Chemistry 1A General Chemistry Instructor: Dr. Orlando E.
Chemistry 1A General Chemistry Instructor: Dr. Orlando E. Raola Santa Rosa Junior College Chapter 5: Properties of Gases 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 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. The three states of matter A mercury barometer Units of pressure Sample Problem 5.1 Converting Units of Pressure 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 = 291.4 mm Hg. Express the CO2 pressure in torr, atmosphere,
and kilopascal. PLAN: Construct conversion factors to find the other units of pressure. SOLUTION: 291.4 mmHg 1torr = 291.4 torr 1 mmHg 291.4 torr 1 atm = 0.3834 atm 760 torr 0.3834 atm 101.325 kPa 1 atm = 38.85 kPa The relationship between the volume
and pressure of a gas. Boyles Law The relationship between the volume and temperature of a gas. Charless Law Boyles Law V a VxP V a V Amontonss Law combined gas law
P a = constant V a T P n and T are fixed V = constant / P T = constant T T P = constant
Charless Law P 1 P and n are fixed V = constant x T T V and n are fixed P = constant x T V = constant x T PV P T
= constant Avogadros Law For a gas at constant temperature and pressure, the volume is directly proportional to the number of moles of gas V = an a = proportionality constant V = volume of the gas (m3) n = chemical amount of gas (mol) Ideal Gas Law An equation of state for a gas. state is the condition of the gas at a given time. PV = nRT -1 -1 R = 8.314 Jmol K (in common units 0.082 atmLmol-1K-1) Standard molar volume.
THE IDEAL GAS LAW PV = nRT R is the universal gas constant IDEAL GAS LAW PV = nRT or V = fixed n and T Boyles Law V= constant P nRT P fixed n and P fixed P and T Charless Law
Avogadros Law V= constant X T V= constant X n Sample Problem 5.2 PROBLEM: Applying the Volume-Pressure Relationship Boyles apprentice finds that the air trapped in a J tube occupies 24.8 cm3 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:
V1 in cm3 1cm3=1mL unit conversion V1 in mL 103 mL=1L V1 in L xP1/P2 gas law calculation P1 = 1.12 atm P2 = 2.64 atm V1 = 24.8 cm3 V2 = unknown L 24.8 cm3 1 mL
1 cm3 103 mL P1V1 V2 in L n1T1 V2 = P1V1 P2 n and T are constant = P2V2 = 0.0248 L P1V1 = P2V2 n2T2 = 0.0248 L
1.12 atm 2.46 atm = 0.0105 L Sample Problem 5.3 PROBLEM: Applying the Temperature-Pressure Relationship A steel tank used for fuel delivery is fitted with a safety valve that opens when the internal pressure exceeds 1.00x103 torr. It is filled with methane at 230C and 0.991 atm and placed in boiling water at exactly 1000C. Will the safety valve open? PLAN: SOLUTION: P1(atm) T1 and T2(0C)
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 55 dm3. When 1.10 mol of He is added to the blimp, the volume is 26.2 dm3. 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 = 1.10 mol
n2 = unknown n2(mol) of He V1 = 26.2 dm3 V2 = 55.0 dm3 subtract n1 mol to be added xM g to be added V1 n1 = V2 n2 n2 = n1 55.0 dm3
P1V1 n1T1 = P2V2 n2T2 V2 V1 4.003 g He = 9.24 g He n2 = 1.10 mol = 2.31 mol 3 26.2 dm mol He Sample Problem 5.5 PROBLEM: PLAN:
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 O2. Calculate the pressure of O2 at 210C. V, T and mass are given. From mass, find amount (n), and use the ideal gas law to find P. SOLUTION: 1m3 V 438 L 3 0.438 m3 10 L T 21C 273.15 294.15 K m = 0.885 kg 1000 g 0.885 kg m 1kg n=
27.7 mol -1 M 32.00 g mol n RT P V n RT 27.7 mol 8.314 kg m2 s 2 294.15 K 5 P 1.55 10 Pa 3 V 0.438 m Relationship between density and molar mass for gases m Since n and PV nRT , then M
mRT PV M The density of a gas is mRT directly proportional to its M molar mass. PV m and because d V dRT finally M P The density of a gas is inversely proportional to the temperature. Sample Problem 5.7 PROBLEM: Calculating Gas Density
