1. AP Chemistry 2014-2015 CH 5 GASES

2.  Pressure is a measure of the force that a gas exerts on its container. It can be described as force per unit area. 5.1 PRESSURE

4.  Gas pressure is measured using barometers, which use mercury (hence the unit mmHg).  A manometer is a device for measuring the pressure of a gas in a container. The pressure of the gas is given by h (the difference in mercury levels).

5.  The most commonly used unit is atmospheres (atm), but others are used as well.  1 atm = 760.00 mmHg = 760.00 torr = 101.325 kPa = 1.013 x 10 3 Pa = standard pressure (sea level)  1 Pa = 1 N/m 2

6. The pressure of a gas is measured as 49 torr. Convert this to both atm and Pa. EXERCISE 5.1 PRESSURE CONVERSIONS 49 torr * (1 atm/760 torr) = 0.064 atm 0.064 atm * (1.013 x 10 3 Pa/1 atm) = 65 Pa

7.  Boyle’s law states that the volume of a confined gas is inversely proportional to the pressure exerted on the gas. Another way to phrase this is “for a quantity of a gas at constant temperature, the product of pressure and volume is a constant”. We call this product the “Boyle’s law constant”.  All gases behave in this manner under ideal conditions—in this case, low pressure. Ideal gases always follow Boyle’s law.  Easiest form to memorize: P 1 V 1 = P 2 V 2 5.2 THE GAS LAWS OF BOYLE, CHARLES, AND AVOGADRO

8. Sulfur dioxide is found in the exhaust of automobiles and power plants. Consider a 1.53 L sample of this gas at a pressure of 5600 Pa. If the pressure is changed to 15000 Pa at a constant temperature, what will be the new volume of the gas? What is the Boyle’s law constant of this gas? EXERCISE 5.2 BOYLE’S LAW P1V1 = P2V2 (5600 Pa)(1.53 L) = (15000 Pa)(V2) V2 = 0.57 L P1V1 = 5600 Pa * 1.53 L = 8568 PaL  8600 PaL

9.  Charles’s law states that if a given quantity of a gas is held at a constant pressure, then its volume is directly proportional to the absolute temperature (meaning, temperature in Kelvin).  Easiest form to memorize V 1 /T 1 = V 2 /T 2 also V 1 T 2 = V 2 T 1

10. A sample of gas at 15°C and 1 atm has a volume of 2.58 L. What volume will this gas occupy at 38°C and 1 atm? EXERCISE 5.3 CHARLES’S LAW P1 = P2 = 1 atm T1 = 15°C + 273 = 288 K V1 = 2.58 L V2 = ? T2 = 38°C + 273 = 311 K (2.58 L/288K) = (V2/311K) V2 = 2.79 L

11.  Gay-Lussac’s law of combining volumes states that volumes of gases always combine with one another in the ratio of small whole numbers, as long as the volumes are measured at the same T and P.  Easiest form to memorize: P 1 /P 2 = T 1 /T 2 ; also P 1 T 2 = P 2 T 1

12.  Avogadro’s hypothesis states that equal volumes of gases under the same conditions of temperature and pressure contain equal numbers of molecules.  Avogadro’s law states that the volume of a gas, at a given T and P, is directly proportional to the quantity of the gas. V 1 /n 1 = V 2 /n 2

13. Combined gas law P 1 V 1 /T 1 = P 2 V 2 /T 2 also P 1 V 1 T 2 = P 2 V 2 T 1  memorize it (“peas and vegetables on the table”)

14. Suppose we have a 12.2 L sample containing 0.50 moles of oxygen gas at a pressure of 1 atm and a temperature of 25°C. If all this oxygen is converted to ozone (O 3 ) at the same temperature and pressure, what would be the volume of the ozone? (answer = 8.1 L) EXERCISE 5.4 AVOGADRO’S LAW 3 O 2 (g)  2 O 3 (g) If you look at the mole ratio of ozone to diatomic oxygen, it’s clear that the ozone would take up 2/3 the volume the oxygen occupies. 12.2 L * 2/3 = 8.1 L, nothing fancy about it.

15.  PV = nRT  R = 0.08206 Latm/molK, though it has other values and units as well; useful only at low pressures and high temperatures 5.3 THE IDEAL GAS LAW

16. A sample of hydrogen gas has a volume of 8.56 L at a temperature of 0°C and a pressure of 1.5 atm. Calculate the moles of hydrogen molecules present in this gas sample. (answer = 0.57 moles) EXERCISE 5.5 IDEAL GAS LAW I PV = nRT n = PV = (1.5 atm)(8.56 L) = 0.57 moles RT(.08206 Latm/molK)(273 K)

17. A sample containing 0.35 moles of argon gas at a temperature of 13°C and a pressure of 568 torr is heated to 56°C and a pressure of 897 torr. Calculate the change in volume that occurs. (answer = decreases by 3 L) EXERCISE 5.6 IDEAL GAS LAW II V = nRT/P We’ll calculate V for both sets of conditions, and then compare. Note that the temperatures need to be converted to Kelvin and that the pressure need to be converted to atmospheres. 286 K 329 K0.747 atm 1.18 atm

