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34 Pages

### Chapter 9 - Part 2

Course: CHEM 1010, Spring 2011
School: UOIT
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Word Count: 1718

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9 Chapter Gases Lecture 20 (CHEM 1010) March 31 Agenda: Ideal Gas Law Applying the Ideal Gas Law Daltons Law of Partial Pressures Kinetic Molecular Theory of Gases Grahams Law (Diffusion and Effusion world gas Real Ideal Gas Law Pressure 1 atm = 760 mmHg = 1 torr = 101.325 kPa Boyles Law V inversely proportional to P Charles Law V directly proportional to T Avogadros Law Ideal Gas Law PV =...

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9 Chapter Gases Lecture 20 (CHEM 1010) March 31 Agenda: Ideal Gas Law Applying the Ideal Gas Law Daltons Law of Partial Pressures Kinetic Molecular Theory of Gases Grahams Law (Diffusion and Effusion world gas Real Ideal Gas Law Pressure 1 atm = 760 mmHg = 1 torr = 101.325 kPa Boyles Law V inversely proportional to P Charles Law V directly proportional to T Avogadros Law Ideal Gas Law PV = nRT R can be V directly proportional to n 8.314 J/(K mol) or 8.314 kg m2/(s2 K mol) 0.08206 L atm/(K mol) Problem 9.68 challenge Pure oxygen gas was first prepared by heating mercury (II) oxide, HgO: 2HgO(s) 2Hg (l) + O2(g) What volume (in liters) of oxygen at STP is released by heating 10.57 g of HgO? Problem 9.69 challenge How many grams of HgO would you need to heat if you wanted to prepare 0.0155 mol of O2 according to 2HgO(s) 2Hg (l) + O2(g). Gas Laws and Chemical Reactions Challenge 9.12 5th Propane gas (C3H8) is used as fuel in rural areas. How many liters of CO2 are formed at STP by the complete combustion of the propane in a container with a volume of 15.0L and a pressure of 4.5 atm at 25C. The unbalance equation is C3H8(g) + O2(g) CO2(g) + H2O(l) Gas Laws and Chemical Reactions Challenge 9.12 5th Propane gas (C3H8) is used as fuel in rural areas. How many liters of CO2 are formed at STP by the complete combustion of the propane in a container with a volume of 15.0L and a pressure of 4.5 atm at 25C. The unbalance equation is C3H8(g) + O2(g) CO2(g) + H2O(l) Gas Laws and Chemical Reactions Calculate the volume of oxygen in liters at 35C and 630.0 mmHg, that could be obtained by heating 10.0 g of potassium chlorate (KClO3). Step #2: # mols O2 = 0.1223 mol of O2 V = ?, R = 0.0821 L atm K-1 mol-1 T = 35 + 273 = 308 K, P = 630 / 760 = 0.82895 atm V = nRT/P V = [(0.1223mol)(0.0821LatmK-1mol-1)(308K)]/0.82895atm = 3.73L = volume of O2 at 35 C & 630 torr (or 630 mm Hg) Daltons Law of Partial Pressures Many chemical problems involve mixtures of gases and not pure gases. In the case of gas mixtures the total gas pressure can be related to the pressures of individual gas components (partial pressures). Daltons Law of Partial pressures states: The total pressure of a mixture of gases is just the sum of the pressures that each gas would exert if it were present alone. PT = P1 + P2 + P3 + Daltons Law of Partial Pressures It is possible given the mole fraction of a gaseous component and the total pressure to determine the individual partial pressures: Pi = X iPT The mole fraction of a component: ni X= Mole fraction (Xi) = the i ratio of the number of moles of nT one component (ni) to the number of moles of all components present (nT). Applying Daltons Law of Partial Pressures A sample of natural gas contains 8.24 moles of methane (CH4), 0.421 mole of ethane (C2H6) and 0.116 mole of propane (C3H8). If the total pressure of the gases is 1.37 atm, what are the partial pressures of the gases? Pi = X iPT ni Xi = nT Applying Daltons Law of Partial Pressures Partial Pressures Challenge Applying Daltons Law of Partial Pressures Q. 9-113 5th When 2.00 mol of NOCl (g) was heated to 225C in a 400.0 L steel reaction vessel, the NOCl partially decomposed according to the equation 2 NOCl (g) 2 NO (g) + Cl2(g). The pressure in the vessel after reaction is 0.246 atm. (a) what is the partial pressure of each gas in the vessel after reaction? (b) What percentage of the NOCl decomposed? Kinetic Molecular Theory of Gases The gas laws allow us to predict the behavior of gases. They do not explain what is happening at the molecular level to cause this behavior. Much of the physical properties displayed by gases can be explained by examining the molecular motion of individual molecules. Molecular motion is a form of energy and can also be defined as the capacity to do work: energy = work done = force x distance The energy of motion is known as Kinetic energy. Kinetic Molecular Theory of Gases Relating the physical properties of gasses to the motion of gas molecules lead to a set of four generalizations about the behavior of gasses. These generalizations are known as the kinetic molecular theory of gases (kinetic theory of gases). 1. Gases consist of very small molecules (atoms). 