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Chapter 10 Gases

Chapter 10 Gases. 1.6. m. x ________. s 2. 10.2 Pressure. Force = mass x acceleration

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Chapter 10 Gases

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  1. Chapter 10Gases

  2. 1.6 m x ________ s2 10.2 Pressure Force = mass x acceleration When you weigh yourself on a scale, you are measuring the force that your body exerts on the scale. Let’s say that your body mass is 68 kg. Also, the acceleration due to gravity that the earth exerts on you is constant 9.8 m/s2. The total force exerted by your body on the scale is F = m x a F = 68 kg x 9.8 m = 670 kg m s2 s2 The SI unit for force is the Newton (N). 1 N = 1 kg m s2 670 N = 150 lbs (4.47 N/pound) 1. Let’s say that you (with your mass of 68 kg) are on the moon, where the acceleration due to gravity is about 1.6 m/s2. What force would you exert on a scale in Newtons and in pounds? 1 lb 68 kg 110 N 108.8 N x ________ 24 lbs = = 4.47 N

  3. Pressure = Force Area If you weigh 670 N on earth, and you are standing on a 0.5 m x 0.5 m square, your area = (0.5m)2 = 0.25 m2, the pressure you exert is P = F = 670 kg m/s2 = 2700 kg or 2700 N A 0.25 m2 m s2 m2 The SI unit for pressure is the Pascal (Pa) 1 Pa = 1 kg = 1 N ms2 m2 2. Let’s say that you are wearing high heels with a total area (for both heels) of 110-4 m2. (Men, make believe that you are doing this.) The force is 670 N. Calculate the pressure that you exert. P = F = A 670 kg m/s2 = 1 x 10-4 m2 6.7 x 106 kg m s2 or 6.7 x 106 Pa

  4. Air exerts pressure as well. The “standard atmosphere” is = 101,325 Pa, but this is an awkwardly large number. We commonly express air pressure in units of atmospheres (atm), torr, and mm Hg. 1 atm = 101,325 Pa 1 atm = 760 torr = 760 mm Hg 3. The pressure exerted by a gas is measured to be 0.985 atm. Convert this pressure to torr and pascals. 760 torr Pa 101325 0.985 atm x ________ 749 torr 0.985 atm x ________ 98300 Pa = = 1 atm 1 atm

  5. V2 = P1V1 ____ V2 = (1.3 atm)(27 L) ____________ P2 3.9 atm 10.3 The Gas Laws of Boyle, Charles, and Avogadro Boyle’s Law- The pressure exerted by a gas is inversely proportional to the volume that the gas occupies. In other words, as you squeeze the gas, it exerts more pressure. As an example, fill a zip-lock freezer bag with air, and seal it. Try to squeeze it (reduce the volume). It is very difficult to squeeze because the pressure of the gas is increasing as you reduce the volume of the freezer bag. Pressure x Volume = Constant PV = K If PV is a constant for a gas at constant temperature, then P1V1 = P2V2 (at constant temp) 4. A gas which has a pressure of 1.3 atm occupies a volume of 27 L. What volume will the gas occupy is the pressure is increased to 3.9 atm at constant temperature? Given V1= 27 L P1= 1.3 atm P2= 3.9 atm V2= ? V2 = 9.0 L • 5. The PV constant in the previous example was 35 L atm (1.3 atm x 27 L). Convert this constant to • a. L Pa, and • b. Fundamental SI units 101325 Pa 35 L atm x ________ 3.5 x106 L Pa = 1 atm 1 kg m-1 s-2 1 x 10-3 m3 x ________ x __________ 3.5 x 103 kg m2 s-2 3.5 x 106 L Pa = 1 L 1 Pa 3.5 x 103 J PV is energy!!!

