BEHAVIOR OF GASES

BEHAVIOR OF GASES • Gases have weight • Gases take up space • Gases exert pressure • Gases fill their containers Gases doing all of these things!

Kinetic Theory of GasesThe basic assumptions of the kinetic molecular theory are: • Gases are mostly empty space • The molecules in a gas are separate, very small and very far apart

Kinetic Theory of GasesThe basic assumptions of the kinetic molecular theory are: • Gas molecules are in constant, chaotic motion • Collisions between gas molecules are elastic (there is no energy gain or loss)

Kinetic Theory of GasesThe basic assumptions of the kinetic molecular theory are: • The average kinetic energy of gas molecules is directly proportional to the absolute temperature • Gas pressure is caused by collisions of molecules with the walls of the container

Measurements of Gases • To describe a gas, its volume, amount, temperature, and pressure are measured. • Volume: measured in L, mL, cm3 (1 mL = 1 cm3) • Amount: measured in moles (mol), grams (g) • Temperature: measured in KELVIN (K) • K = ºC + 273 • Pressure: measured in mm Hg, torr, atm, etc. • P = F / A (force per unit area)

Moderate Force (about 100 lbs) Small Area (0.0625 in2) P = F /A Enormous Pressure (1600 psi)

Bed of Nails Moderate Force Small Pressure P = F / A Large Surface Area (lots of nails)

Units of Pressure • Units of Pressure: • 1 atm = 760 mm Hg • 1 atm = 760 torr • 1 atm = 1.013 x 105 Pa • 1 atm = 101.3 kPa • 1 atm = 1.013 bar

Boyle’s Law For a given number of molecules of gas at a constant temperature, the volume of the gas varies inversely with the pressure. As P, V (when T and n are constant) and vice versa…. INVERSE RELATIONSHIP V  1/P P1V1 = P2V2

Example: A sample of gas occupies 12 L under a pressure of 1.2 atm. What would its volume be if the pressure were increased to 3.6 atm? (assume temp is constant) • P1V1 = P2V2 • (1.2 atm)(12 L) = (3.6 atm)V2 • V2 = 4.0 L

Charles’ Law Jacques Charles (1746-1828) The volume of a given number of molecules is directly proportional to the Kelvin temperature. As T, V  (when P and n are constant) and vice versa…. DIRECT RELATIONSHIP V  T

Example: A sample of nitrogen gas occupies 117 mL at 100.°C. At what temperature would it occupy 234 mL if the pressure does not change? (express answer in K and °C) • V1 / T1= V2 / T2 • (117 mL) / (373 K) = (234 mL) / T2 • T2 = 746 K • T2 = 473 ºC

Combined gas law Combining Boyle’s law (pressure-volume) with Charles’ Law (volume-temp): OR This is for one gas undergoing changing conditions of temp, pressure, and volume.

P1 = 985 torr V1 = 105 L T1 = 27 °C = 300. K P2 = 1 atm = 760 torr V2 = ? T2 = 0 °C = 273 K Example 1:A sample of neon gas occupies 105 L at 27°C under a pressure of 985 torr. What volume would it occupy at standard conditions? P1V1T2 = P2V2T1 (985 torr)(105 L)(273K)= (760torr)(V2)(300K) V2= 124 L

P1 = 80.0 kPa V1 = 10.0 L T1 = 240 °C = 513 K P2 = 107 kPa V2 = 20.0 L T2 = ? Example 2:A sample of gas occupies 10.0 L at 240°C under a pressure of 80.0 kPa. At what temperature would the gas occupy 20.0 L if we increased the pressure to 107 kPa? P1V1T2 = P2V2T1 (80.0kPa)(10.0L)(T2)= (107kPa)(20.0L)(513K) T2= 1372K≈ 1370K

P1 = 1.3 atm = 988 mmHg V1 = 23.2 L T1 = 22.2 °C = 295.2 K P2 = ? V2 = 11.6 L T2 = 12.5 °C = 285.5 K Example 3:A sample of oxygen gas occupies 23.2 L at 22.2 °C and 1.3 atm. At what pressure (in mm Hg) would the gas occupy 11.6 L if the temperature were lowered to 12.5 °C? P1V1T2 = P2V2T1 (988mm Hg)(23.2L)(285.5K)= (P2)(11.6L)(295.2K) P2= 1938 mm Hg ≈ 1900 mmHg

Gases: Standard Molar Volume & The Ideal Gas Law • Avogadro’s Law: at the same temperature and pressure, equal volumes of all gases contain the same # of molecules (& moles). • Standard molar volume = 22.4 L @STP • This is true of “ideal” gases at reasonable temperatures and pressures ,the behavior of many “real” gases is nearly ideal.

Example: 1.00 mole of a gas occupies 36.5 L, and its density is 1.36 g/L at a given temperature & pressure. a) What is its molecular weight (molar mass)?

Example: 1.00 mole of a gas occupies 36.5 L, and its density is 1.36 g/L at a given temperature & pressure. b) What is the density of the gas under standard conditions?

