Gases and the Kinetic Molecular Theory Chapter 5 Gases and the Kinetic Molecular Theory Dr. Wolf’s CHM 101

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 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 Dr. Wolf’s CHM 101

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. Dr. Wolf’s CHM 101

The three states of matter. Figure 5.1 The three states of matter. Dr. Wolf’s CHM 101

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. Dr. Wolf’s CHM 101

A mercury barometer. Figure 5.3 Dr. Wolf’s CHM 101

Two types of manometer Figure 5.4 Dr. Wolf’s CHM 101

Table 5.2 Common Units of Pressure Atmospheric Pressure Scientific Field pascal(Pa); kilopascal(kPa) 1.01325x105Pa; 101.325 kPa SI unit; physics, chemistry chemistry atmosphere(atm) 1 atm 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 Dr. Wolf’s CHM 101

Converting 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 = 291.4mmHg. Calculate the CO2 pressure in torrs, atmospheres, and kilopascals. PLAN: Construct conversion factors to find the other units of pressure. SOLUTION: 291.4mmHg 1torr 1mmHg = 291.4torr 1atm 760torr 291.4torr = 0.3834atm 101.325kPa 1atm 0.3834atm = 38.85kPa Dr. Wolf’s CHM 101

Boyle’s Law - The relationship between volume and the pressure of a gas. (Temperature is kept constant.) Dr. Wolf’s CHM 101

PV = constant V = constant / P Boyle’s Law 1 P V a n and T are fixed Dr. Wolf’s CHM 101

Charles’s Law - The relationship between volume and the temperature of a gas. (Pressure is kept constant.) Dr. Wolf’s CHM 101

When the amount of gas, n, is constant V a 1 P Boyle’s Law n and T are fixed Charles’s Law V a T P and n are fixed V T = constant V = constant x T Amonton’s Law P a T V and n are fixed P T = constant P = constant x T V a T P V = constant x T P PV T = constant combined gas law When the amount of gas, n, is constant Dr. Wolf’s CHM 101

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 n = constant V = constant x n Twice the amount gives twice the volume Dr. Wolf’s CHM 101

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 Dr. Wolf’s CHM 101

PV = nRT THE IDEAL GAS LAW PV nT 1atm x 22.414L 1mol x 273.15K fixed n and T fixed n and P fixed P and T Boyle’s Law Charles’s Law Avogadro’s Law V P V = constant X n = constant V = constant X T THE IDEAL GAS LAW PV = nRT PV nT 1atm x 22.414L 1mol x 273.15K 0.0821atm-L Mol-K R = = = Dr. Wolf’s CHM 101

Applying the Volume-Pressure Relationship Sample Problem 5.2 Applying the Volume-Pressure Relationship 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)? PLAN: SOLUTION: n and T are constant V1 in cm3 P1 = 1.12atm P2 = 2.64atm unit conversion 1cm3=1mL V1 = 24.8cm3 V2 = unknown V1 in mL 1mL 1cm3 L 103mL 103mL=1L 24.8cm3 = 0.0248L = V1 V1 in L gas law calculation xP1/P2 P1V1 n1T1 P2V2 n2T2 P1V1 = P2V2 = R = = V2 in L P1V1 1.12atm 2.46atm V2 = = 0.0105L = 0.0248L P2 Dr. Wolf’s CHM 101

Applying the Temperature-Pressure Relationship 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) P1V1 n1T1 P2V2 n2T2 = P1 T1 P2 T2 = x T2/T1 P2(torr) 0.991atm 1atm 760 torr = 753 torr P2 = P1 T2 T1 = 753 torr 373K 296K = 949 torr Dr. Wolf’s CHM 101

Applying the Volume-Amount Relationship 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 P1V1 n1T1 P2V2 n2T2 = n2(mol) of He V1 = 26.2dm3 V2 = 55.0dm3 subtract n1 V1 n1 V2 n2 = n2 = n1 V2 V1 mol to be added x M n2 = 1.10mol 55.0dm3 26.2dm3 = 2.31mol g to be added 4.003g He mol He n2 - n1 = 2.31 -1.10 = 1.21 mol He = 4.84g He Dr. Wolf’s CHM 101

Solving for an Unknown Gas Variable at Fixed Conditions 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 0.885kg 103g kg mol O2 32.00g O2 = 27.7mol O2 210C + 273.15 = 294K = 27.7mol 294K atm*L mol*K 0.0821 x 438L P = nRT V = 1.53atm Dr. Wolf’s CHM 101

Calculating Gas Density 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 RT M x P mult. both sides by M d = n / V = P / RT Mass = n M SOLUTION: 44.01g/mol x 1atm atm*L mol*K 0.0821 x 273K = 1.96g/L d = (a) 1.96g L mol CO2 44.01g CO2 6.022x1023molecules mol = 2.68x1022molecules CO2/L Dr. Wolf’s CHM 101

