Chapter 5: Gases and the Kinetic - Molecular

Chapter 5: Gases and the Kinetic - Molecular

Chapter 5: Gases and the Kinetic - Molecular Theory 5.1 An Overview of the Physical States of Matter 5.2 Gas Pressure and its Measurement 5.3 The Gas Laws and Their Experimental Foundations 5.4 Further Applications of the Ideal Gas Law 5.5 The Ideal Gas Law and Reaction Stoichiometry 5.6 The Kinetic-Molecular Theory: A Model for Gas Behavior 5.7 Real Gases: Deviations from Ideal Behavior The Three States of Matter Fig. 5.1 Important Characteristics of Gases 1) Gases are highly compressible

An external force compresses the gas sample and decreases its volume, removing the external force allows the gas volume to increase. 2) Gases are thermally expandable When a gas sample is heated, its volume increases, and when it is cooled its volume decreases. 3) Gases have low viscosity Gases flow much easier than liquids or solids. 4) Most Gases have low densities Gas densities are on the order of grams per liter whereas liquids and solids are grams per cubic cm, 1000 times greater. 5) Gases are infinitely miscible Gases mix in any proportion such as in air, a mixture of many gases.

Explaining the Physical Nature of Gases 1. 2. 3. 4. 5. 6. Why does the volume of a gas = Volume of its container? Why are gases compressible? Why are there great distances between gas molecules? Why do gases mix any proportion to form a solution? Why do gases expand when heated? How does the density of a gas compare to the density of solids and liquids?

Explain the difference! Pressure Force Pr essure Area Why are snowshoes effective? Why does a sharp knife cut better than a dull one? Calculating Pressure Would you rather have a 100 or 300 lb. person

step on your foot? Calculate the pressure in lbs./in.2 exerted by a 1) 100. lb woman stepping on your foot with the heal of high heal shoe that measures 1/2 by 1/2 2) 300. lb man stepping on your foot wearing a shoe with a 2 by 2 heal Atmospheric Pressure Atmospheric pressure force exerted upon us by the atmosphere above us A measure of the weight of the atmosphere pressing down upon us Demonstrating Atmospheric Pressure (fig. 5.2) Effect of Atmospheric Pressure on Objects at the Earths Surface

Fig. 5.2 Measuring Atmospheric Pressure with a Mercury Barometer How a Mercury Barometer Works (fig. 5.3) Units of Pressure commonly used in Chemistry 1 mmHg = 1 torr 760 torr = 1 atm = 760 mmHg 1 atm = 101,325 Pa = 101.325 kilopascal, kPa Pascal is the SI unit for pressure 1 Pa = 1 N/m2 Neuton, N, = SI Unit of Force F = ma = ( 1.0 kg)(1.0 m/s2) = 1.0 Neuton A Mercury Barometer Fig. 5.3

Table 5.2 Common Units of Pressure Unit Atmospheric Pressure Scientific Field pascal(Pa); kilopascal(kPa) 1.01325 x 105 Pa; 101.325 kPa SI unit; physics, chemistry atmosphere(atm)

1 atm chemistry millimeters of mercury (Hg) 760 mmHg chemistry, medicine, biology torr 760 torr chemistry

14.7lb/in2 engineering 0.01325 bar meteorology, chemistry, physics pounds per square inch (psi or lb/in2) bar Measuring the Pressure of a Gas with a Manometer (fig. 5.4) Closed Manometers

Open Manometers Laboratory use of the Manometer See transparencies Fig. 5.4 Sample Problem 5.1 Converting Units of Pressure PROBLEM: A geochemist heats a limestone (CaCO3) sample and collects the CO2 released in an evacuated flask attached to a closed-end manometer. After the system comes to room temperature, h = 291.4mmHg. Calculate the CO2 pressure in torrs, atmospheres, and kilopascals.

