Chapter 12. The Gaseous State of Matter

Transcription

1 Chapter 12 The Gaseous State of Matter The air in a hot air balloon expands When it is heated. Some of the air escapes from the top of the balloon, lowering the air density inside the balloon, making the balloon buoyant. Introduction to General, Organic, and Biochemistry 10e John Wiley & Sons, Inc Morris Hein, Scott Pattison, and Susan Arena

2 Chapter Outline 12.1 General Properties 12.2 The Kinetic-Molecular Theory 12.3 Measurement of Pressure 12.4 Dependence of Pressure on Number of Molecules and Temperature 12.5 Boyle s Law 12.6 Charles Law 12.7 Gay-Lussac s Law 12.8 Combined Gas Laws 12.9 Dalton s Law of Partial Pressures Avogadro s Law Mole-Mass-Volume Relationships of Gases Density of Gases Ideal Gas Law Gas Stoichiometry

3 General Properties Gases Have an indefinite volume Expand to fill a container Have an indefinite shape Take the shape of a container Have low densities d = 1.2 g / L at 25 C air d = 1.0 g/ml HO 2 Have high kinetic energies

4 Kinetic Molecular Theory (KMT) Assumptions of the KMT and ideal gases include: 1. Gases consist of tiny particles 2. The distance between particles is large compared with the size of the particles. 3. Gas particles have no attraction for each other 4. Gas particles move in straight lines in all directions, colliding frequently with each other and with the walls of the container.

5 Kinetic Molecular Theory Assumptions of the KMT (continued): 5. Collisions are perfectly elastic (no energy is lost in the collision). 6. The average kinetic energy for particles is the same for all gases at the same temperature. 1 2 KE = where is mass and is velocity 2 mv m v 7. The average kinetic energy is directly proportional to the Kelvin temperature.

6 Diffusion

7 Effusion Gas molecules pass through a very small opening from a container at higher pressure of one at lower pressure. Graham s law of effusion: rate of effusion of gas A density B molar mass B = = rate of effusion of gas B density A molar mass A

8 Your Turn! Which gas will diffuse most rapidly? a. He b. Ne c. Ar d. Kr rate of effusion of gas A density B molar mass B = = rate of effusion of gas B density A molar mass A

9 Measurement of Pressure Force Pressure = Area Pressure depends on the Number of gas molecules Temperature of the gas Volume the gas occupies

10 Atmospheric Pressure Atmospheric pressure is due to the mass of the atmospheric gases pressing down on the earth s surface.

11 Barometer

12 Pressure Conversions Convert 675 mm Hg to atm. Note: 760 mm Hg = 1 atm 1 atm 675 mm Hg = atm 760 mm Hg Convert 675 mm Hg to torr. Note: 760 mm Hg = 760 torr. 760 torr 675 mm Hg = 675 torr 760 mm Hg

13 Your Turn! A pressure of 3.00 atm is equal to a. 819 torr b torr c torr d. 253 torr

14 Dependence of Pressure on Number of Molecules P is proportional to n (number of molecules) at T c (constant T) and V c (constant V). The increased pressure is due to more frequent collisions with walls of the container as well increased force of each collision.

15 Dependence of Pressure on Temperature P is proportional to T at n c (constant number of moles) and V c. The increased pressure is due to more frequent collisions higher energy collisions

16 Your Turn! If you change the temperature of a sample of gas from 80 C to 25 C at constant volume, the pressure of the gas a. will increase. b. will decrease. c. will not change

17 Boyle s Law 1 At Tc and nc : V α or PV = PV P What happens to V if you double P? V decreases by half! What happens to P if you double V? P decreases by half!

18 Boyle s Law A sample of argon gas occupies ml at 920. torr. Calculate the pressure of the gas if the volume is increased to 937 ml at constant temperature. Knowns V 1 = 500 ml P 1 = 920. torr V 2 = 937 ml Set-Up Calculate P 2 P 2 = PV V torr 500. ml = = 491 torr 937 ml

19 Boyle s Law Another approach to the same problem: Since volume increased from 500. ml to 937 ml, the pressure of 920. torr must decrease. Multiply the pressure by a volume ratio that decreases the pressure: P ml = 920. torr = 491 torr 937 ml

20 Your Turn! A 6.00 L sample of a gas at a pressure of 8.00 atm is compressed to 4.00 L at a constant temperature. What is the pressure of the gas? a atm b atm c atm d atm

21 Your Turn! A 400. ml sample of a gas is at a pressure of 760. torr. If the temperature remains constant, what will be its volume at 190. torr? A ml B ml C ml D. 1.60x10 2 ml

