ก ก Gases and the Kinetic- Molecular Theory ก ก ก ก Mc-Graw Hill 1 Gases and the Kinetic Molecular Theory 5.1 An Overview of the Physical States of Matter 5. 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
Table 5.1 Some Important Industrial Gases Name (Formula) Methane (CH 4 ) Ammonia (NH 3 ) Chlorine (Cl ) Oxygen (O ) Ethylene (C H 4 ) Origin and Use natural deposits; domestic fuel from N +H ; fertilizers, explosives electrolysis of seawater; bleaching and disinfecting liquefied air; steelmaking high-temperature decomposition of natural gas; plastics Atmosphere-Biosphere Redox Interconnections 3 5.1 An Overview of the Physical States of Matter The Distinction of Gases from Liquids and Solids 1. Gas volume changes greatly with pressure.. Gas volume changes greatly with temperature. 3. Gases have relatively low viscosity. 4. Most gases have relatively low densities under normal conditions. 5. Gases are miscible. 4
The three states of matter: Bromine 5 5. Gas Pressure and Its Measurement Effect of atmospheric pressure on objects at the Earth s surface. P inside = P outside P inside = 0 6
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. P = F A Manometer - A device to measure gas pressure in a closed container. 7 d A mercury barometer Torricelli 1608-1647 Is h dependent on d? 8
closed-end Two types of manometer open-end 9 Table 5. Common Units of Pressure Unit pascal(pa); kilopascal(kpa) atmosphere(atm) Atmospheric Pressure 1.0135x10 5 Pa; 101.35 kpa 1 atm* Scientific Field SI unit; physics, chemistry chemistry millimeters of mercury(hg) 760 mm Hg* chemistry, medicine, biology torr 760 torr* chemistry pounds per square inch (psi or lb/in ) 14.7lb/in engineering bar 1.0135 bar meteorology, chemistry, physics *This is an exact quantity; in calculations, we use as many significant figures as necessary. 10
Sample Problem 5.1 Converting Units of Pressure PROBLEM: PLAN: A geochemist heats a limestone (CaCO 3 ) sample and collects the CO released in an evacuated flask attached to a closedend manometer. After the system comes to room temperature, h = 91.4 mm Hg. Calculate the CO pressure in torrs, atmospheres, and kilopascals. Construct conversion factors to find the other units of pressure. SOLUTION: 91.4 mmhg 1torr 1 mmhg 91.4 torr 1 atm 760 torr = 91.4 torr = 0.3834 atm 0.3834 atm 101.35 kpa 1 atm = 38.85 kpa 11 5.3 The Gas Laws and Their Experimental Foundations Boyle s Law - The relationship between the volume and pressure of a gas. (Temperature is kept constant.) 1
Charles s Law - The relationship between volume and the temperature of a gas. (Pressure is kept constant.) 13 Amontons Law The relationship between the pressure and temperature of a gas. (Volume is kept constant.) 14
1 Boyle s Law V α P n and T are fixed V x P = constant V = constant / P 167 1691 1837 193 Charles s Law V α T P and n are fixed V T = constant V = constant x T 1746 183 Amontons Law P α T V and n are fixed Combined gas law P T = constant P = constant x T T V α V = constant x T PV P P T 1663 1705 = constant 15 Gay-Lussac s Law Law of Combining Volumes Gay-Lussac 1778 1850 The ratio between the combining volumes of gases and the product, if gaseous, can be expressed in small whole numbers. Gay-Lussac s law was discovered in 1809. This law played a major role in the development of modern gas stoichiometry because in 1811, Avogadro used Gay-Lussac s Law to form Avogadro's hypothesis. 16
Avogadro s Law - The volume of a gas is directly proportionate to the amount of gas. (Pressure and temperature are kept constant.) An experiment to study the relationship between the volume and amount of a gas. 1776 1856 V α n V n = constant V = constant x n 17 If the constant pressure is 1 atm and the constant temperature is 0 o C, 1 mole of any gas has a volume of.4 L. This is known as the Standard Molar Volume 18
