

8. P total = P H 2 O + P gas W hen one collects a gas over water, there is water vapor mixed in with the gas. To find only the pressure of the gas, one must subtract the water vapor pressure from the total pressure. P gas = P total – P H 2 O equalize water level inside & outside =

9. Gas Stoich with Molar Volume Assume reactions below occurred at STP. 1.Calculate the moles of NH 4 CI reacted with Ca(OH) 2 to produce 11.2 L of NH 3 (g). 2 NH 4 Cl + Ca(OH) 2  2 NH 3 + CaCI 2 + 2 H 2 O 0.500 mol 11.2 L NH 3 x 1 mol NH 3 x 22.4 L NH 3 2 mol NH 4 Cl = 2 mol NH 3

10. 2.What volume of O 2 gas is produced from 490 g KClO 3 at 298 K and 1.06 atm? KClO 3 (s)  KClO (s) + O 2 (g) = ____L O 2 490 g KClO 3 x g KClO 3 mol KClO 3 x mol O 2 122.55 = 1 1 1 92.3 L O 2 NOT at STP Molar Mass of KClO 3 is 122.55 g/mol PV = nRT use… 4.00 (1.06 atm) V = (4.00 mol )(0.08206)(298 K) (would be 89.6 L if at STP)

11. Describe the relationship of the molecular velocity (v) of a gas to its molar mass (M) and its temperature (T). ½ m v 2 = ½ m v 2 same T, same KE KE = ½ m v 2  T,  KE ↑ T, ↑ v ↑ M, ↓ v KE = ½ mv 2

12. Less ideal gas: H 2 O (g) or CO 2 (g) F 2 (g) or Br 2 (g) Non-Ideal: (Low T) (High P) Ideal: (High T) (Low P) WHY?(high KE)(high V total ) (low KE)(low V total ) Ideal Gas vs Non-Ideal Gas IMAFS & V particle are negligible At low T, IMAFs not negligible, collisions less frequent and of less force IMAFs At high P, particle volume is not negligible compared to smaller total volume V gas = V container – V particles IMAFS & V particle are NOT negligible …under ideal conditions: PV ( usually) = nRT