  6. V2 = T2V1 ______ V2 = (333 K)(0.842 L) _____________ T1 303 K Charles’s Law – At a constant pressure, the volume of a gas is directly proportional to the temperature (in Kelvins) of the gas. Volume = Constant Temperature Using the same reasoning as we did in Boyle’s Law, if we change the temperature (at constant pressure), the volume will change so that the ratio of volume to temperature will remain constant. V1 = V2 T1 T2 6. A gas at 30.0 C and 1.00 atm occupies a volume of 0.842 L. What volume will the gas occupy at 60.0 C and 1.00 atm? Given V1= 0.842 L T1= 30.0 C + 273 = 303 K T2= 60.0 C + 273 = 333 K V2= ? V2 = 0.925 L

  7. V2 = n2V1 ______ V2 = (1.71 mol)(5.20 L) ______________ n1 0.436 mol Avogadro’s Law- For a gas at constant temperature and pressure the volume is directly proportional to the number of moles of gas. Volume = Constant x number of moles (constant T & P) V = kn (constant T & P) If you triple the number of moles of gas, the volume will also triple. V1 = V2 n1 n2 7. A 5.20 L sample at 18.0 C and 2.00 atm pressure contains 0.436 moles of gas. If we add an additional 1.27 moles of the gas at the same temperature and pressure, what will the total volume occupied by the gas be? Given V1= 5.20 L n1= 0.436 mol n2= 1.27 mol V2= ? = 1.71 mol V2 = 20.3 L + 0.436 mol

  8. P= (0.614 mol) (0.08206 atm L/mol K)(285 K) _______________________________ 12.9 L P = nRT ____ V 10.4 The Ideal Gas Law The ideal gas law – is a combination of Boyle’s, Charles’s, and Avogadro’s laws. It relates pressure, temperature, volume, and the number of moles of a gas. Pressure x Volume = # of moles of gas x constant x temperature PV = nRT R = 0.08206 L atm/ mol K 8. A sample containing 0.614 moles of a gas at 12.0 C occupies a volume of 12.9 L. What pressure does the gas exert? Given P= ? T = 12.0 oC + 273 = 285 K V= 12.9 L R = 0.08206 atm L/mol K n = 0.614 mols P = 1.11 atm

  9. V2 = (.848 atm)(7.0 L)(284 K) V2 = P1V1T2 _____________________ _______ n = PV ____ n= (0.355 atm)(6.8 L) (1.52 atm)(277 K) _______________________ = 0.355 atm P2T1 RT (0.08206 atm L/mol K)(377 K) atm 1 x ________ 760 mm Hg 9. A sample of methane gas (CH4) at 0.848 atm and 4.0 C occupies a volume of 7.0 L. What volume will the gas occupy if the pressure is increased to 1.52 atm and the temperature increased to 11.0 C? Given V1= 7.0 L P1= 0.848 atm T1= 4.0 °C + 273 = 277 K V2= ? P2= 1.52 atm T2= 11.0 C + 273 = 284 K V2 = 4.0 L 10. How many moles of a gas at 104 C would occupy a volume of 6.8 L at a pressure of 270. mm Hg? Given V = 6.8 L T = 104 oC + 273 = 377 K P= 270 mm Hg n = ? R = 0.08206 atm L/mol K n = 0.078 mol

  10. Gas Stoichiometry – Molar mass of a gas (STP) standard temperature and pressure 0 C and 1.000 atm One mole of an ideal gas at STP will occupy 22.4 liters. 11. What volume will 1.18 moles of O2 occupy at STP? 22.4 L 1.18 mol x ________ = 26.4 L 1 mol or use PV = nRT

  11. V= (0.341 mol) (0.08206 atm L/mol K)(295 K) _______________________________ 1.04 atm V = nRT ____ P Many gas law problems involve calculating the volume of a gas produced by the reaction of volumes of other gases. The problem solving strategy that we have used throughout your chemistry course is still the same. ideal gas law stoichiometry ideal gas law Volume of reactants  moles of reactants  moles of products volume of products 12. A sample containing 15.0 g of dry ice (CO2 (s)) is put into a balloon and allowed to sublime according to the following reaction: CO2 (s) CO2 (g) How big will the balloon be (volume) at 22.0 C and 1.04 atm, after all of the dry ice has sublimed ? Given V= ? T = 22.0oC + 273 = 295 K P= 1.04 atm R = 0.08206 atm L/mol K n = 15.0 g CO2(s) 1 mol CO2(s) 1 mol CO2(g) x ___________ V = 7.94 L x __________ = 0.341 mol 44.0 g CO2(s) 1 mol CO2(s)