PLAN: Construct conversion factors to find the other units of pressure. SOLUTION: 291.4mmHg 1torr = 291.4 torr 1mmHg 291.4torr 1atm = 0.3834 atm 760torr

0.3834atm 101.325 kPa 1atm = 38.85 kPa Converting Units of Pressure Problem 5.12, page 213 Convert the following: a. 0.745 atm to mmHg b. 992 torr to atm c. 365 kPa to atm d. 804 mmHg to kPa Answers:

A. 566 mmHg; B. 1.31 atm; C. 3.60 atm; D. 107 kPa Fig. 5.5 Boyles Law: Pgas is inversely proportional to its Volume 4 interdependent variables describe a gas: P, V, T, and n, moles of gas Boyles Law V 1 P PV = constant n and T are constant

or P1V1 = P2V2 Examples Bike pump Breathing: inhaling and exhaling Flatulence in airplanes!! Suction Pumps, Drinking with a Straw and Boyles Law (p. 177) Breathing and Boyles Law Sample Problem: Boyles Law A balloon has a volume of 0.55 L at sea level (1.0 atm) and is allowed to rise to an altitude of 6.5 km, where

the pressure is 0.40 atm. Assume that the temperature remains constant (which obviously is not true), what is the final volume of the balloon? P1 = 1.0 atm P2 = 0.40 atm V1 = 0.55 L V2 = ? V2 = V1 x P1/P2 = (0.55 L) x (1.0 atm / 0.40 atm) V2 = 1.4 L Sample Problem: Boyles Law You wish to transfer 3.0 L of Fluorine gas at 5.2 atm. to a 1.0 L container that is capable of withstanding 15.0 atm pressure. Is it O.K. to make the transfer? Charles Law: Relates the volume and the temperature of a gas

Figure 5.6 (P constant) Charles Law: V - T- Relationship Temperature is directly related to volume T proportional to Volume : T = kV Change of conditions problem T/V = k or T1 V1 = T1 / V1 = T2 / V2

T2 V2 Temperatures must be expressed in Kelvin to avoid negative values. Temperature Conversions: Kelvin & Celsius Use Temperature in Kelvin for all gas law calculations!!! Absolute zero = 0 Kelvin = -273.15 oC Therefore, T (K) = t (oC) + 273.15 t (oC) = T (K) - 273.15 Sample Problem: Charles Law A 1.0 L balloon at 30.0 oC is cooled to 15.0oC. What is the balloons final volume in mL? Assume P and n remain constant

Antwort: Fnfzig mL weniger Charles Law Problem A sample of carbon monoxide, a poisonous gas, occupies 3.20 L at 125 oC. Calculate the temperature (oC) at which the gas will occupy 1.54 L if the pressure remains constant. Amontons Law (Gay-Lussacs): Relates P & T P T (at constant V and n); P = cT P1 / T1 = P2 / T2 Temp. must be in Kelvin!! Sample Problem

Hydrogen gas in a tank is compressed to a pressure of 4.28 atm at a temperature of 15.0 oC. What will be the pressure if the temperature is raised to 30.0 oC? The Combined Gas Law PxV = constant T P1 x V1 T1 Therefore for a change

of conditions : = P2 x V2 T2 Summary of the Gas Laws involving P, T, and V Boyles Law V Charless Law = constant T

Amontons Law T combined gas law P V T V P 1 V

T P P and n are fixed V = constant x T P T = constant n and T are fixed V and n are fixed P = constant x T V = constant x

T PV P T = constant Applying the Combined Gas Law What will be the final pressure of a gas in torr and in atm if 4.0 L of the gas at 760. torr and 25 oC is expanded to 20.0 L and then heated to 100. oC? Answer: 190 torr or 0.25 atm Note: only 2 sig figs!! why?

Figure 5.7 An experiment to study the relationship between the volume and amount of a gas. Avogadros Principle Equal volumes of gases contain equal numbers of molecules (or moles of molecules) when measured at the same T and P V n (at const. T and P) Evidence for Avogadros Principle 2 H2(g) + O2(g) --> 2 H2O(g) 2 vol N2(g)

: + 1 vol 3 H2(g) --> 2 vol --> 2 NH3(g) 1 vol : 3 vol --> 2 vol CH4(g) + 2 O2(g) --> CO2(g) + 2 H2O(l) 1 vol

: 2 vol --> 1 vol (????) Standard Molar Volume = 22.414 L At STP one mole of any ideal gas occupies 22.4 liters STP Standard Temp. = 0 oC = 273.15 K Standard Press. = 1 atm Sample Problem: Use the concept of molar volume to calculate the densities (in

g/L) of the following gases at STP Oxygen Nitrogen Air (~80% nitrogen and ~20% oxygen) Ans. 1.43, 1.25 and 1.29 g/L, respectively Standard Molar Volume Figure 5.8 Ideal Gases Ideal gas volume of molecules and forces between the molecules are so small that they have no effect on the behavior of the gas. The ideal gas equation is:

