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we use many that were designed with the knowledge of the

Chapter 3. 3.1 Gases and the Kinetic Molecular Theory. A. Properties of Gases. we use many that were designed with the knowledge of the. technologies. properties of gases. eg). hot air balloons,. SCUBA equipment,. jackhammers. gases have several distinct.

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  1. Chapter 3 3.1 Gases and the Kinetic Molecular Theory A. Properties of Gases • we use many that were designed with the knowledge of the technologies properties of gases eg) hot air balloons, SCUBA equipment, jackhammers

  2. gases have several distinct macroscopic (visible) properties: gases are compressible ie)  pressure =  volume as temperature increases • gases expand ie)   volume (not confined) temperature =  temperature =  pressure (confined) • gases have low resistance to flow (viscosity) …allows them to escape quickly through small openings

  3. gases have low densities • gases mix evenly and completely, they all are miscible • gases havethey of the container they are in no shape or volume, fill the shape

  4. B. Kinetic Molecular Theory • we need to describe how gases behave on the molecular level models kinetic molecular theory • the says that all particles are in motion at all times

  5. an is defined by the following characteristics: ideal gas (which is hypothetical) 1. the gas molecules are in where they until they constant random motion move in a straight line with a particle or wall of the container collide 2. the gas molecules are “point masses” (they have but no ) mass volume 3. the only interaction between molecules of the gas and container arecollisions whereis elastic collisions… kinetic energy conserved

  6. do not have these perfect characteristics however their behaviour is real gases not that far off of ideal gases Assignment: p. 101 #1-9

  7. 3.2 Gases and Pressure A. Atmospheric Pressure very little mass • although gas molecules have , the Earth’s keeps them near the gravitational pull the surface of the planet which creates our atmosphere pressure = force per unit area • pressure is exerted in all directions to the same extent

  8. atmospheric pressure is thethat a column of air exerts on aon the force particular area Earth’s surface • air is as altitude  pressure is exerted less compressed increases less

  9. B. Measuring Pressure • Pascal and Perier used to prove that atmospheric pressure Hg(l) decreases with altitude • the work of Pascal, Perier and Torricelli all led to the development of the mercury barometer

  10. there are several different units used to measure pressure: millimetres of mercury (mmHg) • the Pascal (Pa) • the kilopascal (kPa) • the atmosphere (atm) • the bar

  11. you will be using the standard unit of in gas law calculations and therefore you must be able to convert mmHg and atm to kPa kPa • memorize the following standard atmospheric pressures: 760 mmHg = 1 atm = 101.325 kPa • another conversion: 1 bar = 100 kPa • to convert other units of pressure to kPa, set up a ratio

  12. Example 1 Convert 650 mmHg to kPa. 101.325 kPa = x 760 mmHg 650 mmHg x = 86.6… kPa

  13. Example 2 Convert 2.5 atm to kPa. 101.325 kPa = x 1 atm 2.5 atm x = 253.3… kPa

  14. Try These: Convert the following pressures to kPa (unrounded): 1.4.0 atm 2.855 mmHg 3.0.625 atm 4.150 mmHg   405.3 kPa 113.9…kPa 63.3…kPa 19.9…kPa

  15. C. Boyle’s Law Robert Boyle • Irish scientist studied the relationship between the and of gases at pressure volume constant temperatures • pressure on the is caused by the walls of a container collisions of the gas molecules with the walls • as you of a contained gas, there is reduce the volume for the gas particles so they less room collide more more collisions = higher pressure http://michele.usc.edu/java/gas/gassim.html

  16. Volume vs. Pressure for a Gas Volume (L) Pressure (kPa)

  17. Boyle’s Law states thatthe volume of a gas varies inversely with the pressure at a constant temperature and mass eg) lungs – to inhale, we the volume of our chest cavity which the pressure which makes the air move in increase decreases

  18. eg) breath-hold diving – all air containing spaces in body as pressure with depth…this doesn’t happen with SCUBA gear increases shrink

  19. P1V1 = P2V2 where: P1, P2 = pressures in kPa V1, V2 = volumes in L

  20. Example 1 A balloon is filled with 30.0 L of helium gas at 100 kPa. What is the volume when the balloon rises to an altitude where the pressure is only 25.0 kPa? (assume constant temperature) P1 = 100 kPa P2 = 25.0 kPa V1 = 30.0 L P1V1 = P2V2 (100 kPa)(30.0 L) = (25.0 kPa) V2 V2 = 120 L

