Report

Chapter 5 Gases and the Kinetic Molecular Theory 5-1 Dr. Wolf’s CHM 101 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 5-2 Dr. Wolf’s CHM 101 An Overview of the Physical States of Matter The Distinction of Gases from Liquids and Solids 1. Gas volume changes greatly with pressure. 2. Gas volume changes greatly with temperature. 3. Gas have relatively low viscosity. 4. Most gases have relatively low densites under normal conditions. 5. Gases are miscible. 5-3 Dr. Wolf’s CHM 101 Figure 5.1 5-4 The three states of matter. Dr. Wolf’s CHM 101 Measuring Gas Pressure Barometer - A device to measure atmospheric pressure. Pressure is defined as force divided by area. The force is the force of gravity acting on the air molecules. Manometer - A device to measure gas pressure in a closed container. 5-5 Dr. Wolf’s CHM 101 Figure 5.3 5-6 Dr. Wolf’s CHM 101 A mercury barometer. Figure 5.4 5-7 Two types of manometer Dr. Wolf’s CHM 101 Table 5.2 Common Units of Pressure Unit Scientific Field pascal(Pa); kilopascal(kPa) 1.01325x105Pa; 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 1.01325 bar meteorology, chemistry, physics pounds per square inch (psi or lb/in2) bar 5-8 Atmospheric Pressure Dr. Wolf’s CHM 101 Sample Problem 5.1 PROBLEM: Converting Units of Pressure A geochemist heats a limestone (CaCO3) sample and collects the CO2 released in an evacuated flask attached to a closedend manometer. After the system comes to room temperature, Dh = 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 = 291.4torr 1mmHg 291.4torr 1atm = 0.3834atm 760torr 0.3834atm 101.325kPa 1atm 5-9 Dr. Wolf’s CHM 101 = 38.85kPa Boyle’s Law - The relationship between volume and the pressure of a gas. (Temperature is kept constant.) 5-10 Dr. Wolf’s CHM 101 Boyle’s Law V a PV = constant 5-11 Dr. Wolf’s CHM 101 1 P n and T are fixed V = constant / P Charles’s Law - The relationship between volume and the temperature of a gas. (Pressure is kept constant.) 5-12 Dr. Wolf’s CHM 101 V a Boyle’s Law = constant T P a T Amonton’s Law P T combined gas law P V a T Charles’s Law V 1 = constant V a n and T are fixed P and n are fixed V = constant x T V and n are fixed P = constant x T T P V = constant x T PV P T When the amount of gas, n, is constant 5-13 Dr. Wolf’s CHM 101 = constant Avogadro’s Law - The volume of a gas is directly proportionate to the amount of gas. (Pressure and temperature kept constant.) An experiment to study the relationship between the volume and amount of a gas. V a n V n = constant V = constant x n Twice the amount gives twice the volume 5-14 Dr. Wolf’s CHM 101 If the constant pressure is 1 atm and the constant temperature is 0o C, 1 mole of any gas has a volume of 22.4 L . This is known as the Standard Molar Volume 5-15 Dr. Wolf’s CHM 101 IDEAL GAS LAW fixed n and T Boyle’s Law V = constant fixed n and P fixed P and T Charles’s Law Avogadro’s Law V= V = constant X T constant X n P THE IDEAL GAS LAW PV = nRT R= 5-16 PV nT Dr. Wolf’s CHM 101 = 1atm x 22.414L 1mol x 273.15K = 0.0821atm-L Mol-K Sample Problem 5.2 PROBLEM: Applying the Volume-Pressure Relationship Boyle’s apprentice finds that the air trapped in a J tube occupies 24.8cm3 at 1.12atm. By adding mercury to the tube, he increases the pressure on the trapped air to 2.64atm. Assuming constant temperature, what is the new volume of air (inL)? PLAN: V1 in cm3 1cm3=1mL V1 in mL unit conversion 103mL=1L V1 in L xP1/P2 P1 = 1.12atm P2 = 2.64atm V1 = 24.8cm3 V2 = unknown 24.8cm3 gas law calculation P1V1 V2 in L P1V1 P2 Dr. Wolf’s CHM 101 1mL L 1cm3 103mL =R= n1T1 V2 = 5-17 n and T are constant SOLUTION: P2V2 P1V1 = P2V2 n2T2 1.12atm = 0.0248L = 0.0248L = V1 2.46atm = 0.0105L Sample Problem 5.3 PROBLEM: Applying the Temperature-Pressure Relationship A 1-L steel tank is fitted with a safety valve that opens if the internal pressure exceeds 1.00x103 torr. It is filled with helium at 230C and 0.991atm and placed in boiling water at exactly 1000C. Will the safety valve open? PLAN: SOLUTION: T1 and T2(0C) P1(atm) 1atm=760torr P1(torr) K=0C+273.15 T1 and T2(K) x T2/T1 0.991atm 5-18 P2 = unknown T1 = 230C T2 = 100oC P1V1 = n1T1 P2(torr) P2 = P1 P1 = 0.991atm Dr. Wolf’s CHM 101 n2T2 760 torr = 753 torr 1atm T2 T1 P2V2 = 753 torr 373K 296K = 949 torr P1 T1 = P2 T2 Sample Problem 5.4 PROBLEM: Applying the Volume-Amount Relationship A scale model of a blimp rises when it is filled with helium to a volume of 55dm3. When 1.10mol of He is added to the blimp, the volume is 26.2dm3. 