The Gas Laws Chapter 10

Our Atmosphere 99% N 2 and O 2 78% N 2 21% O 2 1% CO 2 and the Noble Gases

Pressure Pressure = Force Area (Needles, High Heels, Snow shoes) Caused by the collisions of gases against a container We live at about 1 atmosphere of pressure

Barometer Torricelli (1643) Height of column stayed about 760 mm (760 torr) The higher the elevation, the lower the mercury Weather Rising pressure – calm weather Dropping pressure – storm (fast moving air)

Units of Pressure All of the following are equal: 760 mm Hg (760 torr) 29.9 inches Hg (weather reporting) 1 atmosphere (chemistry) kPa (kiloPascals, physics) 760 mm = 29.9 in = 1 atmosphere = kPa (1 psi = 14.7 atm)

Converting Pressures Examples: 1. Express 485 torr in atmospheres. (0.638 atm) 2. Convert 2.4 atmospheres to mm Hg. (1824 mm Hg) 3. Convert 95.0 kPa to atmospheres and mm Hg. (0.938 atm, 712 mm Hg)

The Ideal Gas Law PV = nRT P = pressure in atmosphere V = volume in Liters n = number of moles T = Temperature in Kelvin R = gas constant R = L-atm / mol-K

The Ideal Gas Law Examples: 1. What is the pressure of a 1.45 mol sample of a gas in a 20.0 L container at 25 o C? (1.77 atm) 2. What volume will 5.00 grams of H 2 occupy at 10.0 o C and 1 atm of pressure? (58.1 L) 3. How many grams of O 2 are needed to occupy a mL aerosol can at 20.0 o C and atmospheres? (0.600 g)

STP Standard Temperature & Pressure Standard Temperature = 0 o C (273 K) Standard Pressure = 1 atm 1 mole of a gas occupies 22.4 L at STP 1 moleor22.4 L 22.4 L1 mole

STP Examples: 1. What volume will moles of nitrogen gas occupy at STP? 2. How many grams of chlorine (Cl 2 ) gas are present in 50.0 L at STP?

12.0 grams of Cl 2 is introduced into a 2.00 L flask at 25 o C. a) Calculate the pressure of the gas b) Convert the pressure to mm Hg. c) Calculate the volume the gas would occupy at STP.

Combined Gas Law P 1 V 1 = n 1 RT 1 P 2 V 2 = n 2 RT 2 Solve both equations for R R = P 1 V 1 R = P 2 V 2 n 1 T 1 n 2 T 2 P 1 V 1 = P 2 V 2 n 1 T 1 n 2 T 2

Boyle’s Law Boyle’s Law Apparatus Demo Boyle’s Law – The pressure and volume of a gas are inversely related Bicycle pump example Piston down – low volume, high pressure Piston up – high volume, low pressure

Boyle’s Law Example: 1. The volume of a car’s cylinder is 475 mL at 1.05 atm. What is the volume when the cylinder is compressed and the pressure is 5.65 atm? P 1 V 1 = P 2 V 2 n 1 T 1 n 2 T 2 (Answer: 88.3 mL)

Boyle’s Law Example: 2. A weather balloon has a volume of 40.0 liters on the surface of the earth at 1.00 atm. What will be the volume at atm as it rises? P 1 V 1 = P 2 V 2 n 1 T 1 n 2 T 2

Charles Law Charles Law – The temperature and volume of a gas are directly related “HOTTER = BIGGER” A gas increases in volume 1/273 rd per degree celsius Can be used to find absolute zero Temperature must be in Kelvin

Charles Law 1. A basketball has a volume of 12.0 L when blown up at o C. What will be the volume if it is taken outside on a day when it is only 5.00 o C? P 1 V 1 = P 2 V 2 n 1 T 1 n 2 T 2

Charles Law Collapses to: V 1 = V 2 T 1 T 2

Charles Law 2. If a tire contains 30.0 L of air at 10.0 o C, what volume will it occupy when it is driven and warms up to 50.0 o C? (34.2 L)

Gay-Lussac’s Law Gay-Lussac’s Law = temperature and pressure of a gas are directly related 1. Gas in a spray can has a pressure of 5.00 atm at 25.0 o C. What will be the pressure at o C? (11.3 atm) P 1 V 1 = P 2 V 2 n 1 T 1 n 2 T 2

Avagadro’s Law Avagadro’s Law = The volume of a gas is directly proportional to the moles present “MORE = BIGGER” 1. A balloon has a volume of 1.00 L when 50.0 grams of N 2 are in the balloon. What is the volume if an additional 25.0 grams of N 2 are added? (1.50 L)