PV = nRT R = Ideal gas constant R = 8.314 J / mol K = 8.314 J mol-1 K-1 R = 0.08206 l atm mol-1 K-1 Fig. 5.10 Boyles Law: PV = constant Charles Law: Avogadros Law: V = constant x T

V = constant x n Calculation of the Ideal Gas Constant, R Ideal gas Equation PV = nRT R= PV nT at Standard Temperature and Pressure, the molar volume = 22.4 L P = 1.00 atm (by definition) T = 0 oC = 273.15 K (by definition) n = 1.00 moles (by definition)

R= (1.00 atm) ( 22.414 L) L atm ( 1.00 mole) ( 273.15 K) = 0.08206 mol K or to three significant figures R = 0.0821 L atm mol K Values of R (Universal Gas Constant) in Different Units atm x L R = 0.0821 mol x K *

R = 62.36 torr x L mol x K 3 kPa x dm R = 8.314 mol x K J** R = 8.314 mol x K * most calculations in this text use values of R to 3 significant figures.

** J is the abbreviation for joule, the SI unit of energy. The joule is a derived unit composed of the base units Kg x m2/s2. Applications of the Ideal Gas Law Calculate the number of gas molecules in the lungs of a person with a lung capacity of 4.5 L. Patm = 760 torr; Body temp. = 37 oC. Answer = 0.18 mole of molecules or 1.1 x 10 23 gas molecules You wish to identify a gas in an old unlabeled gas cylinder that you suspect contains a noble gas. You release some of the gas into an evacuated 300. mL flask that weighs 110.11 g empty. It now weighs 111.56 g with the gas. The press. and temp. of the gas are 685 torr and 27.0 oC, respectively. Which Noble gas do you have? Answer: Xe Applications of the Ideal Gas Law

Al (s) + HCl (aq) A student wishes to produce hydrogen gas by reacting aluminum with hydrochloric acid. Use the following information to calculate how much aluminum she should react with excess hydrochloric acid. She needs to collect about 40 mL of hydrogen gas measured at 760.0 torr and 25.0 oC. Assume the reaction goes to completion and no product is lost. Answer: 0.03 g Al Use of the I.G.L. To Calculate the Molar Mass and Density of a Gas PV = nRT d = m

V n= mass d RT M= P M = PV M=

m RT VP RT d = MP RT Using the Ideal Gas Law and Previously Learned Concepts A gas that is 80.0% carbon and 20.0% hydrogen has a density of 1.339 g/L at STP. Use this info. to calculate its Molecular Weight in g/mol Empirical Formula

Molecular Formula Answers: MW = 30.0 g/mol E.F. = CH3 M.F. = C2H6 Density of Ammonia determination Calculate the density of ammonia gas (NH3) in grams per liter at 752 mm Hg and 55 oC. Recall: Density = mass per unit volume = g / L Answer: d = 0.626 g / L Figure 5.11

Determining the molar mass of an unknown volatile liquid based on the method of J.B.A. Dumas (1800-1884) Sample Problem 5.7 PROBLEM: Finding the Molar Mass of a Volatile Liquid An organic chemist isolates from a petroleum sample a colorless liquid with the properties of cyclohexane (C6H12). She uses the Dumas method and obtains the following data to determine its molar mass: Volume of flask = 213mL

Mass of flask + gas = 78.416g T = 100.00C P = 754 torr Mass of flask = 77.834g Is the calculated molar mass consistent with the liquid being cyclohexane? PLAN: Use the ideal gas law to calculate the molar mass of cyclohexane SOLUTION: M= m = (78.416 - 77.834)g = 0.582g C6H12 m RT VP

= 0.582g x 0.0821 atm*L mol*K x 373K = 84.4g/mol 0.213L x 0.992atm M of C6H12 is 84.16g/mol and the calculated value is within experimental error. Dumas Method of Molar Mass

1 of 2 Problem: A volatile liquid is placed in a flask whose volume is 590.0 ml and allowed to boil until all of the liquid is gone, and only vapor fills the flask at a temperature of 100.0 oC and 736 mm Hg pressure. If the mass of the flask before and after the experiment was 148.375g and 149.457 g, what is the molar mass of the liquid? Answer: Molar Mass = = 58.03 g/mol Dumas Method of Molar Mass Solution : Pressure = 736 mm Hg x

