Gaseous state
Table 5.1 Some Important Industrial Gases Name (Formula) Methane (CH 4 ) Ammonia (NH 3 ) Chlorine (Cl 2 ) Oxygen (O 2 ) Ethylene (C 2 H 4 ) Origin and Use natural deposits; domestic fuel from N 2 +H 2 ; fertilizers, explosives electrolysis of seawater; bleaching and disinfecting liquefied air; steelmaking high-temperature decomposition of natural gas; plastics Atmosphere-Biosphere Redox Interconnections
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. Gases have relatively low viscosity. 4. Most gases have relatively low densities under normal conditions. 5. Gases are miscible.
Three states of the matter
Vacuum Effect of atmospheric pressure on objects at the Earth s surface.
A mercury barometer.
closed-end Two types of manometer open-end
Table 5.2 Common Units of Pressure Unit pascal(pa); kilopascal(kpa) atmosphere(atm) Atmospheric Pressure 1.01325x10 5 Pa; 101.325 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 2 ) 14.7lb/in 2 engineering bar 1.01325 bar meteorology, chemistry, physics *This is an exact quantity; in calculations, we use as many significant figures as necessary.
Converting Units of Pressure PROBLEM: A geochemist heats a limestone (CaCO 3 ) sample and collects the CO 2 released in an evacuated flask attached to a closedend manometer. After the system comes to room temperature, Dh = 291.4 mm Hg. Calculate the CO 2 pressure in torrs, atmospheres, and kilopascals. PLAN: Construct conversion factors to find the other units of pressure. SOLUTION: 291.4 mmhg 1torr 1 mmhg 291.4 torr 1 atm 760 torr = 291.4 torr = 0.3834 atm 0.3834 atm 101.325 kpa 1 atm = 38.85 kpa
The relationship between the volume and pressure of a gas. Boyle s Law
Figure 5.6 The relationship between the volume and temperature of a gas. Charles s Law
Boyle s Law V a 1 P n and T are fixed V x P = constant V = constant / P Charles s Law V a T P and n are fixed V T = constant V = constant x T Amontons s Law P a T V and n are fixed P T = constant P = constant x T combined gas law V a T P V = constant x T P PV T = constant
An experiment to study the relationship between the volume and amount of a gas.
Standard molar volume.
The volume of 1 mol of an ideal gas compared with some familiar objects.
Equal volumes of different gases contain, in the same conditions of P and T, the same number of particles 100 ml H 2 double amount of moles of 50 ml of O 2
Avogadro s Law Gas Volume at certain T and P is straight proportional to the gas quantity, that is to its moles number ( n = g/mm) V n
THE IDEAL GAS LAW R = PV nt R is the universal gas constant PV = nrt 1atm x 22.414L 1mol x 273.15K = IDEAL GAS LAW 0.0821atm*L mol*k 3 significant figures PV = nrt or V = nrt P 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
Applying the Volume-Pressure Relationship PROBLEM: Boyle s apprentice finds that the air trapped in a J tube occupies 24.8 cm 3 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: V 1 in cm 3 1cm 3 =1mL V 1 in ml 10 3 ml=1l V 1 in L xp 1 /P 2 V 2 in L unit conversion gas law calculation V 2 = SOLUTION: P 1 V 1 P 2 P 1 = 1.12 atm V 1 = 24.8 cm 3 24.8 cm 3 L P 1 V 1 P 2 V 2 = n 1 T 1 n 2 T 2 = 0.0248 L P and T are constant 10 3 ml 1.12 atm 2.46 atm P 2 = 2.64 atm V 2 = unknown = 0.0248 L P 1 V 1 = P 2 V 2 = 0.0105 L
Applying the Temperature-Pressure Relationship PROBLEM: 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 23 0 C and 0.991 atm and placed in boiling water at exactly 100 0 C. Will the safety valve open? PLAN: P 1 (atm) T 1 and T 2 ( 0 C) 1atm=760torr P 1 (torr) x T 2 /T 1 P 2 (torr) K= 0 C+273.15 T 1 and T 2 (K) 0.991 atm SOLUTION: P 1 = 0.991atm T 1 = 23 0 C P 1 V 1 P 2 V 2 = n 1 T 1 n 2 T 2 760 torr = 753 torr 1 atm P 2 = unknown T 2 = 100 0 C P 1 P 2 = T 1 T 2 P 2 = P 1 T 2 T 1 = 753 torr 373K 296K = 949 torr
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 26.2 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 2. We have to find n 2 and convert it from moles to grams. n 1 (mol) of He x V 2 /V 1 n 2 (mol) of He subtract n 1 mol to be added x M g to be added SOLUTION: V 1 n 1 = 1.10 mol n 2 = unknown V 1 = 26.2 dm 3 V 2 = 55.0 dm 3 V 2 V = n n 1 n 2 = n 2 1 2 n 2 = 1.10 mol V 1 P and T are constant 55.0 dm 3 g He = 2.31 mol4.003 26.2 dm3 mol He P 1 V 1 P 2 V 2 = n 1 T 1 n 2 T 2 = 9.24 g He
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 2. Calculate the pressure of O 2 at 21 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 = 21 0 C (convert to K) n = 0.885 kg (convert to mol) P = unknown 0.885kg 103 g kg nrt P = V mol O 2 32.00 g O 2 = 27.7 mol O 2 atm*l 24.7 mol x 0.0821 x 294.15K mol*k 438 L 21 0 C + 273.15 = 294.15K = 1.53 atm
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. New figures go here. Which of the following balanced equations describes the reaction? (1) A 2 + B 2 2AB (2) 2AB + B 2 2AB 2 (3) A + B 2 AB 2 (4) 2AB 2 A 2 + 2B 2 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 2 to 1 is equation (3).
