Chapter 5. Gases and the Kinetic-Molecular Theory

Transcription

1 Chapter 5 Gases and the Kinetic-Molecular Theory

2 Macroscopic vs. Microscopic Representation

3 Kinetic Molecular Theory of Gases 1. Gas molecules are in constant motion in random directions. Collisions among molecules are perfectly elastic.

4 Kinetic Molecular Theory of Gases 2. The average kinetic energy of the molecules is proportional to the temperature of the gas in kelvins. Any two gases at the same temperature will have the same average kinetic energy 3. A gas is composed of molecules that are separated from each other by distances far greater than their own dimensions. The molecules can be considered to be points; that is, they possess mass but have negligible volume. 4. Gas molecules exert neither attractive nor repulsive forces on one another. 5. Each gas molecule behaves as if it were alone in container (due to #3 and #4)

5 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(k k ) of the particles is constant.

6 Distribution of molecular speeds at three temperatures.

7 Relationship between molar mass and molecular speed.

8 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.

9 Pressure KMT Viewpoint Origin of Pressure Gas molecules hitting container walls Temp, KE, # collisions, P Volume, # collisions, P

10 Pressure Macroscopic Viewpoint Pressure = Force Area Temp, KE, Force, P Units of Pressure 1 pascal (Pa) = 1 N/m 2 1 atm = 760 mmhg = 760 torr Volume, Area, P 1 atm = 101,325 Pa

11 Common Units of Pressure Unit Atmospheric Pressure Scientific Field pascal(pa); kilopascal(kpa) atmosphere(atm) x10 5 Pa; kpa 1 atm SI unit; physics, chemistry chemistry millimeters of mercury(hg) 760 mmhg chemistry, medicine, biology torr 760 torr chemistry pounds per square inch (psi or lb/in 2 ) 14.7lb/in 2 engineering bar bar meteorology, chemistry, physics

12 Atmospheric Pressure Barometer

13 Two types of manometer

14 Converting Units of Pressure PROBLEM: A geochemist heats a limestone (CaCO 3 ) sample and collects the CO 2 released in an evacuated flask attached to a closedend manometer. After the system comes to room temperature, Dh = 291.4mmHg. Calculate the CO 2 pressure in torrs, atmospheres, and kilopascals. PLAN: Construct conversion factors to find the other units of pressure mmHg 291.4torr 1torr 1mmHg 1atm 760torr = 291.4torr = atm atm kPa 1atm = 38.85kPa

15 Effect of Pressure on Volume Boyle s Law 1 atm 2 atm 5 atm

16 The relationship between volume and the pressure of a gas.

17 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

18 1 Boyle s Law V a P n and T are fixed PV = constant V = constant/p

19 Kinetic Molecular theory of gases and Boyle s Law P a collision rate with wall Collision rate Increases with decreased volume P a 1/V Increase P, decrease volume

20 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 = 726 mmhg P 1 x V 1 = P 2 x V 2 P 2 =? V 1 = 946 ml V 2 = 154 ml P 2 = P 1 x V 1 V mmhg x 946 ml = = 4460 mmhg 154 ml

21 Applying the Volume-Pressure Relationship PROBLEM: PLAN: V 1 in cm 3 1cm 3 =1mL V 1 in ml 10 3 ml=1l V 1 in L xp 1 /P 2 V 2 in L Boyle s apprentice finds that the air trapped in a J tube occupies 24.8cm 3 at 1.12atm. By adding mercury to the tube, he increases the pressure on the trapped air to 2.64atm. Assuming constant temperature, what is the new volume of air (inl)? unit conversion gas law calculation SOLUTION: P 1 = 1.12atm V 1 = 24.8cm cm 3 P 1 V 1 n 1 T 1 = 1mL 1cm 3 P 2 V 2 n 2 T 2 P 1 V V 1 2 = = L P 2 P and T are constant L 10 3 ml P 2 = 2.64atm V 2 = unknown = L P 1 V 1 = P 2 V atm 2.46atm = L

22 Effect of Temperature on Volume (Charles Law) Low Temperature High Temperature

23 V a T V = kt V/T = k V 1 /T 1 = V 2 /T 2 As T increases, V Increases

24 Charles Law Animation

25 Applying the Temperature-Pressure Relationship A 1-L steel tank is fitted with a safety valve that opens if the internal pressure exceeds 1.00x10 3 torr. It is filled with helium at 23 0 C and 0.991atm and placed in boiling water at exactly C. Will the safety valve open? P 1 (atm) T 1 and T 2 ( 0 C) 1atm=760torr P 1 (torr) x T 2 /T 1 P 2 (torr) K= 0 C T 1 and T 2 (K) 0.991atm P 1 = 0.991atm T 1 = 23 0 C P 1 V 1 P 2 V 2 = n 1 T 1 n 2 T 2 760torr 1atm = 753torr P 2 = unknown T 2 = 100 o C P 1 P 2 = T 1 T 2 P 2 = P 1 T 2 T 1 = 753torr 373K 296K = 949torr

