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1Dr. Orlando E. Raola Santa Rosa Junior College Chemistry 1AGeneral ChemistryInstructor:Dr. Orlando E. Raola Santa Rosa Junior CollegeChapter 5: Properties ofGases
2Gases and the Kinetic Molecular Theory 5.1 An Overview of the Physical States of Matter5.2 Gas Pressure and Its Measurement5.3 The Gas Laws and Their Experimental Foundations5.4 Applications of the Ideal Gas Law
3The distinction of gases from liquids and solids Gas volume changes greatly with pressure.Gas volume changes greatly with temperature.Gases have relatively low viscosity.Most gases have relatively low densities under normal conditions.Gases are miscible.
7Sample Problem 5.1Converting Units of PressurePROBLEM: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, Dh = mm Hg. Express the CO2 pressure in torr, atmosphere, and kilopascal.PLAN:Construct conversion factors to find the other units of pressure.SOLUTION:291.4 mmHg1torr1 mmHg= torr1 atm760 torr291.4 torr= atmkPa1 atmatm= kPa
8The relationship between the volume and pressure of a gas. Boyle’s Law
9The relationship between the volume and temperature of a gas. Charles’s Law
10V a1PBoyle’s Lawn and T are fixedV x P= constantV = constant / PCharles’s LawV a TP and n are fixedVT= constantV = constant x TAmontons’s LawP a TV and n are fixedPT= constantP = constant x TV aTPV = constant xTPPVT= constantcombined gas law
11n = chemical amount of gas (mol) Avogadro’s LawFor a gas at constant temperature and pressure, the volume is directly proportional to the number of moles of gasV = a na = proportionality constantV = volume of the gas (m3)n = chemical amount of gas (mol)
12(in common units 0.082 atm∙L∙mol-1·K-1) Ideal Gas LawAn equation of state for a gas.“state” is the condition of the gas at a given time.PV = nRTR = J∙mol-1·K-1(in common units atm∙L∙mol-1·K-1)
14PV = nRT THE IDEAL GAS LAW IDEAL GAS LAW nRT P PV = nRT or V = R is the universal gas constantIDEAL GAS LAWnRTPPV = nRT or V =fixed n and Tfixed n and Pfixed P and TBoyle’s LawCharles’s LawAvogadro’s LawconstantPV =constant X nV =V =constant X T
15Sample Problem 5.2Applying the Volume-Pressure RelationshipPROBLEM:Boyle’s apprentice finds that the air trapped in a J tube occupies 24.8 cm3 at 1.12 atm. By adding mercury to the tube, he increases the pressure on the trapped air to 2.64 atm. Assuming constant temperature, what is the new volume of air (in L)?PLAN:SOLUTION:n and T are constantV1 in cm3P1 = 1.12 atmP2 = 2.64 atmunit conversion1cm3=1mLV1 = 24.8 cm3V2 = unknownV1 in mL1 mL1 cm3L103 mL103 mL=1L24.8 cm3= LV1 in Lgas law calculationxP1/P2P1V1n1T1P2V2n2T2P1V1 = P2V2=V2 in LP1V1P2V2 =1.12 atm2.46 atm=L=L
16Sample Problem 5.3Applying the Temperature-Pressure RelationshipPROBLEM:A steel tank used for fuel delivery is fitted with a safety valve that opens when the internal pressure exceeds 1.00x103 torr. It is filled with methane at 230C and atm and placed in boiling water at exactly 1000C. Will the safety valve open?PLAN:SOLUTION:P1(atm)T1 and T2(0C)P1 = 0.991atmP2 = unknown1atm=760torrK=0CT1 = 230CT2 = 1000CP1(torr)T1 and T2(K)P1V1n1T1P2V2n2T2=P1T1P2T2=x T2/T1P2(torr)0.991 atm1 atm760 torr= 753 torrP2 =P1T2T1= 753 torr373K296K= 949 torr
17Sample Problem 5.4Applying the Volume-Amount RelationshipPROBLEM:A scale model of a blimp rises when it is filled with helium to a volume of 55 dm3. When 1.10 mol of He is added to the blimp, the volume is 26.2 dm3. How many more grams of He must be added to make it rise? Assume constant T and P.PLAN:We are given initial n1 and V1 as well as the final V2. We have to find n2 and convert it from moles to grams.n1(mol) of HeSOLUTION:P and T are constantx V2/V1n1 = 1.10 moln2 = unknownP1V1n1T1P2V2n2T2=n2(mol) of HeV1 = 26.2 dm3V2 = 55.0 dm3subtract n1V1n1V2n2=n2 = n1V2V1mol to be addedx Mn2 = 1.10 mol55.0 dm326.2 dm34.003 g Hemol Heg to be added= 9.24 g He= 2.31 mol
18Sample Problem 5.5Solving for an Unknown Gas Variable at Fixed ConditionsPROBLEM:A steel tank has a volume of 438 L and is filled with kg of O2. Calculate the pressure of O2 at 210C.PLAN:V, T and mass are given. From mass, find amount (n), and use the ideal gas law to find P.SOLUTION:
19Relationship between density and molar mass for gases The density of a gas is directly proportional to its molar mass.The density of a gas is inversely proportional to the temperature.