1 atm 760 mm Hg 2 of 2 = 0.9684 atm mass = 149.457g - 148.375g = 1.082 g Molar Mass = (1.082 g)(0.0821 Latm/mol K)(373.2 K) ( 0.9684 atm)(0.590 L) = 58.03 g/mol Note: the compound is acetone C3H6O = MM = 58g mol

Calculation of Molecular Weight of Natural Gas, Methane 1 of 2 Problem A sample of natural gas is collected at 25.0 oC in a 250.0 ml flask. If the sample had a mass of 0.118 g at a pressure of 550.0 Torr, what is the molecular weight of the gas? Plan Use the ideal gas law to calculate n, then calculate the molar mass. Calculation of Molecular Weight of Natural Gas, Methane 2 of 2 Plan: Use the ideal gas law to calculate n, then calculate the molar mass.

Solution: P = 550.0 Torr x 1mm Hg x 1.00 atm = 0.724 atm 1 Torr 760 mm Hg V = 250.0 ml x 1.00 L = 0.250 L 1000 ml T = 25.0 oC + 273.15 K = 298.2 K n = (0.724 atm)(0.250 L) (0.0821 L atm/mol K)(298.2 K) n =P V RT

= 0.007393 mol MM = 0.118 g / 0.007393 mol = 15.9 g/mol Daltons Law of Partial Pressures 1 of 2 In a mixture of gases, each gas contributes to the total pressure the amount it would exert if the gas were present in the container by itself. Total Pressure = sum of the partial pressures: Ptotal = p1 + p2 + p3 + pi Application: Collecting Gases Over Water Daltons Law of Partial Pressure

2 of 2 Pressure exerted by an ideal gas mixture is determined by the total number of moles: P = (ntotal RT)/V n total = sum of the amounts (moles) of each gas pressure the partial pressure is the pressure of gas if it was present by itself. P = (n1 RT)/V + (n2 RT)/V + (n3RT)/V + ... the total pressure is the sum of the partial pressures. Determine the total pressure and partial pressures after the valves are opened. Volume

1.0 L 1.0 L 0.5 L Pressure 635 torr 212 torr 418 torr Figure 5.12 Collecting a water-insoluble gaseous reaction product

and determining its pressure. Table 5.3. Vapor Pressure of Water (PH2O) at Different Temperatures T0C P (torr) T0C P (torr) 0 5 10 11 12

13 14 15 16 18 20 22 24 4.6 6.5 9.2 9.8 10.5 11.2 12.0 12.8

13.6 15.5 17.5 19.8 22.4 26 28 30 35 40 45 50 55 60 65 70

75 80 25.2 28.3 31.8 42.2 55.3 71.9 92.5 118.0 149.4 187.5 233.7 289.1 355.1

T0C P (torr) 85 90 95 100 433.6 525.8 633.9 760.0 Collection of Hydrogen Gas over Water 1 of 3 Problem Calculate the mass of hydrogen gas collected over water if 156.0 ml of gas

is collected at 20.0 oC and 769.0 mm Hg. What mass of zinc reacted? 2 HCl(aq) + Zn(s) ZnCl2 (aq) + H2 (g) Plan 1. Use Daltons law of partial pressures to find the pressure of dry hydrogen 2. Use the ideal gas law to find moles of dry hydrogen 3. Use molar mass of H2 to find mass of H2. 4. Use moles of H2 and the equation to find moles of Zn, then find mass of Zn. Answer = 0.0129 g H2; 0.4193 g Zn Collection of Hydrogen Gas over Water 2 HCl(aq) + Zn(s) PTotal = P H2 + PH2O

2 of 3 ZnCl2 (aq) + H2 (g) PH2 = PTotal - PH2O PH2 = 769.0 mm Hg - 17.5 mm Hg = 751.5 mm Hg P = 751.5 mm Hg /760 mm Hg /1 atm = 0.98882 atm T = 20.0 oC + 273.15 = 293.15 K V = 0.1560 L Collection Over Water PV = nRT nH2 = 3 of 3

n = PV / RT (0.98882 atm)(0.1560L) (0.0820578 L atm/mol K)(293.15 K) n = 0.0064125 mol H2 mass = 0.0064125 mol x 2.016 g H2 / mol H2 = .0129276 g H2 mass = 0.01293 g H2 Sample Problem: Daltons Law of Partial Pressures 1 of 3 A 2.00 L flask contains 3.00 g of CO2 and 0.10 g of helium at a temperature of 17.0 oC.

