VIEWS: 49 PAGES: 49 POSTED ON: 11/9/2012
Chapter 5: Gases Renee Y. Becker Valencia Community College CHM 1045 1 a) Gas is a large collection of particles moving at random throughout a volume b) Collisions of randomly moving particles with the walls of the container exert a force per unit area that we perceive as gas pressure 2 • Units of pressure: atmosphere (atm) Pa (N/m2, 101,325 Pa = 1 atm) Torr (760 Torr = 1 atm) bar (1.01325 bar = 1 atm) mm Hg (760 mm Hg = 1 atm) lb/in2 (14.696 lb/in2 = 1 atm) in Hg (29.921 in Hg = 1 atm) 3 Boyle’s Law • Pressure–Volume Law (Boyle’s Law): 4 Boyle’s Law • Pressure–Volume Law (Boyle’s Law): 1 Volume Pressure • The volume of a fixed amount of gas maintained at constant temperature is inversely proportional to the gas pressure. • Can use any units for P & V V1P1 k1 P1V1 = P2V2 5 Example 1: Boyle’s Law A sample of argon gas has a volume of 14.5 L at 1.56 atm of pressure. What would the pressure be if the gas was compressed to 10.5 L? (at constant temperature and moles of gas) 6 Charles’ Law • Temperature–Volume Law (Charles’ Law): 7 Charles’ Law • Temperature–Volume Law (Charles’ Law): V T • The volume of a fixed amount of gas at constant pressure is directly proportional to the Kelvin temperature of the gas. • Use Kelvins for temperature!! V1 V1 = V2 T1 =k1 T1 T2 8 Example 2: Charles’ Law A sample of CO2(g) at 35C has a volume of 8.56 x10-4 L. What would the resulting volume be if we increased the temperature to 85C? (at constant moles and pressure) 9 Avogadro’s Law • The Volume–Amount Law (Avogadro’s Law): 10 Avogadro’s Law • The Volume–Amount Law (Avogadro’s Law): V n • At constant pressure and temperature, the volume of a gas is directly proportional to the number of moles of the gas present. • Use any volume and moles V1 k1 V1 = V2 n1 n1 n2 11 Example 3: Avogadro’s Law 6.53 moles of O2(g) has a volume of 146 L. If we decreased the number of moles of oxygen to 3.94 moles what would be the resulting volume? (constant pressure and temperature) 12 Combined Gas Law We can combine Boyle’s and charles’ law to come up with the combined gas law P V1 P2 V2 1 T1 T2 Use Kelvins for temp, any pressure, any volume 13 Example 4:Combined Gas Law Oxygen gas is normally sold in 49.0 L steel containers at a pressure of 150.0 atm. What volume would the gas occupy if the pressure was reduced to 1.02 atm and the temperature raised from 20oC to 35oC? 14 Example 5: Gas Laws An inflated balloon with a volume of 0.55 L at sea level, where the pressure is 1.0 atm, is allowed to rise to a height of 6.5 km, where the pressure is about 0.40 atm. Assuming that the temperature remains constant, what is the final volume of the balloon? 15 The Ideal Gas Law • Ideal gases obey an equation incorporating the laws of Charles, Boyle, and Avogadro. P V n R T • 1 mole of an ideal gas occupies 22.414 L at STP • STP conditions are 273.15 K and 1 atm pressure • The gas constant R = 0.08206 L·atm·K–1·mol–1 – P has to be in atm – V has to be in L – T has to be in K 16 Example 6: Ideal Gas Law Sulfur hexafluoride (SF6) is a colorless, odorless, very unreactive gas. Calculate the pressure (in atm) exerted by 1.82 moles of the gas in a steel vessel of volume 5.43 L at 69.5°C. 17 Example 7: Ideal Gas Law What is the volume (in liters) occupied by 7.40 g of CO2 at STP? 18 The Ideal Gas Law • Density and Molar Mass Calculations: mass n MM P MM d volume V R T • You can calculate the density or molar mass (MM) of a gas. The density of a gas is usually very low under atmospheric conditions. 19 Example 8: Density & MM What is the molar mass of a gas with a density of 1.342 g/L at STP? 20 Example 9: Density & MM What is the density of uranium hexafluoride, UF6, (MM = 352 g/mol) under conditions of STP? 21 Example 10: Density & MM The density of a gaseous compound is 3.38 g/L at 40°C and 1.97 atm. What is its molar mass? 22 Dalton’s Law of Partial Pressures 23 Dalton’s Law of Partial Pressures • In a mixture of gases the total pressure, Ptot, is the sum of the partial pressures of the gases: RT P total V n • Dalton’s law allows us to work with mixtures of gases. • T has to be in K • V has to be in L 24 Example 11: Dalton Exactly 2.0 moles of Ne and 3.0 moles of Ar were placed in a 40.0 L container at 25oC. What are the partial pressures of each gas and the total pressure? 