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An Introduction to Gases Chapter 13

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Kinetic Molecular Theory Postulate #1 –Gases consist of tiny particles (atoms or molecules) Postulate #2 –These particles are so small, compared with the distances between them, that the volume (size) of the individual particles can be assumed to be negligible (zero). –Gases are COMPRESSIBLE

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Kinetic Molecular Theory Postulate #3 –The particles are in constant random motion, colliding with the walls of the container. These collisions with the walls cause the pressure exerted by the gas.

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Kinetic Molecular Theory Postulate #4 –The particles are assumed not to attract or to repel each other. Postulate #5 –The average kinetic energy of the gas particles is directly proportional to the Kelvin temperature of the gas.

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Avogadros Hypothesis At the same temp & pressure, equal volumes of gas hold same number of molecules. V and n are directly related.V and n are directly related. twice as many molecules

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Pressure Pressure of air is measured with a BAROMETER (developed by Torricelli in 1643)

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Pressure What is Pressure?What is Pressure? What tool do we use to measure it?What tool do we use to measure it?

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Pressure 1.mmHg (or Torr) 2.Atmospheres (atm) 3.Pascals (used in physics: 1 pascal = 1 newton per square meter) 4. psi Equivalences: 1 atm = 760 mmHg 1 atm = 101,325 Pa = kPa 1 atm = 14.7 psi

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Pressure Calculation What is 475 mm Hg expressed in atm? 475 mm Hg = 1 atm 760 mm Hg atm

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The pressure of a tire is measured as 294,000 pascals. What is this pressure in atm? Pressure Calculation 294, 000 Pa 101, 325 Pa 1 atm = 2.90 atm

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Daltons Law The Law of Partial Pressure The total pressure of a mixture of gases is the sum of the partial pressures of the gases in the mixture. P total = P A + P B + P C

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Gas Laws Calculations Get out a calculator!!!

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The Gas Law PV=nRT P = pressure ( atm or kPa ) V= volume ( L ) n= number of moles (mol) T= temperature (K)

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R – The Proportionality Constant Value depends on units L (kPa) mol (K) L (atm) mol (K) Or

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The Gas Law – Problem If 7.0 moles of an ideal gas has a volume of 12.0 L with a temperature of 300. K, what is the pressure in kPa? P (12.0 L)=(7.0 mol)(300 K)8.314 L (kPa) mol (K) PV = nRT P = kPa P = 1500 kPa

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Combined Gas Law Lets say we have some O 2 gas AND we change some conditions. Would there be anything similar between the two gases?

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Combined Gas Law – Problem You have 3 moles of a solution at 300. K and 15 atm in a 2 L container. If the container is heated to 350. K and the volume decreased to 1 L, what will the new pressure be? P1P1 15 atmP2P2 want V1V1 2 LV2V2 1 L n1n1 3 molesn2n2 R1R1 constantR2R2 T1T KT2T K

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Combined Gas Law – Problemc P1V1P1V1 = n1R1T1n1R1T1 P2V2P2V2 n2R2T2n2R2T2 P1V1P1V1 = T1T1 P2V2P2V2 T2T2 If we know that R 1 = R 2 and the mass is constant then (15 atm)(2 L) = (300. K) P 2 (1L)(350. K) Replace with numbers

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Combined Gas Law – Problem (15 atm)(2 L) = (300. K) P 2 (1L)(350. K) (15 atm)(2 L)(350. K) = P2P2 (1L)(300. K) P 2 = 35 atm

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Pressure & Volume At constant Temperature Pressure and Volume vary inversely. –Why? –More collisions More pressure P 1 V 1 = P 2 V 2

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P & V – Example Problem If you start with L of a gas at 7.0 atm and you move the gas to a container with 3.5 L available, how much pressure will the gas exert? 7 atm (0.500 L) =P2P2 3.5 L P 1 (V 1 ) = P 2 (V 2 ) 7.0 atm (0.500 L) = P 2 (3.5 L) 1.0 atm = P 2

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Put a few drops of water in a can. Heat the can until the water boils. What is happening to the gas inside? Now flip the can over into cold water. Predict what do you predict will happen? Demo

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On a Larger Scale

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Temperature & Volume At constant Pressure Volume & Temperature vary directly. –Why? –More collisions More Volume V1V1 = V2V2 T1T1 T2T2

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T & V – Example Problem If a gas is in a balloon with a volume of 12.0 L and at a temperature of 300. K, what will the volume be if you place the balloon in a freezer at 250. K? V1V1 = V2V2 T1T1 T2T L = V2V K250. K 12.0 L (250. K) = V2V K 10.0 L= V2V2

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S.T.P. Standard Temperature and PressureStandard Temperature and Pressure These are conditions that are universal Standard Temperature: 0ºC or 273K Standard Pressure: 1atm or kPa

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S.T.P. – Example Problem What is the volume of 1 mole of a gas at STP? P1 atm Vwant n1 mole R (L)(atm)/(K)(mole) T273 K PV = nRT (1atm)V = (1 mole)( [Latm/Kmole])(273K) V= 22.4 L

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