5-1 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Chapter 10 Gases Lecture Presentation

5-2 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Characteristics of Gases Unlike liquids and solids, gases –Expand to fill their containers. –Are highly compressible. –Have extremely low densities.

5-3 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Pressure is the amount of force applied to an area: Pressure P = FAFA

5-4 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Common Units of Pressure Atmospheric PressureUnitScientific Field chemistryatmosphere(atm)1 atm* pascal(Pa); kilopascal(kPa) x10 5 Pa; kPa SI unit; physics, chemistry millimeters of mercury(Hg) 760 mm Hg*chemistry, medicine, biology torr760 torr*chemistry pounds per square inch (psi or lb/in 2 ) 14.7lb/in 2 engineering bar barmeteorology, chemistry, physics *This is an exact quantity; in calculations, we use as many significant figures as necessary.

5-5 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Sample Problem 5.1 Converting Units of Pressure PROBLEM: A geochemist heats a limestone (CaCO 3 ) sample and collects the CO 2 released in an evacuated flask attached to a closed- end manometer. After the system comes to room temperature,  h = mm Hg. Calculate the CO 2 pressure in torrs, atmospheres, and kilopascals. SOLUTION: PLAN: Construct conversion factors to find the other units of pressure mmHg 1torr 1 mmHg = torr torr 1 atm 760 torr = atm atm kPa 1 atm = kPa

5-6 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Standard Pressure Normal atmospheric pressure at sea level is referred to as standard pressure. It is equal to –1.00 atm –760 torr (760 mmHg) – kPa

5-7 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. The Gas Laws Gas experiments revealed that four variables affect the state of a gas: Temperature, T Volume, V Pressure, P Quantity of gas present, n (moles) These variables are related through equations know as the gas laws.

5-8 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Boyle’s Law 1662 The volume occupied by a gas is inversely related to its pressure Boyle’s Law: P 1 x V 1 = P 2 x V 2

5-9 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. A sample of chlorine gas occupies a volume of 946 mL at a pressure of 726 mmHg. What is the pressure of the gas (in mmHg) if the volume is reduced at constant temperature to 154 mL? P 1 x V 1 = P 2 x V 2 P 1 = 726 mmHg V 1 = 946 mL P 2 = ? V 2 = 154 mL P 2 = P 1 x V 1 V2V2 726 mmHg x 946 mL 154 mL = = 4460 mmHg 5.3 P x V = constant

5-10 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Charles’ Law 1787 At constant pressure, the volume occupied by a fixed amount of gas directly proportional to its absolute temperature First conceived by Guillaume Amontons in 1702 and published by Joseph Gay-Lussac Charles’s Law: V 1 /T 1 = V 2 /T 2

5-11 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. A sample of carbon monoxide gas occupies 3.20 L at C. At what temperature will the gas occupy a volume of 1.54 L if the pressure remains constant? V 1 = 3.20 L T 1 = K V 2 = 1.54 L T 2 = ? T 2 = V 2 x T 1 V1V L x K 3.20 L = = 192 K 5.3 V 1 /T 1 = V 2 /T 2 T 1 = 125 ( 0 C) (K) = K

5-12 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Avogadro’s Law At fixed temperature and pressure, equal volumes of any ideal gas contain equal number of particles (or moles) Avogadro’s Law: V 1 /n 1 = V 2 /n 2

5-13 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 5.7 Standard molar volume.

5-14 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Ideal-Gas Equation V  1/P (Boyle’s law) V  T (Charles’s law) V  n (Avogadro’s law) So far we’ve seen that Combining these, we get V V  nT P

5-15 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Ideal-Gas Equation The constant of proportionality is known as R, the gas constant.

5-16 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Ideal-Gas Equation The relationship then becomes nT P V V  nT P V = R or PV = nRT

5-17 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Molecular Mass We can manipulate the density equation to enable us to find the molecular mass of a gas: Becomes  molar mass of a gaseous substance P  RT d = dRT P  =

5-18 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. A 2.10-L vessel contains 4.65 g of a gas at 1.00 atm and C. What is the molar mass of the gas? 5.4 dRT P M = d = m V 4.65 g 2.10 L = = 2.21 g L M = 2.21 g L 1 atm x x K Latm molK M = 54.6 g/mol

5-19 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1)What is the volume (in liters) occupied by 49.8 g of HCl at STP? 2) 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 0 C is heated to 85 0 C at constant volume. What is the final pressure of argon in the lightbulb (in atm)? 5.4

5-20 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. What is the volume (in liters) occupied by 49.8 g of HCl at STP? PV = nRT V = nRT P T = 0 0 C = K P = 1 atm n = 49.8 g x 1 mol HCl g HCl = 1.37 mol V = 1 atm 1.37 mol x x K Latm molK V = 30.6 L 5.4

5-21 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 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 0 C is heated to 85 0 C at constant volume. What is the final pressure of argon in the lightbulb (in atm)? PV = nRT n, V and R are constant nR V = P T = constant P1P1 T1T1 P2P2 T2T2 = P 1 = 1.20 atm T 1 = 291 K P 2 = ? T 2 = 358 K P 2 = P 1 x T2T2 T1T1 = 1.20 atm x 358 K 291 K = 1.48 atm 5.4

5-22 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Dalton’s Law of Partial Pressures Formulated in 1801 The total pressure of a mixture of gases is just the sum of the pressures that each gas would exert if it was alone.

5-23 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Dalton’s Law of Partial Pressures V and T are constant P1P1 P2P2 P total = P 1 + P 2 5.6

5-24 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Consider a case in which two gases, A and B, are in a container of volume V. P A = n A RT V P B = n B RT V n A is the number of moles of A n B is the number of moles of B P T = P A + P B X A = nAnA n A + n B X B = nBnB n A + n B P A = X A P T P B = X B P T P i = X i P T 5.6 mole fraction (X i ) = nini nTnT

5-25 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. A sample of natural gas contains 8.24 moles of CH 4, moles of C 2 H 6, and moles of C 3 H 8. If the total pressure of the gases is 1.37 atm, what is the partial pressure of propane (C 3 H 8 )? P i = X i P T X propane = P T = 1.37 atm = P propane = x 1.37 atm= atm 5.6

5-26 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. In Groups

5-27 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Gas Stoichiometry What is the volume of CO 2 produced at 37 0 C and 1.00 atm when 5.60 g of glucose are used up in the reaction: C 6 H 12 O 6 (s) + 6O 2 (g) 6CO 2 (g) + 6H 2 O (l) g C 6 H 12 O 6 mol C 6 H 12 O 6 mol CO 2 V CO g C 6 H 12 O 6 1 mol C 6 H 12 O g C 6 H 12 O 6 x 6 mol CO 2 1 mol C 6 H 12 O 6 x = mol CO 2 V = nRT P mol x x K Latm molK 1.00 atm = = 4.76 L 5.5