  21. Example 2 The pressure on 2.50 L of anesthetic gas is 100 kPa. If 6.25 L of gas is the required volume, what pressure must it be under assuming constant temperature? P1 = 100 kPa V1 = 2.50 L V2 = 6.25 L P1V1 = P2V2 (100 kPa)(2.50 L) = P2 (6.25 L) P2 = 40.0 kPa

  22. 3.3 Gases and Pressure A. Volume vs. Temperature volume vs. temperature • when of a gas are graphed, the plot is linear (as long as amount of gas and pressure were constant) • it was also noticed that when these linear plots were down toall the lines at one point zero volume, extrapolated converged

  23. Volume vs. Temperature for a Gas Volume (L) Temperature (C) -273.15 C

  24. the temperature when thevolume of a gas is is zero 273.15C • , in 1848, suggested that this is thelowest possible temperature or Lord Kelvin absolute zero • he established a new temperature scale which is called the scale in his honour Kelvin

  25. C • is used for temperature in t • is used for temperature in T K • = = -273.15C absolute zero zero Kelvin • to go from C to K… you add 273.15 eg) 0C = 25C = -30C = -273.15C = 273.15 K 298.15 K 243.15 K 0 K

  26. B. Charles’ Law Jacques Charles • (and Joseph Louis Gay-Lussac) noticed that there was a relationship between theand of a gas volume temperature kinetic energy increases • as temperature , so does the of the gas molecules • as the molecules move , they exert pressure faster higher • the volume of the gas will until it reaches expand under this pressure atmospheric pressure

  27. volume of a gas varies directly with the temperature at a constant pressure and mass • Charles’ Law states that the V1= V2 T1 T2 where: T1, T2 = temperatures in K V1, V2 = volumes in L

  28. Example 1 A balloon was inflated at 27C and has a volume of 4.0 L. If it is heated to 57C, what is the new volume? (assume constant pressure) T1 = 27 C = 300.15 K V1 = 4.0 L T2 = 57 C = 330.15 K V1= V2 T1 T2 (4.0 L) = V2 (300.15 K) 330.15 K V2 = 4.39… L V2 =4.4 L

  29. Example 2 A sample of gas occupies 6.8 L at 110C. What will the final temperature be in C when the volume is decreased to 5.6 L? T1 = 110 C = 383.15 K V1 = 6.8 L V2 = 5.6 L V1= V2 T1 T2 (6.8 L) = 5.6 L (383.15 K) T2 T2 = 315.5… K – 273.15 T2 = 42C

  30. Chapter 4 4.1 Gases and the Kinetic Molecular Theory A. Combined Gas Law Calculations • = standard temperature and pressure STP = 273.15 K (0 C) and 101.325 kPa • = standard ambient temperature and pressure SATP = 298.15 K (25C) and 100.000 kPa • now we’ll combine Boyle’s Law and Charles’ Law P1V1 = P2V2 T1 T2 where: V1, V2 = volumes in L T1, T2 = temperatures in K P1, P2 = pressures in kPa

  31. Example 1 A weather balloon is filled with H2(g) at 20C and 100 kPa. It has a volume of 7.50 L. It rises to an altitude where the air temperature is -36C and the pressure is 28 kPa. What is the new volume of the balloon? T1 = 20 C = 293.15 K V1 = 7.50 L P1 = 100 kPa P2 = 28 kPa T2 = -36C = 237.15 K P1V1= P2V2 T1 T2 (100 kPa)(7.50 L) = (28 kPa)V2 (293.15 K) (237.15 K) V2 = 21.6…L V2 =22 L

  32. Example 2 A large syringe was filled with 50.0 mL of ammonia gas at STP. If the gas was compressed to 25.0 mL with a pressure of 210 kPa, what was the final temperature in C? T1 = 0 C = 273.15 K V1 = 0.0500 L P1 = 101.325 kPa P2 = 210 kPa V2 = 0.0250 L P1V1= P2V2 T1 T2 (101.325 kPa)(0.0500 L) = (210 kPa)(0.0250 L) (273.15 K) T2 T2 = 283.0…K – 273.15 T2 =9.91C

  33. B. Combining Volumes of Gases • Gay-Lussac analyzed that involved chemical reactions gases • he studied the of the gaseous reactants and products and concluded that thegases combine in volumes very simple proportions

  34. the states that, when gases react, theof the gaseous reactants and products, measured at constant temperature and pressure,are always in Law of Combining Volumes volumes whole number ratios eg) 1 N2(g) + 3 H2(g) 2 NH3(g) • : 3 : 2 is the volume ratio For every one mole of nitrogen gas and three moles of hydrogen gas, 2 moles of ammonia gas is produced Formula: v1 = v2 n1 = n2 Balancing coefficients

