Chapter 5 Gases and the Kinetic-Molecular Theory

Chapter 5 Gases and the Kinetic-Molecular Theory Name (Formula) Methane (CH 4 ) Ammonia (NH 3 ) Chlorine (Cl 2 ) Oxygen (O 2 ) Ethylene (C 2 H 4 ) Origin and Use natural deposits; domestic fuel from N 2 + H 2 ; fertilizers, explosives electrolysis of seawater; bleaching and disinfecting liquefied air; steelmaking high-temperature decomposition of natural gas; plastics Atmosphere-Biosphere Redox Interconnections

Figure 5.1 The three states of matter.

An Overview of the Physical States of Gases Note The Distinction of Gases from Liquids and Solids 1. Gas volume changes greatly with pressure. 2. Gas volume changes greatly with temperature. 3. Gases have relatively low viscosity. 4. Most gases have relatively low densities under normal conditions. 5. Gases are miscible. 6. Gases are compressible. 7. Gas particles have negligible attraction for each other (ideal gases).

Figure 5.2 Effect of atmospheric pressure on objects at the Earth s surface.

Variations in pressure, temperature, and composition of the Earth s atmosphere.

Figure 5.3 A mercury barometer.

Figure 5.4 closed-end Two types of manometer open-end

Table 5.2 Common Units of Pressure Unit pascal(pa); kilopascal(kpa) atmosphere(atm) Atmospheric Pressure 1.01325x10 5 Pa; 101.325 kpa 1 atm* Scientific Field SI unit; physics, chemistry chemistry millimeters of mercury(hg) 760 mm Hg* chemistry, medicine, biology torr 760 torr* chemistry pounds per square inch (psi or lb/in 2 ) 14.7lb/in 2 engineering bar 1.01325 bar meteorology, chemistry, physics *This is an exact quantity; in calculations, we use as many significant figures as necessary.

PROBLEM SOLVING 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. Exercise # 8: Do problems 1 & 2.

Boyle s Law V 1 P n and T are fixed V x P = constant V = constant / P Charles s Law V T P and n are fixed V T = constant V = constant x T Amontons s Law P T V and n are fixed P T = constant P = constant x T combined gas law V T P V = constant x T P PV T = constant

Basic Behavior of a Gas Boyle s Law: P 1 V 1 = P 2 V 2 Where T and n are held constant.

Figure 5.5 The relationship between the volume and pressure of a gas. Boyle s Law

Problem Solving Do # s 3 & 4 of Exercise 8

Relationship of Pressure and Temperature Charles Law: V 1 = V 2 T 1 T 2 Where P and n are held constant. **** Temperature must be in Kelvin****

Figure 5.6 The relationship between the volume and temperature of a gas. Charles s Law

Problem Solving Do Problem # s 5 & 6 of Exercise 8

Relationship Between Temperature and Pressure Amonton s Law: P 1 = P 2 T 1 T 2 Where V and n are held constant. **** Temperature must be in Kelvin****

Relationship Between Volume, Temperature, and Pressure The Combined Gas Law: P 1 V 1 = P 2 V 2 T 1 T 2 Where n is held constant. Do problem # s 7 & 8 of Exercise 8

The Relationship Between the Volume and Amount of a Gas.

Avogadro s Law Equal volumes of a gas will contain the same number of particles if P & T are held constant. At standard temperature, 273.15 K and pressure, 1.00 atm (STP) 1 mole of gas = 22.4 L

Figure 5.8 Standard molar volume.

Problem Solving Do problems 9 & 10 of Exercise 8

Figure 5.9 The volume of 1 mol of an ideal gas compared with some familiar objects.

Figure 5.10 THE IDEAL GAS LAW R = PV nt = R is the universal gas constant PV = nrt 1atm x 22.414L 1mol x 273.15K = IDEAL GAS LAW PV = nrt or V = nrt P 0.0821atm*L mol*k 3 significant figures fixed n and T fixed n and P fixed P and T Boyle s Law Charles s Law Avogadro s Law V = constant P V = constant X T V = constant X n

Problem Solving Do problem # s 11 & 12 of Exercise 8

Determination of Density of a Gas Think about what units you will need.? Grams? Volume Gas densities are usually expressed in g/l (m/v) Consider: PV = nrt, then Recall that n (moles) = m(mass)/mm(molecular mass) So PV = m RT and rearranging we get m = d = MM x P MM V RT Do Problem #15 of Exercise 8

