Gases. Announcements KNOW THESE. Gases and the Kinetic Molecular Theory. Chapter 5

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1 Announcements --Exam 2 Sept 5 6:00-7:30M Coverage Chapter 4-6. lease see blog for skipped material. Gases Chapter 5 Quiz 4 (Today or Thursday) Quiz 5 (Thursday Sept 23) 40 MC/2 long question format: 1.5 hrs Bring only a calculator, pen(s) and your head Gases and the Kinetic Molecular Theory 5.1 An Overview of the hysical States of Matter 5.2 Gas ressure 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 Gases have different physical properties compared to liquids and solids 1. Gas are compressible 2. Exert pressure 3. Have low densities and viscosities. 4. Gas volume varies with temperature. 5. Gases readily mix. 6. Gases assume shape of container. Gas Liquid Solid ressure is the force exerted on a unit area of surface. Chemists use different units to describe pressure = Force Area = kg m/s 2 m 2 mg Area mg l w kg = m s 2 = 1 ascal SI Unit of ressure Unit atmosphere (atm) mm of (Hg) torr bar pascal(a) Atmospheric ressure 760 mm Hg* 760 torr* lb/in 2 (psi) 14.7 lb/in bar x 10 5 a KNOW THESE

2 ressure The pressure of the atmosphere is measured with a barometer. Glass tube closed at one end no pressure (vaccum) --filled with mercury --inverted into the same liquid creates a vacuum. The weight of the earth s atmosphere applies a force (F= mg) that supports the weight of the mercury in the column to a height, h. What is the pressure in atmospheres, pascals, mmhg, psi of a barometer that reads 739 torr? # atmospheres = 739 torr X 760 torr 101,325 ascals ascals = atm X psi = atm X = atm = 9.85 X lbs/in 2 = 14.3 psi At sea level = 760 mm Hg = Centuries of experiments have shown that 4 macroscopic variables are needed to completely describe the state of a gas : pressure, volume, temperature, number of moles. Boyle s law: V! 1 (at constant n and T) Charles law: V! T (at constant n and ) Avogadro s law: V! n (at constant and T) Consolidated into 1 equation V = n Ideal Gas Law! The three historical gas laws can be rewritten to solve for volume and combined into 1 equation: Boyle s law: 1 V = k (at constant n and T) Charles law: V = k1 T (at constant n and ) Avogadro s law: V = k2 n (at constant and T) Ideal Gas Law: V = k x k1 x k2 n T ressure (atm) V = n moles Temperature (Kelvin) = R n T Gas Constant = L atm mol -1 K -1 Boyles Law: At constant temperature (in Kelvin) and constant moles of gas, the pressure of a gas is inversely proportional to the gas volume. Final Condition 2 at V2 1 V 1 = 2 V 2 = k Boyles Law: At constant temperature (in Kelvin) and constant moles of gas, the pressure of a gas is inversely proportional to the gas volume. x V = constant.6.3 Intial condition 1 at V1 2 Volume 4 1 V 1 = 2 V 2 = k at constant T and constant n

3 Charles Law: At constant pressure and moles of gas, the volume of a of gas is directly proportional to the absolute temperature in kelvin. V! T V = constant x T Tubing Hg Gas Avogadro s Law: At a specified temperature and pressure, the volume of a gas is proportional to the number of moles of the gas. V! n V = k n V1 = V2 n 1 n 2 T= 25 C Low Temp High Temp Temp ( C) moles Temp ( C) gas Avogadro s Law: At a specified temperature and pressure, the volume of a gas is proportional to the number of moles of the gas. Scientists define a standard laboratory temperature and pressure called ST. Memorize this! Constant temperature Constant pressure V! n V = k n V1 = V2 n 1 n 2 Avogadro s Law: one mole of any gas will occupy the same volume as one mole of any other gas. ST means: Temperature = K (0 C) ressure: = = 760 torr What is the volume occupied by 1 mole of any gas under conditions of ST? Experiments show that at ST, one mole of an ideal gas occupies 22.4 L. V = n V = n V = (1 mol)( L atm mol 1 K 1 )273.15K = 22.4 L