To apply a green chemistry approach, a chemical engineer uses waste CO2 from a manufacturing process, instead of chlorofluorocarbons, as a blowing agent in the production of polystyrene containers. Find the density (in g/L) of CO2 and the number of molecules (a) at STP (00C and 1 atm) and (b) at room conditions (20.0C 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 Avogadros number. MxP d = mass/volume PV = nRT V = nRT/P d = RT SOLUTION: 44.01 g/mol x 1atm = 1.96 g/L d= (a) atm*L 0.0821 x 273.15K
mol*K 1.96 g mol CO2 6.022x1023 molecules = 2.68x1022 molecules CO2/L L 44.01 g CO2 mol Sample Problem 5.6 Calculating Gas Density continued (b) d= 44.01 g/mol x 1 atm 0.0821 1.83g mol CO2
L 44.01g CO2 = 1.83 g/L atm*L x 293K mol*K 6.022x1023 molecules mol = 2.50x1022 molecules CO2/L Determining the molar mass of an unknown volatile liquid. based on the method of J.B.A. Dumas (1800-1884) Sample Problem 5.8 PROBLEM: Finding the Molar Mass of a Volatile Liquid
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 T = 100.00C P = 754 torr Mass of flask + gas = 78.416 g Mass of flask = 77.834 g Calculate the molar mass of the liquid. PLAN: Use unit conversions, mass of gas, and density-M relationship. SOLUTION: M= m = (78.416 - 77.834) g = 0.582 g m RT VP =
0.582 g x 0.0821 atm*L mol*K 0.213 L x 0.992 atm x 373K = 84.4 g/mol Relationship between density and molar mass for gases m Since n and PV nRT , then M mRT PV M The density of a gas is mRT directly proportional to its M molar mass.
PV m and because d V dRT finally M P The density of a gas is inversely proportional to the temperature. Sample Problem 5.7 Calculating Gas Density PROBLEM: To apply a green chemistry approach, a chemical engineer uses waste CO2 from a manufacturing process, instead of chlorofluorocarbons, as a blowing agent in the production of polystyrene containers. Find the density (in g/L) of CO2 and the number of molecules (a) at STP (273.15 K and 1 bar) and (b) at room conditions (20.0C 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 Avogadros number. MxP d = mass/volume PV = nRT V = nRT/P d = RT SOLUTION: -1 (a) 44.01 gmol 1 bar d= =1.94 gL 0.08314barLmol K 275.15 K 1.94 g mol CO2
L 44.01 g CO2 -1 -1 6.022x1023 molecules mol -1 = 2.65x1022 molecules CO2/L Sample Problem 5.6 Calculating Gas Density continued (b) d=
44.01 g/mol x 1 atm 0.0821 1.83g mol CO2 L 44.01g CO2 = 1.83 g/L atm*L x 293K mol*K 6.022x1023 molecules mol = 2.50x1022 molecules CO2/L Determining the molar mass of an unknown volatile liquid.
based on the method of J.B.A. Dumas (1800-1884) Sample Problem 5.8 PROBLEM: Finding the Molar Mass of a Volatile Liquid 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 T = 100.00C P = 754 torr Mass of flask + gas = 78.416 g Mass of flask = 77.834 g Calculate the molar mass of the liquid. PLAN: Use unit conversions, mass of gas, and density-M relationship. SOLUTION: M=
m = (78.416 - 77.834) g = 0.582 g m RT VP = 0.582 g x 0.0821 atm*L mol*K 0.213 L x 0.992 atm x 373K = 84.4 g/mol Sample Problem 5.5 PROBLEM: PLAN: 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 O2. Calculate the pressure of O2 at 210C. V, T and mass are given. From mass, find amount (n), and use the ideal gas law to find P. SOLUTION: 1m3 V 438 L 3 0.438 m3 10 L T 21C 273.15 294.15 K m = 0.885 kg 1000 g 0.885 kg m 1kg n= 27.7 mol
-1 M 32.00 g mol n RT P V n RT 27.7 mol 8.314 kg m2 s 2 294.15 K 5 P 1.55 10 Pa 3 V 0.438 m Mixtures of gases Gases mix homogeneously in any proportions. Each gas in a mixture behaves as if it were the only gas present. Daltons Law of Partial Pressures Ptotal = P1 + P2 + P3 + ...
