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Notes One Unit Five

Characteristics of Gases Pressure of fluids Standard Temperature and Pressure Converting Pressures Gas Laws. Notes One Unit Five. Pages 422-440. Pressure Versus Molecular Collision. Pressure is caused by molecular collision A molecule colliding creates a force.

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Notes One Unit Five

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  1. Characteristics of Gases Pressure of fluids Standard Temperature and Pressure Converting Pressures Gas Laws Notes One Unit Five Pages 422-440

  2. Pressure Versus Molecular Collision • Pressure is caused by molecular collision • A molecule colliding creates a force. • Catching a ball creates a force. • P=F/A • pp 427

  3. Pressure viewed as created in a fluid • Created by the weight • The deeper you go, the more weight . pp 427

  4. Air is a fluid…just like water pp 427

  5. Torricellian Barometer pp 427 Air Pressure 780 torr 760 torr 740 torr Mercury

  6. Pop can Demo

  7. Magdeburg plates Demo

  8. Standard Pressure, Temperature and Volume 1 atm = 14.7 psi 1 atm = 29.92 in Hg 1 atm = 760 mm Hg 1 atm = 760 torr 1 atm = 101,325 Pa 1 atm = 101.325 kPa 1 atm = 1.01325 bar 273K or 0oC K=oC+273 22.4 Liter/mole for any gas at STP pp 427

  9. Converting Pressures • Convert 25 lb/in2 to torr • Convert 75 Kpa to in Hg

  10. Charles’ and Boyle’s Demo Pressure is constant. Temperature is constant. Moles are constant. Moles are constant. P1= 1.0atm P2= 2.0atm 1.0atm P1= 1.0atm P2= V2= 0.50L V1= 1.0L V1= V2= 2.0L 1.0L T1= 273K T2= 546K T1= 273K T2= 273K V1 V2 = P1 x V1 = P2 x V2 T1 T2 1.0ax 1.0L = 2.0ax 0.50L 1.0L 2.0L = pp 433-440 273K 546K

  11. Combined Gas Law Equation • Is…….. pp 433-440

  12. Combined gas Law Problem One • A gas occupies 2.0 m3 at 121.2 K, exerting a pressure of 100.0 kPa. What volume will the gas occupy at 410.0 K if the pressure is increased to 220.0 kPa? Assign variables and calculate V2. (2.0m3 ) (100.0KPa) (220.0KPa ) (V2 ) = (121.2K ) (410.0K ) (2.0m3 ) (100.0KPa) (410.0K ) V2 = (121.2K ) (220.0KPa ) pp 433-440 V2= 3.1M3

  13. Combined gas Law Problem Two 10.0 gram • A 10.0 gram sample of ethane(C2H6) gas is at STP. If the volume is changed to 26.0 liters, what is the new Kelvin temperature of the gas? 1) Calculate Formula mass. Mass E # 24.0 2x 12.0 = C 6.0 H 6x 1.0 = 30.0g/m 2) Calculate V1. 10.0g V1 = V1= 7.46L 30.0g/mol 22.4L 3) Assign variables and calculate T2. (101.325KPa) (26.0L) (101.325KPa) (7.46L) = (T2) (273K) (273K) (101.325KPa) (26.0L) T2= 951K T2 = (101.325KPa) (7.46L) pp 433-440

  14. Notes Two Unit Five Grahams’ Law Calculation Review Mass-Mass Calculation Mass-Volume Calculation @STP Volume-Mass Calculation @STP Pages 441-450

  15. Graham’s Law Demo 17.0g/m 36.5g/m pp 442

  16. Graham’s Law • Describes how speed compares between gas molecules with different masses. • Two different gases: • 1)Same Temperature • 2)Different Masses • Kinetic energy  ½ M1V12= ½ M2V22 pp 442

  17. Graham’s Equation ½ M2V22= ½ M1V12 M2V22 M1V12 ÷ by M1 = M1 M1 M2V22 V12 ÷ by V22 V12 = V22 V22 M1 Square root of both sides pp 442

  18. Grahams’ Law Problem One • At a high temperature molecules of chlorine gas travel 15.90cm. What is the mass of vaporized metal (gas) under the same conditions, if the metal travels 8.97cm? # E Mass Cl 2x 35.5 = 71.0g/m (15.90cm) = 14.9 2 = 8.97cm M2= 223g/m pp 442

  19. Grahams’ Law Problem Two • At a certain temperature molecules of chlorine gas travel at 0.450 km/s. What is the speed of sulfur dioxide gas under the same conditions? Mass # E Cl 2x 35.5 = 71.0g/m Mass E # 32.1 S 1x 32.1 = 32.0 O 2x 16.0 = 64.1g/m = V2 V2= 0.474Km/s pp 442

