Presentation on theme: "1 Warm-Up n What does the motion of gas molecules look like? n Why does a balloon inflate when you blow it up? Why will soda explode from a bottle if opened."— Presentation transcript:
1 Warm-Up n What does the motion of gas molecules look like? n Why does a balloon inflate when you blow it up? Why will soda explode from a bottle if opened after shaking it?
2 Chapter 5 The Gas Laws
3 Section 5.1- Pressure n Force per unit area (P = force/area). n Gas molecules fill container. n Molecules move around and hit sides. n Collisions are the force. n Container is the area. n Measured with a barometer.
4 How Does A Barometer Work? n The pressure of the atmosphere at sea level will cause the column of mercury to rise to 760 mm Hg. n 1 atm = 760 mm Hg Pressure of atmosphere pushes on Hg 760 mm Hg Vacuum As a result, Hg rises up into the glass tube *Hg stops rising when it’s equal to atmospheric pressure
5 Units of pressure n 1 atmosphere = 760 mm Hg n 1 mm Hg = 1 torr n 1 atm = 101,325 Pascals = kPa *The first two are provided on the AP equation sheet. No need to memorize the third- I assume you’ll be given that if you need to use it.
6 THE GAS LAWS OF BOYLE, CHARLES, AND AVOGADRO Section 5.2
7 About the Laws… n You should be aware of the following laws, however we will not focus heavily on them as they can be derived from the ideal gas law. n After briefly going through each of the following laws, we will see how to derive each from the ideal gas law.
8 Boyle’s Law n Pressure and volume are inversely related at constant temperature. n P 1 V 1 = P 2 V 2 As one goes up, the other goes down.As one goes up, the other goes down. n Ex: if P increases (at constant T), V must go down n Further studies show that Boyle’s Law is only true at very low P This will be discussed more in 5.8 This will be discussed more in 5.8 n Gases that obey these laws are called ideal gases.
9 Charles’s Law n Volume of a gas varies directly with the temperature at constant pressure. n V 1 V 2 T 1 T 2 T 1 T 2 n As one goes up/down, so does the other. =
10 Avogadro's Law n At constant temperature and pressure, the volume of gas is directly related to the number of moles. n V 1 V 2 n 1 n 2 n 1 n 2 n As one goes up/down, so does the other. =
11 Gay- Lussac Law n At constant volume, pressure and temperature are directly related. n P 1 P 2 T 1 T 2 T 1 T 2 n As one goes up/down, so does the other. =
12 Combined Gas Law n Combination of Boyle’s Law, Charles’ Law, and Gay-Lussac Law. n Moles of gas remain constant. n P 1 V 1 P 2 V 2 T 1 T 2 T 1 T 2 =
13 Summary n Boyle’s: P 1 V 1 = P 2 V 2 n Charles’: V 1 /T 1 = V 2 /T 2 n Avogadro’s: V 1 /n 1 = V 2 /n 2 n Gay-Lussac: P 1 /T 1 = P 2 /T 2 n Combined: P 1 V 1 /T 1 = P 2 V 2 /T 2 n That’s a lot of laws! Or we can just use the Ideal Gas Law!
14 Combined Gas Law Cont. n Ex: A 2.3L sample of gas has a pressure of 1.2atm at 200.K. If the pressure is raised to 1.4atm and the temperature is increased to 300.K, what is the volume of the gas? n V 2 = P 1 V 1 T 2 T 1 P 2 T 1 P 2 n V 2 = 3.0 L
15Practice n Ex: A 12.2L sample of gas has 0.50mol of O 2 at 1atm and 25°C. How many moles of O 2 would occupy 19.4L at the same temperature and pressure? Solution: V 1 /n 1 = V 2 /n 2 (12.2L)/(0.50mol) = (19.4L)/(n 2 ) n 2 =0.80mol *In other words, 0.80mol of O 2 would be required to fill 19.4L in order to keep the same pressure as 0.50mol of O 2 in 12.2L.
16 AP Practice Question n A sample of argon gas is sealed in a container. The volume of the container is doubled. If the pressure remains constant, what must happen to the temperature? a) It doesn’t change. b) It is halved. c) It is doubled. d) It is squared.