26 Kinetic theory of gases and Charles Law -Average kinetic energy a T -Increase T, Gas Molecules hit walls with greater Force, this Increases the Pressure BUT since pressure must remain constant, and only volume can change -Volume Increase to reduce Pressure -Increase Temperature, Increase Volume

27 Determination of Absolute Zero

28 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

29 V a number of moles (n) V = constant x n V 1 /n 1 = V 2 /n 2 Avogadro s Law 5.3

30 Applying the Volume-Amount Relationship A scale model of a blimp rises when it is filled with helium to a volume of 55dm 3. When 1.10mol of He is added to the blimp, the volume is 26.2dm 3. How many more grams of He must be added to make it rise? Assume constant T and P. We are given initial n 1 and V 1 as well as the final V 2. We have to find n 2 and convert it from moles to grams. n 1 (mol) of He x V 2 /V 1 n 2 (mol) of He subtract n 1 mol to be added x M g to be added V 1 n 1 = 1.10mol n 2 = unknown V 1 = 26.2dm 3 V 2 = 55.0dm 3 V 2 V = n n 1 n 2 = n n 2 = 1.10mol 55.0dm dm 3 V 1 P and T are constant 4.003g He = 2.31mol mol He P 1 V 1 P 2 V 2 = n 1 T 1 n 2 T 2 = 4.84g He

31 Kinetic theory of gases and Avogadro s Law More moles of gas, more collisions with walls of container More collisions, higher pressure BUT since pressure must remain constant and only volume can change Volume increases to decrease pressure to original value

32 Boyle s Law V a 1 P n and T are fixed Charles s Law V a T P and n are fixed V T = constant V = constant x T Amonton s Law P a T V and n are fixed P T = constant P = constant x T combined gas law V a T P V = constant x T P PV T = constant

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34 Gas Law Animation

35 Ideal Gas Equation Boyle s law: V a 1 (at constant n and T) P 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 nt V = constant x P = R nt P R is the gas constant R = L atm / (mol K) PV = nrt

36 Obtaining Other Gas Law Relationship PV = nrt PV nt R PV 1 n T PV 2 n T 2 2 2

37 THE IDEAL GAS LAW PV = nrt R = PV nt = 1atm x L 1mol x K = atm*L mol*k IDEAL GAS LAW PV = nrt or V = nrt P 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

38 PV = nrt nr V P 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 = constant P 2 = T 1 T 2 P 2 = P 1 x T 2 T 1 n, V and R are constant P 1 = 1.20 atm T 1 = 291 K = 1.20 atm x 358 K 291 K P 2 =? T 2 = 358 K = 1.48 atm

39 Types of Problems PV 1 1 PV 2 2 Make Substitution into PV = nrt n T 1 1 n 2 T 2 moles ( n) mass, g MolarMass, g / mole Given initial conditions, determine final conditions; Cancel out what is constant Density mass Volume

40 Using Gas Variables to Find Amount of Reactants and Products PROBLEM: A laboratory-scale method for reducing a metal oxide is to heat it with H 2. The pure metal and H 2 O are products. What volume of H 2 at 765torr and C is needed to form 35.5g of Cu from copper (II) oxide? PLAN: Since this problem requires stoichiometry and the gas laws, we have to write a balanced equation, use the moles of Cu to calculate mols and then volume of H 2 gas. mass (g) of Cu mol of Cu mol of H 2 L of H 2 divide by M molar ratio SOLUTION: 35.5g Cu 0.559mol H 2 x use known P and T to find V mol Cu 63.55g Cu CuO(s) + H 2 (g) Cu(s) + H 2 O(g) 1mol H 2 1 mol Cu = 0.559mol H atm*l x mol*k 1.01atm 498K = 22.6L

41 Solving for an Unknown Gas Variable at Fixed Conditions A steel tank has a volume of 438L and is filled with 0.885kg of O 2. Calculate the pressure of O 2 at 21 0 C. V, T and mass, which can be converted to moles (n), are given. We use the ideal gas law to find P. V = 438L T = 21 0 C (convert to K) n = 0.885kg (convert to mol) P = unknown 0.885kg 103 g kg nrt P = = V mol O g O 2 = 27.7mol O mol x L atm*l mol*k 21 0 C = 294K x 294K = 1.53atm

42 Using the Ideal Gas Law in a Limiting-Reactant Problem PROBLEM: The alkali metals [Group 1A(1)] react with the halogens [Group 7A(17)] to form ionic metal halides. What mass of potassium chloride forms when 5.25L of chlorine gas at 0.950atm and 293K reacts with 17.0g of potassium? PLAN: After writing the balanced equation, we use the ideal gas law to find the number of moles of reactants, the limiting reactant and moles of product. SOLUTION: 2K(s) + Cl 2 (g) 2KCl(s) P = 0.950atm V = 5.25L T = 293K n = PV Cl = 0.950atm x 5.25L = 0.207mol 2 RT atm*l x 293K mol*k 2mol KCl 17.0g mol K 0.207mol Cl = 0.435mol K 2 1mol Cl 39.10g K 2 2mol KCl Cl 2 is the limiting reactant mol K 2mol K 74.55g KCl 0.414mol KCl = 30.9 g KCl mol KCl n = unknown = 0.414mol KCl formed = 0.435mol KCl formed

