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Chapter 5 Gases and the Kinetic Molecular Theory

Chapter 5 Gases and the Kinetic Molecular Theory. 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.

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Chapter 5 Gases and the Kinetic Molecular Theory

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  1. Chapter 5 Gases and the Kinetic Molecular Theory

  2. Gases and the Kinetic Molecular Theory 5.1An Overview of the Physical States of Matter 5.2Gas Pressure and Its Measurement 5.3The Gas Laws and Their Experimental Foundations 5.4Further Applications of the Ideal Gas Law 5.5The Ideal Gas Law and Reaction Stoichiometry 5.6The Kinetic-Molecular Theory: A Model for Gas Behavior 5.7Real Gases: Deviations from Ideal Behavior

  3. 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.

  4. Figure 5.1 The three states of matter.

  5. 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.

  6. A mercury barometer. Figure 5.3

  7. Two types of manometer Figure 5.4

  8. 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 pounds per square inch (psi or lb/in2) 14.7lb/in2 engineering bar 1.01325 bar meteorology, chemistry, physics Table 5.2 Common Units of Pressure Unit Atmospheric Pressure Scientific Field

  9. 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, Dh = 291.4mmHg. Calculate the CO2 pressure in torrs, atmospheres, and kilopascals. PLAN: Construct conversion factors to find the other units of pressure. 1torr 1mmHg 1atm 760torr 101.325kPa 1atm Sample Problem 5.1 Converting Units of Pressure SOLUTION: 291.4mmHg = 291.4torr 291.4torr = 0.3834atm 0.3834atm = 38.85kPa

  10. Boyle’s Law - The relationship between volume and the pressure of a gas. (Temperature is kept constant.)

  11. 1 P Boyle’sLaw V a n and T are fixed PV = constant V = constant / P

  12. Charles’s Law - The relationship between volume and the temperature of a gas. (Pressure is kept constant.)

  13. 1 T T V a P P P = constant P V T T = constant V a V = constant x = constant PV T Boyle’sLaw n and T are fixed Charles’sLaw V a T P and n are fixed V = constant x T Amonton’sLaw P a T V and n are fixed P = constant x T combined gas law When the amount of gas, n, is constant

  14. = constant V n 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 = constant x n Twice the amountgives twice the volume

  15. 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

  16. PV 1atm x 22.414L 0.0821atm-L nT 1mol x 273.15K Mol-K IDEAL GAS LAW V P fixed n and T fixed n and P fixed P and T Boyle’s Law Charles’s Law Avogadro’s Law V = constant X n = constant V = constant X T THE IDEAL GAS LAW PV = nRT R = = =

  17. PROBLEM: 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)? unit conversion 1mL L 1cm3 103mL gas law calculation P2V2 P1V1 n2T2 n1T1 1.12atm V2 = = 0.0105L = 0.0248L 2.46atm Sample Problem 5.2 Applying the Volume-Pressure Relationship PLAN: SOLUTION: n and T are constant V1 in cm3 P1 = 1.12atm P2 = 2.64atm 1cm3=1mL V1 = 24.8cm3 V2 = unknown V1 in mL 103mL=1L 24.8cm3 = 0.0248L = V1 V1 in L xP1/P2 P1V1 = P2V2 = R = = V2 in L P1V1 P2

  18. P1 P2 = = T1 T2 760 torr P2V2 0.991atm P1V1 1atm n2T2 n1T1 T2 373K P2 = P1 = 753 torr T1 296K Sample Problem 5.3 Applying the Temperature-Pressure Relationship PROBLEM: 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: P1(atm) T1 and T2(0C) P1 = 0.991atm P2 = unknown 1atm=760torr K=0C+273.15 T1 = 230C T2 = 100oC P1(torr) T1 and T2(K) x T2/T1 P2(torr) = 753 torr = 949 torr

  19. = = n2 = n1 V1 V2 V2 P2V2 P1V1 n1 V1 n2 n2T2 n1T1 55.0dm3 4.003g He n2 = 1.10mol 26.2dm3 mol He Sample Problem 5.4 Applying the Volume-Amount Relationship PROBLEM: 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 SOLUTION: P and T are constant x V2/V1 n1 = 1.10mol n2 = unknown n2(mol) of He V1 = 26.2dm3 V2 = 55.0dm3 subtract n1 mol to be added x M = 2.31mol g to be added n2 - n1 = 2.31 -1.10 = 1.21 mol He = 4.84g He

  20. 103g mol O2 0.885kg kg 32.00g O2 atm*L nRT 27.7mol x 0.0821 x 294K mol*K V P = = 438L Sample Problem 5.5 Solving for an Unknown Gas Variable at Fixed Conditions PROBLEM: A steel tank has a volume of 438L and is filled with 0.885kg of O2. Calculate the pressure of O2 at 210C. PLAN: V, T and mass, which can be converted to moles (n), are given. We use the ideal gas law to find P. SOLUTION: V = 438L T = 210C (convert to K) n = 0.885kg (convert to mol) P = unknown 210C + 273.15 = 294K = 27.7mol O2 = 1.53atm

