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Instructor: Dr. Orlando E. Raola Santa Rosa Junior College Chapter 5: Properties of Gases Chemistry 1A General Chemistry

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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 Applications of the Ideal Gas Law

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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. The distinction of gases from liquids and solids

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The three states of matter

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A mercury barometer

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Units of pressure

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Sample Problem 5.1 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 closed- end manometer. After the system comes to room temperature, h = 291.4 mm Hg. Express the CO 2 pressure in torr, atmosphere, and kilopascal. SOLUTION: PLAN: Construct conversion factors to find the other units of pressure. 291.4 mmHg 1torr 1 mmHg = 291.4 torr 291.4 torr 1 atm 760 torr = 0.3834 atm 0.3834 atm 101.325 kPa 1 atm = 38.85 kPa

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The relationship between the volume and pressure of a gas. Boyle’s Law

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The relationship between the volume and temperature of a gas. Charles’s Law

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Boyle’s Lawn and T are fixed V 1 P Charles’s Law V T P and n are fixed V T = constant V = constant x T Amontons’s Law P T V and n are fixed P T = constant P = constant x T combined gas law V T P V = constant x T P PV T = constant V x P= constantV = constant / P

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Avogadro’s Law For a gas at constant temperature and pressure, the volume is directly proportional to the number of moles of gas V = a n a = proportionality constant V = volume of the gas (m 3 ) n = chemical amount of gas (mol)

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Ideal Gas Law An equation of state for a gas. “state” is the condition of the gas at a given time. PV = nRT R = 8.314 J∙mol -1 ·K -1 (in common units 0.082 atm∙L∙mol -1 ·K -1 )

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Standard molar volume.

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THE IDEAL GAS LAW PV = nRT IDEAL GAS LAW nRT P PV = nRT or V = Boyle’s Law V = constant P V = Charles’s Law constant X T Avogadro’s Law constant X n fixed n and Tfixed n and Pfixed P and T R is the universal gas constant

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Sample Problem 5.2 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: SOLUTION: V 1 in cm 3 V 1 in mL V 1 in L V 2 in L unit conversion gas law calculation P 1 = 1.12 atmP 2 = 2.64 atm V 1 = 24.8 cm 3 V 2 = unknown n and T are constant 24.8 cm 3 1 mL 1 cm 3 L 10 3 mL = 0.0248 L P1V1P1V1 n1T1n1T1 P2V2P2V2 n2T2n2T2 = P 1 V 1 = P 2 V 2 P1V1P1V1 P2P2 V 2 = =0.0248 L 1.12 atm 2.46 atm =0.0105 L 1cm 3 =1mL 10 3 mL=1L xP 1 /P 2

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Sample Problem 5.3 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:SOLUTION: P 1 (atm)T 1 and T 2 ( 0 C) P 1 (torr)T 1 and T 2 (K) P 1 = 0.991atmP 2 = unknown T 1 = 23 0 CT 2 = 100 0 C P 2 (torr) 1atm=760torr x T 2 /T 1 K= 0 C+273.15 P1V1P1V1 n1T1n1T1 P2V2P2V2 n2T2n2T2 = P1P1 T1T1 P2P2 T2T2 = 0.991 atm 1 atm 760 torr = 753 torr P 2 =P1P1 T2T2 T1T1 = 753 torr 373K 296K = 949 torr

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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 26.2 dm 3. How many more grams of He must be added to make it rise? Assume constant T and P. PLAN: SOLUTION: 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 n 2 (mol) of He mol to be added g to be added x V 2 /V 1 x M subtract n 1 n 1 = 1.10 moln 2 = unknown V 1 = 26.2 dm 3 V 2 = 55.0 dm 3 P and T are constant P1V1P1V1 n1T1n1T1 P2V2P2V2 n2T2n2T2 = V1V1 n1n1 V2V2 n2n2 = n 2 = n 1 V2V2 V1V1 n 2 = 1.10 mol 55.0 dm 3 26.2 dm 3 = 2.31 mol 4.003 g He mol He = 9.24 g He

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Sample Problem 5.5Solving for an Unknown Gas Variable at Fixed Conditions PROBLEM: 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: SOLUTION: V, T and mass are given. From mass, find amount (n), and use the ideal gas law to find P.