Calculating Gas Density Sample Problem 5.6 Calculating Gas Density continued d = 44.01g/mol x 1atm x 293K atm*L mol*K 0.0821 (b) = 1.83g/L 1.83g L mol CO2 44.01g CO2 6.022x1023molecules mol = 2.50x1022molecules CO2/L Dr. Wolf’s CHM 101

Calculating the Molar Mass, M, of a Gas 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 Dr. Wolf’s CHM 101

Finding the Molar Mass of a Volatile Liquid 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: Volume of flask = 213mL Mass of flask + gas = 78.416g T = 100.00C 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 = 0.582g 0.582g atm*L mol*K 0.0821 373K x x m RT P V M = = = 84.4g/mol 0.213L x 0.992atm M of C6H12 is 84.16g/mol and the calculated value is within experimental error. Dr. Wolf’s CHM 101

Dalton’s Law of Partial Pressures 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 c1 = n1 n1 + n2 + n3 +... = n1 ntotal Dr. Wolf’s CHM 101

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. Dr. Wolf’s CHM 101

Applying Dalton’s Law of Partial Pressures Sample Problem 5.8 Applying Dalton’s 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% 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. PLAN: Find the c and P from Ptotal and mol% 18O2. 18O2 = 4.0mol% 18O2 100 mol% 18O2 SOLUTION: = 0.040 c 18O2 divide by 100 c 18O2 P = c x Ptotal = 0.040 x 0.75atm 18O2 = 0.030atm multiply by Ptotal partial pressure P 18O2 Dr. Wolf’s CHM 101

CaC2(s) + 2H2O(l) C2H2(g) + Ca(OH)2(aq) Sample Problem 5.9 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 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? 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. P C2H2 SOLUTION: = (738-21)torr = 717torr Ptotal P C2H2 atm 760torr P = 0.943atm n = PV RT 717torr H2O n C2H2 g C2H2 x M Dr. Wolf’s CHM 101

Calculating the Amount of Gas Collected Over Water Sample Problem 5.9 Calculating the Amount of Gas Collected Over Water continued 0.943atm x 0.523L n C2H2 = = 0.203mol atm*L mol*K 0.0821 x 296K 0.203mol 26.04g C2H2 mol C2H2 = 0.529 g C2H2 Dr. Wolf’s CHM 101

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). amount (mol) of gas A amount (mol) of gas B P,V,T of gas A P,V,T of gas B ideal gas law ideal gas law molar ratio from balanced equation Dr. Wolf’s CHM 101

Using Gas Variables to Find Amount of Reactants and Products Sample Problem 5.10 Using Gas Variables to Find Amount of Reactants and Products 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. mass (g) of Cu CuO(s) + H2(g) Cu(s) + H2O(g) SOLUTION: divide by M mol Cu 63.55g Cu 1mol H2 1 mol Cu 35.5g Cu = 0.559mol H2 mol of Cu molar ratio x 498K atm*L mol*K 0.0821 1.01atm 0.559mol H2 = 22.6L mol of H2 use known P and T to find V L of H2 Dr. Wolf’s CHM 101

Using the Ideal Gas Law in a Limiting-Reactant Problem Sample Problem 5.11 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 5.25L 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. 2K(s) + Cl2(g) 2KCl(s) V = 5.25L T = 293K n = unknown P = 0.950atm SOLUTION: PV RT = 0.950atm atm*L mol*K 0.0821 x 293K x 5.25L n = = 0.207mol Cl2 0.207mol Cl2 2mol KCl 1mol Cl2 = 0.414mol KCl formed 17.0g 39.10g K mol K = 0.435mol K 2mol KCl 2mol K = 0.435mol KCl formed Cl2 is the limiting reactant. 0.435mol K 74.55g KCl mol KCl 0.414mol KCl = 30.9 g KCl Dr. Wolf’s CHM 101

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(Kk) of the particles is constant. Dr. Wolf’s CHM 101

A molecular description of Boyle’s Law 1 P Boyle’s Law V a n and T are fixed Dr. Wolf’s CHM 101

A molecular description of Charles’s Law V a T n and P are fixed Dr. Wolf’s CHM 101

Dalton’s Law of Partial Pressures A molecular description of Dalton’s law of partial pressures. Dalton’s Law of Partial Pressures Ptotal = P1 + P2 + P3 + ... Dr. Wolf’s CHM 101

A molecular description of Avogadro’s Law V a n P and T are fixed Dr. Wolf’s CHM 101

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. Dr. Wolf’s CHM 101

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 Dr. Wolf’s CHM 101

Relationship between molar mass and molecular speed. Dr. Wolf’s CHM 101

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. = rate A rate B ( MB / MA ) ½ Dr. Wolf’s CHM 101

Applying Graham’s Law of Effusion Sample Problem 5.12 Applying Graham’s 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 M of He = 4.003g/mol CH4 He rate = ( 16.04/ 4.003 ) ½ = 2.002 Dr. Wolf’s CHM 101

End of Chapter 5 Dr. Wolf’s CHM 101