The volume of 1 mol of an ideal gas compared with some familiar objects. He.4 L 5-gal fish tank 18.9 L 13-in TV 1.6 L 7.5 L 19 THE IDEAL GAS LAW R = PV = nt R is the universal gas constant PV = nrt 1atm x.414l 1mol x 73.15K = IDEAL GAS LAW PV = nrt or V = nrt P 0.081atm*L mol*k 3 significant figures fixed n and T fixed n and P fixed P and T Boyle s Law Charles s Law Avogadro s Law V = constant P V = constant X T V = constant X n 0
Sample Problem 5. Applying the Volume-Pressure Relationship PROBLEM: Boyle s apprentice finds that the air trapped in a J tube occupies 4.8 cm 3 at 1.1 atm. By adding mercury to the tube, Hg increases the pressure on the trapped air to.64 atm. Assuming constant temperature, what is the new volume of air (in L)? PLAN: V 1 in cm 3 1cm 3 =1mL unit V 1 in ml conversion 10 3 ml=1l V 1 in L xp 1 /P V in L Answer : 0.0105 L gas law calculation V = SOLUTION: P 1 V 1 P P 1 = 1.1 atm V 1 = 4.8 cm 3 4.8 cm 3 1 ml 1 cm 3 L 10 3 ml P 1 V 1 P V = n 1 T 1 n T = 0.048 L P and T are constant P =.64 atm V = unknown = 0.048 L P 1 V 1 = P V 1.1 atm.46 atm = 0.0105 L 1 Sample Problem 5.3 Applying the Temperature-Pressure Relationship PROBLEM: PLAN: A steel tank used for fuel delivery is fitted with a safety valve that opens when the internal pressure exceeds 1.00x10 3 torr. It is filled with methane at 3 0 C and 0.991 atm and placed in boiling water at exactly 100 0 C. Will the safety valve open? SOLUTION: P 1 (atm) T 1 and T ( 0 C) 1atm=760torr P 1 (torr) x T /T 1 P (torr) K= 0 C+73.15 T 1 and T (K) 0.991 atm P 1 = 0.991atm T 1 = 3 0 C P 1 V 1 P V = n 1 T 1 n T 760 torr = 753 torr 1 atm P = unknown T = 100 0 C P 1 P = T 1 T P = P 1 T Answer : 949 torr T 1 = 753 torr 373K 96K = 949 torr
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 55 dm 3. When 1.10 mol of He is added to the blimp, the volume is 6. dm 3. How many more grams of He must be added to make it rise? Assume constant T and P. PLAN: We are given initial n 1 and V 1 as well as the final V. We have to find n and convert it from moles to grams. n 1 (mol) of He SOLUTION: P and T are constant x V /V 1 n (mol) of He subtract n 1 mol to be added x M g to be added Answer : 9.4 g He V 1 n 1 = 1.10 mol n = unknown V 1 = 6. dm 3 V = 55.0 dm 3 V V = n n 1 n = n 1 n = 1.10 mol V 1 55.0 dm 3 6. dm =.31 mol 4.003 g He 3 mol He P 1 V 1 P V = n 1 T 1 n T = 9.4 g He 3 Sample Problem 5.5 PROBLEM: Solving for an Unknown Gas Variable at Fixed Conditions A steel tank has a volume of 438 L and is filled with 0.885 kg of O. Calculate the pressure of O at 1 0 C. 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 = 438 L T = 1 0 C (convert to K) n = 0.885 kg (convert to mol) P = unknown 0.885kg 103 g kg nrt P = = V mol O 3.00 g O = 7.7 mol O atm*l 4.7 mol x 0.081 x 94.15K mol*k 438 L 1 0 C + 73.15 = 94.15K = 1.53 atm Answer : 1.53 atm 4
Sample Problem 5.6 Using Gas Laws to Determine a Balanced Equation PROBLEM: The piston-cylinders below depict a gaseous reaction carried out at constant pressure. Before the reaction, the temperature is 150K; when it is complete, the temperature is 300K. Which of the following balanced equations describes the reaction? (1) A + B AB () AB + B AB (3) A + B AB (4) AB A + B PLAN: SOLUTION: We know P, T, and V, initial and final, from the pictures. Note that the volume doesn t change even though the temperature is doubled. With a doubling of T then, the number of moles of gas must have been halved in order to maintain the volume. Looking at the relationships, the equation that shows a decrease in the number of moles of gas from to 1 is equation (3). 5 5.4 Further Applications of the Ideal Gas Law The Density of a Gas density = m/v n = m/m PV = nrt PV = (m/m)rt m/v = M x P/ RT The density of a gas is directly proportional to its molar mass. The density of a gas is inversely proportional to the temperature. 6
