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Chapter 5 Gases. Objectives Describe the properties of gases Describe the Kinetic Molecular Theory, Ideal Gases Explain air pressure and barometers Convert.

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Presentation on theme: "Chapter 5 Gases. Objectives Describe the properties of gases Describe the Kinetic Molecular Theory, Ideal Gases Explain air pressure and barometers Convert."— Presentation transcript:

1 Chapter 5 Gases

2 Objectives Describe the properties of gases Describe the Kinetic Molecular Theory, Ideal Gases Explain air pressure and barometers Convert pressure units Perform calculations using the ideal gas law

3 Why Study Gases? We deal with gases on a daily basis –Filling your car tires –Barometric Pressure (Weather) –Breathing air Many reactants and products are gases –We need to know how to work with gases –In the lab gases are usually measured in volumes instead of masses

4 Properties of Gases Gases expand to fill their container Gases can be compressed Gases are fluids –Meaning they flow Gases have a low density –Liq. N 2 =.807g/mL at -196ºC –Gas N 2 =.625 g/L at 0ºC –Or about 1000 times different! Gases effuse and diffuse

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7 Kinetic Molecular Theory Particles of a gas are in constant motion Volume of the individual particles is zero Particles colliding with the side of the container cause pressure Particles exert no force on each other –Means = No Intermolecular Forces Temperature of a gas is directly related to its Kinetic Energy

8 What’s Wrong? Gas particles do have volume –However the ratio of the particle volume to container volume is almost zero Gas particles experience intermolecular forces –However, the particles are relatively far apart and free to move so –Forces are weak

9 Ideal Gas Gases that behave according to the kinetic molecular theory (KMT) No such gas exists Good approximation most of the time Simplifies our treatment of gases Corrections for real gases are fairly small

10 Gas Variables Pressure (P) in atm, mmHg, torr, kPa Volume (V) in mL, L Temperature (T) in K Moles (n) in mol These 4 variables can completely describe a gas

11 Pressure Measure of force per unit area –Force/Area –SI Unit is N/m 2 or Pascal (Pa) –1 Newton is about 100 grams –So, 1 Pa is about 100 grams on 1m 2 Or a very small pressure Atmospheric pressure is quite substantial –101,300 Pa or 101.3 kPa

12 Measuring Pressure Pressure is most commonly measured with a barometer Invented by Evangelista Torricelli in approximately 1644 Filled a glass tube with liquid mercury and inverted the tube in a dish of mercury At sea level the column stood at 76 cm When the barometer was taken to higher elevation the level dropped

13 Torricelli’s Explanation Air pressed down on the dish of mercury Mercury was forced up the column Mercury rose until the weight of the mercury equaled the weight of the air

14 Figure 11.404a

15 Pressure Measurements Normal pressure at sea level in a barometer –760. millimeters mercury (mmHg) –1.00 atmospheres (atm) –101.3 kilopascals (kPa) –760. torr (in honor of Torricelli) –14.7 pounds per square inch (psi) –29.9 inches of mercury (inHg) –We will use the first three the most

16 Example Convert 728 mmHg to A) atm B) kPa

17 Temperature Kelvin temperature is the ONLY scale used in gas calculations Used because 0 K is absolute zero Note it is NOT ºK! Converting from Celsius to Kelvin –Temp. in K = Temp in C + 273 0 ºC = 273 K 100 ºC = 373 K Room temp is about 22 ºC or 295 K

18 The Ideal Gas Law Mathematical equation that relates all variables for a gas PV=nRT –P,V,n,T have been discussed –What is R? –Universal Gas Constant Same for ALL gases But can change with pressure unit –Works well at low pressures and high temp.