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.
PROBLEM: Calculating Gas Density To apply a green chemistry approach, a chemical engineer uses waste CO 2 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 2 and the number of molecules (a) at STP (0 0 C and 1 atm) and (b) at room conditions (20. 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 0.0821 x 273.15K mol*k 1.96 g mol CO 2 6.022x10 23 molecules = 2.68x10 22 molecules CO 2 /L L 44.01 g CO 2 mol
Calculating Gas Density continued (b) d = 44.01 g/mol x 1 atm 0.0821 atm*l mol*k x 293K = 1.83 g/l 1.83g L mol CO 2 6.022x10 23 molecules 44.01g CO 2 mol = 2.50x10 22 molecules CO 2 /L
Determining the molar mass of an unknown volatile liquid. based on the method of J.B.A. Dumas (1800-1884)
Finding the Molar Mass of a Volatile Liquid PROBLEM: 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 ml Mass of flask + gas = 78.416 g T = 100.0 0 C Mass of flask = 77.834 g P = 754 torr Calculate the molar mass of the liquid. PLAN: Use unit conversions, mass of gas, and density-m relationship. SOLUTION: m = (78.416-77.834) g = 0.582 g atm*l m RT 0.582 g 0.0821 x 373K M = mol*k VP 0.213 L x 0.992 atm = 84.4 g/mol
Mixtures 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 Pressures P total = P 1 + P 2 + P 3 +... P 1 = c 1 x P total c 1 = n 1 n 1 + n 2 + n 3 +... = n 1 n total
Applying Dalton s Law of Partial Pressures PROBLEM: In a study of O 2 uptake by muscle at high altitude, a physiologist prepares an atmosphere consisting of 79 mol% N 2, 17 mol% 16 O 2, and 4.0 mol% 18 O 2. (The isotope 18 O will be measured to determine the O 2 uptake.) The pressure of the mixture is 0.75atm to simulate high altitude. Calculate the mole fraction and partial pressure of 18 O 2 in the mixture. PLAN: Find the c and P from P total and mol% 18 O 2. 18 O 2 18 O 2 mol% 18 O 2 divide by 100 SOLUTION: c 18 O 2 = 4.0 mol% 18 O 2 100 c 18 O 2 P = c x P total = 0.040 x 0.75 atm = 0.030 atm 18 O 18 2 O 2 multiply by P total
The Molar Mass of a Gas n = mass M = PV RT M = m RT VP d = m V M = d RT P
Table 5.3 Vapor Pressure of Water (P H 2 O ) at Different T T( 0 C) P (torr) T( 0 C) P (torr) 0 5 10 11 12 13 14 15 16 18 20 22 24 26 28 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 25.2 28.3 30 35 40 45 50 55 60 65 70 75 80 85 90 95 100 31.8 42.2 55.3 71.9 92.5 118.0 149.4 187.5 233.7 289.1 355.1 433.6 525.8 633.9 760.0
Collecting a water-insoluble gaseous reaction product and determining its pressure.