18. = 11 L = 8.0 L The change in volume is 8.0 – 11 = -3 L. V1 = nRT1 = (0.35 moles) (.08206 Latm/molK)(286 K) P1 0.747 atm V2 = nRT2 = (0.35 moles) (.08206 Latm/molK)(329 K) P2 1.18 atm

20.  Using PV = nRT to solve for the volume of one mole of gas at STP, we get V/n = RT/P. This is the molar volume of a gas at STP.  At STP, 1 mole of an ideal gas occupies 22.42 L.  Use the ideal gas law to convert quantities that are not at STP. 5.4 GAS STOICHIOMETRY

21. A sample of nitrogen gas has a volume of 1.75 L at STP. How many moles of N 2 are present? (answer =.0781 mol) EXERCISE 5.7 GAS STOICHIOMETRY I At STP, 1 mole of an ideal gas occupies 22.42 L. 1.75 L1 mole= 7.81 x 10 -2 moles nitrogen 22.42 L

22. Quicklime (calcium oxide) is produced by the thermal decomposition of calcium carbonate. Calculate the volume of carbon dioxide at STP produced from the decomposition of 152 g of calcium carbonate. (answer: 34.1 L at STP) EXERCISE 5.8 GAS STOICHIOMETRY II CaCO 3  CaO + CO 2 152 g CaCO 3 1 mole CaCO 3 1 mole CO 2 22.42 L 100.09 g 1 mole CaCO 3 1 mole CO 2 = 34.1 L

23.  “Molecular Mass kitty cat”—all good cats put dirt (dRT) over their pee (P). Ew, but it works.  Remember that the densities of gases are reported in g/L not g/mL. THE DENSITY OF GASES

24. The density of a gas was measured at 1.50 atm and 27°C and found to be 1.95 g/L. Calculate the molar mass of the gas. (answer = 32.0 g/mole) EXERCISE 5.10 GAS DENSITY/MOLAR MASS MM = dRT = (1.95 g/L)(.08206 Latm/molK)(300 K) P1.50 atm = 32.0 g/mole

25.  Dalton’s law states that the pressure of a mixture of gases is the sum of the pressures of the different components of the mixture  P total = P 1 + P 2 + …. + P n  no matter the identity of the gases in a mixture, unless it is stated that a chemical reaction occurs…then you will need to use stoichiometry  This law uses the concept of mole fractions. That’s not on the AP exam anymore, so we’ll leave this as it is.  The partial pressure of each gas in a mixture of gases in a container depends on the number of moles of that gas. 5.5 DALTON’S LAW OF PARTIAL PRESSURES

27. Mixtures of helium and oxygen are used in scuba diving tanks to help prevent “the bends”. For a particular dive, 46 L He at 25°C and 1.0 atm, and 12 L oxygen gas at 25°C and 1.0 atm were pumped into a tank with a volume of 5.0 L. Calculate the partial pressure of each gas and the total pressure in the tank at 25°C. (answer: P He = 9.2 atm, P 2 = 2.4 atm, P total = 11.6 atm) EXERCISE 5.11 DALTON’S LAW First, we need to use Boyle’s Law for each gas (P1V1 = P2V2) to find their individual pressures in the tank. Adding these pressures together gives us the total pressure.

28. Total pressure: P total = P He + P O2 = 9.2 atm + 2.4 atm = 11.6 atm For oxygen: P2 = P1V1 = (1.0 atm)(12 L) = 2.4 atm V2 5.0 L For helium: P2 = P1V1 = (1.0 atm)(46 L) = 9.2 atm V2 5.0 L

29.  It is common to collect a gas by water displacement which means some of the pressure is due to water vapor collected as the gas was passing through. You must correct for this by looking up the partial pressure due to water vapor, which depends on the temperature WATER DISPLACEMENT (APPLICATION OF DALTON’S LAW)

31. A sample of solid potassium chlorate was heated in a test tube and decomposed according to the following equation: 2 KClO 3 (s)  2 KCl(s) + 3 O 2 (g) The oxygen produced was collected by water displacement at 22°C at a total pressure of 754 torr. The volume of the gas collected was 0.650 L. Calculate the partial pressure of oxygen in the gas collected and the mass of potassium chlorate in the sample that was decomposed. (answer = 733 torr, 2.12 g KClO 3 ) EXERCISE 5.12 GAS COLLECTION OVER WATER Look up the partial pressure of water at 22°C, and subtract it from the total pressure to find the partial pressure of oxygen. 754 torr – 19.8 torr = 734 torr (a little different from the given answer; different tables can unfortunately have different values)

