Engineering Chemistry 1B (ENCB101) Gases PDF 2024

DEPARTMENT OF CHEMISTRY ENGINEERING CHEMISTRY 1B (ENCB101) GASES Dr Lethiwe Mthembu 2024 ABOUT ME Dr. Lethiwe Mthembu (Chemistry, Biomass Processing) I enjoy travelling and reading. MUT Alumna (Diploma...

DEPARTMENT OF CHEMISTRY ENGINEERING CHEMISTRY 1B (ENCB101) GASES Dr Lethiwe Mthembu 2024 ABOUT ME Dr. Lethiwe Mthembu (Chemistry, Biomass Processing) I enjoy travelling and reading. MUT Alumna (Diploma Am into photography, capturing those and BTech in Chemistry). good moments. DUT Alumna (Masters I am passionate about Chemistry and PhD in Chemistry). because it solve problems. Wits MBA Candidate. RULES Attending a class is compulsory, If you can’t attend report with a reason via email: [email protected] When writing me an email, you start by writing a subject of the email. The main reason you’re sending me an email, following by greeting and introducing yourself (Name, surname, student number) and the course you’re doing because I have other classes that I teach. Email me during weekdays from 08:00 to 16:00. Not at night and not on weekends. Always check your DUT email address that is how I will communicate with you if I need to talk to you. If it happens you couldn’t write a test or didn’t attend a practical session report as soon as possible with a good excuse (Doctors letter). Tutorial session: You will submit your tutorials answers to me before attending a session via moodle and my email. Physics - study of the properties of matter that are shared by all substances Chemistry - the study of the properties of the substances that make up the universe and the changes that these substances undergo Physical Chemistry - the best of both worlds! PHYSICAL CHEMISTRY In Physical Chemistry matter is referred to as a system. The system may be a: Liquid Solid Gas or any combination of these. PROPERTIES OF THE SYSTEM ▪Pressure ▪Temperature ▪Volume PROPERTIES OF THE SYSTEM These properties may be intensive or extensive. Intensive property – value Extensive property- value of of the property does not the property changes with the change with the quantity of quantity of matter present. the matter present. Examples: volume and mass. Examples: pressure, temperature, and refractive index.  GASES ▪The Gas Laws of Boyle, Charles and Avogadro ▪The Ideal Gas Law ▪Gas Stoichiometry ▪Dalton’s Laws of Partial Pressure ▪The Kinetic Molecular Theory of Gases ▪Effusion and Diffusion ▪Collisions of Gas Particles with the Container Walls ▪Intermolecular Collisions Real Gases ▪Chemistry in the Atmosphere Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia PHYSICAL CHARACTERISTICS OF GASES ▪Gases assume the volume and shape of their containers. ▪Gases are easily compressible. ▪Gases will mix evenly and completely when confined to the same container. ▪Gases have much lower densities than liquids and solids. Elements that exist as gases at 25 0C and 1 atmosphere PRESSURE Pressure is force per unit area Units of Pressure 1 pascal (Pa) = 1 N/m2 Conversion of Pressure units  Current way to PRESSURE GAUGE measure pressure PRESSURE CONVERSION EXERCISE 1 standard atmosphere = 1 atm= 101.3 KPa = 760 mm of Hg = 760 torr A. Convert 135.4 KPa to torr. B. Convert 950 torr to mm Hg. C. Convert 5.76 atm to in of KPa PRESSURE CONVERSION ANSWERS 1 standard atmosphere = 1 atm= 101.3 KPa = 760 mm of Hg = 760 torr A. Convert 135.4 KPa to torr. 101,3 KPa = 760 torr Answer: 1 015,8 torr B. Convert 950 torr to mm Hg. 760 torr = 760 mm Hg Answer: 950 mm Hg C. Convert 5.76 atm to in of Kpa 1 atm = 101,3 Kpa Answer: 583,5 Kpa HOW TO CALCULATE PRESSURE? BOYLE’S LAW: PRESSURE AND VOLUME The product of the pressure and volume, PV, of a sample of gas is a constant at a constant temperature: PV = k = Constant (fixed T,n) BOYLE’S LAW EXAMPLE A sample of chlorine gas occupies a volume of 946 mL at a pressure of 726 mmHg. What is the pressure of the gas (in mmHg) if the volume is reduced at constant temperature to 154 mL? P1 x V1 = P2 x V2 P1 = 726 mmHg V1 = 946 mL P2 = ? V2 = 154 mL SOLUTION P2 = P1 X V1/V2 = 726 mmHg X 946 mL/154 mL = 4 