The IDEAL GAS LAW PV=nRT • Shows the relationship among the pressure, volume, temp. and # moles in a sample of gas. • P = pressure (atm) • V = volume (L) • n = # moles • T = temp (K) • R = universal gas constant = 0.0821 The units of R depend on the units used for P, V & T

P = (1820 torr)(1 atm/760 torr) = 2.39 atm V = ? n = (50.0 g)(1 mol / 30.0 g) = 1.67 mol T = 140 °C + 273 = 413 K Example 1:What volume would 50.0 g of ethane, C2H6, occupy at 140 ºC under a pressure of 1820 torr? PV = nRT (2.39 atm)(V) = (1.67 mol)(0.0821 L·atm/mol·K)(413 K) V= 23.6 L

P = 1.00atm V = 8.96 L n = ? T = 273 K Example 2:Calculate (a) the # moles in, and (b) the mass of an 8.96 L sample of methane, CH4, measured at standard conditions. (a) PV = nRT (1 atm)(8.96 L) = (n)(0.0821 L·atm/mol·K)(273 K) n = 0.400 mol

Or the easier way… Example 2:Calculate (a) the # moles in, and (b) the mass of an 8.96 L sample of methane, CH4, measured at standard conditions. (a)

Convert moles to grams… Example 2:Calculate (a) the # moles in, and (b) the mass of an 8.96 L sample of methane, CH4, measured at standard conditions. (b)

P = ? V = 25.0 L n = (50.0 g)(1 mol / 30.0 g) = 1.67 mol T = 25 °C + 273 = 298 K Example 3:Calculate the pressure exerted by 50.0 g ethane, C2H6, in a 25.0 L container at 25 ºC? PV = nRT (P)(25.0 L) = (1.67 mol)(0.0821 L·atm/mol·K)(298 K) P= 1.63 atm

Determining Molecular Weights & Molecular Formulas of Gases: If the mass of a volume of gas is known, we can use this info. to determine the molecular formula for a compound.

First find # moles Then use mass to determine g/mol… Example 1: A 0.109 g sample of pure gaseous compound occupies 112 mL at 100. ºC and 750. torr. What is the molecular weight of the compound? PV = nRT (0.99 atm)(0.112 L) = (n)(0.0821 L·atm/mol·K)(373 K) n= 0.00362 mol

First find the empirical formula… Example 2: A compound that contains only C and H is 80.0% C and 20.0% H by mass. At STP, 546 mL of the gas has a mass of 0.732 g. What is the molecular (true) formula for the compound? Empirical formula = CH3

Next, determine the MW of the sample… Example 2: A compound that contains only C and H is 80.0% C and 20.0% H by mass. At STP, 546 mL of the gas has a mass of 0.732 g. What is the molecular (true) formula for the compound? 0.024 mol

Finally, determine the molecular formula… Example 2: A compound that contains only C and H is 80.0% C and 20.0% H by mass. At STP, 546 mL of the gas has a mass of 0.732 g. What is the molecular (true) formula for the compound? Empirical formula = CH3  15.0 g/mol True MW = 30.0 g/mol CH3 x 2 = C2H6

Partial Pressures and Mole Fractions • In a mixture of gases each gas exerts the pressure it would exert if it occupied the volume alone. • The total pressure exerted by a mixture of gases is the sum of the partial pressures of the individual gases: • Ptotal = P1 + P2 + P3 + …

Example: If 100.0 mL of hydrogen gas, measured at 25C and 3.00 atm, and 100.0 mL of oxygen, measured at 25C and 2.00 atm, what sould be the pressure of the mixture of gases? • Ptotal = P1 + P2 + P3 + … • PT = 3.00 atm + 2.00 atm • PT = 5.00 atm Notice the two gases are measured at the same temp. and vol.

Vapor Pressure of a Liquid • The pressure exerted by its gaseous molecules in equilibrium with the liquid; increases with temperature

Vapor Pressure of a Liquid • Patm = Pgas + PH2O • or • Pgas = Patm - PH2O

Vapor Pressure of a Liquid See A-2 for a complete table

Example 1: A sample of hydrogen gas was collected by displacement of water at 25 C. The atmospheric pressure was 748 mm Hg. What pressure would the dry hydrogen exert in the same conditions? • PH2 = Patm - PH2O • PH2 =748 mm Hg – 23.76 mm Hg • PH2 = 724.24 mm Hg • PH2724 mm Hg

Example 2: A sample of oxygen gas was collected by displacement of water. The oxygen occupied 742 mL at 27 C. The barometric pressure was 753 mm Hg. What volume would the dry oxygen occupy at STP? • PO2 = Patm - PH2O • PO2 =753 mm Hg – 26.74 mm Hg • PO2 = 726 mm Hg • P1V1T2 = P2V2T1 • (726 mm Hg)(0.742 L)(273K) = (760 mm Hg)(V2)(300K) • V2 = 0.645 L

Example 3: A student prepares a sample of hydrogen gas by electrolyzing water at 25 C. She collects 152 mL of H2 at a total pressure of 758 mm Hg. Calculate: (a) the partial pressure of hydrogen, and (b) the number of moles of hydrogen collected. • PH2 = Patm - PH2O • PH2 =758 mm Hg – 23.76 mm Hg • PH2 = 734 mm Hg

Example 3: A student prepares a sample of hydrogen gas by electrolyzing water at 25 C. She collects 152 mL of H2 at a total pressure of 758 mm Hg. Calculate: (a) the partial pressure of hydrogen, and (b) the number of moles of hydrogen collected. PV = nRT (0.966 atm)(0.152 L) = (n)(0.0821 L·atm/mol·K)(298 K) n= 0.00600 mol H2

Graham’s Law of Diffusion & Effusion • Where, • Rate = rate of diffusion or effusion • MM=molar mass

Stoichiometry of Gaseous Reactions A balanced equation can be used to relate moles or grams of substances taking part in a reaction. (AND VOLUME!)

Example: Hydrogen peroxide is the active ingredient in commercial preparations for bleaching hair. What mass of hydrogen peroxide must be used to produce 1.00 L of oxygen gas at 25 C and 1.00 atm? 2H2O2 O2 + 2H2O PV = nRT (1.00 atm)(1.00 L) = (n)(0.0821 L·atm/mol·K)(298 K) n= 0.0409 mol O2

BEHAVIOR OF GASES