Chapter 5 The Gaseous State Bushra Javed CHM

Chapter 5: The Gaseous State Bushra Javed CHM 2045 1

Contents 1. Gas pressure & it’s units 2. Empirical Gas Laws • Boyle’s. Law • Charles’Law • Combined gas Law • Avogadro’s Law • Dalton’s Law • Ideal Gas Law 2

Contents 3. Kinetic-Molecular Theory 4. Molecular Speeds 5. Diffusion and Effusion Graham’s Law 6. Real Gases 3

The Gaseous State • Gases differ from liquids and solids: • Have low densities, are compressible & can expand • Physical condition of any gas can be defined • by four variables: • Pressure, P • Volume, V • temperature, T • amount or number of moles n. 4

Pressure , P of a Gas Pressure is the force exerted per unit area. It can be given by two equations: The SI unit for pressure is the pascal, Pa. 5

Pressure , P of a Gas Other Units atmosphere, atm mm. Hg torr bar 6

Atmospheric Pressure • Atmospheric pressure results from the exertion & pressure of air molecules in the environment. • A barometer is a device for measuring the pressure of the atmosphere. • A manometer is a device for measuring the pressure of a gas or liquid in a vessel.

Atmospheric Pressure 5|8

Empirical Gas Laws All gases behave quite simply with respect to temperature, pressure, volume, and molar amount. By holding two of these physical properties constant, it becomes possible to show a simple relationship between the other two properties. T 5|9

Boyle’s Law Pressure –Volume Relationship The volume of a sample of gas at constant temperature varies inversely with the applied pressure. The mathematical relationship: In equation form: 5 | 10

Boyle’s Law Pressure –Volume Relationship • When the volume decreases, the gas molecules collide with the container more often and the pressure increases. • When the volume increases, the gas molecules collide with the container less often and the pressure decreases. 11

Boyle’s Law plot of V versus P for 1. 000 g O 2 at 0°C. This plot is nonlinear. 12

Boyle’s Law At one atmosphere the volume of the gas is 100 m. L. When pressure is doubled, the volume is halved to 50 m. L. When pressure is tripled, the volume decreases to onethird, 33 m. L. 5 | 13

Boyle’s Law Example 1 A volume of oxygen gas occupies 38. 7 m. L at 751 mm. Hg and 21°C. What is the volume if the pressure changes to 359 mm. Hg while the temperature remains constant? Vi = 38. 7 m. L Pi = 751 mm. Hg Ti = 21°C Vf = ? Pf = 359 mm. Hg Tf = 21°C 5 | 14

Boyle’s Law Vi = 38. 7 m. L Pi = 751 mm. Hg Ti = 21°C Vf = ? Pf = 359 mm. Hg Tf = 21°C = 81. 0 m. L 5 | 15

Boyle’s Law Example 2 A sample of methane, CH 4, occupies a volume of 244. 0 m. L at 25°C and exerts a pressure of 1135. 0 mm. Hg. If the volume of the gas is allowed to expand to 720. 0 m. L at 298 K, what will be the pressure of the gas? P 2 = ? a. 3350 mm. Hg b. 385 mm. Hg c. 4580 mm. Hg d. 0. 0149 mm. Hg 16

If you plot volume vs. temperature for any gas at constant pressure, the points will all fall on a straight line If the lines are extrapolated back to a volume of “ 0, ” they all show the same temperature, − 273. 15 °C, called absolute zero 17

Absolute Zero The temperature -273. 15°C is called absolute zero. It is the temperature at which the volume of a gas is hypothetically zero. This is the basis of the absolute temperature scale, the Kelvin scale (K). 5 | 18

Charles’s Law The volume of a sample of gas at constant pressure is directly proportional to the absolute temperature (K). The mathematical relationship: In equation form: 19

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. 5 | 20

Charles’s Law Example 3 You prepared carbon dioxide by adding HCl(aq) to marble chips, Ca. CO 3. According to your calculations, you should obtain 79. 4 m. L of CO 2 at 0°C and 760 mm. Hg. How many milliliters of gas would you obtain at 27°C? Vi = 79. 4 m. L Pi = 760 mm. Hg Ti = 0°C = 273 K Vf = ? Pf = 760 mm. Hg Tf = 27°C = 300. K

Charles’s Law Vi = 79. 4 m. L Pi = 760 mm. Hg Ti = 0°C = 273 K Vf = ? Pf = 760 mm. Hg Tf = 27°C = 300. K = 87. 3 m. L