4 Assuming constant T and constant, Avogadro s Law says we can also use volumes as stoichiometric factors in chemical equations 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? NH3(g) + O 2 (g) 4NH (g) 3 + 5O 2 (g) From the equation From the problem NO(g) + H 2 O(g) 4NO(g) + 6H 2 O(g) 4 mole NH 3 4 mole NO At constant T and Avogadro s Law 1 volume NH 3 1 volume NO Advise for doing gas law problems. 1. Memorize: V = n and units 2. Watch for the gottcha sucker traps-- mostly unit conversions traps. 3. You do not need to remember Boyle s Law, Charles Law, Avogadro s law to solve any problem...just V = n that it. 4. When multiple variables are changing, place variables on one side, constants on the other. What is the volume (in liters) occupied by 49.8 g of HCl at ST? V = n V = n T = 0 0 C = K = 1 mol HCl n = 49.8 g x = 1.37 mol g HCl V = Remember that ST means: L atm 1.37 mol x x K mol K V = 30.6 L A bottle of chlorine gas occupies a volume of 946 ml at a pressure of 726 mmhg. What is the pressure of the gas (in mm Hg) if the gas is compressed at constant temperature to 154 ml in this problem n are all constants--put them on the right 2 = 1 x V 1 V 2 1 = 726 mmhg V 1 = 946 ml V = n = k V = constant and thus: 1 V 1 = 2 V 2 2 =? V 2 = 154 ml 726 mmhg x 946 ml = = 4460 mmhg 154 ml A sample of carbon monoxide gas occupies 3.20 L at C. At what temperature will the gas occupy a volume of 1.54 L if the pressure remains constant? Use V = n and move all the identified constants to the right side of the equation. Move all the variables in the problem to the left side. V = n --n, R, = constants with V and T variables V/T = nr/ = k and therefore V 1 T 1 = V 2 T 2 = Constant Now plug and solve the problem using the correct units!

5 We can extend the usefulness of the gas law to density of a gas and to the gas molecular weight! Molar Mass (M ) V = n = MW m MW = m V Density (d) from the Ideal Gas Equation d = m V = MW m/v = d is the density of the gas in g/l m is the mass of the gas in g M is the molar mass of the gas Rate the gases according to density from lowest to highest assuming all of the gases are maintained at the same pressure and temperature. H2 Cl2 CO2 NH3 2.0 g/mol 71. g/mol 44. g/mol 17 g/mol density = m V = MW MW = A 2.10-L vessel contains 4.65 g of a gas at 1.00 atm and C. What is the molar mass of the gas? V = n = m MW MW = ( m V ) = d solving for MW d = density of gas ( 4.65 g 2.10 L )( L atm mol 1 K 1 ) (300.15K) = 54.6 g/mol 1.00 atm Find the density (in g/l) of CO 2 and the number of molecules (a) at ST (0 0 C and ) and (b) at room conditions (20. 0 C and 1.00 atm). V = n = m MW d = m V = (MW)() (44.01 g/mol)() d = L atm mol 1 K 1 = 1.96 g/l (273.15K) 1.96 g molecules CO 2 /L = X X L mol CO x10 23 molecules g CO 2 mol = 2.68x10 22 molecules CO 2 /L Copyright The McGraw-Hill Companies, Inc. ermission required for reproduction or display. Sample roblem 5.7 Calculating Gas Density 5- ROBLEM: To apply a green chemistry approach, a chemical engineer uses waste CO 2 from a manufacturing process, instead of chlorofluorocarbons, as a blowing agent in the production of polystyrene containers. Find the density (in g/l) of CO 2 and the number of molecules (a) at ST (0 0 C and ) and (b) at room conditions (20. 0 C and 1.00 atm). LAN: 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 Avogadro s number. d = mass/volume V = n V = n/ d = M x SOLUTION: (a) g/mol x 1atm d = = 1.96 g/l atm*l x K mol*k 1.96 g mol CO x10 23 molecules L g CO 2 mol = 2.68x10 22 molecules CO 2 /L