Chapter 5 Gases PDF

Gases Chapter 5 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.  Objectives Chapter 5 5.1 Substances That Exist as Gases 5.2 Pressure of a Gas 5.4 The Ideal Gas Equation 5.5 Gas Stoichiometry 5.6...

Gases Chapter 5 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.  Objectives Chapter 5 5.1 Substances That Exist as Gases 5.2 Pressure of a Gas 5.4 The Ideal Gas Equation 5.5 Gas Stoichiometry 5.6 Dalton's Law of Partial Pressures Elements that exist as gases at 250C and 1 atmosphere 5.1 5.1  Physical Characteristics of Gases Gases assume the volume and shape of their containers. Gases are the most compressible state of matter. Gases will mix evenly and completely when confined to the same container. Gases have much lower densities than liquids and solids. 5.1  Force Pressure = Area (force = mass x acceleration) Units of Pressure 1 pascal (Pa) = 1 N/m2 1 atm = 760 mmHg = 760 torr 1 atm = 101,325 Pa Barometer 5.2 16 km 0.2 atm 6 km 0.5 atm Sea level 1 atm 5.2 Manometers Used to Measure Gas Pressures 5.2  Avogadro’s Law V  number of moles (n) Constant temperature V = constant x n Constant pressure V1 / n1 = V2 / n2 5.3 Ammonia burns in oxygen to form nitric oxide (NO) and water vapor. How many volumes of NO are obtained from one volume of ammonia at the same temperature and pressure? 4NH3 + 5O2 4NO + 6H2O 1 mole NH3 1 mole NO At constant T and P 1 volume NH3 1 volume NO 5.3  Ideal Gas Equation 1 Boyle’s law: V (at constant n and T) P Charles’ law: V  T (at constant n and P) Avogadro’s law: V n (at constant P and T) nT V P nT nT V = constant x =R R is the gas constant P P PV = nRT 5.4  The conditions 0 0C and 1 atm are called standard temperature and pressure (STP). Experiments show that at STP, 1 mole of an ideal gas occupies 22.414 L. PV = nRT PV (1 atm)(22.414L) R= = nT (1 mol)(273.15 K) R = 0.082057 L atm / (mol K) R = 8.314 m3 Pa / (mol K) 5.4  Gas Stoichiometry What is the volume of CO2 produced at 37 0C and 1.00 atm when 5.60 g of glucose are used up in the reaction: C6H12O6 (s) + 6O2 (g) 6CO2 (g) + 6H2O (l) g C6H12O6 mol C6H12O6 mol CO2 V CO2 1 mol C6H12O6 6 mol CO2 5.60 g C6H12O6 x x = 0.187 mol CO2 180 g C6H12O6 1 mol C6H12O6 L atm 0.187 mol x 0.0821 x 310.15 K nRT mol K V= = = 4.76 L P 1.00 atm 5.5  What is the volume (in liters) occupied by 49.8 g of HCl at STP? T = 0 0C = 273.15 K P = 1 atm PV = nRT 1 mol HCl nRT n = 49.8 g x = 1.37 mol V= 36.45 g HCl P L atm 1.37 mol x 0.0821 mol K x 273.15 K V= 1 atm V = 30.6 L 5.4  Change in State Pi, Vi, Ti Pf, Vf, Tf Change ni any nf parameter Re-arrange the equation Change in state at constant volume P1 P2 = T1 T2  Argon is an inert gas used in lightbulbs to retard the vaporization of the filament. A certain lightbulb containing argon at 1.20 atm and 18 0C is heated to 85 0C at constant volume. What is the final pressure of argon in the lightbulb (in atm)? PV = nRT n, V and R are constant nR = P = constant P1 = 1.20 atm P2 = ? V T T1 = 291 K T2 = 358 K P1 P2 = T1 T2 T2 P2 = P1 x = 1.20 atm x 358 K = 1.48 atm T1 291 K 5.4 Density (d) Calculations m PM m is the mass of the gas in g d= = V RT M is the molar mass of the gas Molar Mass (M ) of a Gaseous Substance dRT M= d is the density of the gas in g/L P 5.4  A 2.10-L vessel contains 4.65 g of a gas at 1.00 atm and 27.0 0C. What is the molar mass of the gas? dRT m 4.65 g g M= d= = = 2.21 P V 2.10 L L g L atm 2.21 x 0.0821 mol K x 300.15 K L M= 1 atm M = 54.6 g/mol 5.4  Dalton’s Law of Partial Pressures V and T are constant P1 P2 Ptotal = P1 + P2 5.6 Consider a case in which two gases, A and B, are in a container of volume V. nART PA = nA is the number of moles of A V nBRT nB is the number of moles of B PB = V nA nB PT = PA + PB XA = XB = nA + nB nA + n B PA = XA PT PB = XB PT ni Pi = Xi PT mole fraction (Xi) = nT 5.6  A sample of natural gas contains 8.24 moles of CH4, 0.421 moles of C2H6, and 0.116 moles of C3H8. If the total pressure of the gases is 1.37 atm, what is the partial pressure of propane (C3H8)? Pi = Xi PT PT = 1.37 atm 0.116 Xpropane = = 0.0132 8.24 + 0.421 + 0.116 Ppropane = 0.0132 x 1.37 atm = 0.0181 atm 5.6  Bottle full of oxygen gas and water vapor 2KClO3 (s) 2KCl (s) + 3O2 (g) PT = PO2 + PH2 O 5.6 5.6  Chemistry in Action: Scuba Diving and the Gas Laws Depth (ft) Pressure (atm) 0 1 33 2 66 3 P V 5.6 W = - PAL W = - P(VA - Vi)