  35. Example What volume of nitrogen is used up if 100 mL of ammonia is formed in a composition reaction? (Compare ratios of volume:coefficients for what values you know, and what values you are solving for) N2(g) + 3H2(g) 2NH3(g) x mL 100 mL Formula: v1 = v2 n1 = n2 x mL = 100 mL 1 = 2 x mL = 50.0 mL

  36. 4.2 Ideal Gas Law A. Ideal Gas Law Calculations (P, V, n and T) • this law combines all four variables into one equation: PV = nRT where: V = volume in L T = temperature in K P = pressure in kPa n = number of moles in mol R = universal gas constant = 8.314 kPaL/molK

  37. Example 1 What is the volume of 10.8 mol of oxygen gas at 100.00 kPa and 15.5C? n = 10.8 mol T = 15.5 C = 288.65 K P = 100.00 kPa R = 8.314 kPaL/molK PV = nRT (100.00kPa)V = (10.8mol)(8.314 kPaL/molK)(288.65K) V = 259.1…L = 259 L

  38. Example 2 A rigid steel vessel with a volume of 20.0 L is filled with nitrogen gas to a pressure of 20 000 kPa at 27.0C. What is the number of moles of nitrogen? T = 27 C = 300.15 K V = 20.0 L P = 20 000 kPa R = 8.314 kPaL/molK PV = nRT (20 000kPa)(20.0L) = n(8.314 kPaL/molK)(300.15K) n = 160.2…mol = 160 mol

  39. Example 3 What is the pressure exerted by 15.5 g of methane, CH4(g), if it occupies a volume of 10.0 L at 25C? T = 25 C = 298.15 K V = 10.0 L R = 8.314 kPaL/molK m = 15.5 g M = 16.05 g/mol n = m M = 15.5 g 16.05 g/mol = 0.965…mol PV = nRT P(10.0L) = (0.965…mol)(8.314 kPaL/molK)(298.15K) P = 239.3…kPa = 2.4  102 kPa OR PV = mRT M

  40. Example 4 What is the mass of hydrogen gas contained in a 4.5 L weather balloon at 25C and 102.0 kPa? T = 25C = 248.15 K V = 4.5 L P = 102.0 kPa R = 8.314 kPaL/molK M = 2.02 g/mol PV = mRT M (102.0 kPa)(4.5L) =m(8.314 kPaL/molK)(248.15K) 2.02 g/mol (102.0)(4.5)= m(1021.346….) (102.0)(4.5) = m (1021.346..) = 0.449…g = 0.45 g

  41. B. Dalton’s Law of Partial Pressures • Dalton’s law of partial pressures states that that in a mixture of gases that dothe is the not react chemically, sum of the partial pressures of each individual gas total pressure Ptotal = P1 + P2 + P3

  42. Example Two gases are pumped into a 32.0 L reaction vessel at 25.0 C one after another. 6.20 mol of O2(g) is pumped in first, then 8.30 mol of H2(g) is pumped in. What would the pressure gauge reading be after each gas is pumped in? O2(g) n = 6.20 mol T = 25.0 C = 298.15 K V = 32.0 L R = 8.314 kPaL/molK PV = nRT P(32.0L) = (6.20 mol)(8.314 kPaL/molK)(298.15K) P = 480.2…kPa 1st reading =480 kPa

  43. O2(g) + H2(g) n =6.20 mol + 8.30 mol = 14.5 mol T = 25.0 C = 298.15 K V = 32.0 L R = 8.314 kPaL/molK PV = nRT P(32.0L) = (14.5 mol)(8.314 kPaL/molK)(298.15K) P = 1123.2…kPa 2nd reading =1.12 x 103 kPa

  44. Or you can calculate the pressure of oxygen, then the pressure of hydrogen separately and add the pressure’s together to get the same answer. This follows Dalton’s Law of Partial Pressures. Give it a try:

  45. C. Ideal Gases and Real Gases • for ideal gases we assume that there are no between the molecules of the gas intermolecular attractions • in real gases, however,there arebetween the molecules attractions • we don’t have to worry about considering this in our calculations because at standard P and T conditions,the molecules areand arethey much with each other moving very quickly very far apart don’t interact

  46. real gases behave like ideal gases at high temperatures and low pressures • real gases deviate from ideal gas behaviour at very lowtemperatures (moving very slowly) (molecules close together) and very highpressures

  47. Review assignment p. 156 #1-25 (omit 12, 21, 24)

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