P1= c1 x Ptotal where c1 is the mole fraction c1 = n1 n1 + n2 + n3 +... = n1 ntotal Mole fraction and partial pressure For each component we define the mole fraction xB nB nB xB nA nB nC ... na a
and because of Daltons law, we can calculate the partial pressure of each component as pB xB Ptot Sample Problem 5.9 Applying Daltons Law of Partial Pressures PROBLEM: In a study of O2 uptake by muscle at high altitude, a physiologist prepares an atmosphere consisting of 79 mol% N2, 17 mol% 16 O2, 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. PLAN: Find the c 18 and P18 from Ptotal and mol% 18O2. O2 O2 4.0 mol% 18O2 18
mol% O2 SOLUTION: = 0.040 = c 18 O2 100 divide by 100 c 18O2 P18 multiply by Ptotal partial pressure P 18 O2 O2 = c 18 x Ptotal = 0.040 x 0.75 atm = 0.030 atm O2
Table 5.3 Vapor Pressure of Water (P H2O T(0C) 0 5 10 11 12 13 14 15 16 18 20 22 24 26 28 ) at Different T P (torr)
433.6 525.8 633.9 760.0 Collecting a water-insoluble gaseous reaction product and determining its pressure. Sample Problem 5.10 Calculating the Amount of Gas Collected Over Water 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 98.32 kPa and the volume is 523 mL. At the temperature of the gas (230C), the vapor pressure of water is 2.798 kPa. 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. Ptotal P SOLUTION: PC H = (98.32-2.798) kPa =95.52 kPa 2 2 C2H2 P PV H2O n= RT n g C2H2 C2H2 xM Sample Problem 5.9 Calculating the Amount of Gas Collected Over Water continued 4
nC H 2 2 4 3 9.552 10 Pa5.23 10 m 8.314 Jmol K 296K 0.0203mol 1 26.04 g C2H2 mol C2H2 1
0.0203 mol = 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 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 5.11 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 H2 reduces the metal oxide, forming the pure metal and H2O. On a laboratory scale, what volume of H2 at 765 torr and 2250C 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 H2 gas. mass (g) of Cu SOLUTION: divide by M mol of Cu 35.5 g Cu molar ratio mol of H2 mol Cu Cu(s) + H2O(g) 1 mol H2 63.55 g Cu 1 mol Cu 0.559 mol H2 x 0.0821
use known P and T to find V L of H2 CuO(s) + H2(g) atm*L mol*K x 1.01 atm = 0.559 mol H2 498K = 22.6 L Sample Problem 5.12 Using 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? 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: PV n = Cl2 RT 17.0g 2K(s) + Cl2(g) = 0.950 atm x 5.25L 0.0821 mol K 39.10 g K 2KCl(s)
atm*L mol*K Cl2 is the limiting reactant. 74.55 g KCl mol KCl V = 5.25 L T = 293K n = unknown x 293K = 0.435 mol K 0.414 mol KCl = 0.207 mol P = 0.950 atm 0.207 mol Cl2
0.435 mol K = 30.9 g KCl 2 mol KCl = 0.414 mol 1 mol Cl2 KCl formed 2 mol KCl 2 mol K = 0.435 mol KCl formed 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 I2 is charged with 10.0 g of F2 and the reaction proceeds until one of the reactants is completely consumed. After the reaction is complete, the temperature in the flask is 125 C.
a) What is the partial pressure of IF5 in the flask? b) What is the mole fraction of IF5 in the flask? 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(Ek) of the particles is constant. Distribution of molecular speeds as a function of temperature Distribution of molecular speeds as a function of temperature Ek a T Ek constant T
Molecular description of Boyles Law Molecular description of Daltons law of partial pressures Molecular description of Charless law Molecular description of Avogadros law Relationship between molecular speed and mass 1 2 Ek mu 2 urms 3RT M The meaning of temperature 3 R
Ek T 2 NA Absolute temperature is a measure of the average kinetic energy of the molecular random motion Effusion and difussion Effusion: describes the passage of gas through a small orifice into an evacuated chamber. Diffusion: describes the mixing of gases. The rate of diffusion is the rate of gas mixing. Effusion: M2 Rateof effusion for gas1 = Rateof effusion for gas 2 M1
Diffusion: Distance traveled by gas1 M2 = Distance traveled by gas 2 M1 Effusion and KMT Diffusion through space distribution of molecular speeds mean free path collision frequency Sample Problem 5.13 Applying Grahams 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 rate rate He CH4 = 16.04 4.003 = 2.002 M of He = 4.003g/mol Table 5.4 Molar Volume of Some Common Gases at STP (00C and 1 atm)
Gas He H2 Ne Ideal gas Ar N2 O2 CO Cl2 NH3 Molar Volume (L/mol) 22.435 22.432 22.422 22.414 22.397 22.396 22.390 22.388 22.184 22.079
Condensation Point (0C) -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 Effect of molecular attractions on pressure Effect of molecular volume on measured volume van der Waals equation Van der Waals
equation for n moles of a real gas Gas 2 n P+a V-nb =nRT V a (atmL 2mol-2) He Ne Ar Kr Xe H2 N2 O2 Cl2 CO2 CH4
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