19 Universal Gas Constant The units of R can change as the pressure units change R has two values R = 8.314 (L*kPa)/(mol*K) R = 0.08206 (L*atm)/(mol*K) –Use the first if you are in kPa –Use the second if you are in atm –If you are in mmHg convert

20 Units in the Ideal Gas Law PV=nRT P can be in atm or kPa V must be in Liters (L) n must be in moles (mol) T must be in Kelvins (K)

21 Changing the Law The ideal gas law can be manipulated to solve for an unknown variable Often used in stoichiometry problems You will always know R. –It is never a variable Just use algebra to isolate the variable you desire

22 Solve for T

23 Solve for P

24 Example A sample of gas at 25ºC and 3.40 L contains 3.33 moles. What is the pressure (in kPa)?

25 Example 5.69 grams of Oxygen gas at 250.ºC has a pressure of 722mmHg. What is the volume?

26 Changing Gas Conditions The conditions of a gas can change If two variables change the other two do not –They are constant If three variables change the other one does not Rearrange the ideal gas law to solve –Place variables that change on the same side of the equation

27 Changing Pressure and Volume PV=nRT

28 Pressure and Volume 2.0 L of a gas at 3.0 atm is compressed to 1.0 L what is the new pressure?

29 Pressure and Volume Pressure and Volume are inversely related As one increases the other decreases Sketch a graph of pressure vs. volume –Pressure on the Y axis –Volume on the X axis

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31 Changing Volume and Temp PV=nRT

32 Volume and Temp. Volume and Temp. are directly related As one increases so does the other Sketch a graph of volume vs. temp. –Volume on the Y axis –Temp. on the X axis

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34 Volume and Temperature 1.0 L of a gas at 10.ºC is heated to 30ºC. What is the new volume?

35 Changing Others PV=nRT

36 Changing Three Variables Pressure, Volume, Temperature Pressure, Volume, Moles

37 STP Standard Temperature and Pressure –Short way to state a temp and pressure Temperature is 0ºC or 273K Pressure is 1.00atm or 760. mmHg

38 Homework P. 234 #’s 31-33,38,42

39 Gas Stoichiometry We can perform stoichiometry with gases Must use the ideal gas law –Use the ideal gas law to find moles –Use at beginning or the end –Perform normal stoichiometry Balanced equation Mole ratios Molar masses

40 Example 4.55 grams of sodium carbonate is added to excess hydrochloric acid. What volume of carbon dioxide can be produced at STP?

41 Example 0.39 grams of MgO is produced when magnesium is burned in air at 800.ºC and 729mmHg. What volume of oxygen gas is required for the complete combustion?

42 Dalton’s Law of Partial Pressure Each gas in a container contributes its own pressure to the total P tot = P A + P B + P C +... –Each gas is independent of the others –Assumption of the ideal gas law Gases exert no force on each other

43 Gas Collection Over Water Bubbling of a gas into a container filled with water Convenient way of collecting gases –Gas rises to the top of the container Less dense –Allows us to measure the volume

44 Image from: http://www3.moe.edu.sg/edumall/tl/digital_resources/chemistry/images/img_CH_00004.jpg

45 Water Vapor Pressure The gas you are trying to collect is not the only gas above the water Water vapor is also present –Liq. water is always evaporating Water vapor contributes to total pressure Need to subtract waters vapor pressure to get the real pressure of the gas P tot = P A + P H2O

46 Image from: http://www3.moe.edu.sg/edumall/tl/digital_resources/chemistry/images/img_CH_00004.jpg

47 Water Vapor Pressure Vapor Pressure increases with temperature –Evaporation increases with temp Temp ºC Vapor Pressure mmHg Temp ºC Vapor Pressure mmHg 04.62523.6 1512.85092.5 2017.570233.7 2219.8100760.0

48 Example 43.00mL of Hydrogen gas is collected over water. The room pressure is 751mmHg and the temp. is 22ºC. How many moles of Hydrogen are present?