32. To calculate the mass of the potassium chlorate, we will need to 1.Find the number of moles of oxygen gas using PV = nRT Pressure = 734 torr = 0.966 atm Volume = 0.650 L Temperature = 22°C = 295.15 K (or 295 K if you’re lazy) 2.Use stoichiometry to “go from” moles of oxygen gas to moles of potassium chlorate 3.Convert moles to grams of potassium chlorate using its GFM PV = nRT  n = PV/RT Moles oxygen gas = n = (0.966 atm)(0.650 L)_____ (0.08206 Latm/molK)(295.15 K) =0.0259 moles oxygen gas

33. To calculate the mass of the potassium chlorate, we will need to 1.Find the number of moles of oxygen gas using PV = nRT 0.0259 moles oxygen gas 2.Use stoichiometry to “go from” moles of oxygen gas to moles of potassium chlorate 3.Convert moles to grams of potassium chlorate using its GFM 0.0259 moles O 2 2 moles KClO 3 122.55 g KClO 3 3 moles O 2 1 mole KClO 3 2 KClO 3 (s)  2 KCl(s) + 3 O 2 (g) = 2.12 g KClO 3

34.  Assumptions  All particles are in constant, random motion  All collisions between particles are perfectly elastic  The volume of the particles in a gas is negligible  The average kinetic energy of the molecules in a gas it is its Kelvin temperature  These assumptions ignore intermolecular forces.  Reminder…gases expand to fill their containers; they are also compressible.  Distribution of molecular speeds 5.6 THE KINETIC MOLECULAR THEORY OF GASES

35.  The plot of gas molecules having various speeds vs. fraction of molecules with a particular speed is a curve. This equation (Maxwell’s equation) is  U rms = root mean speed  T = temperature, Kelvin  MM = mass of a mole of gas particles in kg (weird, I know— respect the math though)  Use the “energy R” or 8.314510 J/molK for this equation since kinetic energy is involved. 5.7 EFFUSION AND DIFFUSION

37.  Calculate the root mean square velocity for the atoms in a sample of helium gas at 25°C. EXERCISE 5.13 ROOT MEAN SQUARE VELOCITY R = 8.314510 J/molKT = 298.15 K Molar mass = 4.003 g/mole, or 0.004003 kg/mole U rms = √((3 * 8.314510 * 298.15)/(0.004003)) = √(7433/0.004003) = 1360 m/s

38.  The mean free path is the average distance a particle travels between collisions. It’s on the order of a tenth of a micrometer, and very erratic.  The graph on the right shows the effect of temperature on the numbers of particles with a given velocity. It’s no surprise that increasing the temperature increases the mean velocity, but the shape of the curve changes as well. There will always be the odd particle present with zero or near-zero velocity, so as mean temperature increases, the curve gets pulled to the right and squashed. Keep in mind that all three of these curves represent the same number of particles. You may see graphs like this on the AP exam where you have to identify the highest temperature based on the shape of the graph.

39.  Now, for Graham’s Law of Diffusion and Effusion.  Diffusion describes the mixing of gases; the rate of diffusion is the rate of mixing.  Effusion describes the passage of a gas through a tiny orifice into an evacuated chamber. The rate of effusion measures the speed at which the gas is transferred into the chamber.  The rates of effusion of two gases are inversely proportional to the square roots of their molecular masses at the same temperature and pressure.  Remember that rate is a change in a quantity over time.

41. Calculate the ratio of the effusion rates of hydrogen gas and uranium hexafluoride gas. (answer = 13.2) EXERCISE 5.14 EFFUSION RATES I Let’s make hydrogen “gas 1” and uranium hexafluoride “gas 2”, just to keep them straight. Rate H 2 /Rate UF 6 = √MM UF 6 /MM H 2 = √(352.02 g/mole)/(2.02 g/mole) = √174.27 = 13.2

42. A pure sample of methane is found to effuse through a porous barrier in 1.50 minutes. Under the same conditions, an equal number of molecules of an unknown gas effuse through the barrier in 4.73 minutes. What is the molar mass of the unknown gas? (answer = 159 g/mole) EXERCISE 5.15 EFFUSION RATES II Read carefully—the equation we’ve seen uses rates, but this problem describes times instead. The equation must change. Think about it… Time of gas 1 = √MM1/MM2 Time of gas 2

43. Time of gas 1 = √MM1/MM2 Time of gas 2 (Time 1/Time 2) 2 = MM1/MM2 MM1 = (Time 1/Time 2) 2 * MM2 So let’s make gas 1 our unknown, and gas 2 methane (molar mass = 16.04 g/mole MM1 = (4.73 min/1.50 min) 2 * 16.04 g/mole = 9.933 * 16.04 g/mole = 159 g/mole

44. CLASSIC EXAMPLE OF DIFFUSION

45.  Most gases behave ideally until you reach high pressure and low temperature (gases also tend to liquefy at these conditions). 5.8 REAL GASES

46.  The van der Waals equation corrects for the negligible volume of molecules and accounts for inelastic collisions leading to IMFs. a and b are van der Waals constants.  The “a” term increases the pressure to correct for lower pressure due to IMFs.  The “b” term accounts for decreased free volume, since particles take up space.  It is more important to understand the concepts behind this equation than to plug in values and use it.