459.7 mmHg BOYLE’S LAW: THE EFFECT OF PRESSURE ON GAS VOLUME Example The cylinder of a bicycle pump has a volume of 1131 cm3 and is filled with air at a pressure of 1.02 atm. The outlet valve is sealed shut, and the pump handle is pushed down until the volume of the air is 517 cm3. The temperature of the air trapped inside does not change. Compute the pressure inside the pump. SOLUTION V1 = 1 131 cm3 P1 = 1,02 atm V2 = 517 cm3 P2 = ? P1V1= P2V2 P2 = P1V1/V2 P2 = 1,02 atm x 1131 cm3/517 cm3 P2 = 2,23 atm  CHARLES’ LAW: TEMPERATURE VS VOLUME Charles's law, a statement that the volume occupied by a fixed amount of gas is directly proportional to its absolute temperature, if the pressure remains constant. V = kT Where V = volume k = non-zero constant T = temperature EXAMPLE A sample of carbon monoxide gas occupies 3.20 L at 125 0C. At what temperature will the gas occupy a volume of 1.54 L if the pressure remains constant? V1/T1 = V2/T2 V1 = 3.20 L T1 = 398.15 K V2 = 1.54 L T2 = ? SOLUTION T1T2 X V1/T1 = V2/T2 X T1T2 T2 = (V2 X T1)/V1 T2 = (1,54 L X 398.15 k)/3,20 L T2 = 191,6 k AVOGADRO’S LAW For a gas at constant temperature and pressure the volume is directly proportional to the number of moles of gas. V = kn Where: V = volume n = number of moles of gas k = proportionality constant AVOGADRO’S LAW EXAMPLE Ammonia burns in oxygen to form nitric oxide (NO) and water vapor. How many volumes of NO are obtained from one volume of ammonia at the same temperature and pressure? 4NH3 + 5O2 4NO + 6H2O SOLUTION 1 mole NH3 = 1 mole NO At constant T and P 1 volume NH3 = 1 volume NO GAS LAWS EQUATIONS IDEAL GAS EQUATION  IDEAL GAS LAW The conditions 0 0C and 1 atm are called standard temperature and pressure (STP). Experiments show that at STP, 1 mole of an ideal gas occupies 22.414 L. PV = nRT R = PV /nT = (1 atm)(22.414L)/ (1 mol)(273.15 K) R = 0.082057 L atm / (mol K) IDEAL GAS LAW CONSTANTS EXAMPLE At some point during its ascent, a sealed weather balloon initially filled with helium at a fixed volume of 1.0 x 104 L at 1.00 atm and 30 oC reaches an altitude at which the temperature is -10 oC yet the volume is unchanged. Calculate the pressure at that altitude. SOLUTION At some point during its ascent, a sealed weather balloon initially filled with helium at a fixed volume of 1.0 x 104 L at 1.00 atm and 30 oC reaches an altitude at which the temperature is -10 oC yet the volume is unchanged. Calculate the pressure at that altitude. P2 = P1T2/T1 = (1atm)(263K)/(303K) = 0,868 atm EXAMPLE What mass of hydrogen gas is needed to fill a weather balloon to a volume of 10.0 L, 1.00 atm and 30 ̊ C? 1) Use PV = nRT; n=PV/RT 2) Find the number of moles. 3) Use the atomic weight to find the mass. SOLUTION What mass of hydrogen gas is needed to fill a weather balloon to a volume of 10.0 L, 1.00 atm and 30 ̊ C? n = PV/RT = (1.00 atm) (10.0 L)/ (303,15 K)-1 (0.082 L atm mol-1 K- 1)-1 = 0.402 mol Mass = (0.402 mol)(2.0 g mol-1) = 0.804 g LEARNING CHECK What is the volume (in liters) occupied by 49.8 g of HCl at STP? T = 0 0C = 273.15 K P = 1 atm mHCl = 49,8 g STP conditions V=? Answer: 30.7 L SOLUTION n = 49.8 g/36.45 g/mol HCl = 1.37 mol PV = nRT V = nRT/P = 1.37 mol x 0.0821 L. atm. mol-1.k-1 x 273.15 k/1 atm = 30.7 L LEARNING CHECK Argon is an inert gas used in lightbulbs to retard the vaporization of the filament. A certain lightbulb containing argon at 1.20 atm and 18 0C is heated to 85 0C at constant volume. What is the final pressure of argon in the lightbulb (in atm)? P1 = 1.20 atm T1 = 18 oC + 273.15 k = 291.15 k T2 = 85 oC + 273.15 k = 358.15 k V = constant P2= ? In atm Answer: 1.5 atm  SOLUTION PV = nRT n, V and R are constant 𝑛𝑅 𝑃 = = constant 𝑉 𝑇 𝑃1 𝑇1 = 𝑃2 𝑇2 1.2 atm x 358.15 k P2 = = 1.5 atm 291.15 𝑘 CALCULATION FOR GAS DENSITY AND MOLAR MASS Density (d) Calculations: 𝑚 𝑃𝑀 d= = 𝑉 𝑅𝑇 m is the mass of the gas in g M is the molar mass of the gas Molar Mass (M) of a Gaseous Substance 𝑑𝑅𝑇 M= 𝑃 d is the density of the gas in g/L GAS DENSITY AND MOLAR MASS Example: Calculate the density of gaseous hydrogen at a pressure of 1.32 atm and a temperature of -45oC. P = 1.32 atm T = -45 oC + 273.15 k = 228.15 k d =? SOLUTION 𝑃𝑀 1.32 atm x 2.0 g.mol−1 d= = = 0.14 g/L 𝑅𝑇 0.0821 L.atm.mol−1.k−1 x 228.15 𝑘 EXAMPLE A 0.100g sample of a compound of empirical formula CH2F2 occupies 0.0470 L at 298k and 755 mm of Hg. What is the molar mass of the compound? Answer: 52.4 g/mol SOLUTION GAS STOICHIOMETRY Stoichiometry is the quantitative study of the relative amounts of reactants and products in chemical reactions; gas stoichiometry involves chemical reactions that produce gases. Stoichiometry is based on the law of conservation of mass, meaning that the mass of the reactants must be equal to the mass of the products. This assumption can be used to solve for unknown quantities of reactants or products. GAS STOICHIOMETRY Use volumes to determine stoichiometry. The volume of a gas is easier to measure than the mass. GAS STOICHIOMETRY GAS STOICHIOMETRY GAS STOICHIOMETRY DALTON’S LAW OF PARTIAL PRESSURES The pressure exerted by an individual gas in a mixture is known as its partial pressure. Assuming we have a mixture of ideal gases, we can use the ideal gas law to solve problems involving gases in a mixture. Dalton's law of partial pressures states that the total pressure of a mixture of gases is equal to the sum of the partial V and T are constant pressures of the component gases: MIXTURES OF GASES Dalton’s Law of Partial Pressures The total pressure of a mixture of gases equals the sum of the partial pressures of the individual gases.  MOLE FRACTION OF GAS MIXTURE The composition of a gas mixture can be The ratio of the pressure of a gas A to the described by the mole fractions of the total pressure of a gas mixture that contains gases present. A. We can use the ideal gas law to The mole fraction ( χ ) of any component of describe the pressures of both gas A and a mixture is the ratio of the number of the mixture: moles of that component to the total number of moles of all the species present PA=nART/V and Ptot=ntRT/V. The ratio of in the mixture ( ntot ): the two is thus: Rearranging the equation gives: EXAMPLE A sample of natural gas contains 8.24 moles of CH4, 0.421 moles of C2H6, and 0.116 moles of C3H8. If the total pressure of the gases is 1.37 atm, what is the partial pressure of propane (C3H8)? Answer: 0.0181 atm  SOLUTION A sample of natural gas contains 8.24 moles of CH4, 0.421 moles of C2H6, and 0.116 moles of C3H8. If the total pressure of the gases is 1.37 atm, what is the partial pressure of propane (C3H8)? EXAMPLE A solid hydrocarbon is burned in air in a closed container, producing a mixture of gases having a total pressure of 3.34 atm. Analysis of the mixture shows it to contain 0.340 g of water vapor, 0.792 g of carbon dioxide, 0.288 g of oxygen, 3.790 g of nitrogen, and no other gases. Calculate the mole fraction and partial pressure of carbon dioxide in this mixture. SOLUTION THE KINETIC MOLECULAR THEORY OF GASES The Ideal Gas Law is an empirical relationship based on experimental observations. ▪Boyle law ▪Charles law ▪Avogadro law Kinetic Molecular Theory is a simple model that attempts to explain the behavior of gases. THE KINETIC MOLECULAR THEORY OF GASES To apply the kinetic model of gases, five assumptions are made: 1. Negligible particle volume: The volume of the individual gas particles is extremely small compared to the total volume of the gas. Most volume is occupied by a gas is empty space. 2. No attractions or repulsive forces: Gas particles undergo no intermolecular attractions or repulsions. This assumption implies that the particles possess no potential energy and thus their total energy is simply equal to their kinetic energies. 3. Gas particles in constant motion: Gas particles are in continuous, random motion. They move in straight lines until they collide with each other or the walls of their container. 4. Elastic Collisions: Collisions between gas particles are completely elastic. In other words, there is no net loss or gain of kinetic energy when particles collide. 