# Gases: Their Properties & Behavior. Chapter 09 Slide 1

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1 9 Gases: Their Properties & Behavior Chapter 09 Slide 1

2 Gas Pressure 01 Chapter 09 Slide 2

3 Gas Pressure 02 Units of pressure: atmosphere (atm) Pa (N/m 2, 101,325 Pa = 1 atm) Torr (760 Torr = 1 atm) bar ( bar = 1 atm) mm Hg (760 mm Hg = 1 atm) lb/in 2 ( lb/in 2 = 1 atm) in Hg ( in Hg = 1 atm) Chapter 09 Slide 3

4 Boyle s Law 01 Pressure Volume Law (Boyle s Law): Chapter 09 Slide 4

5 Boyle s Law 01 Pressure Volume Law (Boyle s Law): Chapter 09 Slide 5

6 Boyle s Law 02 Pressure Volume Law (Boyle s Law): Volume 1 Pressure The volume of a fixed amount of gas maintained at constant temperature is inversely proportional to the gas pressure. V 1 P 1 = k 1 Chapter 09 Slide 6

7 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 V mmhg x 946 ml = = 4460 mmhg 154 ml 5.3 Chapter 09 Slide 7

8 As T increases V increases Chapter 09 Slide 8

9 Charles Law 01 Temperature Volume Law (Charles Law): T(K)=Temperature in Kelvin T (K) = t (0C) Chapter 09 Slide 9

10 Charles Law 01 Temperature Volume Law (Charles Law): V T The volume of a fixed amount of gas at constant pressure is directly proportional to the Kelvin temperature of the gas. V 1 T 1 =k 1 Chapter 09 Slide 10

11 Variation of gas volume with temperature at constant pressure. P 1 <P 2 <P 3 <P 4 V α T V = constant x T V 1 /T 1 = V 2 /T 2 Temperature must be in Kelvin T (K) = t ( 0 C) Chapter 09 Slide 11

12 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 /T 1 = V 2 /T 2 V 1 = 3.20 L T 1 = K V 2 = 1.54 L T 2 =? T 2 = V 2 x T 1 V L x K = = 192 K 3.20 L Chapter 09 Slide 12

13 Avogadro s Law 01 The Volume Amount Law (Avogadro s Law): Chapter 09 Slide 13

14 Avogadro s Law 01 The Volume Amount Law (Avogadro s Law): V n At constant pressure and temperature, the volume of a gas is directly proportional to the number of moles of the gas present. V 1 = k n 1 1 Chapter 09 Slide 14

15 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? 4NH 3 + 5O 2 4NO + 6H 2 O 1 mole NH 3 1 mole NO At constant T and P 1 volume NH 3 1 volume NO Chapter 09 Slide 15

16 Ideal Gas Equation Boyle s law: V α 1 (at constant n and T) P Charles law: V α T (at constant n and P) Avogadro s law: V α n (at constant P and T) V α nt P nt V = constant x = R nt P P R is the gas constant PV = nrt Chapter 09 Slide 16

17 The conditions 0 0 C and 1 atm are called standard temperature and pressure (STP). Experiments show that at STP, 1 mole of an ideal gas occupies L. PV = nrt R = PV nt = (1 atm)(22.414l) (1 mol)( K) R = L atm / (mol K) Chapter 09 Slide 17

18 The Ideal Gas Law 01 Ideal gases obey an equation incorporating the laws of Charles, Boyle, and Avogadro. P V = n R T R =The gas constant R = L atm K 1 mol 1 Chapter 09 Slide 18

19 What is the volume (in liters) occupied by 49.8 g of HCl at STP? T = 0 0 C = K PV = nrt V = nrt P P = 1 atm n = 49.8 g x 1 mol HCl g HCl = 1.37 mol V = L atm 1.37 mol x x K mol K 1 atm V = 30.6 L 5.4 Chapter 09 Slide 19

20 PV = nrt nr V P 1 T 1 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)? = P T = P 2 T 2 n, V and R are constant = constant P 1 = 1.20 atm T 1 = 291 K P 2 = P 1 x T 2 T 1 = 1.20 atm x 358 K 291 K P 2 =? T 2 = 358 K = 1.48 atm Chapter 09 Slide 20

21 USING GAS DENSITY The density of air at 15 o C and 1.00 atm is 1.23 g/l.. What is the molar mass of air? 1. Calc. moles of air. V = 1.00 L, m=1.23 g, P = 1.00 atm, T = ( )= 288 K n = PV/RT = mol, m=1.23 g 2. Calc. molar mass mass/mol = 1.23 g/ mol = 29.1 g/mol Chapter 09 Slide 21

22 Dalton s Law of Partial Pressures 01 Flash Animation - Click to Continue Chapter 09 Slide 22

23 Dalton s Law of Partial Pressures V and T are constant P 1 P 2 P total = P 1 + P 2 Chapter 09 Slide 23

24 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 Chapter 09 Slide 24

25 P A = n A RT V P B = n B RT V P T = n A RT V n B RT V RT V + = (n A + n B ) P A / P T = n A n A + n B X A = n A n A + n B P A / P T = X A P A = X A P T P B = X B P T X B = n B n A + n B P i = X i P T P T = P A + P B Chapter 09 Slide 25

