Chapter 5 The Gaseous State

Contents and Concepts Gas Laws We will investigate the quantitative relationships that describe the behavior of gases. 1. Gas Pressure and Its Measurement 2. Empirical Gas Laws 3. The Ideal Gas Law 4. Stoichiometry Problems Involving Gas Volumes 5. Gas Mixtures; Law of Partial Pressures Copyright Cengage Learning. All rights reserved. 5 2

Kinetic-Molecular Theory This section will develop a model of gases as molecules in constant random motion. 6. Kinetic Theory of Gases 7. Molecular Speeds; Diffusion and Effusion 8. Real Gases Copyright Cengage Learning. All rights reserved. 5 3

Physical Characteristics of Gases 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. NO 2 gas 5

Some applications How does a pressure cooker work? How is gas pressure applied in spray cans? How does a hot air balloon work? Why do we not want our tires to be full during hot summer days? Why do balloons deflate when left outside on a cold weather?

Pressure = Force Area (force = mass x acceleration) Units of Pressure 1 pascal (Pa) = 1 N/m 2 1 atm = 760 mmhg = 760 torr 1 atm = 101,325 Pa 8

Manometers Used to Measure Gas Pressures closed-tube open-tube 9

Boyle s Law The volume of a sample of gas at constant temperature varies inversely with the applied pressure. The mathematical relationship: V 1 P In equation form: PV constant PV i i P V f f Copyright Cengage Learning. All rights reserved. 5 10

Boyle s Law P a 1/V P x V = constant P 1 x V 1 = P 2 x V 2 Constant temperature Constant amount of gas 11

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 x V = constant P 1 x V 1 = P 2 x V 2 P 1 = 726 mmhg P 2 =? V 1 = 946 ml V 2 = 154 ml P 2 = P 1 x V 1 V 2 726 mmhg x 946 ml = = 4460 mmhg 154 ml 12

1. The pressure on a 2.50 L of anesthetic gas changes from 105 kpa to 40.5 kpa. What will be the new volume if the temperature remains constant? 2. A gas with a volume of 4.00 L at a pressure of 205 kpa is allowed to expand to a volume of 12.0 L. What is the pressure in the container of the temperature remains constant?

Depth (ft) Chemistry in Action: Scuba Diving and the Gas Laws Pressure (atm) 0 1 33 2 66 3 P V 14

Variation in Gas Volume with Temperature at Constant Pressure As T increases V increases 15

Variation of Gas Volume with Temperature at Constant Pressure Charles & Gay-Lussac s Law V a T V = constant x T Temperature must be in Kelvin V 1 /T 1 = V 2 /T 2 T (K) = t ( 0 C) + 273.15 16

A balloon was immersed in liquid nitrogen (black container) and is shown immediately after being removed. It shrank because air inside contracts in volume. As the air inside warms, the balloon expands to its orginial size. Copyright Cengage Learning. All rights reserved. 5 17

Charles Law Volume vs. Temperature at constant pressure Volume is directly proportional to temperature Vα T @ constant P V=kT; V/T= k V 1 /T 1 = V 2 /T 2 1. If a sample of gas occupies 6.80 L at 325 0 C, what will be its volume at 25 0 C if the pressure does not change? Note: convert Celsius to Kelvin scale by adding 273 to the given 0 C.

2. Exactly 5.00 L of air at -50.0 0 C is warmed to 100.0 0 C. What is the new volume if the pressure remains constant?

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 /T 1 = V 2 /T 2 V 1 = 3.20 L T 1 = 398.15 K V 2 = 1.54 L T 2 =? T 1 = 125 ( 0 C) + 273.15 (K) = 398.15 K T 2 = V 2 x T 1 V 1 1.54 L x 398.15 K = = 192 K 3.20 L 20

Gay-Lussac s Law Pressure and Temperature at constant Volume P is proportional to T @ constant volume PαT @ constant V P=kT; P/T=k A gas has a pressure of 6.58 kpa at 539K. What will be the pressure at 211 K if the volume does not change? P 1 /T 1 = P 2 /T 2

The pressure in an automobile tire is 198 kpa at 27 0 C. At the end of a trip on a hot sunny day, the pressure has risen to 225 kpa. What is the temperature of the air in the tire? Helium gas in a 2.00-L cylinder is unter 1.12 atm pressure. At 36.5 0 C, that same gas sample has a pressure of 2.56 atm. What was the initial temperature of the gas?