Sample Problem 5.7 Calculating Gas Density PROBLEM: To apply a green chemistry approach, a chemical engineer uses waste CO from a manufacturing process, instead of chlorofluorocarbons, as a blowing agent in the production of polystyrene containers. Find the density (in g/l) of CO and the number of molecules (a) at STP (0 0 C and 1 atm) and (b) at room conditions (0. 0 C 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. M x P d = mass/volume PV = nrt V = nrt/p d = RT SOLUTION: 44.01 g/mol x 1atm d = = 1.96 g/l (a) atm*l 0.081 x 73.15K mol*k 1.96 g mol CO 6.0x10 3 molecules =.68x10 molecules CO /L L 44.01 g CO mol Answers : (a) 1.96 g/l,.68x10 molecules CO /L 7 Sample Problem 5.7 Calculating Gas Density continued (b) d = 44.01 g/mol x 1 atm 0.081 atm*l mol*k x 93K = 1.83 g/l 1.83g L mol CO 6.0x10 3 molecules 44.01g CO mol =.50x10 molecules CO /L Answers : (b) 1.83 g/l,.50x10 molecules CO /L 8
Calculating the Molar Mass, M, of a Gas 1800-1884 Dumas method - Determining the molar mass of an unknown volatile liquid. 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 9 Sample Problem 5.8 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 (C 6 H 1 ). She uses the Dumas method and obtains the following data to determine its molar mass: Volume of flask = 13mL Mass of flask + gas = 78.416g T = 100.0 0 C 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 = 0.58g M = m RT P V = atm*l 0.58g x 0.081 x 373K mol*k 0.13L x 0.99atm = 84.4g/mol M of C 6 H 1 is 84.16g/mol and the calculated value is within experimental error. Answer : 84.4 g/mol 30
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 P total = P 1 + P + P 3 +... 1766 1844 P 1 = χ 1 x P total where χ 1 is the mole fraction χ 1 = n 1 n 1 + n + n 3 +... = n 1 n total 31 Sample Problem 5.9 Applying Dalton s Law of Partial Pressures PROBLEM: In a study of O uptake by muscle at high altitude, a physiologist prepares an atmosphere consisting of 79 mol% N, 17 mol% 16 O, and 4.0 mol% 18 O. (The isotope 18 O will be measured to determine the O uptake.) The pressure of the mixture is 0.75atm to simulate high altitude. Calculate the mole fraction and partial pressure of 18 O in the mixture. PLAN: Find the χ and P from P total and mol% 18 O. 18 O 18 O mol% 18 O divide by 100 SOLUTION: χ 18 O = 4.0 mol% 18 O 100 = 0.040 χ 18 O P = χ x P total = 0.040 x 0.75 atm = 0.030 atm 18 O 18 O multiply by P total partial pressure P 18 O Answer : 0.030 atm 3
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. 33 Table 5.3 Vapor Pressure of Water (P H O ) at Different T T( 0 C) P (torr) T( 0 C) P (torr) 0 5 10 11 1 13 14 15 16 18 0 4 6 8 4.6 6.5 9. 9.8 10.5 11. 1.0 1.8 13.6 15.5 17.5 19.8.4 5. 8.3 30 35 40 45 50 55 60 65 70 75 80 85 90 95 100 31.8 4. 55.3 71.9 9.5 118.0 149.4 187.5 33.7 89.1 355.1 433.6 55.8 633.9 760.0 34