17 Demonstration Warm-Up! n Observe the demonstration. n Keep in mind the properties of gases we have discussed so far: P, V, T, and n. n Think about these properties before and after imploding the can. Why do you think the can was crushed? n As temperature decreases, so does the pressure and volume. n Remind you of a law we looked at?
18 Sections 1&2 Homework n Pgs #: 2, 6, 34, 35
19 THE IDEAL GAS LAW Section 5.3
20 Ideal Gas Law n PV = nRT n At standard temperature and pressure (STP): V = 22.4L at 1atm, 0ºC, and n = 1mol. These conditions were used to determine R (ideal gas constant): »R = L atm/mol K »= J/mol K »= L torr/mol K n Tells you about a gas NOW. n The other laws tell you about a gas when it changes. Choose R value according to units of P KNOW THIS!
21 Ideal Gas Law Cont. n Looking back at the possible values for R, you will notice that all units for temperature are in K. –When using the ideal gas law for calculations, convert all temperatures to K! –Recall conversion: K = °C (provided on AP equation sheet)
22 Ideal Gas Law Derivation Practice n May be asked to prove one of the laws discussed before! n Strategy: get all constants in the ideal gas law on one side and changing variables on the other. n We will go several of these in class.
23 AP Practice Question A 1.15mol sample of carbon monoxide gas has a temperature of 27°C and a pressure of 0.300atm. If the temperature is lowered to 17°C at constant volume, what is the new pressure? a) 0.290atmc) 0.206atm b) 0.519atmd) 0.338atm
24 Ideal Gas Law- Why ‘Ideal’? n Ideal gases are hypothetical substances. – Gases only approach ideal behavior at low pressure (< 1 atm) and high temperature. –They do not behave exactly according to this law, but they behave closely enough. –Law provides good estimates of gas behavior under these conditions. n Unless told otherwise, assume ideal gas behavior and use the ideal gas law.
25 AP Practice Question n A sample of aluminum metal is added to HCl. How many grams of aluminum metal must be added to an excess of HCl to produce 33.6L of hydrogen gas at STP? a) 18.0g b) 35.0g c) 27.0g d) 4.50g
26 Section 3 Homework n Complete the gas laws worksheet AND #33, 40, 43, 52 on pg
27 GAS STOICHIOMETRY Section 5.4
28 Gases and Stoichiometry n Reactions involve moles of substances. n Recall that at STP (0ºC and 1 atm) 1mol of any gas occupies 22.4 L. –At STP this can be a conversion factor: 1mol/22.4L or 22.4L/1mol n If not at STP, use the ideal gas law to calculate moles or volume of a substance.
29 Section 4 Example n Quicklime (CaO) is produced by the thermal decomposition of calcium carbonate. Calculate the volume of carbon dioxide produced at STP if 152g of calcium carbonate are completely decomposed. CaCO 3 CaO + CO 2 n Convert to moles: 152g x 1mol = 1.52mol g CaCO g CaCO 3 n 1:1 mole ratio of CaCO 3 to CO mol CO 2 n Use STP conditions & stoichiometry: –At STP 1mol = 22.4L –1.52mol x (22.4L/1mol) = 34.1L CO 2 Can double check using ideal gas law
30 Gas Density and Molar Mass n Recall: D = m/V Let m molar stand for molar mass Let m molar stand for molar mass m molar = m/n so n = m/m molar m molar = m/n so n = m/m molar n PV = nRT solve for n: n= PV/RT Thus m/m molar = PV/RT Thus m/m molar = PV/RT Solve for m molar : m molar = mRT/VP Solve for m molar : m molar = mRT/VP n Replace m/V with D: m molar = DRT/P n If density, temperature, and pressure are known, molar mass can be found.