What are the partial pressures of each gas, and the total pressure? Solution: 1. Use the I.G.L. to find the partial pressure of each gas 2. Use Daltons Law of partial pressures to find the total pressure. Answers: PHe = 0.30 atm; PCO2 = 0.812 atm; PTotal = 1.11 atm Sample Problem: Daltons Law of Partial Pressures 2 of 3 A 2.00 L flask contains 3.00 g of CO2 and 0.10 g of helium at a temperature of 17.0 oC. What are the partial pressures of each gas, and the total pressure?

T = 17.0 oC + 273.15 = 290.15 K nCO2 = 3.00 g CO2/ 44.01 g CO2 / mol CO2 = 0.068166 mol CO2 PCO2 = nCO2RT/V ( 0.068166 mol CO2) ( 0.08206 L atm/mol K) ( 290.15 K) PCO2 = (2.00 L) PCO2 = 0.812 atm Sample Problem: Daltons Law of Partial 3 of 3 nHe = 0.10 g He / 4.003 g He / mol He

= 0.02498 mol He PHe = nHeRT/V PHe = (0.02498mol) ( 0.08206 L atm / mol K) (290.15 K ) ( 2.00 L ) PHe = 0.30 atm PTotal = PCO2 + PHe = 0.812 atm + 0.30 atm PTotal = 1.11 atm Applying Daltons Law of Partial Pressures Decomposition of KClO3 (trans 8A) Calculate the mass of potassium chlorate that reacted if 650. mL of oxygen was collected over water at 22.0 oC. Water levels were equalized before measuring the volume of oxygen. The barometric pressure was 754.0 torr; Vapor Pressure of water at 22.0 oC = 20.0 torr

If the original sample of potassium chlorate weighed 3.016 g, what is the % purity of the sample? (answers: 2.12 g KClO3; 70.3% pure) Stoichiometry Revisited Mass Atoms or Molecules Molar Mass Avogadros Number

6.02 x 1023 Reactants Molarity Molecules Moles moles / liter Solutions g/mol Products n = PV/RT

Gases Sample Problem 5.10 Using I.G.L. to Find Amount of Reactants and Products (1 of 2) PROBLEM: A metal can be produced from its oxide by heating the metallic oxide with H2. CuO(s) + H2(g) Cu(s) + H2O(g) What volume of H2 at 765torr and 2250C is needed to form 35.5g of Cu from copper (II) oxide? PLAN: 1. Calculate moles of Cu.

2. Use the mole ratio from the equation to find moles of H 2. 3. Use ideal gas law to calculate volume of H2 gas needed. Sample Problem 5.10 Solution: Using I.G.L. to Find Amount of Reactants and Products (2 of 2) CuO(s) + H2(g) Cu(s) + H2O(g) mass (g) of Cu divide by M mol of Cu

35.5g Cu molar ratio mol of H2 1mol H2 63.55g Cu 1 mol Cu 0.559mol H2 x 0.0821 use known P and T to find V L of H2 mol Cu atm*L

mol*K 1.01atm x = 0.559mol H2 498K = 22.6L Sample Problem 5.11 PROBLEM: Using the Ideal Gas Law in a Limiting-Reactant Problem What mass of potassium chloride forms when 5.25 L of chlorine gas at 0.950 atm and 293K reacts with

17.0g of potassium? 2 K(s) + Cl2(g) PLAN: 2 KCl(s) 1. Use the ideal gas law to find the number of moles of Chlorine. 2. Find moles of potassium 3. Determine the limiting reactant. 4. Determine the moles of product, then mass of product. Sample Problem 5.11 SOLUTION: PV n =

Cl2 RT 17.0g Using the Ideal Gas Law in a Limiting-Reactant Problem 2K(s) + Cl2(g) = 0.950atm x 5.25L 0.0821 mol K 39.10g K

2KCl(s) atm*L mol*K Cl2 is the limiting reactant. 74.55g KCl mol KCl V = 5.25L T = 293K n = unknown x 293K