26 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 Chapter 09 Slide 26

27 Gas Stoichiometry 01 In gas stoichiometry, for a constant temperature and pressure, volume is proportional to moles. Assuming no change in temperature and pressure, calculate the volume of O 2 (in liters) required for the complete combustion of 14.9 L of butane (C 4 H 10 ): 2 C 4 H 10 (g) + 13 O 2 (g) 8 CO 2 (g) + 10 H 2 O(l) Chapter 09 Slide 27

28 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 1 mol C 5.60 g C 6 H 12 O 6 H 12 O 6 6 x 180 g C 6 H 12 O 6 6 mol CO x 2 = mol CO 1 mol C 6 H 12 O 2 6 V = nrt P L atm mol x x K mol K = = 4.76 L 1.00 atm Chapter 09 Slide 28

29 Chapter 09 Slide 29

30 Kinetic Molecular Theory 01 This theory presents physical properties of gases in terms of the motion of individual molecules. Average Kinetic Energy Kelvin Temperature Gas molecules are points separated by a great distance Particle volume is negligible compared to gas volume Gas molecules are in rapid random motion Gas collisions are perfectly elastic Gas molecules experience no attraction or repulsion Chapter 09 Slide 30

31 Kinetic-Molecular Theory Energy can be transferred between molecules during collisions, but the average kinetic energy of the molecules does not change with time, as long as the temperature of the gas remains constant. Chapter 09 Slide 31

32 Kinetic Molecular Theory 03 Average Kinetic Energy (KE) is given by: KE = 1 2 mu2 u 2 = u2 N Chapter 09 Slide 32

33 Kinetic Molecular Theory 03 Average Kinetic Energy (KE) is given by: KE = 1 2 mu2 u 2 = u2 N Chapter 09 Slide 33

34 Kinetic Molecular Theory 05 Maxwell speed distribution curves. Chapter 09 Slide 34

35 Kinetic Molecular Theory 04 The Root Mean Square Speed: is a measure of the average molecular speed. u 2 = 3RT M Taking square root of both sides gives the equation Calculate the root mean square speeds of helium atoms and nitrogen molecules in m/s at 25 C. R=8.314 J/K.mol 1J = 1Kg.m 2 /s 2 3RT u rms = M H m/s Chapter 09 Slide 35

36 Graham s Law 02 Effusion is when gas molecules escape, through a tiny hole into a vacuum. Chapter 09 Slide 36

37 Graham s Law 03 Graham s Law: Rate of effusion is proportional to its rms speed, u rms. Rate α u rms = 3RT M For two gases at same temperature and pressure: Rate 1 Rate 2 = M 2 M 1 = M 2 M 1 Chapter 09 Slide 37

38 A Problem to Consider How much faster would H 2 gas effuse through an opening than methane, CH 4? Rate of Rate of H CH 2 = 4 M m M m (CH (H 2 4 ) ) Rate of H Rate of CH 16.0 g/mol 2.0 g/mol 2 = = So hydrogen effuses 2.8 times faster than CH 4 Chapter 09 Slide 38

39 Behavior of Real Gases 01 Deviations result from assumptions about ideal gases. 1. Molecules in gaseous state do not exert any force, either attractive or repulsive, on one another. 2. Volume of the molecules is negligibly small compared with that of the container. Chapter 09 Slide 39

40 Behavior of Real Gases 02 At higher pressures, particles are much closer together and attractive forces become more important than at lower pressures. As a result, the pressure of real gases will be smaller than the ideal value Chapter 09 Slide 40

41 Behavior of Real Gases 03 The volume taken up by gas particles is actually less important at lower pressures than at higher pressure. As a result, the volume at high pressure will be greater than the ideal value. Chapter 09 Slide 41

42 Behavior of Real Gases 04 Test of ideal gas behavior. Chapter 09 Slide 42

43 Behavior of Real Gases 05 Test of ideal gas behavior. Chapter 09 Slide 43

44 Behavior of Real Gases 05 Corrections for non-ideality require van der Waals equation. P + a n V 2 2 ( V n b) = nrt Intermolecular Attractions Excluded Volume Chapter 09 Slide 44

45 A Problem to Consider If sulfur dioxide were an ideal gas, the pressure at 0 o C exerted by mol occupying L would be atm. Use the van der Waals equation to estimate the real pressure. Use the following values for SO 2 a = L 2. atm/mol 2 b = L/mol Chapter 09 Slide 45

46 A Problem to Consider First, let s rearrange the van der Waals equation to solve for pressure. 2 P = nrt V - nb - n V R= L. atm/mol. K T = K a = L 2. atm/mol 2 V = L b = L/mol a 2 Chapter 09 Slide 46

47 A Problem to Consider P = nrt V - nb - 2 n a 2 V P 2 (1.000 mol) = - 2 L atm (1.000 mol)( mol K)(273.2K) L - (1.000 mol)( L/mol) P = atm (6.865 (22.41 L) L atm 2 mol 2 ) The real pressure exerted by 1.00 mol of SO 2 at STP is slightly less than the ideal pressure. Chapter 09 Slide 47

48 Thermochemical Equations The physical states of all reactants and products must be specified in thermochemical equations. H 2 O (s) H 2 O (l) ΔH = 6.01 kj H 2 O (l) H 2 O (g) ΔH = 44.0 kj How much heat is evolved when 266 g of white phosphorus (P 4 ) burn in air? P 4 (s) + 5O 2 (g) P 4 O 10 (s) ΔH = kj 1 mol P g P 4 x g P 4 x 3013 kj 1 mol P 4 = 6470 kj Chapter 09 Slide 48

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