1. The volume of mol of a gas is mL at 1.00 atm and 20.0 o C. What is the volume at 2.00 atm and 30.0 o C? (259 mL)

Gas Density and Molar Mass Remember D = massMolar Mass = mass volume moles

Ex 1 What is the density of carbon tetrachloride vapor at 714 torr and 125 o C? (HINT: Pretend 1 L, solve for n) (4.43 g/L)

Ex 2 The average molar mass of atmosphere of Titan (Saturn’s largest moon) is 28.6 g/mol. If the surface temperature is 95 K and the pressure 1.6 atm, calculate the gas density of Titan’s atmosphere? (ANS: 5.9 g/L)

Ex 3 A 936 mL flask masses g empty. When it is filled with gas to a pressure of 735 torr at 31.0 o C, it is found to mass g. What is the molar mass of the gas?

n = (0.967 atm)(0.936 L) ( L-atm/mol-K)(304 K) n = mol mass = g – g = 2.89 g MM = 2.89 g=79.6 g/mol mol

Ex 4 Calculate the average molar mass of dry air if it has a density of 1.17 g/L at 21 o C and torr. ANS: 29.0 g/mol

Calculate the molar mass of a gas whose density is 2.59 g/L at STP.

Gases and Reaction Stoichiometry: Ex 1 1. What mass of Al is needed to produce 50.0 L of H 2 at STP? 2Al(s) + 6HCl(aq)  2AlCl 3 (aq) + 3H 2 (g) (ANS: 40.2 g Al)

Gases and Reaction Stoichiometry: Ex 2 2. What volume of NO gas measured at atm and 25 o C will be produced from the reaction of 19.5 g of O 2 ? 4NH 3 (g) + 5O 2 (g)  4NO(g) + 6H 2 O(l) (Ans: 16.4 L)

Gases and Reaction Stoichiometry: Ex 3 3. Car safety bags are inflated through the decomposition of NaN 3. How many grams of NaN 3 are needed to produce 36.0 L of N 2 at 1.15 atm and 26.0 o C? 2NaN 3 (s)  2Na(s) + 3N 2 (g) (Ans: 73.1 g)

Gases and Reaction Stoichiometry: Ex 4 4. How many liters of H 2 and N 2 at 1.00 atm and 15.0 o C are needed to produce grams of NH 3 ? N 2 (g) + 3H 2 (g)  2NH 3 (g)

Dalton’s Law of Partial Pressures Dalton’s Law – the total pressure of a gas is equal to the sum of the partial pressures P tot = P A + P B + P C + P D +….. P atm = P N2 + P O2 + P rest 1 atm = 0.78atm atm atm

Dalton’s Law of Partial Pressures 1. Three gases are mixed in a 5.00 L container. In the container, there are 255 torr of Ar, 228 torr of N 2, and 752 torr of H 2. What is the total pressure? (1.63 atm)

Dalton’s Law of Partial Pressures 2. On a humid day, the partial pressure of water in the atmosphere is 18.0 torr. a) If the total pressure is 766 torr, what are the pressures of all of the other gases? b) If the atmosphere is 78.0% N 2 and 21.0% O 2, what are their pressures on this humid day?

Dalton’s Law of Partial Pressures 3. What is the total pressure (in atm) exerted by a mixture of 12.0 g of N 2 and 12.0 g of O 2 in a 2.50 L container at 25.0 o C? (HINT: Calculate the moles of each gas, then use PV=nRT twice). (ANS: 7.87 atm)

Mole Fraction Mole fraction = moles gas A = X A total moles P A = X A P tot

Mole Fraction: Ex 1 A gas mixture contains mol of oxygen and mole of nitrogen. If the total pressure is 745 torr, what is the partial pressure of the two gases? X O2 =0.200 mol= mol X N2 =0.500 mol= mol

P O2 = X O2 P tot P O2 = (0.286)(745 torr) = 213 torr P N2 = X N2 P tot P N2 = (0.714)(745 torr) = 532 torr

Ex 2 The atmosphere of Titan is 82 mol % nitrogen, 12 mol % argon, and 6 mol % methane. Calculate the partial pressure of each gas if the total pressure on Titan is 1220 torr. P N2 = (0.82)(1220 torr) =1000 torr P Ar = (0.12)(1220 torr) = 150 torr P CH4 = (0.06)(1220 torr) = 73 torr