31 AP Practice Question Determine the formula for a gaseous silane (Si n H 2n+2 ) if it’s density is 5.47g/L at 0ºC and 1.00atm. *There are several ways to solve! a) SiH 4 b) Si 2 H 6 c) Si 3 H 8 d) Si 4 H 10
32 Section 4 Homework n Pg #51, 54, 57, 63, 64
33 DALTON’S LAW OF PARTIAL PRESSURES Section 5.5
34 Dalton’s Law of Partial Pressures n The total pressure in a container is the sum of the pressure each gas would exert if it were alone in the container. n Total pressure = sum of partial pressures. n P tot = P 1 + P 2 + P –P 1, P 2, P 3 are individual gases n From the ideal gas law: P Total = (n Total )RT V
35 Partial Pressures Cont. n What does Dalton’s Law tell us about ideal gases? n Total # of gas particles, not their identities, is important. –V of individual gas particles doesn’t affect the total P. –Forces between gas particles doesn’t affect the total P. n If these were important, the different identities of gas particles would affect the total P differently.
36 AP Practice Question A gaseous mixture at 25°C contained 1mol CH 4 and 2mol O 2, and P = 2atm. The gases underwent the following reaction: CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O(g) What is the P in the container after the reaction goes to completion and the T is allowed to return to 25°C? a) 1atm b) 2atm c) 3atm d) 4atm
37 AP Practice Question n A sealed, rigid container is filled with three identical gases: A, B, and C. The partial pressure of each gas is known as well as T and V. What additional information is needed to find the masses of the gases in the container? a) average distance travelled between molecular collisions b) the intermolecular forces c) the molar masses of the gases d) the total pressure
38 The mole fraction n Ratio of moles of a substance to the total moles. symbol is Greek letter chi symbol is Greek letter chi = n 1 = P 1 = n 1 = P 1 n tot P tot n tot P tot n Mole fractions have no units!
39 AP Practice Question n A reaction makes a mixture of CO 2, CO, and H 2 O. The gaseous products contained 0.60mol CO 2, 0.30mol CO, and 0.10mol H 2 O. If the total P is 0.80atm, what is the partial P of CO? a) 0.24atm b) 0.34atm c) 0.080atm d) 0.13atm
40 Vapor Pressure n Water evaporates! n When water evaporates, the resulting water vapor has a pressure. –Vapor pressure changes with T- must be looked up. n Gases are often collected over water so the vapor pressure of water must be subtracted from the total pressure. n Vapor pressure must be given.
41 AP Practice Question A sample of methane gas was collected over water at 35°C. The sample had a total pressure of 756mm Hg. Determine the partial pressure of methane gas in the sample. (Vapor pressure of water at 35°C is 41mm Hg.) a) 760mm Hg b) 41mm Hg c) 715mm Hg d) 797mm Hg
42 Section 5 Homework n Pg #65, 67, 69, 72
43 Collapsing Can Demo n Watch the demonstration. n Why did the can collapse? -The heat vaporized the water, which in turn increased P and pushed air out of the can. -When the can was inverted the water vapor quickly cooled. This caused a quick drop in P (created a partial vacuum because essentially no air was left to maintain P). -The atmospheric P outside of the can was much greater than P inside of the can, which allowed the can to be crushed.
44 THE KINETIC MOLECULAR THEORY OF GASES Section 5.6
45 Kinetic Molecular Theory (KMT)- Explains Behavior & Properties of Gases 1. Gases are made up of molecules or atoms. 2. V of particles can be ignored (very small in comparison to distance b/t particles). 3. Particles constantly move and collide with each other and the walls of the container. Collisions with the walls of the container cause P of the gas. 4. Particles don’t attract or repel each other; when they collide, it’s elastic (no KE is lost- it’s transferred). 5. The average KE is proportional to the Kelvin T.
46 KMT Cont. n Assumes gases are ideal. n BUT no gases are truly ideal- they only approach ideal behavior (specifically nonpolar gases at low P and high T). n In reality, gases DO have V (although small), and they CAN interact with each other. n Even so, assuming ideal behavior gives us good enough answers about properties of gases.
47KMT n #3 describes motion; let’s quantify it: u rms = √(3RT/m molar ) u rms = √(3RT/m molar ) –u rms is root mean square velocity –R value used is 8.314J/molK –molar mass in kg/mol (b/c J = kgm 2 /s 2 ) n #5: KE per mole (average KE) = 3/2 RT -Recall definition of T! Directly related! -Units: J/mol n KE per molecule = ½ mv 2 this is the only equation given on AP exam! - Units: J Large! For H 2 at 20°C = 2,000m/s
48 Root Mean Square Velocity Example n What is the root mean square velocity for the atoms in a sample of He gas at 25°C? n Convert T to K: = 298K n M = 4.00g/mol kg/mol n u rms = 136m/s
49 Range of velocities n The average distance a molecule travels between collisions with another gas particle is called the mean free path and is small (near ) –Results in a range of velocities. n Temperature is an average. There are molecules of many speeds in the average. n This is shown on a graph called a velocity distribution.