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The density of a gas is directly proportional to its molar mass. The density of a gas is inversely proportional to the temperature. Relationship between density and molar mass for gases

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Sample Problem 5.7 Calculating Gas Density PROBLEM: 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: SOLUTION: 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 = mass/volumePV = nRTV = nRT/P d= RT M x P d = 44.01 g/molx 1atm atm*L mol*K 0.0821x 273.15K = 1.96 g/L 1.96 g L mol CO 2 44.01 g CO 2 6.022x10 23 molecules mol = 2.68x10 22 molecules CO 2 /L (a)

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Sample Problem 5.6 Calculating Gas Density continued (b)= 1.83 g/L d = 44.01 g/molx 1 atm x 293K atm*L mol*K 0.0821 1.83g L mol CO 2 44.01g CO 2 6.022x10 23 molecules mol = 2.50x10 22 molecules CO 2 /L

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Determining the molar mass of an unknown volatile liquid. based on the method of J.B.A. Dumas (1800-1884)

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Sample Problem 5.8 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: PLAN: SOLUTION: Use unit conversions, mass of gas, and density- M relationship. 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. m = (78.416 - 77.834) g= 0.582 g M = m RT VP = 0.582 g atm*L mol*K 0.0821373K x x 0.213 Lx0.992 atm =84.4 g/mol

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The density of a gas is directly proportional to its molar mass. The density of a gas is inversely proportional to the temperature. Relationship between density and molar mass for gases

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Sample Problem 5.7 Calculating Gas Density PROBLEM: 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 (273.15 K and 1 bar) and (b) at room conditions (20. 0 C and 1.00 atm). PLAN: SOLUTION: 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 = mass/volumePV = nRTV = nRT/P d= RT M x P 1.94 g L mol CO 2 44.01 g CO 2 6.022x10 23 molecules mol = 2.65x10 22 molecules CO 2 /L (a)

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Sample Problem 5.6 Calculating Gas Density continued (b)= 1.83 g/L d = 44.01 g/molx 1 atm x 293K atm*L mol*K 0.0821 1.83g L mol CO 2 44.01g CO 2 6.022x10 23 molecules mol = 2.50x10 22 molecules CO 2 /L

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Determining the molar mass of an unknown volatile liquid. based on the method of J.B.A. Dumas (1800-1884)

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Sample Problem 5.8 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: PLAN: SOLUTION: Use unit conversions, mass of gas, and density- M relationship. 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. m = (78.416 - 77.834) g= 0.582 g M = m RT VP = 0.582 g atm*L mol*K 0.0821373K x x 0.213 Lx0.992 atm =84.4 g/mol

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Sample Problem 5.5Solving for an Unknown Gas Variable at Fixed Conditions PROBLEM: 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: SOLUTION: V, T and mass are given. From mass, find amount (n), and use the ideal gas law to find P.

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Dalton’s Law of Partial Pressures P total = P 1 + P 2 + P 3 +... P 1 = 1 x P total where 1 is the mole fraction 1 = n1n1 n 1 + n 2 + n 3 +... = n1n1 n total Gases mix homogeneously in any proportions. Each gas in a mixture behaves as if it were the only gas present. Mixtures of gases

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Mole fraction and partial pressure For each component we define the mole fraction x B and because of Dalton’s law, we can calculate the partial pressure of each component as

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Sample Problem 5.9 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: SOLUTION: Find the and P from P total and mol% 18 O 2. 18 O 2 mol% 18 O 2 18 O 2 partial pressure P 18 O 2 divide by 100 multiply by P total 18 O 2 = 4.0 mol% 18 O 2 100 = 0.040 = 0.030 atm P = x P total = 0.040 x 0.75 atm 18 O 2

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Table 5.3 Vapor Pressure of Water (P ) at Different T H2OH2O 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 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 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

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Collecting a water-insoluble gaseous reaction product and determining its pressure.