Determination of The Molar Mass of a Gas from its Density n = mass M = PV RT M = m RT VP d = m V M = d RT P

Determination of the Molar Mass of a Gas Think about what units you will need to determine.? Grams? Moles Do Problem # s 13 and 17 of Exercise 8

Figure 5.11 Determining the molar mass of an unknown volatile liquid. based on the method of J.B.A. Dumas (1800-1884)

Problem Solving An organic chemist isolates a colorless liquid from a petroleum sample. She uses the Dumas method and obtains the following data: Volume of flask = 213 ml Mass of flask + gas = 78.416 g T = 100.0 0 C Mass of flask = 77.834 g P = 754 torr Calculate the molar mass of the liquid.

Mixtures of Gases Gases mix homogeneously in any proportions. Each gas in a mixture behaves as if it were the only gas present. Dalton s Law of Partial Pressures P total = P 1 + P 2 + P 3 +... P 1 = 1 x P total where 1 is the mole fraction 1 = n 1 n 1 + n 2 + n 3 +... = n 1 n total

Problem Solving Do problem # 21 of Exercise 8

Collecting a water-insoluble gaseous reaction product and determining its pressure (Week 8 Lab).

Table 5.3 Vapor Pressure of Water (P H 2 O ) at Different T T( 0 C) P (torr) T( 0 C) P (torr) 0 5 10 11 12 13 14 15 16 18 20 22 24 26 28 4.6 6.5 9.2 9.8 10.5 11.2 12.0 12.8 13.6 15.5 17.5 19.8 22.4 25.2 28.3 30 35 40 45 50 55 60 65 70 75 80 85 90 95 100 31.8 42.2 55.3 71.9 92.5 118.0 149.4 187.5 233.7 289.1 355.1 433.6 525.8 633.9 760.0

1 Water-insoluble gaseous product bubbles through water into collection vessel P total P atm

1 Water-insoluble gaseous product bubbles through water into collection vessel 2 P gas adds to vapor pressure of water (P H2 O) to give P total- As shown P total < P atm P total P atm

1 Water-insoluble gaseous product bubbles through water into collection vessel 2 P gas adds to vapor pressure of water (P H2 O) to give P total- As shown P total < P atm 3 P total is made equal to P atm by adjusting height of vessel until water level equals that in beaker P total P atm P total P atm

1 Water-insoluble gaseous product bubbles through water into collection vessel 2 P gas adds to vapor pressure of water (P H2 O) to give P total- As shown P total < P atm 3 P total is made equal to P atm by adjusting height of vessel until water level equals that in beaker P total P gas P atm P total P atm P total = P H2 O

1 Water-insoluble gaseous product bubbles through water into collection vessel 2 P gas adds to vapor pressure of water (P H2 O) to give P total- As shown P total < P atm 3 P total is made equal to P atm by adjusting height of vessel until water level equals that in beaker P total P atm P P atm = total P total P gas P H2 O 4 P total equals P gas plus P H2 O at temperature of experiment. Therefore, P gas = P total P H2 O

Draw Diagram of Lab Apparatus P T = P H2O + P gas P T = barometric pressure P H2O = vapor pressure of H 2 O at a particular temperature

Problem Solving Do Problem # 24 of Exercise 8

Figure 15.13 Summary of the stoichiometric relationships among the amount (mol,n) of a reactant or product and the gas variables of pressure (P), volume (V), and temperature (T). P,V,T of reactant A ideal gas law amount (mol) of reactant A molar ratio from balanced equation amount (mol) of gas B ideal gas law P,V,T of gas B

Gas Stoichiometry Problem Solving Do problem # s 18,19, 20, and 25 of Exercise 8

Postulates of the Kinetic-Molecular Theory Postulate 1: Particle Volume Because the volume of an individual gas particle is so small compared to the volume of its container, the gas particles are considered to have mass, but no volume. Postulate 2: Particle Motion Gas particles are in constant, random, straight-line motion except when they collide with each other or with the container walls. Postulate 3: Particle Collisions Collisions are elastic therefore the total kinetic energy(e k ) of the particles is constant.

Figure 5.14 Distribution of molecular speeds at three temperatures.

A molecular description of Boyle s Law.

A molecular description of Dalton s law of partial pressures.

A molecular description of Charles s Law.

Figure 5.18 A molecular description of Avogadro s Law.

