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It’s a bird, it’s a plane, no it’s a GAS!!!

It’s a bird, it’s a plane, no it’s a GAS!!!. Notes One Unit Five. Characteristics of Gases Pressure of fluids Standard Temperature and Pressure Converting PressuresMilwaukie High Ideal Gas Law. Pages 422-440. Importance of Gases. Airbags fill with N 2 gas in an accident

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It’s a bird, it’s a plane, no it’s a GAS!!!

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  1. It’s a bird, it’s a plane, no it’s a GAS!!!

  2. Notes One Unit Five Characteristics of Gases Pressure of fluids Standard Temperature and Pressure Converting PressuresMilwaukie High Ideal Gas Law Pages 422-440

  3. Importance of Gases Airbags fill with N2 gas in an accident Gas is generated by the decomposition of sodium azide, NaN3 2 NaN3 (s) → 2 Na (s) + 3 N2 (g)

  4. Physical Characteristics of Gases There is a lot of “free” space in a gas Gases can be expanded infinitely Gases assume the volume and shape of their containers Gases have much lower densities than liquids and solids Gases will mix evenly and completely when confined to the same container

  5. Properties of Gases • Gas properties can be modeled using math. Model depends on: • P= pressure (atmosphere) • V= volume of the gas (L) • T= temperature (K) • n= amount, quantity of gas (moles)

  6. Pressure • Caused by the collision of molecules with the walls of a container • A molecule colliding creates force • Catching a ball create force • Pressure= Force/ Area

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

  8. Air is a fluid….. just like water

  9. Measuring Pressure • The first device for measuring atmospheric pressure was developed by Evangelista Torricelli during the 17th century • The device was called a “barometer” • Baro = weight • Meter = measure Air Pressure 760 torr Mercury

  10. Standard Pressure, Temperature and Volume 1.00 atm = 14.7 psi 1.00 atm = 29.92 in Hg 1.00 atm = 760 mm Hg 1.00 atm = 760 torr 1.00 atm = 101,325 Pa 1.00 atm = 101.325 kPa 1.00 atm = 1.01325 bar Standard Temperature 273K or 0oC K=oC+273 Standard Volume: 22.4 Liter/mole for any gas at STP pp 427

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

  12. Starter: Pressure Conversions The pressure of a gas is measured as 49 torr. Represent this pressure in atmospheres, Pascals, and mmHg.

  13. The Ideal Gas Law PV = nRT

  14. Ideal Gases An “ideal” gas exhibits certain theoretical properties. Specifically, an ideal gas … • Obeys all of the gas laws under all conditions. • Does not condense into a liquid when cooled. • Shows perfectly straight lines when its V and T & P and T relationships are plotted on a graph. In reality, there are no gases that fit this definition perfectly. We assume that gases are ideal to simplify our calculations.

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

  16. Ideal Gas Equation Universal Gas Constant Volume(L) PV = nRT Pressure(Torr, kPa, or Atm Temperature(oK) No. of moles(mol) R = 0.0821 atm L / mol K R = 8.314 kPa L / mol K R = 62.4 Torr L / mol K

  17. nRT 2.00mol x 62.4 Torr-L/molK x 573. oK V V = = P 740. Torr PV=nRT Calcuation • What is the volume that 2.00 moles of iodine will occupy under the conditions: Temp = 300.oC and Pressure = 740. Torr ? • Step 1) Write down given information. n = 2.oo moles iodine T = 300.oC = 573 K P = 740. Torr R = 62.4 Torr . L / mol . K Step 2) Equation: PV = nRT Step 3) Solve for variable Step 4) Substitute in numbers and solve = 96.6 L

  18. Allows us to examine the relationships between volume, temperature, pressure, and amount of gas Unit 5Notes #2Gas Laws

  19. Boyle’s Law: P and V • P α 1/V • inversely proportional if moles and temperature are constant ( do not change). • as one increases, the other decreases • pressure is caused by moving molecules hitting container walls • If V is decreased and the # of molecules stays constant, there will be more molecules hitting the walls per unit

  20. Boyle’s Law: P and V

  21. Boyle’s Law

  22. Example: Boyle’s Law Consider a 1.53-L sample of gaseous SO2 at a pressure of 5.6 x 103 Pa. If the pressure is changed to 1.5 x 104 Pa at constant temperature, what will be the new volume of the gas?

  23. Charles’ Law: V and T If n and P are constant, then V α T When T increases, gas molecules move faster and collide with the walls more often and with greater force. The gas will expand To keep the P constant, the V must increase

  24. Charles’ Law: V and T

  25. Charles’ Law • discovered by French physicist, Jacques Charles in 1787 • first person to fill balloon with hydrogen gas and make solo balloon flight

  26. Charles’ Law: V and T • Problem: if we use Celsius, we could end up with negative values from calculations in gas laws for volumes • we need a T system with no negative values: Kelvin Temperature Scale • starts at -273 ° C = absolute zero= 0 K • lowest possible temperature balloon going into liquid nitrogen

  27. Example: Charles’ Law & Temp. A sample of gas at 15°C and 1 atm has a volume of 2.58 L. What volume will this gas occupy at 38°C and 1 atm?

  28. The CombinedGas Law

  29. V1 P1 V2 P2 P1V1 = P2V2 = = T1 T1 T2 T2 P1V1 P2V2 T1 T2 = Combining the gas laws • So far we have seen two gas laws: Robert Boyle Jacques Charles Joseph Louis Gay-Lussac These are all subsets of a more encompassing law: the combined gas law Read pages 437, 438.