5. Average kinetic energy proportional to temperature: The average kinetic energy is the same for all gases at a given temperature, regardless of the identity of the gas. Furthermore, this kinetic energy is proportional to the absolute temperature of the gas. TEMPERATURE AND KMT The last assumption can be written in equation form as: Where, kB is Boltzmann's constant (kB = 1.38 ×10-23 m2 kg s-2 K-1) and T is the absolution temperature (in Kelvin) The equation implies that the speed of gas particles is related to their absolute temperature. In other words, as their temperature increases, their speed increases, and finally their total energy increases as well. However, it is impossible to define the speed of any one gas particle. As such, the speeds of gases are defined in terms of their root-mean-square speed. EXAMPLE At a certain speed, the root-mean-square-speed of the molecules of hydrogen in a sample of gas is 1055 ms-1. Compute the root-mean square speed of molecules of oxygen at the same temperature. EXAMPLE GASEOUS DIFFUSION AND EFFUSION Gaseous particles are in constant random motion. Gaseous particles tend to undergo diffusion because they have kinetic energy. Diffusion is faster at higher temperatures because the gas molecules have greater kinetic energy. Effusion refers to the movement of gas particles through a small hole. Graham’s Law states that the effusion rate of a gas is inversely proportional to the square root of the mass of its particles. DIFFUSION According to Kinetic Molecular Theory, gaseous particles are in a constant state of motion, moving at random speeds and in many different directions. Because of their kinetic energy at temperatures above absolute zero, all particles undergo diffusion. Diffusion refers to the process of particles moving from an area of high concentration to one of low concentration. The rate of this movement is a function of temperature, viscosity of the medium, and the size (mass) of the particles. Diffusion results in the gradual mixing of materials, and eventually, it forms a homogeneous mixture. EFFUSION Not only do gaseous particles move with high kinetic energy, but their small size enables them to move through small openings as well; this process is known as effusion. For effusion to occur, the hole’s diameter must be smaller than the molecules’ mean free path (the average distance that a gas particle travels between successive collisions with other gas particles). The opening of the hole must be smaller than the mean free path because otherwise, the gas could move back and forth through the hole. Effusion is explained by the continuous random motion of particles; over time, this random motion guarantees that some particles will eventually pass through the hole. GRAHAM’S LAW Scottish chemist Thomas Graham experimentally determined that the ratio of the rates of effusion for two gases is equal to the square root of the inverse ratio of the gases’ molar masses. This is written as follows: where M represents the molar mass of the molecules of each of the two gases. The gases’ effusion rate is directly proportional to the average velocity at which they move; a gas is more likely to pass through an orifice if its particles are moving at faster speeds. EXAMPLE What is the ratio of the rate of effusion of ammonia, NH3, to that of hydrogen chloride, HCl? The NH3 molecules effuse at a rate 1.46 times faster than HCl molecules. DERIVATION OF GRAHAM’S LAW Graham’s Law can be understood as a consequence of the average molecular kinetic energy of two different gas molecules (marked 1 and 2) being equal at the same temperature. (Recall that a result of the Kinetic Theory of Gases is that the temperature, in degrees Kelvin, is directly proportional to the average kinetic energy of the molecules.) Therefore, equating the kinetic energy of molecules 1 and 2, we obtain: EXAMPLE A gas mixture contains equal numbers of molecules of N2 and SF6. A small portion of it is passed through a gaseous diffusion apparatus. Calculate how many molecules of N2 are present in the product of gas for every 100 molecules of SF6. SOLUTION REFERENCES Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia THE END