Presentation on theme: "Sit down and get ready for the test, I will hand them out when the bell rings. You have until 8:10 DMA 3/21/11."— Presentation transcript:
1Sit down and get ready for the test, I will hand them out when the bell rings. You have until 8:10 DMA 3/21/11
2You can have a few minutes to finish your test, if you are already done, work on something quietly DMA 3/21/11
3When you are done with your test, begin the following: In your book, read (it is important that you actually read this) Ch 13-1 (pg ) and take notes which include the following info:The 6 physical properties that are common to all gasesWhat it means when the terms “atom” “diatomic molecule” or “polyatomic molecule” are used when taking about gasesThe 6 postulates of the Kinetic-molecular theory of gasesAnswer question #5 on pg. 423This is homework if you don’t get it done today
4Name 4 physical properties of gases DMA 3/22/11Name 4 physical properties of gases
5The Nature of Gases-physical properties Gases expand to fill their containersGases are fluid – they flowGases have low density1/1000 the density of the equivalent liquid or solidGases are compressibleGases effuse and diffuse
6Kinetic Molecular Theory Gases are made of particles, which have massParticles of matter are ALWAYS in motionVolume of individual particles is zero.Collisions of particles with container walls cause pressure exerted by gas.Particles exert no forces on each other.Average kinetic energy proportional to temperature in Kelvin of a gas.
7Measuring GasesWhen you are measuring a gas, you want to know four things about it:1. n --Amount of the gas-this means in moles represented by the letter n2. V—volume3. T—temperature4. P—pressure
8Temperature When using the gas laws, temperature is measured in Kelvin To find Kelvin:Add 273 to Celcius.T(K)=T(oC) + 273
9Converting Celsius to Kelvin Gas law problems involving temperature requirethat the temperature be in KELVINS!Kelvins = C + 273°C = Kelvins - 273
11Converting Temperature Convert the following temperatures to the Kelvin scale.20 C85 C–15 C–190 C
12An Early Barometer Mercury Barometer Atmospheric pressure– the pressureexerted by the air in the atmosphere.Barometer-measures air pressureThe normal pressure due tothe atmosphere at sea levelcan support a column ofmercury that is 760 mm high.
14Standard Temperature and Pressure “STP” P = 1 atmosphere (760 torr)T = 0°C (273 Kelvins)The molar volume of an ideal gas is liters at STP
15PressureIs caused by the collisions of molecules with the walls of a containeris equal to force/unit areaSI units = Newton/meter2 = 1 Pascal (Pa)1 standard atmosphere = 101,325 Pa= kPa1 standard atmosphere = 1 atm =760 mm Hg = 760 torr
16Converting Pressure Units Change 5 atmospheres (atm) into millimeters mercury (mm Hg).1 atm = 760 mm Hg5 atm x 760 mm Hg =1 atm3800 mm Hg
17Converting Pressure Units Change 1900 mm mercury (mm Hg) into atmospheres (atm).1 atm = 760 mm Hg1900 mm Hg x 1 atm =760 mm Hg2.5 atm
18Converting Pressure Units Change 560 kilopascals into millimeters mercury (mm Hg).101.3 kPa = 760 mm Hg560 kPa x 760 mm Hg =101.3 kPa4201 mm Hg
19Converting Pressure Units Change 1013 kilopascals into millimeters mercury (mm Hg).1 atm = kPa1013 atm x 1 atm =101.3 kPa10 atm
20Boyle’s Law* Pressure is inversely proportional to volume when temperature is held constant.Pressure ´ Volume = ConstantP1V1 = P2V2 (T = constant)
21If pressure increases then volume decreases. A Graph of Boyle’s LawIf pressure increases then volume decreases.If pressure decreases, then volume increases.Inversely proportional.
22Charles’s LawThe volume of a gas is directly proportional to temperature, and extrapolates to zero at zero Kelvin. (P = constant)
23Practice Problem P1 = 710 mm Hg, V1 = 2.40 L, P2 = 75 mm Hg V2 = ? A gas at constant temperature occupies a volume of 2.40 L and exerts a pressure of 710 mm Hg What volume will the gas occupy at a pressure of 75 mm Hg?P1 = 710 mm Hg, V1 = 2.40 L, P2 = 75 mm HgV2 = ?P1V1 = P2V2 (Boyle’s Law)710 mm Hg x 2.40 L = 75 mm Hg x V2V2 = 22.7 LThe new volume will be 22.7 liters.