  30. P1T2V1 P1T2V1 P2T1V2 P1T2V1 P2T1V2 P2T1V2 P2 T1 T2 V1 P1 V2 = = = = = = P2V2 T1V2 P1V1 P2T1 T2P1 T2V1 Combined Gas Law Equations

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

  32. 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

  33. Combined gas Law Problem Two 1) Calculate Formula mass. • 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? 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

  34. 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

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

  36. 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

  37. 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

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

  39. Notes Two Unit Five Review Mass-Mass Calculation Mass-Volume Calculation @STP Volume-Mass Calculation @STP Pages 441-450

  40. Let’s Put All The Steps Together! (Review of Stoich!) • Factor label Method: it uses UNITS to cancel and work your way through the problem Given Info (g) 1 Mole Given (mol)Moles RequestedMM Requested (g) of Requested (g) MM Given (g) Moles Given 1 Mole Requested = Given Mole to Mole Requested The given converts grams of given to moles of given. The middle converts moles of given to moles of requested. The requested converts moles of requested to grams of requested. All in one multistep process !!! THIS IS A MASS MASS PROBLEM.

  41. 8 mol CO2 44.0 g CO2 1 mol C8H8 1 mol C8H8 1 mol CO2 104.0 g C8H8 = 20.3 g CO2 Mass-Mass Calculation 1C8H8(l)+ 15O2(g) 8 CO2(g) + 4 H2O(g) • How many grams of carbon dioxide will be made by burning 6.00 grams of Isovanillin? 6.00 g C8H8

  42. How many grams of silver will be made by decomposing 8.00 grams of Silver oxide? 4 mol Ag 107.9 g Ag 1 mol AgO 2 mol AgO 1 mol Ag 231.8 g AgO = 7.45 g Ag Mass-Mass Calculation 2Ag2O(s) 1O2(g)+ 4Ag(s) 8.00 g AgO

  43. Molar Volume @STP 1molN2(g) = 28.0g/mol 22.4L/mol 32.0g/mol 22.4L/mol 1molO2(g) = 1molAr(g)= 39.9g/mol 22.4L/mol

  44. 2 mol NH3 22.4 l NH3 1 mol N2 1 mol N2 1 mol NH3 28.0 g N2 = 89.6 L NH3 Mass-Volume Calculation • How many liters of ammonia will be made reacting 56.0 grams of nitrogen with hydrogen @ STP? 1N2(g) + 3 H2(g) 2 NH3(l) 56.0 g N2

  45. 2 mol NH3 22.4 l NH3 1 mol N2 1 mol N2 1 mol NH3 22.4 l N2 = 67.2 L NH3 Volume-Volume Calculation 1N2(g) + 3 H2(g) 2NH3(l) • How many liters of ammonia will be made reacting 33.6 liters of nitrogen with hydrogen @STP? 33.6 l N2

  46. Notes Three Unit Five • Molar volume @ Non-STP Conditions • R is Universal Gas Constant Pages 452-459

  47. 1 mol Zn 1 mol H2 28.1 L H2 1 mol Zn 65.4 g Zn 1 mol H2 = 5.59 L H2 Gas Stoich Probs NON-STP • How many liters of hydrogen gas at 650. torr and 20. C will be formed by reacting 13.0 g of zinc with excess HCl? 1Zn + 2HCl  1 ZnCl2 + 1 H2 Use: PV = nRT to determine the volume of one mole of any gas will occupy at this temperature and Pressure. V = nRT = (1.00 mol)(62.4 L-torr/molK)(293K) = 28.1 L P 650. Torr 13.0 g Zn

  48. 1 mol Fe 3 mol H2 48.2 L H2 2 mol Fe 55.8 g Fe 1 mol H2 = 145 L H2 Gas Stoich Probs NON-STP • What volume of hydrogen gas at 1000. torr and 500. C would be needed to form 112 g iron? • 1Fe2O3 + 3H2  2 Fe + 3H2O Use: PV = nRT to determine the volume of one mole of any gas will occupy at this temperature and Pressure. V = nRT = (1.00 mol)(62.4 L-torr/molK)(773K) = 48.2 L P 1000. Torr 112 g Fe

  49. 1 mol KClO3 3 mol O2 19.1 L O2 2 mol KClO3 122.6 g KClO3 1 mol O2 = 2.34 L O2 Gas Stoich Probs NON-STP • What volume of oxygen gas at 1.25 atm and 18 C would be formed by heating 10.0 g of potassium chlorate? 2KClO3  2 KCl + 3O2 Use: PV = nRT to determine the volume of one mole of any gas will occupy at this temperature and Pressure. V = nRT = (1.00 mol)(0.0821 L-atm/molK)(291K) = 19.1 L P 1.25. Atm 10.0 g KClO3

  50. 1 mol CH3Br 1 mol CH4 25.1 L CH4 1 mol CH3Br 25.1 L CH3Br 1 mol CH4 = 9.58 L CH4 Gas Stoich Probs NON-STP • What volume of methane gas (CH4) at 98.5 kPa and 24 C would be needed to form 9.58 L of methyl bromide (CH3Br)? 1CH4 + 1 Br2  1 CH3Br + 1HBr Use: PV = nRT to determine the volume of one mole of any gas will occupy at this temperature and Pressure. V = nRT = (1.00 mol)(8.314 L-kpa/molK)(297K) = 25.1 L P 98.5 kpa 9.58 LCH3Br

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