Charles’s Law Example 4 The volume of a sample of gas measured at 10. 0°C and 1. 00 atm pressure is 6. 00 L. What must the final temperature be in order for the gas to have a final volume of 7. 00 L at 1. 00 atm pressure? T 2 = ? Convert °C into Kelvin then back into °C. a. – 30. 4°C b. 8. 6°C c. 11. 7°C d. 57. 2°C 23

Combined Gas Law The volume of a sample of gas at constant pressure is inversely proportional to the pressure and directly proportional to the absolute temperature. The mathematical relationship: In equation form:

Combined Gas Law Example 5 Divers working from a North Sea drilling platform experience pressure of 5. 0 × 101 atm at a depth of 5. 0 × 102 m. If a balloon is inflated to a volume of 5. 0 L (the volume of the lung) at that depth at a water temperature of 4°C, what would the volume of the balloon be on the surface (1. 0 atm pressure) at a temperature of 11°C? Vf or V 2 = ? 5 | 25

Vi = 5. 0 L Pi = 5. 0 × 101 atm Ti = 4°C = 277 K = 2. 6 102 L Vf = ? Pf = 1. 0 atm Tf = 11°C = 284 K

Combined Gas Law Example 6 If 7. 75 L of radon gas is at 1. 55 atm and – 19 °C, what is the volume at STP? Standard Temperature and Pressure (STP) The reference condition for gases, chosen by convention to be exactly 0°C and 1 atm pressure (a)4. 65 L (b)5. 37 L (c)8. 33 L (d)12. 9 L 27

Avogadro’s Law Amedeo Avogadro (1776– 1856) • Volume directly proportional to the number of gas molecules ü V = constant x n ü constant P and T ü more gas molecules = larger volume • Count number of gas molecules by moles • Equal volumes of gases contain equal numbers of molecules ü the gas doesn’t matter 28

Avogadro’s Law Example 6 If 1. 00 mole of a gas occupies 22. 4 L at STP, what volume would 0. 750 moles occupy? Ans: 16. 8 L 29

Standard Conditions • Because the volume of a gas varies with pressure and temperature, chemists have agreed on a set of conditions to report our measurements so that comparison is easy – we call these standard conditions – STP • Standard pressure = 1 atm • Standard temperature = 273 K – 0 °C 30

Molar Volume • Solving the ideal gas equation for the volume of 1 mol of gas at STP gives 22. 4 L – 6. 022 x 1023 molecules of gas – notice: the gas is immaterial • We call the volume of 1 mole of gas at STP the molar volume – it is important to recognize that one mole measures of different gases have different masses, even though they have the same volume 31 Tro: Chemistry: A Molecular Approach, 2/e

Molar Volume 32 Tro: Chemistry: A Molecular Approach, 2/e

Ideal Gas Law/Ideal gas equation Combining the three Laws V α 1/ P (Boyle’s Law) V α T (Charles’ Law) V αn (Avogadro’s Law) V α T. n/P or V = R. T. n / P Where R = molar gas constant 33

Ideal Gas Equation • Rearranging the above equation gives: • PV = n. RT R = molar gas constant = 0. 0821 L. atm/ (mol. K) R = 0. 0821 atm. L/(K. mol) R = 8. 3145 J/(K. mol) R = 8. 3145 kg. m 2 /(. K. mol) 34

Ideal Gas Law Equation Example 7 A 50. 0 -L cylinder of nitrogen, N 2, has a pressure of 17. 1 atm at 23°C. What is the mass of nitrogen in the cylinder? Hint: Find mol first. V = 50. 0 L P = 17. 1 atm T = 23 C = 296 K Ans mass = 986 g 5 | 35

Density at Standard Conditions • • Density is the ratio of mass to volume Density of a gas is generally given in g/L The mass of 1 mole = molar mass The volume of 1 mole at STP = 22. 4 L 36 Tro: Chemistry: A Molecular Approach, 2/e

Density at Standard Conditions Example 8: Calculate the density of N 2(g) at STP 37

Gas Density and Molar Mass Gas Density Using the ideal gas law, it is possible to calculate the moles in 1 L at a given temperature and pressure. The number of moles can then be converted to grams (per liter). Molar Mass To find molar mass, find the moles of gas, and then find the ratio of mass to moles. 38

Gas Density and Molar Mass Example 8 What is the density of methane gas (natural gas), CH 4, at 125°C and 3. 50 atm? Note: Since density of gases is measured in g/L, you can assume the volume is 1 L. Hint: Find mols first, then convert to grams ; Mm = 16. 04 g/mol a) b) c) d) 1. 71 g/L 1. 72 g/L 4. 5 g/L 0. 04 g/L 39