24Practice Problem P1 = 3.4 atm, V1 = 14 L, P2 = 5 atm V2 = ? Sulfur dioxide gas at a pressure of 3.4 atm and a volume of 14 liters increases pressure to 5 atm. What is the resulting volume?P1 = 3.4 atm, V1 = 14 L, P2 = 5 atmV2 = ?P1V1 = P2V2 (Boyle’s Law)3.4 atm x 14 L = 5 atm x V2V2 = 9.5 LThe new volume will be 9.5 liters.
25Practice Problem P1 = 11.8 kPa, V1 = 420 ml, V2 = 630 ml P2 = ? Carbon monoxide at a pressure of kPa in a vessel of 420 ml expands to a new volume of 630 ml. What is the new pressure in kPa?P1 = 11.8 kPa, V1 = 420 ml, V2 = 630 mlP2 = ?P1V1 = P2V2 (Boyle’s Law)11.8 kPa x 420 ml = P2 x 630 mlP2 = 7.9 kPaThe new pressure is 7.9 kilopascals.
26Practice ProblemGiven 90 ml of H2 gas collected when the temperature is 27 C, how many ml will H2 occupy at 42 C?V1 = 90 ml, T1 = 27 C, T2 = 42 C, V2 = ?T1 = 27 C = 300 KT2 = 42 C = 315 KV1 = V2 (Charles’ Law)T1 T290 ml = V2 V2 = 90 ml x 315 K300 K K KV2 = 94.5 ml
27Why Don’t I Get a Constant Value for PV = k? Air is not madeof ideal gases2. Real gasesdeviate fromideal behaviorat high pressure
31Gay Lussac’s Law The pressure and temperature of a gas are directly related, provided that the volumeremains constant.
32The Combined Gas Law The combined gas law expresses the relationship between pressure, volume and temperature of afixed amount of gas.Why would you need this law when you already have all the others? What is different here?
33Combined Gas LawIf you should only need one of the other gas laws, you can cover up the item that is constant and you will get that gas law!=P1V1P2Boyle’s LawCharles’ LawGay-Lussac’s LawV2T1T2
34Combined Gas Law Problem A sample of helium gas has a volume of L, a pressure of atm and a temperature of 29°C. What is the new temperature(°C) of the gas at a volume of mL and a pressure of 3.20atm?What do you need to do first?Set up Data TableP1 = atm V1 = 180 mL T1 = 302 KP2 = atm V2= 90 mL T2 = ??
35Calculation = 604 K P1 = 0.800 atm V1 = 180 mL T1 = 302 K P1 V P2 V2= P1 V1 T2 = P2 V2 T1T T2T2 = P2 V2 T1P1 V1T2 = 3.20 atm x mL x 302 K atm x mLT2 = 604 K = °C= K
36Learning CheckA gas has a volume of 675 mL at 35°C and atm pressure. What is the temperature in °C when the gas has a volume of L and a pressure of 802 mm Hg?
37One More Practice Problem A balloon has a volume of 785 mL on a fall day when the temperature is 21°C. In the winter, the gas cools to 0°C. What is the new volume of the balloon?
38To Do Today Finish Boyle’s Law and Charles’ Law homework-due today! Combined gas law practice problems-due MondayIn book-review Ch 13 to help your understanding
39DMA 3/28/11A balloon has a volume of 785 mL on a fall day when the temperature is 21°C. In the winter, the gas cools to 0°C. What is the new volume of the balloon?
41Dalton’s Law of Partial Pressures For a mixture of gases in a container,PTotal = P1 + P2 + PThis is particularly useful in calculating the pressure of gases collected over water.
42Dalton’s Law of Partial Pressures 2 H2O2 (l) ---> 2 H2O (g) + O2 (g)0.32 atm atmWhat is the total pressure in the flask?Ptotal in gas mixture = PA + PB + ...Therefore,Ptotal = PH2O + PO2 = atmDalton’s Law: total P is sum of PARTIAL pressures.