PROBLEM: Calculating the Amount of Gas Collected Over Water Acetylene (C 2 H 2 ), an important fuel in welding, is produced in the laboratory when calcium carbide (CaC 2 ) reaction with water: CaC 2 (s) + 2H 2 O(l) C 2 H 2 (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 (23 0 C), 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 C 2 H 2. The ideal gas law will allow us to find n. Converting n to grams requires the molar mass, M. P P SOLUTION: total P C2 H = (738-21)torr = 717torr 2 C2 H 2 H 2 O n = PV atm 717torr = 0.943atm RT 760torr n g C2 H 2 C2 H 2 x M
Calculating the Amount of Gas Collected Over Water continued n C2 H 2 = = 0.0203mol 0.0203mol 26.04g C 2 H 2 mol C 2 H 2 = 0.529 g C 2 H 2
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
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 2 reduces the metal oxide, forming the pure metal and H 2 O. On a laboratory scale, what volume of H 2 at 765 torr and 225 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 2 gas. mass (g) of Cu SOLUTION: CuO(s) + H 2 (g) Cu(s) + H 2 O(g) divide by M mol Cu 1 mol H mol of Cu 35.5 g Cu 2 = 0.559 mol H 2 63.55 g Cu (da 1 mol Cu molar ratio vcorrggere!!!) 0.559 mol H mol of H 2 x 0.0821 atm*l x 498K = 22.6 L 2 mol*k use known P and T to find V 1.01 atm L of H 2
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 5.25 L of chlorine gas at 0.950 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) + Cl 2 (g) 2KCl(s) P = 0.950 atm V = 5.25 L PV 0.950 atm X 5.25 L T = 293K n = unknown = = 0.207 mol Cl 2 RT atm*l 0.0821 x 293K mol*k 2 mol KCl 17.0g mol K 0.207 mol Cl = 0.435 mol K 2 = 0.414 mol 1 mol Cl 39.10 g K 2 KCl formed 2 mol KCl Cl 2 is the limiting reactant. 0.435 mol K = 0.435 mol 2 mol K 74.55 g KCl KCl formed 0.414 mol KCl = 30.9 g KCl mol KCl
Diffusion of a gas
Effusion of a gas It is the movement of gas through a tiny opening in a container into another containerwhere the pressure is very low.
Graham law Rate of effusion of gas1 (molar mass of gas 2) 1/2 ------------------------------ = ------------------------------------- Rate of effusion of gas2 (molar mass of gas 1) 1/2 By considering a constant path size for both gases: t2 (molar mass of gas2) 1/2 -------- = ------------------------------- t1 (molar mass of gas1) 1/2 where t = effusion time
Kinetic molecular theory of gases It describes the behavior of matter at the molecular or atomic level as concerns the gaseous state. The molecular model allows us to interpret the macroscopic evidences by a statistical mathematical model.
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(e k ) of the particles is constant.
Distribution of molecular speeds at three temperatures.
A molecular description of Boyle s Law.
A molecular description of Dalton s law of partial pressures.
A molecular description of Charles s Law.
Avogadro s Law V a n E k = 1/2 mass x speed 2 E k = 1/2 mass x u 2 u 2 is the root-mean-square speed u rms = 3RT M R = 8.314Joule/mol*K 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
A molecular description of Avogadro s Law.
Relationship between molar mass and molecular speed.
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 rate He CH 4 = 16.04 4.003 = 2.002
Diffusion of a gas particle through a space filled with other particles. distribution of molecular speeds mean free path collision frequency
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 2 Ne Ideal gas Ar N 2 O 2 CO Cl 2 NH 3 22.435 22.432 22.422 22.414 22.397 22.396 22.390 22.388 22.184 22.079-268.9-252.8-246.1 --- -185.9-195.8-183.0-191.5-34.0-33.4
The behavior of several real gases with increasing external pressure.
The effect of intermolecular attractions on measured gas pressure.
The effect of molecular volume on measured gas volume.
Table 5.5 Van der Waals Constants for Some Common Gases Van der Waals equation for n moles of a real gas (P n2 a )(V nb) nrt 2 V Gas a atm*l 2 mol 2 b L mol He Ne Ar Kr Xe H 2 N 2 O 2 Cl 2 CO 2 CH 4 NH 3 H 2 O 0.034 0.211 1.35 2.32 4.19 0.244 1.39 1.36 6.49 3.59 2.25 4.17 5.46 0.0237 0.0171 0.0322 0.0398 0.0511 0.0266 0.0391 0.0318 0.0562 0.0427 0.0428 0.0371 0.0305