V a number of moles (n) V = constant x n Avogadro s Law Constant temperature Constant pressure V 1 / n 1 = V 2 / n 2 23

Avogadro s Principle Volume is proportional to number of moles V α n where n is the number of moles More moles occupy greater volume

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 25

Combined Gas Law P α T V P=k T V PV = k T At 0.00 0 C and 1.00 atm pressure a sample of gas occupies 30.0 ml. if the temperature is increased 30.0 0 C and the gas sample is transferred to 20.0 ml container, what is the gas pressure?

Standard Temperature and Pressure (STP) The reference condition for gases, chosen by convention to be exactly 0 C and 1 atm pressure. The molar volume, V m, of a gas at STP is 22.4 L/mol. The volume of the yellow box is 22.4 L. To its left is a basketball. Copyright Cengage Learning. All rights reserved. 5 27

STP At STP, T=0 0 C or 273 K P= 1atm At Standard temperature and pressure, molar volume is 22.4 L Meaning, 1 mole of any gas occupies 22.4 liters at STP

Ideal Gas Law Ideal gas behaves as if there is no IMF present among the molecules of gas. PV = R nt R = 1atm (22.4L) 1 mole ( 273) R=.0821 atm-l/mol-k

Ideal Gas Equation Boyle s law: P a 1 (at constant n and T) V Charles law: V a T (at constant n and P) Avogadro s law: V a n (at constant P and T) V a nt P V = constant x nt P = R nt P R is the gas constant PV = nrt 30

What is the volume (in liters) occupied by 49.8 g of HCl at STP? T = 0 0 C = 273.15 K PV = nrt V = nrt P P = 1 atm n = 49.8 g x 1 mol HCl 36.45 g HCl = 1.37 mol V = L atm 1.37 mol x 0.0821 x 273.15 K mol K 1 atm V = 30.7 L 31

Sample problems for Ideal gas law If the pressure exerted by the gas at 25 0 C in a volume of 0.044 L is 3.81 atm, how many moles of gas are present? Determine the celsius temperature of 2.49 moles of gas contained in a 1.00-L vessek at a pressure of 143 kpa.

Density (d) Calculations d = m V = PM 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 M = drt P d is the density of the gas in g/l 33

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? M = drt P d = m V 4.65 g = = 2.21 g 2.10 L L M = 2.21 g L x 0.0821 1 atm L atm mol K x 300.15 K M = 54.5 g/mol 34

Density of gas from ideal gas equation D=M/V V=M/D PV=nRT P(Mass/D)= nrt Since n= mass/mw Then, P(Mass/D)=(Mass/MW)RT Therefore: P(MW)=DRT D=P(MW) RT What is the density of a gas at STP that has a molar mass of 44.0 g/mol? D=1atm(44.0g/mol).0821 (273K) D= 1.96 g/l

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 x 1 mol C 6 H 12 O 6 180 g C 6 H 12 O 6 6 mol CO x 2 = 0.187 mol CO 1 mol C 6 H 12 O 2 6 V = nrt P L atm 0.187 mol x 0.0821 x 310.15 K mol K = = 4.76 L 1.00 atm 37

Gas Mixtures Dalton found that in a mixture of unreactive gases, each gas acts as if it were the only gas in the mixture as far as pressure is concerned. Copyright Cengage Learning. All rights reserved. 5 38

Dalton s Law of Partial Pressures V and T are constant P 1 P 2 P total = P 1 + P 2 39

Consider a case in which two gases, A and B, are in a container of volume V. P A = n ART V P B = n BRT 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 = n A n A + n B X B = n B n A + n B P A = X A P T P B = X B P T P i = X i P T mole fraction (X i ) = n i n T 40

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 = P T = 1.37 atm 0.116 8.24 + 0.421 + 0.116 = 0.0132 P propane = 0.0132 x 1.37 atm = 0.0181 atm 41

A 100.0-mL sample of air exhaled from the lungs is analyzed and found to contain 0.0830 g N 2, 0.0194 g O 2, 0.00640 g CO 2, and 0.00441 g water vapor at 35 C. What is the partial pressure of each component and the total pressure of the sample? Copyright Cengage Learning. All rights 5 42

