AP CHEMISTRY Chapter 5 Gases Introduction We live in a solution of gases; nitrogen, oxygen and other gases surround and support us. In this chapter we will focus on the behavior of

gases and the laws that govern that behavior. 5.1 Pressure Properties of Gases Gases are compressible Gases uniformly fill any container Gases mix easily with other gases Barometer In

1643, Evangelista Torricelli invented the barometer, an instrument for measuring atmospheric pressure. Atmospheric pressure results from gravitys pull on air masses. dish filled with mercury with a closed tube

outside pressure causes mercury level to rise and fall The Manometer Manometer- instrument used for measuring pressure Gas pressure less than atmospheric pressure gas pressure = atm pressure h Gas pressure greater than atmospheric pressure gas pressure= atm pressure + h a) Gas pressure = atmospheric pressure h

b) Gas pressure = atmospheric pressure + h Units of Pressure mm Hg is also called torr 760mm Hg = 760 torr = 1 atm SI unit of pressure is N/m2 or Pascal (Pa) 1 atm = 101,325 Pa or 101.325 kPa

oyle B We and s a e p les! b a t

vege Boyles Law 1st quantitative study of gases, 1600s. Pressure and volume are inversely related. P1 = V2 P1V1= P2V2 P2 V1 Ideal gas- gas that obeys Boyles law An ideal gas is expected to

have a constant value of PV, as shown by the dotted line. CO2 shows the largest change in PV, but this change is actually quite small. PV changes from about 22.42

Latm at 0.00 atm to 22.26 Latm at 1.00 atm. A Charlie Brown Christmas is on TV. Notice how the balloon shrinks in the cold water and expands in the warm water. Charles Law 1700s -volume of a gas is directly proportional to

Kelvin temperature V1 = V2 Temp must be in Kelvin!!!!!! T1 T2 The volume of a gas at absolute zero is zero. Avogadros Law -equal volumes of gases at the same temperature and pressure contain the same # of particles

-for a gas at constant temp. and pressure, the volume is directly proportional to the # of moles of gas Gay-Lussacs Law s e v ri er. d le ruis y Ga PT c

a Pressure of a gas is directly proportional to Kelvin temperature P1 = P2 Temp must be in Kelvin!!!!!! T1 T2 Combined Gas Law P1V1 = P2V2 T1 T2 Peas and Vegetables on the Table

The Ideal Gas Law Ideal Gas Law Combining Boyles, Charles, and Avogadros laws we get PV= nRT. R = 0.08206 L.atm/K.mol (proportionality constant)

Most gases behave ideally at pressures less than 1 atm. PV=nRT We can use the ideal gas law for all gas law problems by putting changing variables on one side and the constant on the other. Ex. If P&V change w/ others constant: P1V1 = nRT and P2V2 = nRT so P1V1 = P2V2 Ex.

If V&T change with others constant: V1 = nR and V2 = nR so V1 = V2 T1 P T2 P T1 T2 Ex. The gas pressure inside an aerosol can is 1.5 atm at 25oC. Assuming that the gas is ideal, what would the pressure be if the

can were heated to 450oC? If the volume is held constant, then the formula P1V1/T1 = P2V2/T2 shortens down to P1/T1 = P2/T2 . 1.5atm = x atm 298K 723K X = 3.6 atm Ex. A quantity of helium gas occupies a volume of 16.5 L at 78oC and 45.6

atm. What is the volume at STP? We use P1V1/T1 = P2V2/T2 and solve for V2. (45.6atm)(16.5L) = (1 atm)(x L) (351K) (273K) X = 585 L We will use PV = nRT and solve for n or moles so that we can convert to mass. (15.3 atm)(50.0L) = (n)(0.08206L.atm/mol.K)(296K) n = 31.5 mol O2 31.5 mol O2

