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Chapter 5: Gases

Chapter 5: Gases. Remember the conversions for pressure: 760 mm Hg = 760 torr 1 atm = 760 mm Hg 760 mm Hg = 101.325 Pa Convert 2.3 atm to torr: Example ……………………………………… 2.0 atm x 760 torr = 1520 torr 1 atm . The Manometer . In an open manometer, the P gas = P atm – h

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Chapter 5: Gases

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  1. Chapter 5: Gases Remember the conversions for pressure: 760 mm Hg = 760 torr 1 atm = 760 mm Hg 760 mm Hg = 101.325 Pa Convert 2.3 atm to torr: Example ……………………………………… 2.0 atm x 760 torr = 1520 torr 1 atm

  2. The Manometer • In an open manometer, the Pgas= Patm –h • In a closed manometer, the Pgas = the height above the mercury on the gas valve side

  3. Gas Laws: Boyles; Charles; Combined; Dalton • Boyles Law: one common use is to predict the new volume of gas when the pressure is changed or vice versa; P1V1= P2V2 • Example: • A 1.5 L sample of SO2 gas at 5.6 x 103 Pa is changed to a pressure of 1.5 x 104 Pa, what will the volume be? • (1.53 L ) (5.6 x 103 Pa) = V2 (1.5 x 104 Pa) • V2 = 0.57 L

  4. Charles Law: describes the dependence of the volume of gas on temperature at constant pressure • V1 = V2 Pressure is constant T1 T2 • Remember for this to work, temperature must be in degrees Kelvin (K = ºC + 273) • Example: A sample of gas at 15 ºC and 1 atm has a volume of 2.58 L what volume will it have at 38 ºC and 760 mmHg? • 1 atm= 760 mmHg pressure is constant • Convert all temperatures to K: • T1 = 273 + 15 = 283 K and T2 = 273 + 38 = 311 K • 2.58 L = V2 283 K 311 K • V2 = 2.79 L

  5. What do you need to memorize? • The combined gas law: P1V1 = P2V2 T1 T2 • This way, you can cancel any variable that is constant in the problem • Ideal Gas Law: PV = nRT • Where P = pressure v = volume n = mols of gas R = universal gas constant, 0.08206 L atm K mol T = temperature in K

  6. The ideal gas law applies best at high temperatures and low pressures; unless you are given other information, you should assume ideal gas behavior • In the upper atmosphere oxygen is converted to ozone by the equation: O2(g)  O3(g) if .5 mol of oxygen at 1 atm and 25 ºC were converted to ozone at STP, what volume of ozone would be produced? • Remember STP is 1 atm and 0 ºC • Calculate the volume of oxygen at the given conditions: P= 1 atm n= .5 mol R= .08206 atm L/K mol T= 25 + 273=298 k (1 atm) V = (.5 mol) (.08206 atm L/K mol) (298 K) V1= 12.2 L • Use Combined gas to calculate volume of ozone at given conditions V1 = 12.2 L V2 = ? P1V1 = P2V2 T1= 298 T2= 0+ 273= 273 T1 T2 P1= 1 atm P2= 1 atm (1atm) (12.2L) = (1 atm) V2 V2 = 11.2 L 298 K 273 K

  7. Gas Stoichiometry • Key point to remember: At STP (0 ºC and 1 atm) the volume of 1 mol of any gas is 22.4L • You will need to use varied combinations of the ideal gas law, combined gas law, and stoichiometry to solve these types of problems • The presence of mols means we need to be able to predict the products and balance the equations

  8. Ex: a 2.80 L sample of methane gas at 25 ºC and 1.65 atm is mixed with a 3.8 L sample of oxygen at 31 ºC and 1.25 atm then this mixture is ignited , calculate the volume of CO2 formed at a pressure of 2.50 atm and a temperature of 125 ºC • Write the equation: CH4 + O2 CO2 + H2O • Balance it: CH4 + 2 O2 CO2 + 2 H2O • Calculate mols of O2 and mols of CH4 n= PV/RT no2= (1.25 atm)(3.8L) nCH4= (1.65 atm)(2.80L) (0.08206)(298) (0.08206)(304) no2= 1.75 mols O2 nCH4= 0.189 mol CH4 • Determine limiting reactant: mole ratio of methane: oxygen from balanced equation: 1:2 oxygen is excess, methane is limiting • So mols of CO2 produced = 0.189 CH4 x 1 mol CO2 = 0.189 CO2 1 mol CH4

  9. Final calculation • Remember the final conditions of the products are 125 ºC (398 K); 2.5 atm; and now 0.189 mols CO2 so we must use the ideal gas law to complete the calculation: • PV=nRT (2.5 atm) V = (0.189 mols) (0.08206 atm L)(398 K) K mol V= 2.47 L