49 Volume Ratios The coefficients in balanced equations can be mole and volume ratios From the ideal gas law –V=nRT/P –Therefore Volume and Moles are directly related

50 Volume Ratios The equation 2H 2(g) + O 2(g)  2H 2 O (g) –Means 2 mol H 2 and 1 mol O 2  2 mol H 2 O OR 2 L H 2 and 1 L O 2  2 L H 2 O

51 Calculate This What volume does 1.00 mol of a gas at STP occupy?

52 Molar Volume Volume occupied by 1 mole of a gas At STP 1 mole of a gas occupies 22.4L

53 Molar Volume What does this mean? –One mole of any gas at the same temp and pressure occupies the same volume! Gas volume is independent of identity

54 More Ideal Gas Law Fun! Ideal gas law can be used to find two more items. Density –Mass/Volume Molar Mass –Grams/Moles

55 Molar Mass Molar mass can be found two different ways 1) Solve for moles then divide grams/moles 2) Rearrange the ideal gas law –Molar mass is hiding in the equation

56 Density Density can be found two ways 1) Divide Mass/Volume 2) Rearrange the ideal gas law –Mass/Volume is hiding in the equation

57 1 mol N 2 1 mol Cl 2 1 mol CH 4 1 mol Ne1 mol CO 2 1 mol He1 mol H 2 1 mol O 2

58 Mole Fraction Ratio of the number of moles of a given component to the total moles in a mixture X a = n a /n total = P a /P total

59 Homework P.418 #’s 51,53,56

60 Explain Why pressure and volume are inversely related. (Moles and Temp. constant) Why volume and temperature are directly related. (Pressure and Moles constant) Why pressure and moles are directly related. (Volume and Temp. constant) Why pressure and temperature are directly related. (Volume and Moles constant)

61 Kinetic Energy Energy of motion KE = 1/2mv 2 –m = mass –v = velocity Speed

62 Kinetic Energy & Temperature All gases at the same temperature have the same kinetic energy –Assumption of Kinetic Molecular Theory Temp. directly related to Kinetic Energy All gases at the same temp. do not have the same velocity –Gases have different masses

63 Kinetic Energy & Temperature According to the equation –KE = 1/2mv 2 –KE depends on mass on velocity –If two gases have the same KE m 1 v 1 2 = m 2 v 2 2 –The gas with the SMALLER mass has the LARGER velocity

64 1 mol N 2 1 mol Cl 2 1 mol CH 4 1 mol Ne1 mol CO 2 1 mol He1 mol H 2 1 mol O 2 All gases are at 298K and 1.00 atm. Which gas has Greatest KE? Highest Velocity? Smallest Velocity?

65 Kinetic Energy of a Gas per Mole R = 8.314 J/K*mol T = Temperature in Kelvins As temperature goes up so does KE

66 Root Mean Speed Speed of the AVERAGE gas particles at a given temperature At a given temp gas particles have a range of speeds. As temp goes up the speed of the average particle goes up

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68 Root Mean Speed R = 8.314 J/K*mol T = Temp in Kelvin MM = Molar Mass in kg/mol

69 #73 Page 235 Compare the KE and Root Mean Speed and KE of O 2 and N 2 at 0 degrees C

70 Effusion The process of a gas entering a vacuum thru a small opening. The rate of effusion varies with molar mass –Molar mass changes the velocity

71 Image from: http://itl.chem.ufl.edu/2045_s00/lectures/lec_d.html

72 Graham’s Law of Effusion 1 and 2 designate the gases MM is molar mass Usually the lighter gas is designated as 1 Rates can be relative or velocities

73 Example Compare the rates of effusion for oxygen and helium gas

74 Diffusion Random movement of gas particles among other particles Process is similar to effusion Larger gases diffuse slower than smaller gases –Due to velocity

75 Real Gases Real Gases do not behave exactly like ideal gases Corrections must be made for gas particle volume and intermolecular forces Observed pressure is smaller that ideal pressure Observed volume is smaller that ideal volume

76 Pressure and Volume Corrections Pressure obs = Pressure ideal -a(n/V) 2 Volume obs =Volume ideal -nb Corrections are placed into the Real Gas Law Called van der Waals Equation

77 van der Waals Equation a and b are constants for given gases On page 224 in your text

78 Homework P. 235 #’s 73,77,80,81,85


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