  21. M x P RT 44.01g/mol x 1atm x 273K atm*L 0.0821 mol*K 1.96g mol CO2 6.022x1023molecules L 44.01g CO2 mol Sample Problem 5.6 Calculating Gas Density PROBLEM: 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). 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 Avogardro’s number. d = mass/ V PV = nRT mult. both sides by M d = n / V = P / RT Mass = n M SOLUTION: = 1.96g/L d = (a) = 2.68x1022molecules CO2/L

  22. 44.01g/mol x 1atm d = x 293K atm*L 0.0821 mol*K 1.83g mol CO2 6.022x1023molecules L 44.01g CO2 mol Sample Problem 5.6 Calculating Gas Density continued (b) = 1.83g/L = 2.50x1022molecules CO2/L

  23. Calculating the Molar Mass, M, of a Gas Since PV = nRT Then n = PV / RT And n = mass / M So m / M = PV / RT AndM= mRT / PV OrM= d RT / P

  24. Volume of flask = 213mL T = 100.00C P = 754 torr Mass of flask + gas = 78.416g Mass of flask = 77.834g atm*L 0.0821 m RT 0.582g 373K x mol*K = 84.4g/mol P V 0.213L x 0.992atm Sample Problem 5.7 Finding the Molar Mass of a Volatile Liquid PROBLEM: 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: 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 x M = = M of C6H12 is 84.16g/mol and the calculated value is within experimental error.

  25. n1 n1 = ntotal n1 + n2 + n3 +... c1 = 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

  26. 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.

  27. PROBLEM: 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. Find the c and P from Ptotal and mol% 18O2. 18O2 18O2 4.0mol% 18O2 = c 18O2 100 P = c x Ptotal = 0.040 x 0.75atm 18O2 18O2 partial pressure P 18O2 Sample Problem 5.8 Applying Dalton’s Law of Partial Pressures PLAN: mol% 18O2 SOLUTION: = 0.040 divide by 100 c 18O2 = 0.030atm multiply by Ptotal

  28. PROBLEM: 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? P P g C2H2 C2H2 C2H2 PV atm P n = RT 760torr n C2H2 Sample Problem 5.9 Calculating the Amount of Gas Collected Over Water 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. SOLUTION: = (738-21)torr = 717torr Ptotal = 0.943atm 717torr H2O x M

  29. n = C2H2 26.04g C2H2 0.203mol mol C2H2 atm*L 0.0821 mol*K Sample Problem 5.9 Calculating the Amount of Gas Collected Over Water continued 0.943atm x 0.523L = 0.203mol x 296K = 0.529 g C2H2

  30. 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 B amount (mol) of gas A P,V,T of gas A P,V,T of gas B ideal gas law ideal gas law molar ratio from balanced equation

  31. PROBLEM: 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) Cu(s) + H2O(g) mol Cu 1mol H2 63.55g Cu 1 mol Cu atm*L 0.0821 x mol*K x 498K 1.01atm Sample Problem 5.10 Using Gas Variables to Find Amount of Reactants and Products mass (g) of Cu SOLUTION: divide by M 35.5g Cu = 0.559mol H2 mol of Cu molar ratio 0.559mol H2 = 22.6L mol of H2 use known P and T to find V L of H2

  32. 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. P = 0.950atm V = 5.25L 2K(s) + Cl2(g) 2KCl(s) 0.950atm T = 293K n = unknown = = 0.207mol PV x 293K RT 2mol KCl 2mol KCl atm*L 0.207mol Cl2 17.0g = 0.435mol K 0.0821 1mol Cl2 2mol K mol*K mol K 39.10g K 74.55g KCl mol KCl Sample Problem 5.11 Using the Ideal Gas Law in a Limiting-Reactant Problem SOLUTION: x 5.25L n = Cl2 = 0.414mol KCl formed = 0.435mol KCl formed Cl2 is the limiting reactant. 0.435mol K 0.414mol KCl = 30.9 g KCl

  33. Postulate 1: Particle Volume Postulate 2: Particle Motion Postulate 3: Particle Collisions Postulates of the Kinetic-Molecular Theory 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. Gas particles are in constant, random, straight-line motion except when they collide with each other or with the container walls. Collisions are elastic therefore the total kinetic energy(Kk) of the particles is constant.

  34. 1 P A molecular description of Boyle’s Law Boyle’sLaw V a n and T are fixed

  35. A molecular description of Charles’s Law Charles’sLaw V a T n and P are fixed

  36. A molecular description of Dalton’s law of partial pressures. Dalton’s Law of Partial Pressures Ptotal = P1 + P2 + P3 + ...

  37. A molecular description of Avogadro’s Law Avogadro’s Law V a n P and T are fixed

  38. 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.

  39. 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

  40. Relationship between molar mass and molecular speed.

  41. rate A = rate B 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. ( MB / MA ) ½

  42. 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. rate He rate CH4 Sample Problem 5.12 Applying Graham’s Law of Effusion SOLUTION: M of CH4 = 16.04g/mol M of He = 4.003g/mol = ( 16.04/ 4.003 ) ½ = 2.002

  43. End of Chapter 5

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