  10. Try some gas law problems • The density of liquid nitrogen is 0.808 g/ml at -196 ºC. What volume of nitrogen gas at STP must be liquefied to make 10.0 L of liquid nitrogen? • 0.365 mol of water is decomposed to hydrogen gas and oxygen gas by electrolysis. After the gases are dried and collected at 24.5 ºC and 757 torr, what are the volumes of each? • A weather balloon is inflated with 0.295 m3 of helium on the ground at 18 ºC and 756 torr. What will the volume beat an altitude of 10km where the temp is -48 ºC and pressure is 0.14 atm? • A 50.0 L cylinder of chlorine gas at 20.0 ºC and a pressure of 103401 torr springs a leak. The following day the pressure is measured at 41361 torr. How many mols of chlorine gas escaped? • How many cm3 of carbon tetrachloride are produced when 8.0L of chlorine are reacted with 0.75 L of methane at STP?(the other product is HCl) hint cm3 =ml

  11. answers • 6.46 x 103 L • 8.95 L hydrogen gas; 4.47L oxygen gas • 1.6 m3 • 170 mols chlorine gas • 3000 cm3

  12. Real Gases vs Ideal Gases • No gas actually follows the ideal laws, many gases come very close at low pressures and temperatures • Assumptions we make when we explain the properties of an ideal gas: a. Particles are so small that their volumes can be considered negligible b. Particles are in constant motion, collisions of particles with the sides of the container are the cause of pressure c. Particles are assumed to exert no pressure on each other d. Average kinetic energy of a collections of gas is assumed to be directly proportional to the Kelvin temperature of the gas.

  13. How do Real gases behave • Real gas particles actually do take up some volume in the container so that not all volume in the container is available for expansion. • Real gas particles actually do interact with each other and this lowers the total pressure that the real gas exerts on the container. • Pobs = nRT - a (n/v)2 V-nb Pobs = observed pressure n= mols R= constant T= Temp. K A(n/V)2 = pressure correction nb= volume correction

  14. Gas Stoichiometry • 1.0 L of hydrogen gas is collected at 308 K and a pressure of 728 torr. How many grams of iron are required to react with excess HCl to produce this volume of oxygen gas? The products are iron (III) chloride and hydrogen gas. • 2Fe + 6HCl 2 FeCl3 +3 H2 • Convert to STP so we can use 22.4L • 728/760= .958 atm (.958) (1.0) = (1) V V= .849 L H2 308 273 .849LH2 x 1mol H2 x 2 mol Fe x 55.85 g Fe = 1.41 g Fe 22.4 LH2 3 molH2 1 mol Fe

  15. Dalton’s law: states that for a mixture of gases in a container , the total pressure is the sum of each of the individual pressures. • P total = P1 +P2+…Pn= n1+ n2+…nn (RT/V) • 7.8g of solid CO2 is placed in a 4.0 L container at 27 ºC, what is the pressure of the container after all of the CO2 has sublimed? • P (4.0) = (7.8/44)(.08206)(300) P= 1.1 atm • If the same solid CO2 were placed in the same container but it already contained air at 740 torr, what would be the partial pressure of the CO2 and the total pressure of the container? • P CO2 = 1.1 atm • Ptot.= 1.1 + (740/760)= 2.1 atm

  16. What if… • Same solid ofCO2 but now it is placed in a different container filled with .6 mols of air at a volume of 6.3 L and temperature of 290 K. • 1st: calculate pressure at these conditions; this will be the P air (partial pressure) • PV = nRT p(6.3) = .6 (.08206)(290) Pair = 2.27 atm • 2nd: Convert the original CO2 to the new conditions and solve for pressure: can do it either way: PV = nRT • (P)(6.3) = (.18)(.08206)(290) • PCO2= .68 atm • Ptot= .68 + 2.27 = 2.95 atm

  17. Mole fraction and partial pressure • Using the number of moles present in the previous example calculate the mole fraction for each gas • Xair = nair/ ntot = .6/ .78 = .77 • XCO2 = nCO2/ntot = .18/.78 = .23 • This can also be calculated if all you know is the partial pressure of each gas. • Xair= pair/ptot = 2.27/ 2.95 = .77 • X air = p co2 / ptot = .68/2.95 = .23

  18. What do density and molar mass have to do with Gases? • Remember that density is m/v • And if we write the ideal gas law with molar mass it looks like this:PV = (g/MM)(RT) • So if we solve for MM we get MM= g R T = d R T V P P • You can use this information to solve for density and MM as needed:

  19. Diffusion/ Effusion(diffusion is the mixing of gases and effusion is gases passing through a hole into an empty container) • Kinetic energies of two gases are the same at a given temperature so those gases with greater molar masses will move slower with the same energy. • Which gas will effuse through a hole faster, we can use the equations: • Rate gas1 = √MM gas2 Rate gas2 MM gas1 So it follows that the heavier gas will move more slowly out of the hole (Effusion) It also follows that the heavier gas will mix more slowly with other less heavy gases. (Diffusion)

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