P N2 P O2 P CO2 P H2 O 0.0830 g N 2 0.0194 g O 0.00640 2 g CO 0.00441 g H 1mol N 2 28.01 g N 2 0.08206 1L 100.0 ml 3 10 ml 2 1mol 32.00 O 2 g O 2 100.0 ml 3 10 ml 2 1mol 44.01 g CO L atm 308 K mol K Copyright Cengage Learning. All rights reserved. L atm 0.08206 308 K mol K 1L CO 2 0.08206 1L 100.0 ml 3 10 ml 1mol H2O O 0.08206 18.01 g H2O 1L 100.0 ml 3 10 ml 2 L atm 308 K mol K L atm 308 K mol K 0.749 atm 0.153 atm 0.0368 atm 0.0619 atm 5 43

P P P N O 2 2 CO 2 P H2 O 0.749 atm 0.153 atm 0.0368 atm 0.0619 atm P P P P P N2 O2 CO2 H2O P = 1.00 atm Copyright Cengage Learning. All rights 5 44

The partial pressure of air in the alveoli (the air sacs in the lungs) is as follows: nitrogen, 570.0 mmhg; oxygen, 103.0 mmhg; carbon dioxide, 40.0 mmhg; and water vapor, 47.0 mmhg. What is the mole fraction of each component of the alveolar air? P N 570.0 mmhg P 2 103.0 mmhg O P 2 CO 40.0 mmhg 2 P H2 O 47.0 mmhg Copyright Cengage Learning. All rights reserved. 5 45

Mole fraction of N 2 570.0 mmhg 760.0 mmhg Mole fraction of CO 2 40.0 mmhg 760.0 mmhg Copyright Cengage Learning. All rights 5 47 Mole fraction of O 2 103.0 mmhg 760.0 mmhg Mole fraction of H 2 O 47.0 mmhg 760.0 mmhg Mole fraction N 2 = 0.7500 Mole fraction O 2 = 0.1355 Mole fraction CO 2 = 0.0526 Mole fraction O 2 = 0.0618

Collecting a Gas over Water 2KClO 3 (s) 2KCl (s) + 3O 2 (g) P T = P O + P H O 2 2 48

Collecting Gas Over Water Gases are often collected over water. The result is a mixture of the gas and water vapor. The total pressure is equal to the sum of the gas pressure and the vapor pressure of water. The partial pressure of water depends only on temperature and is known (Table 5.6). The pressure of the gas can then be found using Dalton s law of partial pressures. Copyright Cengage Learning. All rights reserved. 5 49

Vapor of Water and Temperature 50

? You prepare nitrogen gas by heating ammonium nitrite: NH 4 NO 2 (s) N 2 (g) + 2H 2 O(l) If you collected the nitrogen over water at 23 C and 727 mmhg, how many liters of gas would you obtain from 5.68 g NH 4 NO 2? P = 727 mmhg P vapor = 21.1 mmhg P gas = 706 mmhg T = 23 C = 296 K Molar mass NH 4 NO 2 = 64.06 g/mol Copyright Cengage Learning. All rights reserved. 5 52

P = 727 mmhg P vapor = 21.1 mmhg P gas = 706 mmhg T = 23 C = 296 K Molar mass NH 4 NO 2 = 64.06 g/mol nrt V P 1mol NH NO 1mol N 4 2 2 5.68 g NH4NO2 64.04 g NH 4 NO 2 1mol NH 4 NO 2 = 0.8887 mol N 2 gas Copyright Cengage Learning. All rights reserved. 5 53

P = 727 mmhg P vapor = 21.1 mmhg P gas = 706 mmhg T = 23 C = 296 K n = 0.0887 mol V V nrt L atm 0.0887 mol 0.08206 (296 K) mol K 1atm 706 mmhg 760 mmhg P = 2.32 L of N 2 (3 significant figures) Copyright Cengage Learning. All rights reserved. 5 54

Kinetic-Molecular Theory (Kinetic Theory) A theory, developed by physicists, that is based on the assumption that a gas consists of molecules in constant random motion. Kinetic energy is related to the mass and velocity: 1 E 2 K mv 2 m = mass v = velocity Copyright Cengage Learning. All rights reserved. 5 55

Postulates of the Kinetic Theory 1. Gases are composed of molecules whose sizes are negligible. 2. Molecules move randomly in straight lines in all directions and at various speeds. 3. The forces of attraction or repulsion between two molecules (intermolecular forces) in a gas are very weak or negligible, except when the molecules collide. 4. When molecules collide with each other, the collisions are elastic. 5. The average kinetic energy of a molecule is proportional to the absolute temperature. Copyright Cengage Learning. All rights reserved. 5 56