  12. V= (0.0223 mol) (0.08206 atm L/mol K)(623 K) _______________________________ 1.00 atm V = nRT ____ P 13. 0.500 L of H2(g) are reacted with 0.600 L of O2(g) at STP according to the equation 2 H2(g) + O2(g) 2 H2O (g) What volume will the H2O occupy at 1.00 atm and 350C ? Given V= ? T = 350oC + 273 = 623 K P= 1.00 atm R = 0.08206 atm L/mol K n = .500 L H2 .600L O2 1 mol H2 2 mol H2O V = 1.14 L x ___________ x __________ = 0.0223 mol 22.4 L H2 2 mol H2 Limiting  mol O2 1 2 mol H2O x ___________ x __________ = 0.0536 mol 22.4 L O2 1 mol O2

  13. m V P RT d = = m V P RT = n V P RT = 10.5 Densities of Gases If we divide both sides of the ideal-gas equation by V and by RT, we get • We know that • moles  molecular mass = mass n  = m • So multiplying both sides by the molecular mass ( ) gives • Mass  volume = density • So, • Note: One only needs to know the molecular mass, the pressure, and the temperature to calculate the density of a gas.

  14. P RT dRT P dRT P d =  =  = Molecular Mass We can manipulate the density equation to enable us to find the molecular mass of a gas: Becomes

  15. MM= (3.40 g/L) (0.08206 atm L/mol K)(307 K) _______________________________ 1.75 atm MM = dRT ____ P 14. A gas at 34.0 C and 1.75 atm has a density of 3.40 g/L. Calculate the molar mass of the gas. Given MM = ? T = 34.0oC + 273 = 307 K P= 1.75 atm R = 0.08206 atm L/mol K d = 3.40 g/L M = 48.9 g/mol

  16. PHe = (0.092 mol) (0.08206 atm L/mol K)(273 K) _______________________________ 4.5 L PHe = nRT nHe = PV ____ ____ nHe = (1.2 atm)(2.0 L) _______________________ Ptotal = PN2 + PHe RT V Ptotal = 1.00 atm + .46 atm (0.08206 atm L/mol K)(319 K) 10.6 Dalton’s Law of Partial Pressure Dalton’s Law of partial pressure – for a mixture of gases in a container, the total pressure is the sum of the pressures that each gas would exert if it were alone. Ptotal = P1 + P2 + … + Pn = (n1 + n2 + … + nn) Key problem solving strategy is the use the ideal gas law to interconvert between pressure and moles of each gas. 15. A volume of 2.0 L of He at 46.0 C and 1.2 atm pressure was added to a vessel that contained 4.5 L of N2 at STP. What is the total pressure and partial pressure of each gas at STP after the He is added? nHe = 0.092 mol Given THe = 46.0°C + 273 = 319 K PHe = 1.2 atm VHe = 2.0 L nHe = ? PHe tank= ? Ptotal tank = ? Vtank = 4.5 L PN2 = 1.00 atm TN2 = 273 K PHe = .46 atm Ptotal = 1.46 atm

  17. ntotal = .20 mol + .092 mol = .292 mol Mole fraction The key idea here is that once you have either the number of moles or the pressure of each component of your system, you can calculate the mole fraction. 16. a. Calculate the number of moles of N2 present in the previous problem. b. Calculate the mole fractions of N2 and He given the mole and pressure data from the previous problem. 1 mol 4.5 L x ________ = .20 mol 22.4 L .69 .69 = = .31 .31 = = 17. The vapor pressure of water in air at 28.0 C is 28.3 torr. Calculate the mole fraction of water in a sample of air at 28.0 C and 1.03 atm pressure. .036 =

