Importance of Gases Airbags fill with N gas in an accident. Gas is generated by the decomposition of sodium azide, NaN.

Transcription

1 Gas Laws

2 Importance of Gases Airbags fill with N 2 gas in an accident. Gas is generated by the decomposition of sodium azide, NaN 3. 2 NaN 3 (s) 2 Na (s) + 3 N 2 (g) 2

3 Importance of Gases C 6 H 12 O 6 + O 2 CO 2 + H 2 O Glucose Cellular respiration is going on throughout your body! It involves two gases: Oxygen Carbon Dioxide You breathe in oxygen and breathe out carbon dioxide. 3

4 Kinetic Molecular Theory Gas molecules are always moving. Moving molecules have kinetic energy (KE). At the same temperature, all gases have the same average KE. As temperature goes up, average KE also increases. 4

5 Velocity of Gas Molecules Molecules of a given gas have a range of speeds. 5

6 General Properties of Gases There is a lot of free space in a gas. Gases can be expanded infinitely. Gases occupy containers uniformly and completely. Gases diffuse and mix rapidly. 6

7 More Properties of Gases tiny particles that are far apart relative to their size constant, random motion elastic collisions (with no loss of kinetic energy) the average kinetic energy depends upon temperature 7

8 Pressure Pressure of air is measured with a BAROMETER developed in 1643 by Evangelista Torricelli 8

9 Pressure Hg rises in tube until force of Hg (down) balances the force of atmosphere (pushing up). P of Hg pushing down is related to Hg density column height 9

10 Pressure Column height measures pressure of atmosphere 1 standard atm = 760 mm Hg = 760 torr (named after Torricelli) SI unit is PASCAL, Pa 1 atm = kpa 10

11 Boyle s s Law For a sample of gas at a constant temperature: PV = k As P goes up, V goes down. As V goes up, P goes down. Comparing a sample of gas at two different pressures: Robert Boyle ( ) 14 th child of Earl of Cork, Ireland. P 1 V 1 = P 2 V 2 11

12 Boyle s Law: PV=k At constant T and n, Volume is inversely proportional to Pressure. 12

13 Boyle s Law Example #1 A gas occupies 4.0 L at 760 torr; what volume will it occupy at 380 torr? P 1 V 1 = P 2 V 2 V 2 = P 1 V 1 P 2 ( )( 4.0 L) = 760torr 380 torr V 2 = 8.0 L 13

14 Boyle s Law Example #2 A gas occupies 12 L at 1.2 atm; what volume will it occupy at 2.4 atm? P 1 V 1 = P 2 V 2 V 2 = P 1 V 1 P 2 = ( 1.2 atm) ( 12 L) 2.4 atm V 2 = 6.0 L 14

15 Temperature For gas laws, we must convert temperature to Kelvin: K = C The symbol is K, not K F Fahrenheit C Celsius K Kelvin H 2 0 boils Room temperature H 2 0 freezes Absolute Zero

16 Charles s s Law For a sample of gas under constant pressure: V = kt V and T (in Kelvin) are directly proportional: As V goes up, T goes up. As V goes down, T goes down. For a sample of gas at two different temperatures: V 1 T 1 = V 2 T 2 Jacques Charles ( ) Isolated boron. Studied gases. Balloonist. 16

17 Charles s Law: V=kT At constant P and n, Volume is directly proportional to Temperature. 17

18 Charles Law Example A gas occupies 117 ml at 100 C. At what temperature will it have a volume of 234 ml? V 1 T 1 = V 2 T 2 T 2 = V 2T 1 V 1 ( )( 373 K) = 234mL 117 ml T 2 = 746K = ( ) C = 473 C 18

19 Lussac s Law For a sample of gas in a fixed volume: P = kt Pressure and Temperature (in K) are directly proportional: Joseph Gay-Lussac ( ) As P goes up, T goes up. As P goes down, T goes down. P 1 T 1 = P 2 T 2 19

20 Lussac s Law: P=kT At constant V and n, Pressure and Temperature (in K) are directly proportional. 20

21 Robert Boyle says Do Ws #1 & Ws #2

22 Combined Gas Law P V 1 1 T 1 = P V 2 2 T 2 Temperature must be in Kelvin. 22

23 Combined Law Example V 1 = 105 L, V 2 =? P 1 = 985 torr, P 2 = 760 torr T 1 = 27 C, T 2 = 273 K V 2 = P 1V 1 T 2 P 2 T 1 = 985torr "105L " 273K 760torr " 300K V 2 =124L 23

24 Standard Temperature and Pressure STP 1.00 atm 273 K (0 C) 24

25 STP Problem (Ws #3: Question #2) A ml gas sample at STP is compressed to a volume of ml, and the temperature is increased to 30.0 C. What is the new pressure of the gas in kpa? P 1 V 1 T 1 = P 2V 2 T 2 T 1 = T STP = 0 C = ( )K = 273K P 1 = P STP =1atm =101.3kPa T 2 = 30 C = ( )K = 303K P 2 = P 1V 1 T 2 T 1 V 2 = 101.3kPa " 700.0mL " 303K 273K " 200.0mL P 2 = 394kPa 25

26 Jacques Charles says Now do Ws #3

27 IDEAL GAS LAW P V = n R T Combines the Gas Laws. Can be derived from experiment and theory. 27

28 P V = n R T The Ideal Gas Law holds for all gases and mixtures of gases, as long as they are ideal gases. P = Pressure (atm, kpa, torr, mm Hg) V = Volume (L) n = Number of moles (mol) T = Temperature (K) R = Gas Law Constant 28