Molar Mass of a Gas • One of the methods chemists use to determine the molar mass of an unknown substance is to heat a weighed sample until it becomes a gas, measure the temperature, pressure, and volume, and use the ideal gas law 40 Tro: Chemistry: A Molecular Approach, 2/e

Gas Density and Molar Mass Example 9 An unknown gas occupies a volume of 4. 75 L at 1227 °C and 5. 00 atm. If the mass is 5. 45 g, what is the molar mass of gas? (R = 0. 0821 atm • L/mol • K) Hint: Find mols first n = 0. 1928 (a)21. 5 g/mol (b) 23. 8 g/mol (c )28. 3 g/mol (d)141 g/mol 41

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. 5 | 42

Originally (left), flask A contains He at 152 mm. Hg and flask B contains O 2 at 608 mm. Hg. Flask A is then filled with oil forcing the He into flask B (right). The new pressure in flask B is 760 mm. Hg. 5 | 43

Dalton’s Law of Partial Pressures Partial Pressure The pressure exerted by a particular gas in a mixture. Dalton’s Law of Partial Pressures The sum of the partial pressures of all the different gases in a mixture is equal to the total pressure of the mixture: P = PA + PB + PC +. . . 5 | 44

Dalton’s Law of Partial Pressures Example 10 A 100. 0 -m. L 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? Hint: Convert grams to mols Then use ideal gas equation, PV= n. RT to find partial pressure of each gas. 5 | 45

5 | 46

Dalton’s Law of Partial Pressures P = 1. 00 atm 5 | 47

Dalton’s Law of Partial Pressures 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. 48

Dalton’s Law of Partial Pressures The partial pressure of water depends only on temperature. (See Table 5. 6). The pressure of the gas can then be found using Dalton’s law of partial pressures. 49

Collecting Gas by Water Displacement 50 Tro: Chemistry: A Molecular Approach, 2/e

Dalton’s Law of Partial Pressures Example 10 If carbon dioxide gas is collected over water at 25 °C and 775 torr, what is the partial pressure of the CO 2? The vapor pressure of water at 25 °C is 23. 8 torr. a) 23. 8 torr b) 750 torr C) 751 torr d) 775 torr e) 799 torr 51

mole fraction(X ) The concentration of any individual gas in a gas mixture can be expressed as a mole fraction(X ) Mole fraction(X) = Moles of component Total Moles in the mixture Mole fraction of component 1, for example , is X = n 1 / n 1 + n 2 + n 3 +. . . = n 1 / n total 52

mole fraction(X ) Example 11 The partial pressure of air in the alveoli (the air sacs in the lungs) is as follows: nitrogen, 570. 0 mm. Hg; oxygen, 103. 0 mm. Hg; carbon dioxide, 40. 0 mm. Hg; and water vapor, 47. 0 mm. Hg. What is the mole fraction of each component of the alveolar air? 5 | 53

mole fraction(X ) 570. 0 mm. Hg 103. 0 mm. Hg 40. 0 mm. Hg 47. 0 mm. Hg P = 760. 0 mm. Hg 5 | 54

Mole fraction of N 2 Mole fraction of. CO 2 Mole fraction of H 2 O 5 | 55

mole fraction(X ) X N 2 = 0. 7500 X O 2 = 0. 1355 X CO 2 = 0. 0526 X H 2 O = 0. 0618 5 | 56

Kinetic-Molecular 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: m = mass v = velocity 5 | 57

Kinetic-Molecular Theory 58

Kinetic-Molecular Theory 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. . 5 | 59

Kinetic-Molecular Theory 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 60

Kinetic-Molecular Theory • An elastic collision occurs when the two objects "bounce" apart when they collide. • In an elastic collision, both momentum and kinetic energy are conserved. Almost no energy is lost to sound, heat, or deformation. 61

Kinetic-Molecular Theory The Kinetic-Molecular Theory explains Boyle’s Law • Compressing a gas makes the V smaller but does not alter the KE avg of the molecules since T is constant. • Though the speed of the particles remains constant, the frequency of collisions increases because the container is smaller. • Therefore, P increases as V decreases. 62