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 He P and T are constant SOLUTION: x V2/V1 n1 = 1.10mol n2 = unknown n2(mol) of He V1 = 26.2dm3 V2 = 55.0dm3 subtract n1 V1 mol to be added n1 = V2 n2 xM n2 = 1.10mol g to be added n2 = n1 55.0dm3 5-19 n1T1 = P2V2 n2T2 V2 V1 = 2.31mol 26.2dm3 n2 - n1 = 2.31 -1.10 = 1.21 mol He Dr. Wolf’s CHM 101 P1V1 4.003g He mol He = 4.84g He Sample Problem 5.5 Conditions A steel tank has a volume of 438L and is filled with 0.885kg of O2. Calculate the pressure of O2 at 210C. PROBLEM: PLAN: Solving for an Unknown Gas Variable at Fixed V, T and mass, which can be converted to moles (n), are given. We use the ideal gas law to find P. SOLUTION: 0.885kg V = 438L T = 210C (convert to K) n = 0.885kg (convert to mol) P = unknown 103g mol O2 kg 32.00g O2 = 27.7mol O2 27.7mol x 0.0821 P= 5-20 nRT V atm*L mol*K = Dr. Wolf’s CHM 101 210C + 273.15 = 294K x 294K = 1.53atm 438L Sample Problem 5.6 Calculate the density (in g/L) of carbon dioxide and the number of molecules per liter (a) at STP (00C and 1 atm) and (b) at ordinary room conditions (20.0C and 1.00atm). PROBLEM: PLAN: Calculating Gas Density 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 Avogardro’s number. d = mass/ V PV = nRT mult. both sides by M n / V = P / RT Mass = n M 44.01g/mol SOLUTION: (a) x 1atm atm*L MxP RT = 1.96g/L d= 0.0821 d = x 273K mol*K 5-21 1.96g mol CO2 L 44.01g CO2 Dr. Wolf’s CHM 101 6.022x1023molecules mol = 2.68x1022molecules CO2/L Sample Problem 5.6 Calculating Gas Density continued (b) 44.01g/mol d= 0.0821 5-22 x 1atm 1.83g mol CO2 L 44.01g CO2 Dr. Wolf’s CHM 101 = 1.83g/L atm*L x 293K mol*K 6.022x1023molecules mol = 2.50x1022molecules CO2/L Calculating the Molar Mass, M, of a Gas Since PV = nRT Then n = PV / RT And n = mass / M So m / M = PV / RT And M = mRT / PV Or M = d RT / P 5-23 Dr. Wolf’s CHM 101 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 T = 100.00C Mass of flask + gas = 78.416g Mass of flask = 77.834g P = 754 torr Is the calculated molar mass consistent with the liquid being cyclohexane? PLAN: Use unit conversions, mass of gas and density-M relationship. SOLUTION: m = (78.416 - 77.834)g = 0.582g 0.582g x 0.0821 M= m RT = PV 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. 5-24 Dr. Wolf’s CHM 101 Partial Pressure of a Gas in a Mixture of Gases Gases mix homogeneously. Each gas in a mixture behaves as if it is the only gas present, e.g. its pressure is calculated from PV = nRT with n equal to the number of moles of that particular gas......called the Partial Pressure. The gas pressure in a container is the sum of the partial pressures of all of the gases present. Dalton’s Law of Partial Pressures Ptotal = P1 + P2 + P3 + ... P1= c1 x Ptotal where c1 is the mole fraction c1 = 5-25 Dr. Wolf’s CHM 101 n1 n1 + n2 + n3 +... = n1 ntotal Collecting Gas over Water Often in gas experiments the gas is collected “over water.” So the gases in the container includes water vapor as a gas. The water vapor’s partial pressure contributes to the total pressure in the container. 5-26 Dr. Wolf’s CHM 101 Sample Problem 5.8 PROBLEM: Applying Dalton’s Law of Partial Pressures In a study of O2 uptake by muscle at high altitude, a physiologist prepares an atmosphere consisting of 79 mol% N2, 17 mol% 16O and 4.0 mol% 18O . (The isotope 18O will be measured to 2, 2 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. 18 PLAN: Find the c 18Oand P18O from Ptotal and mol% O2. 2 mol% 18O 2 SOLUTION: 2 divide by 100 c 18O P18 2 multiply by Ptotal partial pressure P 18O 5-27 2 Dr. Wolf’s CHM 101 O2 c 18O = 2 4.0mol% 18O2 = 0.040 100 = c18 x Ptotal = 0.040 x 0.75atm = 0.030atm O2 Sample Problem 5.9 PROBLEM: Calculating the Amount of Gas Collected Over Water 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 738torr and the volume is 523mL. At the temperature of the gas (230C), the vapor pressure of water is 21torr. 