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Mark S. Cracolice Edward I. Peters www.cengage.com/chemistry/cracolice Chapter 14 The Ideal Gas Law and Its Applications Mark S. Cracolice • The University of Montana Avogadro’s Law Avogadro’s Law Equal volumes of gases at the same temperature and pressure contain the same number of molecules. Volume of a gas is proportional to number of moles: V µ n V=k×n Molar Volume The molar volume is the volume of one mole of gas. The molar volume of an ideal gas depends on the temperature and pressure. One mole of any ideal gas occupies 22.7 liters at 0 0C and 1 bar (STP). The Ideal Gas Law PV = n RT P = pressure of the gas V = volume of the gas T = temperature, in Kelvin n = number of moles R = universal gas constant R = 0.0821 L.atm/mol.K R = 62.4 L torr/mol.K The Ideal Gas Law PV = n RT Since the number of moles is equal to the mass m divided by the molar mass MM n = m / MM PV = (m/MM) x RT Determination of a single variable What volume will be occupied by 0.393 mole of nitrogen at 0.971 atm and 240C? PV = nRT V = nRT/P = 0.393 mol x (0.0821 L.atm/mol.K) x (24+273)K / 0.971 atm = 9.87 L Determination of molar mass 1.67 grams of an unknown liquid are vaporized at a temperature of 125 0C. Its volume is measured as 0.421 liter at 749 torr. Calculate the molar mass. PV = (m/MM) x RT Molar mass= MM = mRT/PV = 1.67g x (62.4 L.torr/mol.K) x (125+273)K/(749 torr x 0.421L) = 132g/mol Gas Density mass Density º volume D m V m PV = RT MM mRT m RT RT = ´ = D ´ (MM) V V MM MM RT P = D ´ MM P = D = (MM) P R T Gas Density Density depends on temperature, pressure and molar mass. Hydrogen and helium have lowest densities. What is the density of nitrogen at 44°C and 0.632 atm? . (MM) P D = RT = 28.02 g 0.632 atm mol mol K 0.0821 L atm 1 (44 + 273) K = 0.680 g/L Molar Mass from Gas Density Find the molar mass of an unknown gas if its density is 3.97g/L at 1250C and 749 torr. (MM) P D = RT Molar mass= MM = D x RT/P = (3.97g/L) x (62.4 L.torr/mol.K) x (125+273)K/(749 torr) • = 132g/mol Molar Volume Molar Volume of a Gas The volume occupied by one mole of gas molecules. V In symbols, MV º n PV = nRT V = nRT P V RT = = MV n P Molar Volume Example: What is the molar volume of a ideal gas at 11°C and 1.33 atm? Solution: Solve with algebra. V RT 0.0821 L × atm MV = = = ´ n P mol × K 1 (11 + 273) K ´ = 17.5 L/mol 1.33 atm Gas Stoichiometry at STP Example: What volume of hydrogen, measured at STP, is released when a 42.7 g Zinc is added to a hydrochloric acid solution? Zn + 2 HCl Given: 42.7 g Zn → H2 + ZnCl2 wanted: volume H2 Molar volume at STP is 22.7 L/mol. Gas Stoichiometry at STP Zn + 2 HCl → H2 + ZnCl2 Mass Zn mol Zn mol of H2volume H2 42.7 g Zn x ( 1 mol Zn/ 65.38 g Zn ) x (1 mol H2 / 1 mol Zn ) x ( 22.7 L H2 / 1 mol H2 ) = 14.8 L H2 Stoichiometry: Molar Volume Method Solving a Gas Stoichiometry Problem using Molar Volume Method Step 1: Use the ideal gas equation to find the molar volume at the given temperature and pressure: V/n = RT/P. Step 2: Use the molar volume to calculate the wanted quantity by all three steps of the stoichiometry path. Stoichiometry: Molar Volume Method What volume of CO2, measured at 131°C and 744 torr, is produced when 16.2 g of C4 H10 is burned completely? Step 1 is to find the molar volume at the given temperature and pressure. Given: 131°C; 744 torr Wanted: MV Molar volume = RT/P = (62.4 L torr/mol. K) x ( 131 + 273) / 744 torr = 33.9 L/mol Stoichiometry: Molar Volume Method Step 2 is to use the molar volume to calculate the wanted quantity by the stoichiometry path. 2 C4H10 + 13 O2 → 10 H2O + 8 CO2 gram of C4H10 → mol C4H10→ mol CO2 → L CO2 16.2 gram of C4H10 x (1 mol.C4H10/ 58.12 g C4H10 ) x(8 mol.CO2/ 2 mol.C4H10 ) x (33.9 L CO2 / mol. CO2) = 37.8 L CO2 Volume–Volume Gas Stoichiometry Volume–Volume Gas Stoichiometry • At constant temperature and pressure, the gas volume is directly proportional to the number of moles. • This means that at the same temperature and pressure the ratio of gas volumes is equal to moles ratio. Volume–Volume Gas Stoichiometry Hydrogen and nitrogen gases react to form gaseous ammonia. How many liters of ammonia can be produced from 5.5 L of nitrogen? Both gases are measured at same temperature and pressure. 3 H2 + N2 → 2 NH3 3 mol of H2 react with 1 mol N2 to form 2 mol NH3 At same T and P, same volumes of gases will have same number of moles. 3 L of H2 react with 1L N2 to form 2 L of NH3 Volume–Volume Gas Stoichiometry N2 + 3 H2 GIVEN: 5.5 L N2 5.5 L N 2 → 2 NH3 WANTED: L NH3 2 L NH 3 1L N2 = 11 L NH 3 Volume–Volume Gas Stoichiometry It is found that 1.75 L of oxygen, measured at 24 0C and 755 torr, is used in burning sulfur. The sulfur dioxide produced is at 165 0C and 785 torr. Find the volume of sulfur dioxide at those condition. O2 + 24 0C,755 torr S → SO2 165 0C,785 torr Volume–Volume Gas Stoichiometry First calculate the volume of O2 at same T and P as SO2 (165 0C, 785 torr) Volume of O2 at 165 0C and 785 torr = 1.75 L x ( 755 torr/785 torr) x ( 438 K / 297 K) = 2.48 L O2 Volume–Volume Gas Stoichiometry Next calculate the volume of sulfur dioxide from the volume of oxygen O2 + S → SO2 165 0C & 785 torr 165 0C & 785 torr Litters SO2 (165 0C & 785 torr) = 2.48 L O2 (165 0C & 785 torr) x (1L SO2/1 L O2 ) = 2.48 L SO2 Homework • Homework: 3, 5, 19, 21, 23, 29, 35, 43, 48, 51, 53