50 number of particles Molecular Velocity 273 K 1273 K 2273 K Notice that with higher T, average velocities increase and so does the velocity range. Maxwell-Boltzmann Distribution
51 AP Practice Question Two balloons are at the same T and P. One contains 14g of nitrogen and the other contains 20.0g of argon. Which of the following is true? a) D of N 2 > D of Ar b) Average speed of N 2 > average speed of Ar molecules c) Average KE of N 2 molecules > average KE of Ar molecules d) V of N 2 container < V Ar
52 AP Practice Question Increasing the T of an ideal gas from 50°C to 75°C at constant V will cause which of the following to increase for the gas? a) average molecular mass of the gas b) average distance between molecules c) average speed of the molecules d) density of the gas
53 Section 6 Homework n Pg #78, 79, 82, 83
54 EFFUSION AND DIFFUSION Section 5.7
55 Effusion n Passage of gas through a small hole, into a vacuum. n Effusion rate = speed at which the gas is transferred into the vacuum. n Graham’s Law - the relative rates of effusion are inversely proportional to the square roots of the molar masses of the gas particles.
56 Diffusion n The spreading of a gas through a room (mixing of gases). n Slow considering molecules move at hundreds of meters per second. –Slower movement is caused by collisions with other molecules in the air. n Best estimate is Graham’s Law. –Ratio is actually less. –More complex analysis required.
57 Section 7 Homework n Pg. 223 #86, 88
58 REAL GASES Sections 5.8 & 5.9
59 Real Gases n Real molecules do take up space and they do interact with each other (especially polar molecules). n Need to add correction factors to the ideal gas law to account for these. n a = correction factor for pressure n b = correction factor for volume
60 Volume Correction n The actual volume free to move in is less because particles do take up some of the volume. n More molecules will have more effect (taking up more space). n Corrected volume V’ = V - nb –b is a constant that differs for each gas. n P’ = nRT (V-nb)
61 Pressure Correction n Molecules are attracted to each other- pressure on the container will be less than ideal gases. n Size of correction factor depends on the # of molecules per liter (conc. of gas). n More molecules = closer together and more likely to interact/attract. n Since two molecules interact, the effect must be squared. P observed = P’ - a 2 () V n a = proportionality constant
62 All Together n P obs = nRT - a n 2 V-nb V n Called the Van der Waal’s equation if rearranged: n Corrected Corrected Pressure Volume () NOT given on AP Equation sheet!
63 Graphing Real Gases n For ideal gases PV/nRT should be 1 (since both are equal according to ideal gas law). n Not seen for real gases. n Notice the effect of T on ideal gas behavior.
64 Graphing Real Gases n Deviation from ideal behavior depends on identity of the gas too. n Smaller, nonpolar gases exhibit more ideal behavior.
65 Where Do Constants Come From? n a and b are experimentally determined. n Different for each gas. n Bigger molecules have larger b. n a depends on both size and polarity. n Note: table of constants for some gases is on pg. 210 in the book.
66 Graphing Real Gases n Take a closer look at H 2 on the graph. –Most ideal behavior, so it has lowest ‘a’ value of the gases shown for Van der Waals equation. –Lower a means less correction needed. –Thus it must have weak intermolecular forces. n Real gas behavior can tell us how big of a role intermolecular forces play in attraction between gas molecules.
67 AP Practice Question The true volume of a real gas is smaller than that calculated from the ideal gas equation. This occurs because the ideal gas equation does not consider which of the following? a) Attraction between molecules b) Shapes of molecules c) Volume of molecules d) Mass of molecules
68 AP Practice Question Which of the following gases probably shows the greatest deviation from ideal gas behavior? a) He b) O 2 c) SF 4 d) SiH 4