43Avogadro’s Hypothesis Equal volumes of gases at the same T and P have the same number of molecules.V = n (RT/P) = knV and n are directly related.twice as many molecules
44Avogadro’s Hypothesis and Kinetic Molecular Theory The gases in this experiment are all measured at the same T and V.P proportional to n
45P V = n R T IDEAL GAS LAW Brings together gas properties. Can be derived from experiment and theory.BE SURE YOU KNOW THIS EQUATION!
46Using PV = nRT L • atm Mol • K P = PressureV = VolumeT = Temperature R =n = number of molesR is a constant, called the Ideal Gas ConstantInstead of learning a different value for R for all the possible unit combinations, we can just memorize one value and convert the units to match R.L • atmMol • K
47Using PV = nRTHow much N2 is required to fill a small room with a volume of 960 cubic feet (27,000 L) to 745 mm Hg at 25 oC?Solution1. Get all data into proper unitsV = 27,000 LT = 25 oC = 298 KP = 745 mm Hg (1 atm/760 mm Hg) = 0.98 atmAnd we always know R, L atm / mol K
48Using PV = nRT 2. Now plug in those values and solve for the unknown. How much N2 is req’d to fill a small room with a volume of 960 cubic feet (27,000 L) to P = 745 mm Hg at 25 oC?Solution2. Now plug in those values and solve for the unknown.PV = nRTRT RTn = 1.1 x 103 mol (or about 30 kg of gas)
49Learning CheckDinitrogen monoxide (N2O), laughing gas, is used by dentists as an anesthetic. If mol of gas occupies a 20.0 L tank at 23°C, what is the pressure (mm Hg) in the tank in the dentist office?
50Try This OneA sample of neon gas used in a neon sign has a volume of 15 L at STP. What is the volume (L) of the neon gas at 2.0 atm and –25°C?
51DMA 3/29/11A sample of neon gas used in a neon sign has a volume of 15 L at STP. What is the volume (L) of the neon gas at 2.0 atm and –25°C?The first thing you need to do is decide which law to use!
53At STP For example What is the volume at STP of 4.00 g of CH4? At STP determining the amount of gas required or produced is easy.22.4 L = 1 moleFor exampleWhat is the volume at STP of 4.00 g of CH4?How many grams of He are present in 8.0 L of gas at STP?
54Gas Stoichiometry: Practice! How many liters of O2 at STP are required to produce 20.3 g of H2O?
55Gases and Stoichiometry 2 H2O2 (l) ---> 2 H2O (g) + O2 (g)Decompose 1.1 g of H2O2 in a flask with a volume of 2.50 L. What is the volume of O2 at STP?Bombardier beetle uses decomposition of hydrogen peroxide to defend itself.
56Gases and Stoichiometry 2 H2O2 (l) ---> 2 H2O (g) + O2 (g)Decompose 1.1 g of H2O2 in a flask with a volume of 2.50 L. What is the volume of O2 at STP?Solution1.1 g H2O mol H2O mol O L O234 g H2O mol H2O mol O2= 0.36 L O2 at STP
57Not At STP Chemical reactions happen in MOLES. If you know how much gas - change it to molesUse the Ideal Gas Law n = PV/RTIf you want to find how much gas - use moles to figure out volume V = nRT/P
58Example #1 HCl(g) can be formed by the following reaction 2NaCl(aq) + H2SO4 (aq)2HCl(g) + Na2SO4(aq)What mass of NaCl is needed to produce 340 mL of HCl at 1.51 atm at 20ºC?
59Example #2 2NaCl(aq) + H2SO4 (aq) 2HCl(g) + Na2SO4 (aq) What volume of HCl gas at 25ºC and 715 mm Hg will be generated if 10.2 g of NaCl react?
60GAS DIFFUSION AND EFFUSION diffusion is the gradual mixing of molecules of different gases.effusion is the movement of molecules through a small hole into an empty container.
61DiffusionDiffusion: describes the mixing of gases. The rate of diffusion is the rate of gas mixing.
62EffusionEffusion: describes the passage of gas into an evacuated chamber.