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): 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 • 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 PV=nRT 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 = 1.00 atm 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 25C and 3.00 atm, and 100.0 mL of oxygen, measured at 25C 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 • PH2724 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
Real Gases & Ideal Gases
The “Ideal” in the ideal gas law refers to the gas in question. The assumption is that all gasses behave in ideal ways. In real life, such is not the case.
For the most precise work, we need to correct for this deviation from ideal behavior.
Deviation from Ideal Behavior The attractive forces among molecules operate at relatively short distances. For a gas at atmospheric pressure, the molecules are far apart and these attractive forces are negligible. At high pressures, the density of the gas increases, and the molecules are much closer to one another. Then intermolecular forces can become significant enough to affect the motion of the molecules, and the gas will no longer behave ideally.
Another way to observe the nonideality of gases is to lower the temperature. Cooling a gas decreases the molecules’ average kinetic energy, which in a sense deprives molecules of the drive they need to break away from their mutual attractive influences. Hot is more ideal!
The nonideality of gases requires some modifications to the ideal gas law. Such modifications were first made by me. J.D. van der Walls in 1873. Dutch Physicist
Net Force The speed of a molecule that is moving toward the container wall is reduced by the attractive forces exerted by its neighbors. Consequently, the impact this molecule will make on the wall is not as great as it would be if no intermolecular forces were present.
The measured gas pressure is always lower than the pressure the gas would have if it behaved ideally.