6/22/20161 Chap 12: Behavior of Gases Real life: How will an airbag protect us? –The compressible nature of gases allows the energy of the collision to be absorbed by the air bag—gas particles are forced closer together— instead of your head.

6/22/20162 Kinetic Theory- particles always in motion 1. The individual volumes of the particles is assumed as __________; gas is easily ____________ because of the space between the particles 2. Collisions are ______- no net loss of energy 3. No ________ or ________ forces exist btwn the particles; gas ________ until it takes the shape and volume of its container 4.Gas particles move _______ in constant ________ motion; changes in direction only occur when particles _______ with each other or with another object 5.Average kinetic energy depends on ___________. –Ideal gas- imaginary perfect gas fitting the theory Properties of Gases: random compressed attractiverepulsive expands rapidly insignificant collide elastic temperature

Question: 6/22/20163

4 Visual Concepts Kinetic Molecular Theory Chapter 12

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6 Visual Concepts Comparing Real and Ideal Gases Chapter 12

6/22/20167 Variables That Describe a Gas _________; P = kPa, atm, mm Hg __________; V = liter ______________; T = kelvin (˚C + 273) _________; n = moles ______________; R = L·atm mol·K STP- standard temperature and pressure; 0˚ C and 1 atm; 22.4 L = 1 mol, known as molar volume pressure volume temperature moles Ideal gas constant

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9 Factors Affecting Gas Pressure: Amount of Gas –By adding gas, you increase the number of _________, thus increasing the number of _________, which explains why _________ increases. –ex. pumping air into a balloon (more molecules  more collision  greater pressure  volume expands) particles collisionspressure

6/22/ There you were just cruisin’ down the street…

6/22/ Blow out! high pressure more gas particles inside low pressure fewer gas particles inside

6/22/ Volume: Gas pressure can also be increased by __________ the volume 100 kPa 100 kPa ? kPa ? kPa V = 1.0 LV = 0.5 L reducing

6/22/ Real life: How does an aerosol spray work? –The difference in pressure btwn the inside of the spray can and the air outside allows them to work. The pressure inside is high, while the outside is low. The spray stops working when they are equal.

6/22/ Temperature : recall temp = average KE, the energy of motion –_________ the temperature provides yet another way to increase gas pressure. (So many varieties) 300 K 600 K 100 kPa ? kPa So what effect would tripling the # of particles have on the pressure? –The pressure would triple. increasing

Barometer- device used to measure atmospheric pressure At standard pressure or one atmosphere pressure a column of mercury is 760 mm high or Torr. Dish of Mercury Column of Mercury 1 atm Pressure

Another unit of pressure is the Pascal.Another unit of pressure is the Pascal kPa = 1 atm = 760 mm Hg = 760 Torr kPa = 1 atm = 760 mm Hg = 760 Torr. Be able to convert between these!Be able to convert between these! 760 mm kPa Pressure

6/22/ Visual Concepts Atmospheric Pressure Chapter 12

6/22/ The volume of the gas varies ________ with pressure; That means: as pressure ________, volume ________ P 1 x V 1 = P 2 x V 2 (it does not matter if you mix “ 1 ” s w/ “ 2 ” s but keep the conditions together) Gas Laws: Boyle’s Law inversely increases decreases

6/22/ Visual Concepts Equation for Boyle ’ s Law Chapter 12

6/22/ Boyle’s Law… Ex. A weather balloon contains 30.0L of helium gas at 103 kPa. a.What is the volume when the balloon rises to an altitude where the pressure is only 25.0 kPa? a.P 1 = 103 kPa b.V 1 = 30.0 L c.P 2 = 25.0 kPa d.V 2 = ? L P 1 x V 1 = P 2 x V 2 or P 1 x V 1 = V 2 P 2 =124 L

6/22/ Boyle’s Law… Ex. A weather balloon contains 30.0L of helium gas at 103 kPa. b.What is the volume of this balloon at sea level? a.P 1 = 103 kPa b.V 1 = 30.0 L c.P 2 = kPa d.V 2 = ? L P 1 x V 1 = P 2 x V 2 or P 1 x V 1 = V 2 P L

6/22/ Boyle’s Law…

6/22/ The volume of the gas is ________ proportional to its temperature That is: as temperature ________, volume ________ V 1 = V 2 T 1 T 2 (T must be in K) Gas Laws: Charles’ Law directly increases

6/22/ Since volume of a gas is dependent on the temperature, if there were no KE (absolute zero), the volume would be zero also. Charles’ Law….

6/22/ Visual Concepts Equation for Charles ’ s Law Chapter 12

6/22/ Ex. A balloon has a volume of 4.0 L at 24°C. What would be its volume if heated to 58°C? Change temperature into the K scale –It does not matter whether 4.0 L is V 1 or V 2, as long as the conditions stay together. Conventionally the “1” are the knowns, though. V 1 = 4.0 L T 1 = 24°C = 297 K V 2 = ? L T 2 = 58°C = 331 K V 1 = V 2 when you solve for V 2 : V 2 = V 1 T 2 T 1 T 2 T 1 Answer is L = 4.5 L Charles’ Law….