20Sample Problem 5.7Calculating Gas DensityPROBLEM:To apply a green chemistry approach, a chemical engineer uses waste CO2 from a manufacturing process, instead of chlorofluorocarbons, as a “blowing agent” in the production of polystyrene containers. Find the density (in g/L) of CO2 and the number of molecules (a) at STP (00C and 1 atm) and (b) at room conditions (20.0C and 1.00 atm).PLAN:Density is mass/unit volume; substitute for volume in the ideal gas equation. Since the identity of the gas is known, we can find the molar mass. Convert mass/L to molecules/L with Avogadro’s number.d=RTM x Pd = mass/volumePV = nRTV = nRT/PSOLUTION:44.01 g/molx 1atmatm*Lmol*K0.0821x K= 1.96 g/Ld =(a)1.96 gLmol CO244.01 g CO26.022x1023 moleculesmol= 2.68x1022 molecules CO2/L
21Sample Problem 5.6Calculating Gas Densitycontinuedd =44.01 g/molx 1 atmx 293Katm*Lmol*K0.0821(b)= 1.83 g/L1.83gLmol CO244.01g CO26.022x1023 moleculesmol= 2.50x1022 molecules CO2/L
22Determining the molar mass of an unknown volatile liquid. based on the method of J.B.A. Dumas ( )
23Sample Problem 5.8Finding the Molar Mass of a Volatile LiquidPROBLEM:An organic chemist isolates a colorless liquid from a petroleum sample. She uses the Dumas method and obtains the following data:Volume of flask = 213 mLMass of flask + gas = gT = CMass of flask = gP = 754 torrCalculate the molar mass of the liquid.PLAN:Use unit conversions, mass of gas, and density-M relationship.SOLUTION:m = ( ) g= g0.582 gatm*Lmol*K0.0821373Kxm RTVPxM ===84.4 g/mol0.213 Lx0.992 atm
24Relationship between density and molar mass for gases The density of a gas is directly proportional to its molar mass.The density of a gas is inversely proportional to the temperature.