Summary: A Molecular View of the Gas Laws Particles exert a force when they collide with the walls of a container. The force is proportional to the pressure. More particles more collisions more pressure Greater T particles move faster more collisions more pressure. Smaller volume more collisions more pressure.

Graham s Law of Effusion Graham s Law of Effusion The rate of effusion of a gas is inversely related to the square root of its molar mass. rate of effusion 1 M

Recall that KE = ½ mv 2 If at some temperature a heavy particle s KE = a light particle s KE Then the heavy gas must be moving slower than the lighter gas. Then KE A = KE B or (1/2 mv 2 ) A = (1/2 mv 2 ) B (m A /m B ) 1/2 = v B /v A Where m = molar mass

Relative number of molecules with a given speed Relationship Between Molar Mass and Molecular Speed H 2 (2) Molecular speed at a given T

Relative number of molecules with a given speed He (4) H 2 (2) Molecular speed at a given T

Relative number of molecules with a given speed H 2 O (18) He (4) H 2 (2) Molecular speed at a given T

Relative number of molecules with a given speed N 2 (28) H 2 O (18) He (4) H 2 (2) Molecular speed at a given T

Relative number of molecules with a given speed O 2 (32) N 2 (28) H 2 O (18) He (4) H 2 (2) Molecular speed at a given T

Effusion: Gaseous particles passing through a tiny hole into an evacuated space. Rate effusion is proportional to 1/(MW) 1/2 Thus..., Rate A = (MW B ) 1/2 Rate B = (MW A ) 1/2 Do Problem # 28 of Exercise 8. Do Problem # 29 of Exercise 8.

Diffusion: Gaseous particles passing through other gaseous particles. Rate diffusion is proportional to 1/(MW) 1/2 Thus..., Rate A = (MW B ) 1/2 Rate B = (MW A ) 1/2

Figure 5.20 Diffusion of a gas particle through a space filled with other particles. distribution of molecular speeds mean free path collision frequency

Real Gases At high temperatures and low pressures most gases behave ideally. Consider gases at high pressure.

Figure 5.23 The effect of molecular volume on measured gas volume.

Figure 5.22 The effect of intermolecular attractions on measured gas pressure.

Real (Non-Ideal) Gases At high pressures we can no longer say the attractions between gases are negligible. We can no longer say the volume of the gaseous particles themselves is negligible. Thus our ideal gas law equation, PV = nrt has to have some correction factors, a and b. ( P + n 2 / V)( V + nb ) = nrt

The behavior of several real gases with increasing external pressure. 2.0 1.5 PV RT 1.0 0.5 0.0 0 200 400 600 800 1000 P ext (atm)

PV RT 1.0 2.0 0 10 20 P ext (atm) 1.5 PV RT 1.0 0.5 0.0 0 200 400 600 800 1000 P ext (atm)

PV RT 1.0 Ideal gas 2.0 0 10 20 P ext (atm) 1.5 PV RT 1.0 Ideal gas 0.5 0.0 0 200 400 600 800 1000 P ext (atm)

PV RT 1.0 H 2 He Ideal gas 2.0 0 10 20 P ext (atm) H 2 He PV RT 1.5 1.0 PV/RT > 1 Effect of molecular volume predominates Ideal gas 0.5 0.0 0 200 400 600 800 1000 P ext (atm)

PV RT 1.0 H 2 He Ideal gas CH 4 CO 2 CH 4 2.0 0 10 20 P ext (atm) H 2 CO 2 He PV RT 1.5 1.0 0.5 PV/RT > 1 Effect of molecular volume predominates PV/RT < 1 Effect of intermolecular attractions predominates Ideal gas 0.0 0 200 400 600 800 1000 P ext (atm)

Table 5.5 Van der Waals Constants for Some Common Gases Van der Waals equation for n moles of a real gas (P n2 a )(V nb) nrt 2 V adjusts P up adjusts V down Gas a atm*l 2 mol 2 b L mol He Ne Ar Kr Xe H 2 N 2 O 2 Cl 2 CO 2 CH 4 NH 3 H 2 O 0.034 0.211 1.35 2.32 4.19 0.244 1.39 1.36 6.49 3.59 2.25 4.17 5.46 0.0237 0.0171 0.0322 0.0398 0.0511 0.0266 0.0391 0.0318 0.0562 0.0427 0.0428 0.0371 0.0305