29 R is the Gas Law Constant P V = n R T (L atm)/(k mol) or 8.31 (L kpa)/(k mol) The units of pressure (atm or kpa) indicate which value of R to use. 29

30 STP Problem What is the volume of 1.00 mol He at STP? L # atm V = nrt 1.00mol " P = mol # K " 273K 1atm V = 22.4L Remember this number! 30

31 1 mole of any gas at STP has the same volume: STP 22.4 Liters 1 atm, 273 K 22.4 L 1 mole of any gas at STP 1 mol H 2 at STP = 22.4 L 1 mol N 2 at STP = 22.4 L 1 mol Ar at STP = 22.4 L 1 mol He at STP = 22.4 L 1 mol Air at STP = 22.4 L 31

32 Ideal Gas Equation and Molecular Weight PV = nrt n=g/mw PV=gRT/MW density = g/v = PMW/RT 32

33 Using PV = nrt How much N 2 would fill a small room with a volume of 960. cubic feet (27,000 L) to P = 745 mm Hg at 25 o C? R = L atm/k mol Solution 1. Get all data into proper units V = 27,000 L T = 25 o C = 298 K 1atm P = 745mmHg " 760mmHg P = 0.980atm 33

34 Using PV = nrt How much N 2 would fill a small room with a volume of 960. cubic feet (27,000 L) to P = 745 mm Hg at 25 o C? R = L atm/k mol Solution 1. Get all data into proper units 2. Now calculate n = PV / RT n = atm " L L # atm K # mol " 298 K n = 1.1 x 10 3 mol (or about 30 kg of gas) 34

35 PV=nRT Example #2 What volume would 4.00 mol of He occupy at a pressure of 748 torr and a temperature of 30.0 C? P = 748torr " V = nrt P = 1atm 760torr = 0.984atm # ( 4.00 mol) L " atm & % ( 303 K $ K " mol' atm ( ) ( ) V = 101L 35

36 PV=(g/MW)RT Calculate the MW (Molecular Weight) of a compound if g occupied 112 ml at C and 750. torr? MW = grt PV = # ( 0.109g) L " atm & % ( 373K $ K " mol' # 750 % $ 760 atm & (( 0.112L) ' ( ) MW = 30.2g /mol 36

37 Gas Density Low density High density d = m V = PMW RT 37

38 Density What is the density of NO 2 if it occupies 1.00L at 1.24 atm and 50 C? d = g V = MWP RT MW NO2 = (2 "16.0) = 46.0g/mol T = 50 C = ( )K = 323K d = " 46.0 g % $ '( 1.24atm) # mol& " L ( atm % $ ' 323K # K ( mol& ( ) d = 2.15 g L 38

39 Joseph Gay-Lussac says Do Ws #4

40 Dalton s Law The partial pressure of a gas in a mixture is simply the fraction of the total pressure which that gas exerts. Component (%) (torr) Nitrogen (N 2 ) Oxygen (O 2 ) Argon (Ar) Carbon dioxide (CO 2 ) Other gases Trace 0.13 Total Pressure

41 Gas Mixtures and Partial Pressures P total = P 1 + P 2 + n total = n A + n B +... P P total total = = n n RT V RT + V total A n RT V B

42 Partial Pressures Calculate the pressure of a 10.0 L container with mole methane, mole H 2 and mole N 2 at 298K. Total _ moles_of _ gas = = 0.900mol P = nrt V = # ( 0.900mol) L " atm & % ( 298K $ K " mol' 10.0L ( ) ( ) P = 2.20atm 42

43 4 Tools for Solving Gas Stoichiometry Problems 1) Use the Molar Mass to convert: Mass (g) Moles Moles Mass (g) 2) Use the Balanced Chemical Equation to convert: Moles of X Moles of Y 3) At STP: 1 mole of gas = 22.4 L 4) If not at STP, use PV=nRT 43

44 Gas Stoichiometry #1 How many liters of oxygen will be formed at STP from the decomposition of 112g of KClO 3? ( ) L O2 = 112 g KClO 3 2 KClO 3 2 KCl + 3 O 2 Method #1: using 1mol gas at STP = 22.4 L " mol KClO 3 %" 3mol O 2 %" $ ' $ ' 22.4L O % 2 $ ' = 30.7L O 2 # 122.6g KClO 3 &# 2mol KClO 3 &# 1mol O 2 & Method #2: using PV=nRT " 1mol KClO mol O2 = ( 112 g KClO 3 ) 3 %" 3mol O 2 % $ ' $ ' =1.37 molo 2 # 122.6g KClO 3 &# 2mol KClO 3 & V = nrt P = # ( 1.37mol) L " atm & % ( 273K $ K " mol' 1atm ( ) V = 30.7 L 44

45 Gas Stoichiometry #2 How many g of NaN 3 are required to fill an air bag with a volume of 45.5 L at 22.0ºC and a pressure of 828 torr? n = PV RT = 2NaN 3 2Na + 3N 2 " 828torr 1atm % $ ' 45.5L # 760torr& " L ( atm % $ '( 295K) # K ( mol& ( ) = 2.05molN 2 " 2mol NaN g NaN 3 = 2.05mol N 3 %" 2 $ ' 65.02g NaN % 3 $ # 3mol N 2 &# mol NaN 3 & ' g NaN 3 = 88.9g 45

46 Amedeo Avogadro says Do Ws #5

47 Avogadro s Hypothesis Equal volumes of gases at the same T and P have the same number of molecules. V = kn V and n are directly proportional. V 1 n 1 = V 2 n 2 twice the moles means twice the volume Amedeo Avogadro ( ) 47

48 Avogadro s Hypothesis: V=kn At constant P and T, Volume is directly proportional to n (moles). 48