25Sample Problem 5.7Calculating Gas DensityPROBLEM:To apply a green chemistry approach, a chemical engineer uses waste CO2 from a manufacturing process, instead of chlorofluorocarbons, as a “blowing agent” in the production of polystyrene containers. Find the density (in g/L) of CO2 and the number of molecules (a) at STP ( K and 1 bar) and (b) at room conditions (20.0C and 1.00 atm).PLAN:Density is mass/unit volume; substitute for volume in the ideal gas equation. Since the identity of the gas is known, we can find the molar mass. Convert mass/L to molecules/L with Avogadro’s number.d=RTM x Pd = mass/volumePV = nRTV = nRT/PSOLUTION:(a)1.94 gLmol CO244.01 g CO26.022x1023 moleculesmol= 2.65x1022 molecules CO2/L
26Sample Problem 5.6Calculating Gas Densitycontinuedd =44.01 g/molx 1 atmx 293Katm*Lmol*K0.0821(b)= 1.83 g/L1.83gLmol CO244.01g CO26.022x1023 moleculesmol= 2.50x1022 molecules CO2/L
27Determining the molar mass of an unknown volatile liquid. based on the method of J.B.A. Dumas ( )
28Sample Problem 5.8Finding the Molar Mass of a Volatile LiquidPROBLEM:An organic chemist isolates a colorless liquid from a petroleum sample. She uses the Dumas method and obtains the following data:Volume of flask = 213 mLMass of flask + gas = gT = CMass of flask = gP = 754 torrCalculate the molar mass of the liquid.PLAN:Use unit conversions, mass of gas, and density-M relationship.SOLUTION:m = ( ) g= g0.582 gatm*Lmol*K0.0821373Kxm RTVPxM ===84.4 g/mol0.213 Lx0.992 atm
29Sample Problem 5.5Solving for an Unknown Gas Variable at Fixed ConditionsPROBLEM:A steel tank has a volume of 438 L and is filled with kg of O2. Calculate the pressure of O2 at 210C.PLAN:V, T and mass are given. From mass, find amount (n), and use the ideal gas law to find P.SOLUTION:
30Mixtures of gases Gases mix homogeneously in any proportions. Each gas in a mixture behaves as if it were the only gas present.Dalton’s Law of Partial PressuresPtotal = P1 + P2 + PP1= c1 x Ptotalwhere c1 is the mole fractionc1 =n1n1 + n2 + n3 +...=n1ntotal
31Mole fraction and partial pressure For each component we define the mole fraction xBand because of Dalton’s law, we can calculate the partial pressure of each component as
32Sample Problem 5.9Applying Dalton’s Law of Partial PressuresPROBLEM:In a study of O2 uptake by muscle at high altitude, a physiologist prepares an atmosphere consisting of 79 mol% N2, 17 mol% 16O2, and 4.0 mol% 18O2. (The isotope 18O will be measured to determine the O2 uptake.) The pressure of the mixture is 0.75atm to simulate high altitude. Calculate the mole fraction and partial pressure of 18O2 in the mixture.PLAN:Find the c and P from Ptotal and mol% 18O2.18O2=4.0 mol% 18O2100mol% 18O2SOLUTION:= 0.040c18O2divide by 100c 18O2P = c x Ptotal = x 0.75 atm18O2= atmmultiply by Ptotalpartial pressure P18O2
33Table 5.3 Vapor Pressure of Water (P ) at Different T H2OT(0C)P (torr)T(0C)P (torr)5101112131415161820222426222.214.171.124.810.511.212.012.813.615.517.519.822.425.228.3303540455055606570758085909510031.842.255.371.992.5118.0149.4187.5233.7289.1355.1433.6525.8633.9760.0
34Collecting a water-insoluble gaseous reaction product and determining its pressure.
35CaC2(s) + 2H2O(l) C2H2(g) + Ca(OH)2(aq) Sample Problem 5.10Calculating the Amount of Gas Collected Over WaterPROBLEM:Acetylene (C2H2), an important fuel in welding, is produced in the laboratory when calcium carbide (CaC2) reaction with water:CaC2(s) + 2H2O(l) C2H2(g) + Ca(OH)2(aq)For a sample of acetylene that is collected over water, the total gas pressure (adjusted to barometric pressure) is kPa and the volume is 523 mL. At the temperature of the gas (230C), the vapor pressure of water is kPa. How many grams of acetylene are collected?PLAN:The difference in pressures will give us the P for the C2H2. The ideal gas law will allow us to find n. Converting n to grams requires the molar mass, M.PC2H2PtotalPC2H2SOLUTION:= ( ) kPa =95.52 kPaPn =PVRTH2OnC2H2gC2H2x M
36Sample Problem 5.9Calculating the Amount of Gas Collected Over Watercontinued0.0203mol26.04 g C2H2mol C2H2= g C2H2