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 5 | 63

Kinetic Energy and Molecular Velocities • Average kinetic energy of the gas molecules depends on the average mass and velocity – KE = ½mv 2 • Gases in the same container have the same temperature, therefore they have the same average kinetic energy • If they have different masses, the only way for them to have the same kinetic energy is to have different average velocities – lighter particles will have a faster average velocity than more massive particles 64

Molecular Speed vs. Molar Mass • To have the same average kinetic energy, heavier molecules must have a slower average speed 65

Temperature and Molecular Velocities _ • KEavg = ½NAmu 2 – NA is Avogadro’s number • KEavg = 1. 5 RT – R is the gas constant in energy units, 8. 314 J/mol∙K • 1 J = 1 kg∙m 2/s 2 • Equating and solving we get – NA∙mass = molar mass in kg/mol • As temperature increases, the average velocity increases 66

Molecular Velocities • All the gas molecules in a sample can travel at different speeds • However, the distribution of speeds follows a statistical pattern called a Boltzman distribution • Ee talk about the “average velocity” of the molecules, but there are different ways to take this kind of average • The method of choice for our average velocity is called the root-mean-square method, where the rms average velocity, urms, is the square root of the average of the sum of the squares of all the molecule velocities 67

Temperature vs. Molecular Speed • As the absolute temperature increases, the average velocity increases – the distribution function “spreads out, ” resulting in more molecules with faster speeds 68

Molecular Speeds When using the equation R = 8. 3145 J/(mol · K). T must be in Kelvins Mm must be in kg/mol 69

Molecular Speeds Example 12 What is the rms speed of carbon dioxide molecules in a container at 23°C? T = 23°C = 296 K CO 2 molar mass = 0. 04401 kg/mol 5 | 70

5 | 71

root-mean-square (rms) Speed Example 13 Calculate the root-mean-square velocity for the O 2 molecules in a sample of O 2 gas at 22. 5°C. (R = 8. 3145 J/Kmol) a) 132. 4 m/s b) 15. 18 m/s c) 479. 9 m/s d) 277. 1 m/s 72

Diffusion and Effusion • The process of a collection of molecules spreading out from high concentration to low concentration is called diffusion • The process by which a collection of molecules escapes through a small hole into a vacuum is called effusion • The rates of diffusion and effusion of a gas are both related to its rms average velocity • For gases at the same temperature, this means that the rate of gas movement is inversely proportional to the square root of its molar mass 73 Tro: Chemistry: A Molecular Approach, 2/e

Effusion 74 Tro: Chemistry: A Molecular Approach, 2/e

Graham’s Law of Effusion Thomas Graham (1805– 1869) • For two different gases at the same temperature, the ratio of their rates of effusion is given by the following equation: 75 Tro: Chemistry: A Molecular Approach, 2/e

Graham’s Law of Effusion Example 14 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? Hydrogen will diffuse more quickly by a factor of 1. 4. 5 | 76

Graham’s Law of Effusion Example 15 Which of the following gases will have the slowest rate of effusion at constant temperature? a)CF 4 b)F 2 c)H 2 d)Ne 77

Ideal vs. Real Gases • Real gases often do not behave like ideal gases at high • pressure or low temperature Ideal gas laws assume 1. no attractions between gas molecules 2. gas molecules do not take up space Ø based on the kinetic-molecular theory • At low temperatures and high pressures these assumptions are not valid 78 Tro: Chemistry: A Molecular Approach, 2/e

Real Gases 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. 5 | 79

The Effect of Molecular Volume Johannes van der Waals (1837– 1923) • At high pressure, the amount of space occupied by the molecules is a significant amount of the total volume • The molecular volume makes the real volume larger than the ideal gas law would predict • van der Waals modified the ideal gas equation to account for the molecular volume – b is called a van der Waals constant and is different for every gas because their molecules are different sizes 80 Tro: Chemistry: A Molecular Approach, 2/e

Real Gas Behavior • Because real molecules attract each other, the molar volume of a real gas is smaller than predicted by the ideal gas law at low temperatures 81

The Effect of Intermolecular Attractions • At low temperature, the attractions between the molecules is significant • The intermolecular attractions makes the real pressure less than the ideal gas law would predict • van der Waals modified the ideal gas equation to account for the intermolecular attractions – a is another van der Waals constant and is different for every gas because their molecules have different strengths of attraction 82

Van der Waals Equation Example 16 a)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. b)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 5 | 83

Van der Waals Equation n = 2. 00 mol V = 10. 0 L T = 25°C = 298 K Ideal gas law: = 4. 89 atm 5 | 84

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 Pactual = 4. 79 atm 5 | 85