2. The combined volume of the atoms is negligible compared with the empty space between them. 3. Gas molecules or atoms have no attractive or repulsive forces between them. Kinetic Molecular Theory of Gases Kinetic molecular theory of gasses continued: 4. Gas molecules are in constant random motion and often collide. All collisions between gas molecules are perfectly elastic (no overall energy loss). 6. The average Kinetic Energy of the gas molecules is proportional to the temperature in Kelvin. The molecular Kinetic theory of gasses allows us to explain a number of macroscopic properties in molecular terms. Kinetic Molecular Theory of Gases Gas pressure (Boyles Law): Caused by the collision between molecules and the walls of their container. Magnitude of gas pressure depends on: The frequency of collision per unit area. How hard the molecules strike the walls. Gas temperature (Charles Law): A measure of the average kinetic energy of the gas molecules. More energetic random motion = higher temperature. Kinetic Molecular Theory of Gases Compressibility of Gases (Avogadros Law): Molecules in the gas phase are widely separated (assumption #1). Therefore gases are more easily compressed to occupy less volume than a liquid or solid. Kinetic Energy Relationships Relationship between T and kinetic energy (EK) total kinetic energy for a mole of gas particles 3RT/2 kinetic energy per particle (includes NA) 3RT/2NA What is the average speed of a gas particle? EK = 3RT/2NA = mu2 m is the mass of the particle and u is the average speed solving for u = 3RT/M (where M is the molar mass) Kinetic Energy Relationships u = 3RT/M molar mass average speed Kinetic Energy Challenge Calculate the average speed of a nitrogen molecule (in meters per second) on a hot day in summer (T = 37C) and on a cold day in winter. (T = 25C). Given: MN2 = 28.01 g/ mol, K = C + 273.15, R = 8.314 kg m2 K-1 mol-1 s-2 u = 3RT/M Grahams Law Diffusion and Effusion of Gases Diffusion is the mixing of gas molecules by random motion with frequent molecular collisions. Effusion is the escape of a gas through a pinhole into a vacuum without molecular collisions with other gas molecules. Grahams Law T influences EK of gas (but not chemical composition) All gases at the same T will have the same EK mu2 = 3RT/2NA ( mu2)gas 1 = ( mu2)gas 2 (uga1)2/(ugas2)2 = m2/m1 for all gases For diffusion , there are more complex interactions occurring (we are not in a vaccum ie for effusion) but Grahams law is a good approximation for the mixing of the gases Effusion challenge 9.20 b 5th Which gas in the following pair diffuses more rapidly and what are the relative rates of diffusion? (b) N2 and acetylene (C2H2) Ugas1/ugas2 = m2/m1 N = 14.01 amu C = 12.01 amu H= 1.01 amu The Behavior of Real Gases: Volume issue The volume of a real gas is larger than predicted by the ideal gas law. The Behavior of Real Gases: Attraction issue Attractive forces between particles become more important at higher pressures. Intermolecular forces start to become very important when molecules are about 10 molecular diameters apart or less. The Behavior of Real Gases van der Waals equation Correction for intermolecular attractions. an2 P+ V2 V - n b = nRT Correction for molecular volume. van der Waals Equation Challenge 9.22 5th Assume that you have 0.500 mol of N2 in a volume of 0.600L at 300K. Calculate the pressure in atm using both the ideal gas law and the van der Waals equation. For N2 a = 1.35 (L2atm)/mol2 and b = 0.0387 L/mol PV = nRT an2 P+ V2 V - n b = nRT Chapter 9 Appendix Kinetic Molecular Theory of Gases Boyles Law: Pressure results from the impact of gas molecules on the walls of a container. The collision rate is a reflection of the number of molecules of gas per unit volume (number density). Decreasing the volume increases the number density and the collision rate. Therefore pressure will be inversely proportional to volume (Boyles Law). 1 V when n and T are constant P Kinetic Molecular Theory of Gases Charles Law: The average kinetic energy of gas molecules is proportional to its temperature (assumption #5). Raising the temperature of a gas will raise the average kinetic energy of a gas. As a result molecules in the gas will collide with the container walls more frequently and harder and the pressure will increase. In order to maintain a constant pressure the volume of the gas must increase proportionally with temperature (Charles Law). Kinetic Molecular Theory of Gases Avogadros Law: At constant T and P adding more particles will result in an increase in volume. If volume did not increase the number of collisions of gas particles with the walls of the container would increase (increasing P). To avoid this increase, the volume increases. Kinetic Molecular Theory of Gases Daltons Law of Partial Pressures: Gas molecules do not attract or repel one another (assumption #4). If this is true then the pressure caused by one type of gas molecule will not be affected by the presence of another type of gas molecule. So the pressure of each gas present can be calculated separately and the total pressure is the sum of these individual pressures (Daltons Law of Partial Pressures).
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