Sample Problem 5.10 Calculating the Amount of Gas Collected Over Water PROBLEM: Acetylene (C H ), an important fuel in welding, is produced in the laboratory when calcium carbide (CaC ) reaction with water: CaC (s) + H O(l) C H (g) + Ca(OH) (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 53mL. At the temperature of the gas (3 0 C), the vapor pressure of water is 1torr. How many grams of acetylene are collected? PLAN: The difference in pressures will give us the P for the C H. The ideal gas law will allow us to find n. Converting n to grams requires the molar mass, M. SOLUTION: P C H = (738-1) torr = 717 torr P total P C H atm P H O n = PV 717 torr = 0.943 atm 760 torr RT 0.943atm x 0.53L n n = = 0.003 mol C H g C H C H atm*l 0.081 x 96K x M mol*k 6.04g C H 0.003mol = 0.59 g C H35 Answer : 0.943 atm, 0.59 g C H mol C H 5.5 The Ideal gas Law and Reaction Stoichiometry 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 amount (mol) of gas A amount (mol) of gas B P,V,T of gas B ideal gas law molar ratio from balanced equation ideal gas law 36
Sample Problem 5.11 Using Gas Variables to Find Amount of Reactants and Products PROBLEM: 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 H reduces the metal oxide, forming the pure metal and H O. On a laboratory scale, what volume of H at 765 torr and 5 0 C 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 H gas. mass (g) of Cu SOLUTION: CuO(s) + H (g) Cu(s) + H O(g) divide by M mol Cu 1 mol H mol of Cu 35.5 g Cu = 0.559 mol H 63.55 g Cu 1 mol Cu molar ratio mol of H 0.559 mol H x 0.081 atm*l x 498K =.6 L mol*k use known P and T to find V 1.01 atm L of H Answer :.6 L 37 Sample Problem 5.11 PROBLEM: PLAN: Using the Ideal Gas Law in a Limiting-Reactant 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 5.5L of chlorine gas at 0.950atm and 93K reacts with 17.0g of potassium? 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. K(s) + Cl (g) KCl(s) P = 0.950atm V = 5.5L SOLUTION: T = 93K n = unknown n = PV x 5.5L = 0.950atm = 0.07mol Cl RT atm*l 0.081 x 93K mol KCl mol*k 0.07mol Cl = 0.414mol 17.0g mol K 1mol Cl KCl formed = 0.435mol K 39.10g K Cl mol KCl is the limiting reactant. 0.435mol K = 0.435mol 74.55g KCl mol K KCl formed 0.414mol KCl = 30.9 g KCl mol KCl Answer : 0.414 mol KCl formed 38
5.6 The Kinetic-Molecular Theory 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 : 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 (E k ) of the particles is constant. Molecular View of the Gas Laws 39 Kinetic Molecular Theory of Gases Bernoulli 1700 178 The kinetic molecular theory of gases proposed in 1734 by Bernoulli. The pressure exerted by a gas on the walls of its container is the sum of the many collisions by individual molecules, all moving independently of each other. This basic law leads to the equation of state of real gases discovered by van der Waals a century later. 40
A molecular description of Boyle s Law 1 Boyle s Law V α P n and T are fixed 41 A molecular description of Charles s Law Charles s Law V α T n and P are fixed 4
A molecular description of Dalton s law of partial pressures Dalton s Law of Partial Pressures P total = P 1 + P + P 3 +... 43 A molecular description of Avogadro s Law Avogadro s Law V α n P and T are fixed 44
Maxwell-Boltzmann Distribution Law of Molecular Velocity N( v) = 4π N( m π k B T ) 3/ v e mv k T B 1831 1879 1844 1906 N is the total number of molecules and N(v)dv is the number of molecules with speeds between v and v+dv. v k BT m p = = 1.41 k BT m most probable speed v = v 8 k T π m 3kBT m B = rms = = 1.73 1.60 kbt m k B T m mean speed root - mean -square speed N ( v) dv = 0 N 45 Average velocity is inversely proportional to the square root of molar mass applies to other transport phenomena such as diffusion, thermal conduction, and nonturbulant flow. Example 46