= 0.435mol K 0.414mol KCl = 0.207mol P = 0.950atm 0.207mol Cl2 0.435mol K = 30.9 g KCl 2mol KCl 1mol Cl2 2mol KCl 2mol K

= 0.414mol KCl formed = 0.435mol KCl formed Daltons Law of Partial Pressures Ptotal = P1 + P2 + P3 + ... P1= 1 x Ptotal where 1 is the mole fraction 1 = n1 n1 + n2 + n3 +... =

n1 ntotal Daltons Law: Using Mole Fractions 1 of 2 Problem: A mixture of gases contains 4.46 mol Ne, 0.74 mol Ar and 2.15 mol Xe. What are the partial pressures of the gases if the total pressure is 2.00 atm ? Solution: The partial pressure of each gas depends on the its mole fraction. Find mole fraction of each gas. Then what??? Answers: PNe = 1.21 atm; PAr = 0.20 atm; PXe = 0.586 atm Daltons Law: Using Mole Fractions 2 of 2 A mixture of gases contains 4.46 mol Ne, 0.74 mol Ar and 2.15 mol

Xe. What are the partial pressures of the gases if the total pressure is 2.00 atm ? Total # moles = 4.46 + 0.74 + 2.15 = 7.35 mol XNe = 4.46 mol Ne / 7.35 mol = 0.607 PNe = XNe PTotal = 0.607 ( 2.00 atm) = 1.21 atm for Ne XAr = 0.74 mol Ar / 7.35 mol = 0.10 PAr = XAr PTotal = 0.10 (2.00 atm) = 0.20 atm for Ar XXe = 2.15 mol Xe / 7.35 mol = 0.293 PXe = XXe PTotal = 0.293 (2.00 atm) = 0.586 atm for Xe Diffusion vs Effusion Diffusion Spreading out of molecules from a region where their concentration is high to a region where their concentration is low For Gases: from high partial pressure to low partial pressure e.g. Perfume, flatulence, etc.

Effusion The diffusion of a gas through a tiny hole (or holes) e.g. Gradual deflation of a balloons, tires, etc. Fig. 5.20 Quantifying Effusion: Grahams Law of Effusion Rate of Effusion 1 MW of Gas OR Rate of Gas A

MW of Gas B Rate of Gas B MW of Gas A Diffusion of NH3 gas and HCl gas NH3 (g) + HCl(g) NH4Cl (s) HCl = 36.46 g/mol NH3 = 17.03 g/mol RateNH3 = RateHCl x ( 36.46 / 17.03 )1/ 2 RateNH3 = RateHCl x 1.463 Sample Problem 5.12 Applying Grahams Law of Effusion

PROBLEM: Calculate the ratio of the effusion rates of helium and methane (CH4). PLAN: The effusion rate is inversely proportional to the square root of the molar mass for each gas. Find the molar mass of both gases and find the inverse square root of their masses. SOLUTION: M of CH4 = 16.04g/mol rate rate He CH4 =

16.04 4.003 = 2.002 M of He = 4.003g/mol Sample Effusion Problems Which effuses faster out of a cars tire, oxygen or nitrogen? How much faster? Answer: 1.069 times or 6.9% faster Suppose that there are two leaky cylinders in a lab, one containing chlorine gas, the other hydrogen cyanide, HCN. Which gas will reach you first?

How many times faster will its molecules diffuse across the room? 1.62 times faster Gaseous Diffusion Separation of Uranium - 235 / 238 235UF6 vs 238UF6 (238.05 + (6 x 19))0.5 Separation Factor = S = 0.5 (235.04 + (6 x 19)) after two runs

S = 1.0086 after approximately 2000 runs 235UF6 is > 99% Purity !!!!! Y - 12 Plant at Oak Ridge National Lab Using the Kinetic Theory to Explain the Gas Laws Kinetic Theory of Gases 1. Gases consist of an exceptionally large number of extremely small particles in random and constant motion 2. The Volume that gas particles themselves occupy is much less than the Volume of their container i.e. The distance between gas particles is vast 3. Collisions are elastic and molecular motion is

linear.....WHY? Postulates of the Kinetic-Molecular Theory Postulate 1: Particle Volume Because the volume of an individual gas particle is so small compared to the volume of its container, the gas particles are considered to have mass, but no volume. Postulate 2: Particle Motion Gas particles are in constant, random, straight-line motion except when they collide with each other or with the container walls. Postulate 3: Particle Collisions Collisions are elastic therefore the total kinetic energy(Kk) of the particles is constant. Figure 5.14