37molar ratio from balanced equation Summary of the stoichiometric relationships among the amount (mol,n) of gaseous reactant or product and the gas variables pressure (P), volume (V), and temperature (T).amount (mol)of gas Aamount (mol)of gas BP,V,Tof gas AP,V,Tof gas Bideal gas lawideal gas lawmolar ratio from balanced equation
38Sample Problem 5.11Using Gas Variables to Find Amount of Reactants and ProductsPROBLEM:Dispersed copper in absorbent beds is used to react with oxygen impurities in the ethylene used for producing polyethylene. The beds are regenerated when hot H2 reduces the metal oxide, forming the pure metal and H2O. On a laboratory scale, what volume of H2 at 765 torr and 2250C is needed to reduce 35.5 g of copper(II) oxide?PLAN:Since this problem requires stoichiometry and the gas laws, we have to write a balanced equation, use the moles of Cu to calculate mols and then volume of H2 gas.mass (g) of CuSOLUTION:CuO(s) + H2(g) Cu(s) + H2O(g)divide by Mmol Cu63.55 g Cu1 mol H21 mol Cu35.5 g Cu= mol H2mol of Cumolar ratiox498Katm*Lmol*K0.08211.01 atm0.559 mol H2= 22.6 Lmol of H2use known P and T to find VL of H2
39Sample Problem 5.12Using the Ideal Gas Law in a Limiting-Reactant ProblemPROBLEM:The alkali metals (Group 1) react with the halogens (Group 17) to form ionic metal halides. What mass of potassium chloride forms when L of chlorine gas at atm and 293K reacts with 17.0 g of potassium?PLAN:After writing the balanced equation, we use the ideal gas law to find the number of moles of reactants, the limiting reactant and moles of product.SOLUTION:2K(s) + Cl2(g) KCl(s)V = 5.25 LT = 293Kn = unknownP = atmPVRT=0.950 atmatm*Lmol*K0.0821x293Kx5.25Ln ==0.207 molCl20.207 mol Cl22 mol KCl1 mol Cl217.0g39.10 g Kmol K= molKCl formed=0.435 mol K2 mol KCl2 mol KCl2 is the limiting reactant.0.435 mol K= molKCl formed74.55 g KClmol KCl0.414 mol KCl= 30.9 g KCl
40ProblemGaseous iodine pentafluoride can be prepared by the reaction between solid iodine and gaseous fluorine. A 5.00-L flask containing 10.0 g of I2 is charged with 10.0 g of F2 and the reaction proceeds until one of the reactants is completely consumed. After the reaction is complete, the temperature in the flask is 125 ºC.What is the partial pressure of IF5 in the flask?What is the mole fraction of IF5 in the flask?
41Postulates of the Kinetic-Molecular Theory Postulate 1: Particle VolumeBecause 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 MotionGas particles are in constant, random, straight-line motion except when they collide with each other or with the container walls.Postulate 3: Particle CollisionsCollisions are elastic therefore the total kinetic energy(Ek) of the particles is constant.
42Distribution of molecular speeds as a function of temperature
43Distribution of molecular speeds as a function of temperature
49The meaning of temperature Absolute temperature is a measure of the average kinetic energy of the molecular random motion
50Effusion and difussion Effusion: describes the passage of gas through a small orifice into an evacuated chamber.Diffusion: describes the mixing of gases. The rate of diffusion is the rate of gas mixing.
53Diffusion through space distribution of molecular speedsmean free pathcollision frequency
54√ Sample Problem 5.13 Applying Graham’s 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/molM of He = 4.003g/molCH4Herate=√16.044.003= 2.002
55(00C and 1 atm) Table 5.4 Molar Volume of Some Common Gases at STP (L/mol)Condensation Point(0C)GasHeH2NeIdeal gasArN2O2COCl2NH322.43522.43222.42222.41422.39722.39622.39022.38822.18422.079-268.9-252.8-246.1----185.9-195.8-183.0-191.5-34.0-33.4
56The behavior of several real gases with increasing external pressure
59van der Waals equation Gas a (atm·L2·mol-2) b (L·mol-1) Van der Waals equation for nmoles of a real gasGas a (atm·L2·mol-2) b (L·mol-1)0.0340.2111.352.324.190.2441.391.366.493.5126.96.36.199HeNeArKrXeH2N2O2Cl2CO2CH4NH3H2O0.02370.01710.03220.03980.05110.02660.03910.03180.05620.04270.04280.03710.0305