Kinetic-Molecular Theory Gas particles are in motion and have a molecular speed, u. But they are moving at various speeds, some very slow, some very fast, but most near the average (mean) speed of all of the particles, u (avg). Since kinetic energy is defined as ½ mass x (speed), we can define the average kinetic energy, E k(avg) = ½mu (avg) E k = ½mu < E k > = ½m< u > Since energy is a function of temperature, the average energy and, hence, average molecular speed will increase with temperature. 47 An increase in temperature results in an increase in average molecular kinetic energy. Average kinetic energy, E k(avg) = ½mu (avg) The relationship between average kinetic energy and temperature is given as, E k(avg) = 3/ (R/N A ) x T (where R is the gas constant in energy units, 8.314 J mol -1 K -1 N A 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, u (rms), is the speed where a molecule has the average kinetic energy. E k(avg) = ½mu (avg) = 3/ (R/N A ) x T (u (avg) )½ = u (rms) The relationship between u (rms) and molar mass is: u (rms) = (3RT/M ) ½ So the speed (or rate of movement) is: rate α 1 / (M ) ½ 48
Relationship between molar mass and molecular speed 49 Effusion Effusion and Diffusion The process by which a gas escapes from its container through a tiny hole into an evacuated space. Diffusion The movement of one gas through another. 50
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. 1805 1869 Rate of effusion α 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 = ( M B / M A ) ½ 51 Sample Problem 5.13 Applying Graham s Law of Effusion PROBLEM: Calculate the ratio of the effusion rates of helium and methane (CH 4 ). 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 CH 4 = 16.04g/mol M of He = 4.003g/mol rate He rate CH 4 = 16.04 4.003 =.00 Answer :.00 5
Mean Free Path and Collision Frequency Two spherical molecules of diameter d will collide if their paths are within a distance d of each other. The mean distance traveled by molecules between collisions is called the mean free path (λ). The average number of collisions per second that each molecule undergoes is called the collision frequency (z). Mean free path = 1 π d n V n V is the number of molecules per unit volume (N/V). mean speed Collision frequency = = mean free path πd νn V 53 Relative mean speed, v rel speed one molecule approaches another Range of approaches, but typical approach is from side v v rel rel rel = v 8kT = πµ A 0.5 k = Boltzmann constant = µ = reduced mass = If m v = m, µ = B m 0.5 8RT = = v π M A R N mamb m + m A B 54
5.7 Real Gases: Deviations from Ideal Gases Table 5.4 Molar Volume of Some Common Gases at STP (0 0 C and 1 atm) Gas Molar Volume (L/mol) Condensation Point ( 0 C) He H Ne Ideal gas Ar N O CO Cl NH 3.435.43.4.414.397.396.390.388.184.079-68.9-5.8-46.1 --- -185.9-195.8-183.0-191.5-34.0-33.4 55 Effects of Extreme Conditions on Gas Behavior PV = nrt PV nrt =1 56
The effect of intermolecular attractions on measured gas pressure 57 The effect of molecular volume on measured gas volume 58
Van der Waals Equation (P + n a )(V nb) = nrt V adjusts P up adjusts V down 1837 193 Table 5.5 Van der Waals Constants for Some Common Gases Gas a atm*l mol b L mol He Ne Ar Kr Xe H N O Cl CO CH 4 NH 3 H O 0.034 0.11 1.35.3 4.19 0.44 1.39 1.36 6.49 3.59.5 4.17 5.46 0.037 0.0171 0.03 0.0398 0.0511 0.066 0.0391 0.0318 0.056 0.047 0.048 0.0371 0.0305 59 P Ideal V Ideal PV = n = PReal + a V = V Real nb nrt n P + a = V nrt n P = a V nb V ( V nb) nrt Effect of stickiness (i.e. probability of collision square of number density). Effect of finite molecular volume Molecular volume Molecular interactions 60
n P + a = V ( V nb) nrt nrt n RT a P = a = V nb V V b V m m 8.314 8.314 61 6
T/T c = 1.5 Critical isotherm T/T c = 1.0 63 Chemistry in Planetary Science Variations in pressure, temperature, and composition of the Earth s atmosphere. 64
65 66