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5-1 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Today’s Quiz 10-11-10 1.What is a gas (in chemistry)? 2.Mention.

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Presentation on theme: "5-1 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Today’s Quiz 10-11-10 1.What is a gas (in chemistry)? 2.Mention."— Presentation transcript:

1 5-1 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Today’s Quiz 10-11-10 1.What is a gas (in chemistry)? 2.Mention 3 properties of a gas. What is the formula for pressure?

2 5-2 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Chapter 5 Gases

3 5-3 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 5.1 The three states of matter.

4 5-4 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 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 Physical Characteristics of Gases

5 5-5 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Properties of Gases Gases have relatively low viscosity –Viscosity - measure of the resistance of a fluid to deform under shear stress Most gases have relatively low density –Expressed as g/L Gases are miscible –Can mix with other gases in any proportion to form a homogeneous solution

6 5-6 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Pressure is the force acting on an object per unit area: Gravity exerts a force on the earth’s atmosphere 14.7 lbs/in 2 (psi) Molecules in a gas are always in motion and move in every direction Exert pressure on a surface uniformly Pressure

7 5-7 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Atmospheric Pressure and The Barometer Atmospheric pressure is measured with a barometer. Standard atmospheric pressure is the pressure required to support 760 mm of Hg in a column.

8 5-8 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 5.3 A mercury barometer.

9 5-9 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Closed End Manometer

10 5-10 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Open End Manometers

11 5-11 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Table 5.1 Common Units of Pressure Atmospheric PressureUnitScientific Field chemistryatmosphere(atm)1 atm* pascal(Pa); kilopascal(kPa) 1.01325x10 5 Pa; 101.325 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 bar1.01325 barmeteorology, chemistry, physics *This is an exact quantity; in calculations, we use as many significant figures as necessary.

12 5-12 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 = 291.4 mm Hg. Calculate the CO 2 pressure in torrs, atmospheres, and kilopascals. SOLUTION: PLAN: Construct conversion factors to find the other units of pressure. 291.4 mmHg 1torr 1 mmHg = 291.4 torr 291.4 torr 1 atm 760 torr = 0.3834 atm 0.3834 atm 101.325 kPa 1 atm = 38.85 kPa

13 5-13 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.

14 5-14 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:

15 5-15 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

16 5-16 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 1802. Charles’s Law:

17 5-17 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. A sample of carbon monoxide gas occupies 3.20 L at 125 0 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 = 398.15 K V 2 = 1.54 L T 2 = ? T 2 = V 2 x T 1 V1V1 1.54 L x 398.15 K 3.20 L = = 192 K 5.3 V 1 /T 1 = V 2 /T 2 T 1 = 125 ( 0 C) + 273.15 (K) = 398.15 K

18 5-18 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:

19 5-19 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Standard Temperature and Pressure - STP 0°C or 273.15K and 1 atm (760 torr) Under these conditions, the volume of 1 mole of an ideal gas is standard molar volume = 22.4L

20 5-20 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 5.7 Standard molar volume.

21 5-21 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. THE IDEAL GAS LAW PV = nRT R = PV nT = 1atm x 22.414L 1mol x 273.15K = 0.0821atm*L mol*K Figure 5.9 R is the universal gas constant 3 significant figures

22 5-22 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Groups

23 5-23 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

24 5-24 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 = 273.15 K P = 1 atm n = 49.8 g x 1 mol HCl 36.45 g HCl = 1.37 mol V = 1 atm 1.37 mol x 0.0821 x 273.15 K Latm molK V = 30.6 L 5.4

25 5-25 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

26 5-26 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Today’s Quiz 10-13-10 What is molar mass?: What is the formula of molar mass? What does the law of partial pressures say?

27 5-27 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Density (d) Calculations d = m V = PMPM RT m is the mass of the gas in g M is the molar mass of the gas Molar Mass ( M ) of a Gaseous Substance dRT P M = d is the density of the gas in g/L 5.4

28 5-28 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 27.0 0 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 0.0821 x 300.15 K Latm molK M = 54.6 g/mol

29 5-29 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 2 5.60 g C 6 H 12 O 6 1 mol C 6 H 12 O 6 180 g C 6 H 12 O 6 x 6 mol CO 2 1 mol C 6 H 12 O 6 x = 0.187 mol CO 2 V = nRT P 0.187 mol x 0.0821 x 310.15 K Latm molK 1.00 atm = = 4.76 L 5.5

30 5-30 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.

31 5-31 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

32 5-32 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

33 5-33 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. A sample of natural gas contains 8.24 moles of CH 4, 0.421 moles of C 2 H 6, and 0.116 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 = 0.116 8.24 + 0.421 + 0.116 P T = 1.37 atm = 0.0132 P propane = 0.0132 x 1.37 atm= 0.0181 atm 5.6


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