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. P SOLUTION: C2H2 = (738-21)torr = 717torr Ptotal P C2H2 atm P = 0.943atm PV 717torr H2O n= 760torr RT n g C2H2 C2H2 5-28 Dr. Wolf’s CHM 101 xM Sample Problem 5.9 Calculating the Amount of Gas Collected Over Water continued n C2H2 0.943atm x = 0.0821 atm*L 0.523L = 0.203mol x 296K mol*K 0.203mol 26.04g C2H2 mol C2H2 5-29 Dr. Wolf’s CHM 101 = 0.529 g C2H2 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). P,V,T of gas A ideal gas law 5-30 Dr. Wolf’s CHM 101 amount (mol) amount (mol) P,V,T of gas A of gas B of gas B molar ratio from balanced equation ideal gas law Sample Problem 5.10 PROBLEM: Using Gas Variables to Find Amount of Reactants and Products A laboratory-scale method for reducing a metal oxide is to heat it with H2. The pure metal and H2O are products. What volume of H2 at 765torr and 2250C is needed to form 35.5g of Cu from 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. CuO(s) + H2(g) mass (g) of Cu divide by M mol of Cu SOLUTION: mol Cu 1mol H2 35.5g Cu = 0.559mol H2 63.55g Cu 1 mol Cu molar ratio mol of H2 0.559mol H2 x 0.0821 use known P and T to find V L of H2 5-31 Dr. Wolf’s CHM 101 Cu(s) + H2O(g) atm*L mol*K 1.01atm x 498K = 22.6L Sample Problem 5.11 Using the Ideal Gas Law in a Limiting-Reactant Problem PROBLEM: The alkali metals [Group 1A(1)] react with the halogens [Group 7A(17)] to form ionic metal halides. What mass of potassium chloride forms when 5.25L of chlorine gas at 0.950atm and 293K reacts with 17.0g 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. 2K(s) + Cl2(g) 2KCl(s) P = 0.950atm V = 5.25L SOLUTION: T = 293K n = unknown PV 0.950atm x 5.25L n = = = 0.207mol Cl2 RT atm*L 0.0821 x 293K 2mol KCl mol*K 0.207mol Cl2 = 0.414mol 1mol Cl2 KCl formed 17.0g mol K = 0.435mol K 39.10g K 2mol KCl = 0.435mol Cl2 is the limiting reactant. 0.435mol K KCl formed 2mol K 74.55g KCl 0.414mol KCl = 30.9 g KCl mol KCl 5-32 Dr. Wolf’s CHM 101 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. 5-33 Dr. Wolf’s CHM 101 A molecular description of Boyle’s Law Boyle’s Law 5-34 Dr. Wolf’s CHM 101 V a 1 P n and T are fixed A molecular description of Charles’s Law Charles’s Law 5-35 Dr. Wolf’s CHM 101 V a T n and P are fixed A molecular description of Dalton’s law of partial pressures. Dalton’s Law of Partial Pressures Ptotal = P1 + P2 + P3 + ... 5-36 Dr. Wolf’s CHM 101 A molecular description of Avogadro’s Law Avogadro’s Law 5-37 Dr. Wolf’s CHM 101 V a n P and T are fixed Kinetic-Molecular Theory Gas particles are in motion and have a molecular speed, . But they are moving at various speeds, some very slow, some very fast, but most near the average speed of all of the particles, (avg) . Since kinetic energy is defined as ½ mass x (speed)2, we can define the average kinetic energy, Ek(avg) = ½m 2 (avg) Since energy is a function of temperature, the average energy and, hence, average molecular speed will increase with temperature. 5-38 Dr. Wolf’s CHM 101 An increase in temperature results in an increase in average molecular kinetic energy. The relationship between average kinetic energy and temperature is given as, Ek(avg) = 3/2 (R/NA) x T (where R is the gas constant in energy units, 8.314 J/mol-K NA is Avogadro’s number, and T temperature in K.) To have the same average kinetic energy, heavier atoms must have smaller speeds. The root-mean-square speed, (rms) , is the speed where a molecule has the average kinetic energy. The relationship between (rms) and molar mass is: (rms) = (3RT/M ) ½ So the speed (or rate of movement) is: rate a 1 / (M ) ½ 5-39 Dr. Wolf’s CHM 101 Relationship between molar mass and molecular speed. 5-40 Dr. Wolf’s CHM 101 Graham’s Law of Effusion Effusion is the process by which a gas in a closed container moves through a pin-hole into an evacuated space. This rate is proportional to the speed of a molecule so..... Graham’s Law of Effusion The rate of effusion of a gas is inversely related to the square root of its molar mass. Rate of effusion a 1 / (M ) ½ So doing two identical effusion experiments measuring the rates of two gases, one known, one unknown, allows the molecular mass of the unknown to be determined. rate A rate B 5-41 Dr. Wolf’s CHM 101 = ( MB / MA ) ½ Sample Problem 5.12 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/mol M of He = 4.003g/mol rate He rate 5-42 CH4 Dr. Wolf’s CHM 101 = ( 16.04/ 4.003 ) ½ = 2.002 End of Chapter 5 5-43 Dr. Wolf’s CHM 101