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

44 Standard Molar Volume

45 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

46 Finding the Molar Mass of a Volatile Liquid A chemist isolates from a petroleum sample a colorless liquid with the properties of cyclohexane (C 6 H 12 ). The Dumas method is used to obtain the following data to determine its molar mass: Volume of flask = 213mL Mass of flask + gas = g T = C Mass of flask = g P = 754 torr Is the calculated molar mass consistent with the liquid being cyclohexane? Use unit conversions, mass of gas and density-m relationship. M = m = ( )g = 0.582g m RT = 0.582g atm*l x x 373K mol*k VP 0.213L x 0.992atm = 84.4g/mol M of C 6 H 12 is 84.16g/mol and the calculated value is within experimental error.

47 Calculating Gas Density Calculate the density (in g/l) of carbon dioxide and the number of molecules per liter (a) at STP (0 0 C and 1 atm) and (b) at ordinary room conditions (20. 0 C and 1.00atm). Density is mass/unit volume; substitute for volume in the ideal gas equation. Since the identity of the gas is known, we can find the molar mass. Convert mass/l to molecules/l with Avogardro s number. M x P d = mass/volume PV = nrt V = nrt/p d = RT 1.96g L (a) d = 44.01g/mol x 1atm = 1.96g/L atm*l mol*k x 273K mol CO x10 23 molecules 44.01g CO 2 mol = 2.68x10 22 molecules CO 2 /L

48 Calculating Gas Density continued (b) d = 44.01g/mol x 1atm atm*l mol*k x 293K = 1.83g/L 1.83g L mol CO x10 23 molecules 44.01g CO 2 mol = 2.50x10 22 molecules CO 2 /L

49 The Molar Mass of a Gas n = mass M = PV RT M = m RT VP d = m V M = d RT P

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

51 Dalton s Law of Partial Pressures P total = P 1 + P 2 + P P 1 = c 1 x P total where c 1 is the mole fraction c 1 = n 1 n 1 + n 2 + n = n 1 n total

52 Applying Dalton s Law of Partial Pressures PROBLEM: In a study of O 2 uptake by muscle at high altitude, a physiologist prepares an atmosphere consisting of 79 mol% N 2, 17 mol% 16 O 2, and 4.0 mol% 18 O 2. (The isotope 18 O will be measured to determine the O 2 uptake.) The pressure of the mixture is 0.75atm to simulate high altitude. Calculate the mole fraction and partial pressure of 18 O 2 in the mixture. PLAN: Find the c and P from P total and mol% 18 O O 2 18 O 2 mol% 18 O 2 divide by 100 SOLUTION: c 18 O 2 = 4.0mol% 18 O = c 18 O 2 P = c x P total = x 0.75atm = 0.030atm 18 O 18 2 O 2 multiply by P total partial pressure P 18 O 2

53 Kinetic theory of gases and 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

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55 Calculating the Amount of Gas Collected Over Water PROBLEM: PLAN: Acetylene (C 2 H 2 ), an important fuel in welding, is produced in the laboratory when calcium carbide (CaC 2 ) reaction with water: CaC 2 (s) + 2H 2 O(l) C 2 H 2 (g) + Ca(OH) 2 (aq) For a sample of acetylene that is collected over water, the total gas pressure (adjusted to barometric pressure) is 738torr and the volume is 523mL. At the temperature of the gas (23 0 C), the vapor pressure of water is 21torr. How many grams of acetylene are collected? The difference in pressures will give us the P for the C 2 H 2. The ideal gas law will allow us to find n. Converting n to grams requires the molar mass, M. P C2 H = (738-21)torr = 717torr 2 P total P C2 H 2 P H2 O n = PV RT n g C2 H 2 C2 H 2 717torr atm 760torr = 0.943atm x M

56 Calculating the Amount of Gas Collected Over Water continued n C2 H 2 = 0.943atm x 0.523L atm*l x 296K mol*k = 0.203mol 0.203mol 26.04g C 2 H 2 mol C 2 H 2 = g C 2 H 2

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58 Gas diffusion is the gradual mixing of molecules of one gas with molecules of another by virtue of their kinetic properties. NH 4 Cl NH 3 17 g/mol HCl 36 g/mol

59 Avogadro s Law V a n E k = 1/2 mass x speed 2 E k = 1/2 mass x u 2 u 2 is the root-mean-square speed u rms = 3RT M R = 8.314Joule/mol*K 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 a 1 M

60 Applying Graham s Law of Effusion PROBLEM: Calculate the ratio of the effusion rates of helium and methane (CH 4 ). PLAN: The effusion rate is inversely proportional to the square root of the molar mass for each gas. Find the molar mass of both gases and find the inverse square root of their masses. SOLUTION: M of CH 4 = 16.04g/mol M of He = 4.003g/mol rate rate He CH 4 = = 2.002

61 Real Gases Nonideal Conditions - when gas gets close to conditions where it will liquify Lower Temperature Higher Pressure P an V 2 2 V nb nrt

62 Real Gases Effect of Intermolecular Forces Corrected Pressure, a Corrected Volume, b