6/22/ Charles’ Law

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6/22/ Pressure of a gas is ______ proportional to the (K) temperature That is: as temperature ________, pressure ________ P 1 = P 2 T 1 T 2 (b/c this is a directly proportional relationship, when graphed out, it will look just like the Charles’ Law.) Gas Laws: Gay- Lussac’s Law directly increases

6/22/ Visual Concepts Gay-Lussac ’ s Law Regarding Combining Volumes of Gases Chapter 12

6/22/ Gay-Lussac’s Law

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6/22/ Literally combines the 3 laws (Please Ve on Time) P 1 V 1 = P 2 V 2 T 1 T 2 Do you see the 3 laws mixed in here? Boyle’s Law: –Cancel out the temperature b/c it stays constant Charles’ Law: –Cancel out the pressure Gay-Lussac’s Law: –cancel the volume Gas Laws: Combined Gas Law

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6/22/ Ideal Gases Ideal Gas Law: –particles exhibit no forces –volume of particles is negligible –Condition: Low pressure and/or high temperature What happens to real gases under other conditions?

6/22/ Visual Concepts Ideal Gas Law Chapter 12

6/22/ Ideal Gas Law: PV = nRT P = pressure; kPa, atm, mm Hg V = volume; liter T = temperature; K n = moles; mol R = ideal gas constant; 8.31 L· kPa K·mol

6/22/ PV = nRT –P = nRT / V –V = nRT / P –n = PV / RT –T = PV / nR –R = Ideal gas constant: R = 8.31 LkPa/ Kmol (Yes! Always…) –Ex.12-5 pg 342 HW: pg 342 # 22-30

6/22/ Visual Concepts Properties of Gases Chapter 12

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6/22/ Avogadro’s Hypothesis –Equal V of gases have equal number of particles at constant T & P –Recall: STP, what is the molar volume? Dalton’s Law:Dalton’s Law –Total pressure is equal to the sum of partial pressures of the component gases (constant V, T) –P total = P 1 + P 2 + P 3 + P 4 +…

6/22/ Visual Concepts Avogadro ’ s Law Chapter 12

Avogadro’s Hypothesis 2 Liters of Helium 2 Liters of Oxygen Has the same number of particles as

Examples What is the volume of 4.59 moles of CO 2 gas at STP? (Use “spider”). Answer liters. (4.59 mol)(22.4 L /mol) How many moles is 5.67 L of O 2 at STP? (5.67 L)(1 mol/22.4 L) = moles O 2 What is the volume of 8.8 g of CH 4 gas at STP? (Watch significant figures) (8.8g)(1 mol/16 g)(22.4 L/mol) = 12 L

6/22/ Visual Concepts Dalton's Law of Partial Pressures Chapter 12

Examples What is the total pressure in a balloon filled with air (O 2 & N 2 ) if the pressure of the oxygen is 170 mm Hg and the pressure of nitrogen is 620 mm Hg? mm Hg. (170 mm Hg mm Hg)

Examples In a second balloon the total pressure is 1.3 atm. What is the pressure of oxygen if the pressure of nitrogen is 720 mm Hg? Steps..In a second balloon the total pressure is 1.3 atm. What is the pressure of oxygen if the pressure of nitrogen is 720 mm Hg? Steps.. P Total = P 1 + P 2 + P 3...P Total = P 1 + P 2 + P atm = Po mm Hg1.3 atm = Po mm Hg Convert to same units (we’ll use mm Hg)Convert to same units (we’ll use mm Hg) 1.3 atm(760 mm Hg/1 atm) = Po mm Hg1.3 atm(760 mm Hg/1 atm) = Po mm Hg 988 mm Hg = Po mm Hg so, mm Hg = Po mm Hg so,... Po 2 = 988 mm Hg mm Hg = 268 mm HgPo 2 = 988 mm Hg mm Hg = 268 mm Hg Can also express as 0.35 atm.Can also express as 0.35 atm.

6/22/ Graham’s Law: –Difussion: Molecules move toward areas of lower concentration until the concentration is uniform –Effusion: Process where gas escapes through a hole Gases w/lower MM effuse faster Eq’ns: KE = ½ mv 2 Ex. P 1 +He vs methane

6/22/ Visual Concepts Equation for Graham ’ s Law of Effusion Chapter 12

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Effusion Passage of gas through a small hole, into a vacuum. The effusion rate measures how fast this happens. Graham’s Law: The rate of effusion is inversely proportional to the square root of the mass of its particles.

6/22/ The Effusion of a Gas into an Evacuated Chamber

Example Nitrogen effuses 1.7 times faster than another gaseous element. Estimate the element’s molar mass and determine its probable identity. (Graham’s Law)(Graham’s Law) Rate of effusion of A = √M B Rate of effusion of B √M A Rate N 2 = 1.7 = √ M X Rate X 1 √M N 2 (1.7)/1 = √M X / √28 √M X = 9, so M X = 81 g/mol Since X is an element, look on period table.. X = Kr

6/22/ Example What is the ratio of the average velocity of hydrogen molecules to neon atoms at the same temperature and pressure? We are not looking for absolute numbers, only the ratio. (Graham’s Law)(Graham’s Law) Rate of effusion of A = √M B Rate of effusion of B √M A Rate H 2 /Rate Ne = √20.18/ √2.02 = 3.16 So, hydrogen effuses 3.16 times faster than neon. End, Chapter 11