22 Charles Law The volume of an ideal gas at absolute zero (-273 C) is zero. Real gases condense at their boiling point so it is not possible to have a gas with zero volume. The gas laws are based on Kelvin temperature. All gas law problems must be worked in Kelvin! At P and n : c V α T or c V T = V T

23 Charles Law A 2.0 L He balloon at 25 C is taken outside on a cold winter day at -15 C. What is the volume of the balloon if the pressure remains constant? Knowns V 1 = 2.0 L T 1 = 25 C= 298 K T 2 = -15 C = 258 K Set-Up rearranged gives V 2 = VT T Calculate V 2 (2.0 L)(258 K) = = 1.7 L 298 K

24 Charles Law Another approach to the same problem: Since T decreased from 25 C to -15 C, the volume of the 2.0L balloon must decrease. Multiply the volume by a Kelvin temperature ratio that decreases the volume: P 2 258K = 2.0L = 1.7L 298K

25 Your Turn The volume of a gas always increases when a. Temperature increases and pressure decreases b. Temperature increases and pressure increases c. Temperature decreases and pressure increases d. Temperature decreases and pressure decreases

26 Your Turn! A sample of CO 2 has a volume of 200. ml at 20.0 C. What will be its volume at 40.0 C, assuming that the pressure remains constant? a ml b ml c. 213 ml d ml

27 Your Turn! A sample of gas has a volume of 3.00 L at 10.0 C. What will be its temperature in C if the gas expands to 6.00 L at constant pressure? a C b. 293 C c. 566 C d. 142 C

28 Gay-Lussac s Law At V and n : P α T or c c P T = P T

29 Combined Gas Laws Used for calculating the results of changes in gas conditions. PV T = PV T Boyle s Law where T c Charles Law where P c Gay Lussacs Law where V c PV V T P T = = V T = P T PV P 1 and P 2, V 1 and V 2 can be any units as long as they are the same. T 1 and T 2 must be in Kelvin.

30 Combined Gas Law If a sample of air occupies 500. ml at STP, what is the volume at 85 C and 560 torr? STP: Standard Temperature 273K or 0 C Standard Pressure 1 atm or 760 torr Knowns V 1 = 500. ml T 1 =273K P 1 = 760 torr T 2 = 85 C = 358K P 2 = 560 torr Set-Up Calculate V 2 2 = V = PV 1 1T2 TP 1 2 (760 torr)(500. ml)(358k) = 890. ml (273K)(560 torr)

31 Combined Gas Law A sample of oxygen gas occupies ml at 722 torr and 25 C. Calculate the temperature in C if the gas has a volume of 2.53 L at 491 mmhg. Knowns V 1 = 500. ml T 1 = -25 C = 248K P 1 = 722 torr V 2 = 2.53 L = 2530 ml P 2 = 560 torr Set-Up Calculate T 2 2 TPV = PV ( 491 torr)( 2530 ml)( 248K) ( 722 torr)( ml) T = =853K = 580 C

32 Your Turn! A sample of gas has a volume of 8.00 L at 20.0 C and 700. torr. What will be its volume at STP? a L b L c L d L

33 Dalton s Law of Partial Pressures The total pressure of a mixture of gases is the sum of the partial pressures exerted by each of the gases in the mixture. P Total = P A + P B + P C +. Atmospheric pressure is the result of the combined pressure of the nitrogen and oxygen and other trace gases in air. P = P + P + P + P + P +. Air N O Ar CO H O

34 Collecting Gas Over Water Gases collected over water contain both the gas and water vapor. The vapor pressure of water is constant at a given temperature Pressure in the bottle is equalized so that the P inside = P atm P = P + P atm gas H O 2

35 Your Turn! A sample of oxygen is collected over water at 22 C and 762 torr. What is the partial pressure of the dry oxygen? The vapor pressure of water at 22 C is 19.8 torr. a. 742 torr b. 782 torr c. 784 torr d torr

36 Avogadro s Law Equal volumes of different gases at the same T and P contain the same number of molecules. The ratio is the same: 1 volume 1 molecule 1 mol 1 volume 1 molecule 1 mol 2 volumes 2 molecules 2 mol

37 Mole-Mass-Volume Relationships Molar Volume: One mole of any gas occupies 22.4 L at STP. Determine the molar mass of a gas, if 3.94 g of the gas occupied a volume of 3.52 L at STP. Knowns m = 3.94 g V = 3.52 L T = 273 K P = 1 atm Set-Up 22.4 L 1 mol = 22.4 L so the conversion factor is 1mol Calculate 3.94 g 1.52 L 22.4 L 1 mol = 58.1g/mol

38 Your Turn! What is the molar mass of a gas if 240. ml of the gas at STP has a mass of grams? a g b g c g d g

39 Density of Gases mass g d = = volume L Calculate the density of nitrogen gas at STP. d STP = molar mass 1 mol 22.4 L g 1 mol d STP = = 1.25g/L 1 mol 22.4 L Note that densities are always cited for a particular temperature, since gas densities decrease as temperature increases.