  18. 10.7 The Kinetic Molecular Theory of Gases Kinetic Molecular Theory – is a model that attempts to explain the properties of an ideal gas. • The volume of the individual particles of a gas can be assumed to be negligible. • The particles are in constant motion. The collisions of the particles with the walls of the container are the cause of the pressure exerted by the gas. • The particles are assumed to exert no forces on each other. • The average kinetic energy of a collection of gas particles is assumed to be directly proportional to the Kelvin temperature of the gas. Temperature is a measure of the average kinetic energy of a gas. (KE)average = 3/2 RT Root Mean Square Velocity R = 8.3145 J/ K mol = 8.3145 kg m2/ s2 because 1J = 1kg m2/s2 K mol T = temperature in Kelvin M= mass of a mole of the gas in kilograms (kg/mol)

  19. 18. Calculate the root mean square velocity for the atoms in a sample of oxygen gas at a. 0 C b. 300 C  = 668 m/s  = 461 m/s Particles collide with one another and exchange energy after traveling a very short distance. This distance between collisions is called the mean free path and is typically about 10-8 m. Because the exchange of energy happens at different times, particles are speeding up and slowing down. They have an average rms velocity, but rarely have precisely that velocity. Therefore, particleshave a large range of velocities, the average of which is the rmsvelocity.

  20. 10.8 Effusion and Diffusion diffusion – relates to the mixing of gases effusion – relates to the passage of a gas through an orifice into an evacuated chamber Graham’s Law of Effusion The heavier the molecule, the slower the rate of effusion through a small orifice. 19. How many times faster would He effuse compared to NO2? MHe = 4.0g/mol MNO2 = 46.0 g/mol = 3.39 so He would effuse 3.39 times as fast as NO2 Diffusion

  21. (P + ) (V − nb) = nRT n2a V2 10.9 Real Gases We know that no gas behaves in a truly ideal fashion. This section is devoted to determining the deviation of a real gas from ideality. Models of behavior are based on the best available data, are necessarily approximations, and when they fail, can help us learn more about our system. In this case, our “ideal gas” model fails under two important conditions: high pressure and low temperature. This forces corrections to the volume and pressure terms in the ideal gas equation. Volume correction: Because gas molecules do take up space, the free volume of the container is not as large as it would be if it were empty.  correction factor Volumeavailable = Vcontainer - nb  # moles of gas Pressure correction: Because gas molecules can interact with each other, they do not collide with the walls of the container (“exert pressure”) to as great an extent as when there is no intramolecular interaction (as with an ideal gas). a = a constant which depends upon the gas n = the number of moles of the gas Combining Pobs and Vcorrected into the ideal gas equation, the CORRECTED equation is called van der Waal’s equation and is given by

  22. P= (.3000 mol) (0.08206 atm L/mol K)(248 K) _______________________________ .2000 L P = nRT ____ V The constants a and b have been tabulated for different gases, and are given in your textbook. Though the numbers get a bit messy, all you are really doing is determining corrected values for P and V. Interactions between molecules are greatest at low temperatures (low rms velocities), high pressuresand low volumes. • 20. Calculate the pressure exerted by 0.3000 mol of He in a 0.2000 L container at –25.0 C • a. using the ideal gas law • b. using van der Waal’s equation Given P= ? T = -25.0 oC + 273 = 248 K V= 0.2000 L R = 0.08206 atm L/mol K n = 0.3000 mol = 30.53 atm

  23. = (.3000 mol) (0.08206 atm L/mol K)(248 K) _______________________________ n2a V2 )(V − nb) = nRT (P + (.3000 mol)2(.0341 atm L2/mol2) (0.2000 L)2 Given P= ? T = -25.0 oC + 273 = 248 K V= 0.2000 L R = 0.08206 atm L/mol K n = 0.3000 mol • 20. Calculate the pressure exerted by 0.3000 mol of He in a 0.2000 L container at –25.0 C • a. using the ideal gas law • b. using van der Waal’s equation a = 0.0341 atm L2/mol2 b = 0.02370 L/mol P + [.2000 L – (.3000 mol)(.02370 L/mol)] = 31.57 atm

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