32.00g O2 = 1.01 x 103 g O2 1 mol O2 Molar Volume = 22.42 L of an ideal gas at STP (Some gases behave more ideally than others.) STP = 0oC and 1 atm Ex. CaH2 reacts with H2O to produce H2

gas. CaH2(s) + 2H2O(l) 2H2(g) + Ca2+(aq) + 2OH-(aq) Assuming complete rxn with water, how many grams of CaH2 are required to fill a balloon to a total pressure of 1.12 atm at 15oC if its volume is 5.50 L? In this problem, we must first find the mol of hydrogen gas produced (this gas causes the pressure in the balloon) and then convert to grams of CaH2. PV=nRT (1.12 atm)(5.50L) = (n)(0.08206L.atm/mol.K)(288K) N = 0.261 mol H2

0.261 mol H2 CaH2 1 mol CaH2 42.1 g CaH2 = 5.49g Ex. How many liters of N2 are required to produce 115 g of NH3 at STP? First, we must write the balanced equation.

N2 + 3H2 2NH3 Now, we convert, using 1 mol=22.4L as a conversion factor because this reaction is occurring at STP. 115 g NH3 1 mol NH3 17.0g NH3 1 mol N2 2 mol NH3 22.4L N2 = 1 mol N2

75.8L N2 Molecular Weight and Density of a Gas n = mass so P = m R T MW (MW)V Since m/V = density (g/L), P = dRT MW MW = dRT

P Molecular Weight Kitty Cat Ex. Calculate the molecular weight of a gas if 0.608g occupies 750 mL at 385 mm Hg and 35oC. MW = dRT/P d= mass/volume = 0.608g/0.750L = 0.811 g/L T = 35C + 273.15 = 308 K P = 385mmHg 1 atm = 0.507 atm

760 mm Hg MW = [(0.811g/L)(0.08206L.atm/mol.K)(308 K)]/0.507 atm MW = 40.4 g/mol Daltons Law of Partial Pressures For a mixture of gases in a container, the total pressure exerted is the sum of the pressures that each gas would exert if it were alone. Ptot = P1 + P2 + P3 + Ptot

Ptot = n1RT + n2RT + n3RT + V V V = ntotal (RT) (It doesnt matter what the gas is.) V Things to Note. If

you do not have a college board account, you need to go to www.collegeboard.org and create an account. You will get reminders about upcoming exams and important information regarding AP testing. Save for exams - with the jump in costs this year for AP exams you need to save your money! Exams will cost $90 each. Free and reduced exams will still be $20 each. "Like" the CSHS facebook page and "Follow" CSHS on Twitter. We will make announcements and reminders on both forums. Upcoming

dates to note.... January 22-23 - all deposits due ($20 each test) February 19-20 - final payments due Partial Pressures of Gases The partial pressure of a particular component of a gaseous mixture is the mole fraction of that component times the total pressure.

P1 = 1 (Ptotal) When gases are collected over water, we must adjust for the pressure of the water vapor. PH2O + Pgas = Ptotal Ex. If a 0.20 L sample of O2 at 0oC and 1.0 atm pressure and a 0.10 L sample of N2 at 0oC and 2.0 atm pressure are both placed in a 0.40 L container at 0oC, what is the total pressure in the container?

In this problem, the total volume should be the sum of the partial pressures of each gas. However, we must take into account the pressure of each gas as it fills the new container. Therefore, we will use P1V1=P2V2 to find each samples partial pressure. Oxygen (1.0atm)(0.20L)=(P2)(0.40L) P2 = 0.50 atm Nitrogen (2.0atm)(0.10L)=(P2)(0.40L) P2 = 0.50 atm Ptotal = PO2 + PN2 = 0.50 + 0.50 = 1.0 atm

Mole Fraction The ratio of the number of moles of a given component in a mixture to the total number of moles in the mixture. is used to symbolize mole fraction 1 = n1 n1 + n2 + n3 +

5.6 Kinetic Molecular Theory of Gases The Kinetic Molecular Theory of Gases is a simple model that attempts to explain properties of an ideal gas. It states that gases consist of particles which have the following properties:

The particles are so small compared to the distances between them that the volume of the individual particles can be assumed to be negligible (zero). The particles are in constant motion. The collisions of the particles with the walls of the container are the cause of the pressure exerted by the gas. The particles are assumed to exert no forces on each other; they are assumed neither to attract nor to repel each other. The average kinetic energy of a collection of gas particles is

assumed to be directly proportional to the Kelvin temperature of the gas. Temperature and Particle Motion Real gases dont conform to these assumptions! Kelvin temperature is an index of the random motions of the particles of a gas, with higher temperatures meaning greater motion. KEavg = 3/2 RT R

= 8.3148 J/K mol Units are J/mol Root mean square velocity You are a Mess urms = u2 = is the average of the squares of the Where M = mass of a mole of gas particles velocities

= root mean square in kg The units of urms are m/s. velocity = urms At any given temperature, lighter molecules have higher root mean square velocities. At any given temp., lighter molecules have higher root mean square velocities.

Ex. Calculate the root mean square velocity in m/s of O2 molecules at 27oC. 3RT 3(8.3145)(300) urms 483.6m / s M 0.03200kg Ex. Calculate the root mean square velocity in m/s of O2 molecules at 27oC.

Calculate the avg. KE of the same molecules. KE = 3/2RT KE = 3/2 (8.3145J/K.mol)300K = 3740 J/mol Mean Free Path Real gases have many collisions between particles.

The average distance a particle travels between collisions in a particular gas sample is called the mean free path. These collisions produce a huge variation in velocities. As temperature increases, the range of velocities is greater. Path of One Particle in a Gas

A Plot of the Relative Number of O2 Molecules That Have a Given Velocity at STP A Plot of the Relative Number of N2 Molecules That Have a Given Velocity at Three Temperatures

Effusion and Diffusion Diffusion- mixing of gases Effusion- the passage of a gas through a tiny orifice into an evacuated chamber Grahams Law of Effusion The rate of effusion of a gas is inversely proportional to the square root of the

mass of its particles. 1 2 = 2 1 MW and MW2 represent the molar masses of the gases. 1 -diffusion rates can be calculated the same way

-lighter gases effuse & diffuse faster than heavier gases HCI(g) and NH3(g) Meet in a Tube Real Gases No gas exactly follows the ideal gas law. A real gas exhibits behavior closest to ideal behavior at low pressures and high

temperatures. Real Gases At high temperatures, there is less interaction between particles because they are moving too fast. At high concentrations, gases have much greater attractive forces between particles. This causes particles to hit the walls of the container with less force (producing less pressure than expected).

Real Gases At high pressure (small volume), the volume of the particles becomes significant, so that the volume available to the gas is significantly less than the container volume. Attractive forces are greatest for large, complex molecules and polar molecules. We can use the Van der Waals equation to adjust for departures

from ideal conditions. PV = nRT becomes: [Pobs + a(n/V)2]V-nb = nRT corrected corrected pressure volume a = correction for pressure that takes into account the intermolecular attractions between molecules. It increases with an increase in MW and an increase in molecular complexity. b= correction for the finite volume of

the gas molecules. It is a measure of the actual volume occupied by the gas molecules. It increases with an increase in mass of the molecule or in the complexity of the structure. We can look up values of a&b for common gases in Table 5.3. Pg 224 Chemistry in the Atmosphere -major gases are N2 and O2

-heaviest gases are concentrated closest to the surface The Ozone Layer Most air pollution centers around NO, Nitric oxide, which is emitted into the air from exhaust of engines. uv NO + O2 NO2 NO2(g)

NO + O O + O 2 O3 Ozone reacts with unburned hydrocarbons in polluted air to produce irritating chemicals. This process creates photochemical smog. S (from coal) + O2 SO2

2SO2 + O2 2SO3 SO3 + H2O H2SO4 Read about scrubbers, etc. A Schematic Diagram of the Process for Scrubbing Sulfur Dioxide from Stack Gases in Power Plants

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