Distribution of molecular speeds at three temperatures. Important Points At a given temperature, all gases have the same 1) molecular kinetic energy distributions 2) average molecular kinetic energy Kinetic Energy: Ek = 1/2 mv2 Figure 5.19 Relationship between molar mass and molecular speed. Ek 1/2 mv2 V mass

Explain Each Gas Law in Terms of the Kinetic Theory Boyles Law (P-V Law) trans 10 Amontons Law (P-T Law) trans 9 &11 Chucks Law (V-T Law) trans 12 Grahams Law of Effusion Hint: KE = 1/2mv2 and gases A and B are at the same temperature. Figure 5.15 A molecular description of Boyles Law

Figure 5.16 A molecular description of Daltons law of partial pressures. Figure 5.17 A molecular description of Charless Law Figure 5.18 A molecular description of Avogadros Law Real Gases: Deviations from the Ideal Gas Law Ideal Gases Obey the gas laws exactly Are Theoretical Gases

Infinitely Small with no intermolecular attractions However, all matter occupies space have I.M.F.s of attraction!! Real Gases Do not follow the gas laws exactly Molecules DO occupy space Molecules DO have attractions causes gases to condense into liquids At high Pressures and low Temps molecular volume and attractions become significant Molar Volume of Some Common Gases at STP (00C and 1 atm) Gas Molar Volume (L/mol)

He H2 Ne Ideal Gas Ar N2 O2 CO Cl2 NH3 Table 5.4 (p. 207) 22.435 22.432 22.422 22.414 22.397

22.396 22.390 22.388 22.184 22.079 Condensation Point (0C) -268.9 -252.8 -246.1 -----185.9 -195.8 -183.0 -191.5 -34.0 -33.4

The Behavior of Several Real Gases with Increasing External Pressure Fig. 5.21 Molecular volume increases measured volume IMFs decrease Pressure The Effect of Molecular Volume on Measured Gas Volume Fig. 5.23 The van der Waals Equation 2 n a (V-nb) = nRT

P+ V2 Gas 2 atm L a mol2 He Ne Ar Kr Xe H2

N2 O2 Cl2 CO2 NH3 0.034 0.211 1.35 2.32 4.19 0.244 1.39 6.49 3.59 2.25 4.17

b L mol 0.0237 0.0171 0.0322 0.0398 0.0511 0.0266 0.0391 0.0318 0.0562 0.0428 0.0371

Van der Waals Calculation of a Real gas 1 of 2 Problem A tank of 20.0 liters contains chlorine gas at a temperature of 20.000C at a pressure of 2.000 atm. If the tank is pressurized to a new volume of 1.000 L and a temperature of 150.000C, calculate the new pressure using the ideal gas equation, and the van der Waals equation. Plan 1. Use ideal gas law to calculate moles of gas under initial cond. 2. 3. Use I.G.L. to calculate ideal pressure under the new cond. Use Van der Waals equation to calculate the real pressure.

Van der Waals Calculation of a Real gas 2 of 2 Solution n = PV RT P= = (2.000 atm)(20.0L) = 1.663 mol (0.08206 Latm/molK)(293.15 K) nRT

(1.663 mol)(0.08206 Latm/molK)(423.15 K) = = 57.745 atm V (1.000 L) nRT n2a P= - 2 (V-nb) V (1.663 mol)(0.08206 Latm/molK)(423.15 K) = (1.00 L) - (1.663 mol)(0.0562) (1.663 mol)2(6.49) = 63.699 - 17.948 = 45.751 atm (1.00 L)2

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    Bug - Variable Scope - variable declared only in one branch of if. We need to make sure that we have a single variable, called result, that is alive:. at the if part.
  • Refinement Planning: Status and Prospectus

    Refinement Planning: Status and Prospectus

    That Fahiem Bacchus and Jonas Kvarnstrom are way better at coming up with domain knowledge for blocks world (and other competition domains) than Dana Nau? Questions worth asking in KB planner comparisons (IMHO) How easy/natural (for humans) is the language...