25 Example 12: Dalton A sample of natural gas contains 6.25 moles of methane (CH4), 0.500 moles of ethane (C2H6), and 0.100 moles of propane (C3H8). If the total pressure of the gas is 1.50 atm, what are the partial pressures of the gases? 26 Dalton’s Law of Partial Pressures • For a two-component system, the moles of components A and B can be represented by the mole fractions (XA and XB). nA Mole fraction is related to XA the total pressure by: nA nB Pi X i Ptot nB XB nA nB X A X B 1 27 Example 13: Mole Fraction What is the mole fraction of each component in a mixture of 12.45 g of H2, 60.67 g of N2, and 2.38 g of NH3? 28 Example 14: Partial Pressure On a humid day in summer, the mole fraction of gaseous H2O (water vapor) in the air at 25°C can be as high as 0.0287. Assuming a total pressure of 0.977 atm, what is the partial pressure (in atm) of H2O in the air? 29 Gas Stoichiometry & Example • In gas stoichiometry, for a constant temperature and pressure, volume is proportional to moles. Example: Assuming no change in temperature and pressure, calculate the volume of O2 (in liters) required for the complete combustion of 14.9 L of butane (C4H10): 2 C4H10(g) + 13 O2(g) 8 CO2(g) + 10 H2O(l) 30 Example 15: All of the mole fractions of elements in a given compound must add up to? 1. 100 2. 1 3. 50 4. 2 31 Example 16: Hydrogen gas, H2, can be prepared by letting zinc metal react with aqueous HCl. How many liters of H2 can be prepared at 742 mm Hg and 15oC if 25.5 g of zinc (MM = 65.4 g/mol) was allowed to react? Zn(s) + 2 HCl(aq) H2(g) + ZnCl2(aq) 32 Kinetic Molecular Theory • This theory presents physical properties of gases in terms of the motion of individual molecules. • Average Kinetic Energy Kelvin Temperature • Gas molecules are points separated by a great distance • Particle volume is negligible compared to gas volume • Gas molecules are in rapid random motion • Gas collisions are perfectly elastic • Gas molecules experience no attraction or repulsion 33 Kinetic Molecular Theory 34 • Average Kinetic Energy (KE) is given by: U = average speed of a 1 KE mu 2 gas particle 2 R = 8.314 J/K mol m = mass in kg 3RT 3RT MM = molar mass, in u kg/mol mNA MM NA = 6.022 x 1023 35 • The Root–Mean–Square Speed: is a measure of the average molecular speed. 3RT 3RT u 2 u rms MM MM Taking square root of both sides gives the equation 36 Example 17: Calculate the root–mean–square speeds of helium atoms and nitrogen molecules in m/s at 25°C. 37 Kinetic Molecular Theory • Maxwell speed distribution curves. 38 Graham’s Law • Diffusion is the mixing of different gases by random molecular motion and collision. 39 Graham’s Law • Effusion is when gas molecules escape without collision, through a tiny hole into a vacuum. 40 Graham’s Law • Graham’s Law: Rate of effusion is proportional to its rms speed, urms. 3RT Rate u rms MM • For two gases at same temperature and pressure: Rate1 MM 2 MM 2 Rate2 MM1 MM1 41 Example 18: Under the same conditions, an unknown gas diffuses 0.644 times as fast as sulfur hexafluoride, SF6 (MM = 146 g/mol). What is the identity of the unknown gas if it is also a hexafluoride? 42 Example 19: Diffusion • What are the relative rates of diffusion of the three naturally occurring isotopes of neon: 20Ne, 21Ne, and 22Ne? 43 Behavior of Real Gases • Deviations result from assumptions about ideal gases. 1. Molecules in gaseous state do not exert any force, either attractive or repulsive, on one another. 2. Volume of the molecules is negligibly small compared with that of the container. 44 Behavior of Real Gases • At higher pressures, particles are much closer together and attractive forces become more important than at lower pressures. 45 Behavior of Real Gases • The volume taken up by gas particles is actually less important at lower pressures than at higher pressure. As a result, the volume at high pressure will be greater than the ideal value. 46 Behavior of Real Gases • Corrections for non-ideality require van der Waals equation. n 2 P a 2 V – n b nRT V Excluded Volume Intermolecular Attractions 47 Example 20: Ideal Vs. Van Der Waals Given that 3.50 moles of NH3 occupy 5.20 L at 47°C, calculate the pressure of the gas (in atm) using (a) the ideal gas equation (b) the van der Waals equation. (a = 4.17, b = 0.0371) 48 Example 21: Ideal Vs. Van Der Waals Assume that you have 0.500 mol of N2 in a volume of 0.600 L at 300 K. Calculate the pressure in atmospheres using both the ideal gas law and the van der Waals equation. • For N2, a = 1.35 L2·atm mol–2, and b = 0.0387 L/mol. 49