  20. How many grams of oxygen will react with 5.00 grams of hydrogen to make water? ___g 42 5.0g ÷2.0g/m 2 H2(g) + O2(g)  H2O(l) 1 2 Review Mass-Mass Calculation X32.0g/m ___m 1.3 ___m 2.5 1) grams H2 to moles H2 5.00g÷ 2) moles H2to moles O2 2.5m H2x 3) moles O2 to grams O2 1.3mO2x 2.0gH2 /m= 2.5m H2 Mass # E H 2.0g/m (1mO2) 2 x 1.0 = =1.3mO2 (2mH2) Mass # E O 2x 16.0 = 32.0g/m 32.0g/m = 42gO2

  21. How many liters of oxygen will react with 5.00 grams of hydrogen to make water? ___L 29 5.0g ÷2.0g/m pp 449 2 H2(g) + O2(g)  H2O(l) 1 2 Mass-Volume Calculation @STP X22.4g/m ___m 1.3 ___m 2.5 1) grams H2 to moles H2 5.00g÷ 2) moles H2to moles O2 2.5m H2x 3) moles O2 to liters O2 1.3mO2x 2.0gH2 /m= 2.5m H2 Mass # E H 2.0g/m (1mO2) 2 x 1.0 = =1.3mO2 (2mH2) Mass # E O 2x 16.0 = 32.0g/m 22.4L/m = 29LO2

  22. How many grams of oxygen will react with 56.0 liters of hydrogen to make water? ____g 40.0 56.0L ÷22.4L/m pp 449 2 1 2 H2(g) + O2(g)  H2O(l) Volume-Mass Calculation @STP ____m 1.25 ____m 2.50 X32.0g/m 1) Liters H2 to moles H2 56.0L÷ 2) moles H2to moles O2 2.50m H2x 3) moles O2 to grams O2 1.25mO2x 22.4LH2 /m= 2.50m H2 Mass # E H 2.0g/m (1mO2) 2 x 1.0 = =1.25mO2 (2mH2) Mass # E O 2x 16.0 = 32.0g/m 32.0g/m = 40.0gO2

  23. Notes Three Unit Five • Kinetic theory of gases • Molar volume @ Non-STP Conditions • R is Universal Gas Constant Pages 452-459

  24. THE KINETIC THEORY OF GASES 6.022x1023atoms/mole • Large number of particles pp 426

  25. THE KINETIC THEORY OF GASES For a collision KEBefore=KEAfter • Large number of particles • Elastic collisions pp 426

  26. THE KINETIC THEORY OF GASES • Large number of particles • Elastic collisions • No external forces pp 433-440

  27. THE KINETIC THEORY OF GASES 1.6x1011 times diameter • Large number of particles • Elastic collisions • No external forces • Separated by largedistances pp 426

  28. THE KINETIC THEORY OF GASES • Large number of particles • Elastic collisions • No external forces • Separated by largedistances • No forces between particles pp 426

  29. Finding volumes @ Non-STP Conditions PV=nRT Ideal Gas Equation What is… P? V? moles n? T? R? Universal Gas Constant (22.4L) (101.325kpa) V P = R R = (1 m) (273.15K) n T - L Kpa = (8.314 ) R - m K pp 446

  30. ____L 9.33 15.00g ÷18.0g/m 2H2O(l) 2H2(l)+ 1O2(l) Finding Volume at Non-STP ______m 0.833 _____m X22.4L/m 0.417 a) What volume of oxygen would be made at STP? 1) grams H2Oto moles H2O 0.833m 15.00g÷ 18.0g/m= Oxygen is made reacting 15.00g water at 209.0Kpa and 20.0oC. 2) moles H2O to moles O2 ( 1mO2) 0.833m x =0.417mO2 (2m H2O) 3) moles O2 to litersO2 # Mass E 0.417mO2x 22.4L/m = 9.33LO2 H 2.0 2 x 1.0 = @STP O 16.0 1 x 16.0 = 18.0g/m

  31. pp 450 b) What is the volume of gas at reaction conditions? PV=nRT (0.417m) (8.314 L•Kpa•m-1•K-1) (293.2K) V= n R T (209.0KPa) V = Finding Volume at Non-STP P 4.86L V= c) How many moles O2 gas are produced? 0.417mO2 n= d) How many grams O2 gas are produced? n x g/m= g Mass # E 0.417mx 32.0g/m= 13.3gO2 O 2x 16.0 = 32.0g/m

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