Ex 3 What is the mole fraction and mole percent of oxygen in exhaled air if P O2 is 116 torr and the P total is 760 torr? P O2 = X O2 P tot X O2 = P O2 /P tot X O2 = 116 torr/760 torr = (15.3%)

Ex 4 A mixture contains 2.15 g H 2 and 34.0 g of O 2. Calculate the partial pressure of each gas if the total pressure is 2.05 atm. ANS: 1.03 atm H 2 and 1.02 atm O 2

Gas Collection Over Water P tot = P gas + P H2O

Ex 1 A sample of KClO 3 is decomposed as shown. If 250 mL of gas are collected at 26 o C and 765 torr total pressure, calculate the partial pressure of O 2. 2KClO 3 (s)  2KCl(s) + 3O 2 (g)

P tot = P O2 + P H2O P O2 = P tot - P H2O P O2 = 765 torr – 25 torr = 740 torr (0.974 atm) How many moles of gas were collected? n = PV/RT n = (0.974 atm)(0.250 L)= mole ( L-atm/mol-K)(299K)

How many grams of KClO 3 were decomposed? 2KClO 3 (s)  2KCl(s) + 3O 2 (g) mol ANS: g KClO 3

Ex 2 When a sample of NH 4 NO 2 is decomposed, 511 mL of N 2 are collected over water at 26 o C and 745 torr total pressure. How many grams of NH 4 NO 2 were decomposed? NH 4 NO 2 (s)  N 2 (g) + 2H 2 O(g) ANS: 1.26 g

Root Mean Square Speed of atoms/molecules  = (3RT/ M ) 1/2 M = molar mass (kg/mol) R = J/mol-K Calculate the rms speed of NH 3 and HCl (25 o C).

Graham’s Law of Effusion – the higher the molar mass of a gas, the slower it moves v 1 =m 2 v 2 m 1

Graham’s Law Example At the same temperature, how much faster does an He atom move than an N 2 molecule? (Ans: 2.65 times faster)

Graham’s Law Example Which is faster (and by how much): Cl 2 or O 2 ? (Ans: O 2 is about 1.5 times faster)

Ideal Gas (Kinetic Molecular Theory) Real Gases (Van der Waals Equation) Compressible (1000X less dense than liquids) Rapid Constant Motion Temp  KE (1/2mv 2 ) Elastic Collisions No Volume Volume of molecules – Important at high pressures No Attraction Molecular attraction – Important at low temperatures (colder, “stickier”)

Real Gases 1. Would the ideal gas law work better on Mars (0.6 kPa pressure) or Venus (9300 kPa)? Explain. 2. Would the ideal gas law work better for H 2 O or Ar? Explain.

1. A gas has a volume of mL at °C and torr. What would the volume of the gas be at °C and torr of pressure? 2. What is the volume at STP of 22 grams of CO 2 ? g of XeF 4 gas is placed into an evacuated 3.00 liter container at 80°C. What is the pressure in the container?

The atmosphere of Jupiter is composed almost entirely of hydrogen (H 2 ) and helium (He). If the average molar mass of Jupiter’s atmosphere is g/mole, calculate the percent composition. (ANS: 87.3% H 2, 12.7% He)

The atmosphere of Mars is composed of CO 2, N 2 and 1.6% Ar. If the average molar mass of the gases in Mars’ atmosphere is g/mole, calculate the percentages of CO 2 and N 2.

20. a) 646 torrb) 105 kPac) atm d) atme) 2.53 bar 22. a) X Earth atm b) 9,100 kPa 26. a) 2.31 Lb) 6.67 L 34. a) 33.4 Lb) 1170 K c) 3.81 atm d) mol g Ne X O 3 molecules 40. a) a) 5.07 atmb) 1.17 Lc) 5.61 atm 42. a) 13.9 kg b) 9760 Lc) 273 K d) 1.96 X 10 4 kPa.

46. CO 2 < SO 2 < HBr 50. a) 5.63 g/Lb) 171 g/mol g CaH kg Fe 62. P tot = 23.3 atm 66. P N2 =0.389 atm, P H2 =0.968, P NH3 =0.496 atm 68. a) X O2 =0.149, X N2 = 0.239, X H2 =0.612 b) P O 2 =0.303atm P N2 =0.488 atm P H2 =1.25atm 70. a) atm b) atm c) P t = atm

76. a) Same # moleculesb) N 2 more dense c) Ave KE are equald) CH4 effuses faster 78. a) SF 6 < HBr < Cl 2 < H 2 S < CO b) 517 m/s (CO)325 m/s (Cl 2 )