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Sample Problem 5.10 Calculating the Amount of Gas Collected Over Water PLAN: SOLUTION: 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. PROBLEM: 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 98.32 kPa and the volume is 523 mL. At the temperature of the gas (23 0 C), the vapor pressure of water is 2.798 kPa. How many grams of acetylene are collected? P total P C2H2C2H2 n C2H2C2H2 g C2H2C2H2 P H2OH2O n = PV RT x M P C2H2C2H2 = (98.32-2.798) kPa =95.52 kPa

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Sample Problem 5.9 Calculating the Amount of Gas Collected Over Water continued 0.0203mol 26.04 g C 2 H 2 mol C 2 H 2 = 0.529 g C 2 H 2

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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 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).

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Sample Problem 5.11Using 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? SOLUTION: 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 mol of Cu mol of H 2 L of H 2 divide by M molar ratio use known P and T to find V CuO( s ) + H 2 ( g ) Cu( s ) + H 2 O( g ) 35.5 g Cu mol Cu 63.55 g Cu 1 mol H 2 1 mol Cu = 0.559 mol H 2 0.559 mol H 2 x498K atm*L mol*K 0.0821x 1.01 atm = 22.6 L

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Sample Problem 5.12Using the Ideal Gas Law in a Limiting-Reactant Problem PROBLEM: The alkali metals (Group 1) react with the halogens (Group 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? SOLUTION: 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 ) + Cl 2 ( g ) 2KCl( s ) V = 5.25 L T = 293Kn = unknown P = 0.950 atm n = PV RT Cl 2 x5.25L = 0.950 atm atm*L mol*K 0.0821x293K =0.207 mol 17.0g 39.10 g K mol K =0.435 mol K 0.207 mol Cl 2 2 mol KCl 1 mol Cl 2 = 0.414 mol KCl formed 0.435 mol K 2 mol KCl 2 mol K = 0.435 mol KCl formed Cl 2 is the limiting reactant. 0.414 mol KCl 74.55 g KCl mol KCl = 30.9 g KCl

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Problem Gaseous iodine pentafluoride can be prepared by the reaction between solid iodine and gaseous fluorine. A 5.00-L flask containing 10.0 g of I 2 is charged with 10.0 g of F 2 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. a) What is the partial pressure of IF 5 in the flask? b) What is the mole fraction of IF 5 in the flask?

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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(E k ) of the particles is constant. Postulate 1: Particle Volume Postulate 2: Particle Motion Postulate 3: Particle Collisions

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Distribution of molecular speeds as a function of temperature

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Molecular description of Boyle’s Law

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Molecular description of Dalton’s law of partial pressures

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Molecular description of Charles’s law

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Molecular description of Avogadro’s law

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Relationship between molecular speed and mass

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The meaning of temperature Absolute temperature is a measure of the average kinetic energy of the molecular random motion

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Effusion and difussion Effusion: describes the passage of gas through a small orifice into an evacuated chamber. Diffusion rate Diffusion: describes the mixing of gases. The rate of diffusion is the rate of gas mixing.

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Effusion : Diffusion :

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Effusion and KMT

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distribution of molecular speeds mean free path collision frequency Diffusion through space

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Sample Problem 5.13 Applying Graham’s Law of Effusion PROBLEM: Calculate the ratio of the effusion rates of helium and methane (CH 4 ). SOLUTION: 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. M of CH 4 = 16.04g/mol M of He = 4.003g/mol CH 4 He rate = √ 16.04 4.003 = 2.002

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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

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The behavior of several real gases with increasing external pressure

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Effect of molecular attractions on pressure

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Effect of molecular volume on measured volume

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van der Waals equation 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 He Ne Ar Kr Xe H 2 N 2 O 2 Cl 2 CO 2 CH 4 NH 3 H 2 O 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 Van der Waals equation for n moles of a real gas Gas a (atm·L 2 ·mol -2 ) b (L·mol -1 )

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