An elastic collision is one in which no kinetic energy is lost. The collision on the left causes the ball on the right to swing the same height as the ball on the left had initially, with essentially no loss of kinetic energy. Copyright Cengage Learning. All rights reserved. 5 57

Kinetic theory of gases and Compressibility of Gases Boyle s Law P a collision rate with wall Collision rate a number density Number density a 1/V P a 1/V Charles Law P a collision rate with wall Collision rate a average kinetic energy of gas molecules Average kinetic energy a T P a T 58

Kinetic theory of gases and Avogadro s Law P a collision rate with wall Collision rate a number density Number density a n P a n Dalton s Law of Partial Pressures Molecules do not attract or repel one another P exerted by one type of molecule is unaffected by the presence of another gas P total = SP i 59

Molecular Speeds According to kinetic theory, molecular speeds vary over a wide range of values. The distribution depends on temperature, so it increases as the temperature increases. Root-Mean Square (rms) Molecular Speed, u A type of average molecular speed, equal to the speed of a molecule that has the average molecular kinetic energy 3RT u M m Copyright Cengage Learning. All rights reserved. 5 60

The distribution of speeds of three different gases at the same temperature The distribution of speeds for nitrogen gas molecules at three different temperatures u rms = 3RT M 61

Gas diffusion is the gradual mixing of molecules of one gas with molecules of another by virtue of their kinetic properties. r 1 r 2 M 2 M 1 = molecular path NH 4 Cl NH 3 17 g/mol HCl 36 g/mol 65

Diffusion The process whereby a gas spreads out through another gas to occupy the space uniformly. Below NH 3 diffuses through air. The indicator paper tracks its progress. Copyright Cengage Learning. All rights reserved. 5 66

Graham s Law of Effusion At constant temperature and pressure, the rate of effusion of gas molecules through a particular hole is inversely proportional to the square root of the molecular mass of the gas. rate of effusion of molecules 1 M m Copyright Cengage Learning. All rights reserved. 5 68

? Both hydrogen and helium have been used as the buoyant gas in blimps. If a small leak were to occur, which gas would effuse more rapidly and by what factor? 1 Rate H 2 2.016 Rate He 1 4.002 4.002 2.016 Hydrogen will diffuse more quickly by a factor of 1.4. Copyright Cengage Learning. All rights reserved. 5 69

Deviations from Ideal Behavior 1 mole of ideal gas PV = nrt n = PV RT = 1.0 Repulsive Forces Attractive Forces 70

At high pressure, some of the assumptions of the kinetic theory no longer hold true: 1. At high pressure, the volume of the gas molecule (Postulate 1) is not negligible. 2. At high pressure, the intermolecular forces (Postulate 3) are not negligible. Copyright Cengage Learning. All rights reserved. 5 72

Effect of intermolecular forces on the pressure exerted by a gas. 73

Van der Waals Equation An equation that is similar to the ideal gas law, but which includes two constants, a and b, to account for deviations from ideal behavior. The term V becomes (V nb). The term P becomes (P + n 2 a/v 2 ). Values for a and b are found in Table 5.7 P n V 2 2 a V nb nrt Copyright Cengage Learning. All rights reserved. 5 74

Van der Waals equation nonideal gas ( P + an 2 )(V nb) = nrt V 2 } corrected pressure } corrected volume 75

Use the van der Waals equation to calculate the pressure exerted by 2.00 mol CO 2 that has a volume of 10.0 L at 25 C. Compare this with value with the pressure obtained from the ideal gas law. n = 2.00 mol V = 10.0 L T = 25 C = 298 K For CO 2 : a = 3.658 L 2 atm/mol 2 b = 0.04286 L/mol Copyright Cengage Learning. All rights reserved. 5 76

n = 2.00 mol V = 10.0 L T = 25 C = 298 K For CO 2 : a = 3.658 L 2 atm/mol 2 b = 0.04286 L/mol P P nrt n 2 V nb 2 2.00 L atm 2 L atm 3.658 2 mol K mol mol 0.08206 298 K 2.00 mol 10.0 L L 2.00 mol 0.04286 mol V P actual = 4.79 atm (3 significant figures) P 4.933 atm 2 a 0.146 atm 10.0 L 2 Copyright Cengage Learning. All rights reserved. 5 78