40 Your Turn! Which of the following gases is the most dense? a. H 2 b. N 2 c. CO 2 d. O 2 Carbon dioxide fire extinguishers can be used to put out fires because CO 2 is more dense than air and can be used to push oxygen away from the fuel source.

41 Ideal Gas Law PV = nrt where R = Latm mol K Calculate the volume of 1 mole of any gas at STP. Knowns Set-Up Calculate V n = 1 mole T = 273K P = 1 atm nrt = P Molar volume! L atm (1 mol)( )(273 K) V = mol K = 22.4 L (1 atm)

42 Ideal Gas Law How many moles of Ar are contained in 1.3L at 24 C and 745 mm Hg? Knowns Set-Up Calculate PV = nrt where Latm R = mol K n n V = 1.3 L T = 24 C = 297 K P = 745 mm Hg = atm PV = RT (0.980 atm)(1.3 L) = =0.052 mol L atm ( )(297 K) mol K

43 Ideal Gas Law Calculate the molar mass (M) of an unknown gas, if 4.12 g occupy a volume of 943mL at 23 C and 751 torr. Knowns m =4.12 g V = 943 ml = L T = 23 C = 296 K P = 751 torr = atm Set-Up n g g = so = M PV M RT M = g RT PV Calculate M = L atm (4.12 g)( )(296 K) mol K (0.988 atm)(0.943 L) =107 g/mol

44 Your Turn! What is the molar mass of a gas if 40.0 L of the gas has a mass of 36.0 g at 740. torr and 30.0 C? a g b g c g d. 333 g

45 Gas Stoichiometry Convert between moles and volume using the Molar Volume if the conditions are at STP : 1 mol = 22.4 L. Use the Ideal Gas Law if the conditions are not at STP.

46 Gas Stoichiometry Calculate the number of moles of phosphorus needed to react with 4.0L of hydrogen gas at 273 K and 1 atm. P 4(s) + 6H 2(g) à 4PH 3(g) Knowns V = 4.0 L T = 273 K P = 1 atm Solution Map L H 2 à mol H 2 à mol P 4 Calculate 4.0 L H2 1 mol H2 22.4L 1 mol P 6 mol H 4 2 = mol P4

47 Gas Stoichiometry What volume of oxygen at 760 torr and 25 C are needed to react completely with 3.2 g C 2 H 6? 2 C 2 H 6(g) + 7 O 2(g) 4 CO 2(g) + 6 H 2 O (l) Knowns m = 3.2 g C 2 H 6 T = 25 C = 298K P = 1 atm Solution Map m C 2 H 6 à mol C 2 H 6 à mol O 2 à volume O 2 Calculate 3.2g C2H 1 mol C2H 6 7 mol O2 6 = 0.37mol O g C2H6 2 mol C2H6 L atm (0.37 mol)( )(298 K) V = mol K = 9.1 L (1 atm)

48 Your Turn! How many moles of oxygen gas are used up during the reaction with 18.0 L of CH 4 gas measured at STP? CH 4(g) + 2 O 2(g) CO 2(g) + 2 H 2 O (l) a moles b moles c moles d moles

49 Volume-Volume Calculations Calculate the volume of nitrogen needed to react with 9.0L of hydrogen gas at 450K and 5.00 atm. N 2(g) + 3H 2(g) à 2NH 3(g) Knowns V = 9.0 L T = 450K P = 5.00 atm Solution Map Assume T and P for both gases are the same. Use volume ratio instead of mole ratio! L H 2 à L N 2 Calculate 9.0 L H2 1 L N 3 L H 2 2 = 3.0 L N2

50 Your Turn! What volume of sulfur dioxide gas will react when 12.0 L of oxygen is consumed at constant temperature and pressure? 2 SO 2 + O 2 2 SO 3 a L b L c L d L

51 Real Gases Most real gases behave like ideal gases under ordinary temperature and pressure conditions. Conditions where real gases don t behave ideally: At high P because the distance between particles is too small and the molecules are too crowded together. At low T because gas molecules begin to attract each other. High P and low T are used to condense gases.