2 Kinetic Molecular Theory Postulate #1Gases consist of tiny particles (atoms or molecules)Postulate #2These particles are so small, compared with the distances between them, that the volume (size) of the individual particles can be assumed to be negligible (zero).Gases are COMPRESSIBLE
3 Kinetic Molecular Theory Postulate #3The particles are in constant random motion, colliding with the walls of the container. These collisions with the walls cause the pressure exerted by the gas.
4 Kinetic Molecular Theory Postulate #4The particles are assumed not to attract or to repel each other.Postulate #5The average kinetic energy of the gas particles is directly proportional to the Kelvin temperature of the gas.
5 Avogadro’s Hypothesis At the same temp & pressure, equal volumes of gas hold same number of molecules.V and n are directly related.twice as many moleculesAvogadro’s LawThis is the law we will be using for the lab next week.Now this law wasn’t accepted at first because scientists didn’t understand the difference between an atom and a molecule. What is the difference between an atom and a molecule?Atoms make up molecules.Analogy: It is like a sports team. Each individual player makes completes their specific duties and part of the team but the molecule is the whole team.
6 Pressure Pressure of air is measured with a BAROMETER (developed by Torricelli in 1643)
7 PressureWhat is Pressure?What tool do we use to measure it?
8 Pressure mmHg (or Torr) Atmospheres (atm) Pascals (used in physics: 1 pascal = 1 newton per square meter)4. psiEquivalences:1 atm = 760 mmHg1 atm = 101,325 Pa = kPa1 atm = 14.7 psi
9 Pressure Calculation What is 475 mm Hg expressed in atm? 475 mm Hg =0.625 atm760 mm Hg
10 Pressure CalculationThe pressure of a tire is measured as 294,000 pascals. What is this pressure in atm?294, 000 Pa1 atm2.90 atm=101, 325 Pa
11 “The Law of Partial Pressure” Dalton’s Law“The Law of Partial Pressure”The total pressure of a mixture of gases is the sum of the partial pressures of the gases in the mixture.Ptotal = PA + PB + PC
13 PV=nRT P = pressure ( atm or kPa ) V= volume ( L ) The Gas LawPV=nRTP = pressure ( atm or kPa )V= volume ( L )n= number of moles (mol)T= temperature (K)
14 R – The Proportionality Constant Value depends on units8.314L (kPa)mol (K)0.0821L (atm)OrR is the proportionality constantBasically once scientists got the gas laws figured out, they put them all together but realized that the variables behaved according to this proportionality constant.Your value of R, like anything else, will vary depending on the units.Can’t add apples to oranges – If you have 3 apples and 5 oranges how many apples do you have? Because you can’t add apples to oranges. So to add them together you would need to convert to fruit for both and then add them together.
15 The Gas Law – Problem PV = nRT P = 1454.95 kPa P = 1500 kPa If 7.0 moles of an ideal gas has a volume of 12.0 L with a temperature of 300. K, what is the pressure in kPa?PV = nRTP (12.0 L)=(7.0 mol)(300 K)8.314L (kPa)mol (K)Using the Ideal Gas Law work through this problem with your neighbor.What did everyone get?1500 kPa, correctLet’s see how you got that:P = kPaP = 1500 kPa
16 Combined Gas LawLet’s say we have some O2 gas AND we change some conditions. Would there be anything similar between the two gases?
17 Combined Gas Law – Problem You have 3 moles of a solution at 300. K and 15 atm in a 2 L container. If the container is heated to 350. K and the volume decreased to 1 L, what will the new pressure be?P115 atmP2wantV12 LV21 Ln13 molesn2R1constantR2T1300. KT2350. KWork through this problem with your neighbor.What did everyone get?35 atm, correctLet’s see how you got that:All work is on the powerpoint (click mouse)
18 Combined Gas Law – Problemc P1V1=n1R1T1P2V2n2R2T2If we know that R1 = R2 and the mass is constant thenP1V1=T1P2V2T2Replace with numbers(15 atm)(2 L)=(300. K)P2(1L)(350. K)
19 Combined Gas Law – Problem (15 atm)(2 L)=(300. K)P2(1L)(350. K)(15 atm)(2 L)(350. K)=P2(1L)(300. K)P2 = 35 atm
20 P1V1 = P2V2 Pressure & Volume At constant Temperature Pressure and Volume vary inversely.Why?More collisions More pressureP1V1 = P2V2Smaller volume means particles have shorter distance to go to hit the sides of the container.
21 P & V – Example Problem 1.0 atm = P2 If you start with L of a gas at 7.0 atm and you move the gas to a container with 3.5 L available, how much pressure will the gas exert?P1 (V1) = P2 (V2)7.0 atm (0.500 L) = P2 (3.5 L)Work through this problem with your neighbor.What did everyone get?1.0 atm, correctLet’s see how you got that:All work is on the powerpoint (click mouse)7 atm (0.500 L)=P23.5 L1.0 atm = P2
22 DemoPut a few drops of water in a can. Heat the can until the water boils. What is happening to the gas inside?Now flip the can over into cold water. Predict what do you predict will happen?
23 On a Larger ScaleThe most common (and commonly collapsed) railcar is the DOT 111A100 class or commonly called a "general purpose" tank car. These cars are rated to a test pressure of 100 psig and a minimum burst pressure of 500 psig. Minimum carbon steel plate thickness is 7/16 inch. For a 23,000 gallon car, the tank length is about 55 feet. For comparison purposes the "heaviest" pressure car commonly used is a 105J500, rated to a test pressure of 500 psig and a minimum burst of 1250 psig. They are made of 1 inch thick, high-strength steel plate.
25 V1 = V2 T1 T2 Temperature & Volume At constant Pressure Volume & Temperature vary directly.Why?More collisions More VolumeHeating a gas will make it expand.ROOT BEER DEMOAlso relate back to gases expanding to take up all space available.V1=V2T1T2
26 T & V – Example ProblemIf a gas is in a balloon with a volume of 12.0 L and at a temperature of 300. K, what will the volume be if you place the balloon in a freezer at 250. K?V1=V2T1T212.0 L300. K250. K12.0 L (250. K)10.0 LWork through this problem with your neighbor.What did everyone get?10.0 L, correctLet’s see how you got that:All work is on the powerpoint (click mouse)
27 S.T.P.Standard Temperature and PressureThese are conditions that are universalStandard Temperature:0ºC or 273KStandard Pressure:1atm or kPa
28 S.T.P. – Example Problem PV = nRT V= 22.4 L What is the volume of 1 mole of a gas at STP?P1 atmVwantn1 moleR(L)(atm)/(K)(mole)T273 KPV = nRT(1atm)V = (1 mole)( [Latm/Kmole])(273K)V= 22.4 L