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Chapter 5: Gases & Kinetic-Molecular Theory

by: Amelia Notetaker

Chapter 5: Gases & Kinetic-Molecular Theory CH 121

Marketplace > University of Alabama - Huntsville > Chemistry > CH 121 > Chapter 5 Gases Kinetic Molecular Theory
Amelia Notetaker
GPA 3.88
GENERAL CHEMISTRY I - 90514 - CH 121 - 02
Pamela D Twigg (P)

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Chapter 5 notes.
GENERAL CHEMISTRY I - 90514 - CH 121 - 02
Pamela D Twigg (P)
Class Notes
Chemistry, twigg, uah, gases, gas, kinetic, Molecular, Theory, theories, notes
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This 0 page Class Notes was uploaded by Amelia Notetaker on Tuesday December 1, 2015. The Class Notes belongs to CH 121 at University of Alabama - Huntsville taught by Pamela D Twigg (P) in Fall 2015. Since its upload, it has received 21 views. For similar materials see GENERAL CHEMISTRY I - 90514 - CH 121 - 02 in Chemistry at University of Alabama - Huntsville.


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Date Created: 12/01/15
Chapter 5 Lecture Notes General properties of gases 0 Unlike like liquids and solids they 0 Expand to ll their containers Highly compressible Extremely low density 0 There is a lot of quotfreequot space in a gas 0 Gases can be expanded in nitely o Gases occupy containers uniformly and completely 0 Gases diffuse and mix rapidly Properties of gases 0 V volume L o T temperature K o N amount moles o P pressure atm Pressure 0 The amount of force applied to an area P FA o Atmospheric pressure is the weight of air per unit of area 0 Measured with a barometer developed by Torricelli in 1643 Hg rises in tub until force of Hg down balances force of atmosphere pushing up P of Hg pushing down is related to Hg density 0 Column height 0 1 standard atm 760 mm Hg 299 in 34 feet of water 0 SI unit is Pascal Pa where 1 atm 101325 Pa Units of pressure 0 Mm Hg or torr Literally the difference between the heights measured in mm h of 2 connected columns of Hg 0 Atmosphere 0 10 atm 760 torr 760 mm Hg 0 At 29000 ft Mt Everest atm pressure 270 mm Hg 0 Pascals 1 Pa 1 Nmquot2 0 Bar 1 bar 10quot5 Pa 100 kPa Standard pressure 0 Normal atmospheric pressure at sea level 0 Equal to o 100 atm 760 torr 760 mm Hg 0 101325 kPa 1013 bar Manometer 0 Used to measure the difference in pressure between atmospheric pressure and that of a gas vessel 0 Ideal gas law 0 PW nRT o Brings together gas properties 0 Can be derived from experiment and theory Boyle39s law 0 If n and T are constant then PV nRT k c This means P goes up as V goes down 0 The volume of a xed molar quantity of gas at constant temperature is inversely proportional to the pressure 0 Charles39 law 0 If n and P are constant then V nRPT kT V and T are directly related o The volume of a xed molar amount of gas at constant pressure is directionally proportional to its absolute temperature Avogadro39s hypothesis 0 Equal volume of gases at the same T and P have the same number of molecules 0 VnRTPkn V and n are directly related Idealgas equation 0 Combining the previous equations we get V nTP R is proportionately constant 0 Using PVnRT o How much N2 is required to ll a small room with a volume of 960 cubic feet 27000 L to P 745 mm Hg at 25 degrees C o R 082057 LatmKmo 0 Get all data into proper units V 27000 L T 298 K P 745 mm 1 atm H9 760 mm Hg 0 98 atm 0 Calculate n PVRT 98 atm27 x 10quot4 L 0821 LatmKmo 298 K o 11 X 10quot3 mol Gases and stoichiometry o 2 H202 l gt 2 H20 g 02 o Decompose 11 g of H202 in a ask with a volume of 25 L 11 g 1 mol H202 340 9 o 032 mol 032 mol 1 mol 02 H202 2 mol H202 016 mol 02 o 016 mol0821298 K 250 L o 16 atm o What is the pressure of 02 at 25 degrees C Dalton39s law of partial pressures o The total of a mixture of gases equals the sum of the pressures that each would exert if it were present alone Ptotal P1 P2 P3 Densities of gases 0 nV PRT o mV PMRT o D mV PMRT 0 One only needs to know the molar mass the pressure and the temperature to calculate the density of gas Molar mass 0 We can manipulate the density equation to enable us to nd the moalr mass of a gas o D PMRT becomes M Using gas density 0 The density of air at 15 degrees C and 100 atm is 123 gL what is the molar mass of air 123082057288100 291gmol Kinetic molecular theory KMT 0 Theory used to explain gas laws KMT assumptions 0 O Gases consist of molecules or atoms in constant random motion 0 Pressure P arises from collisions with container walls No attractive or repulsive forces between molecules coisions elastic total kinetic energy remains the same Volume of molecules is negligible compared to the total volume of the sample We assume molecules are in motion they have a kinetic energy KE 12 mass velocityquot2 For a given gas xed mass as T goes up KE also increases and so doesspeed At the same T a gases have the same average KE Urms sqrt uquot2 sqrt 3RTM Where uquot2 is the average of the squares of the speeds and M is the molar mass Speed increases with T Speed decreased with M Velocity of gas molecules 0 O Molecules of a given gas have a range of speeds At a given temperature average velocity decreases with increasing mass 0 Main Tenets of Kineticmolecular theory 0 Energy can be transferred between molecules during collisions but the average kinetic energy of the molecules does not change with time as long as the temperature of the gas remains constant The average kinetic energy of the molecules is proportional to the absolute temperature KE 32 RNAT R gas constant NA avogadro39s number Deriving the rootmeansquare speed of particles in a sample 0 KE 12 mass velocityquot2 or KE 12 m uquot2 o If we set both equations equal to each other 0 12 m uquot2 32 RTNA NA m M c 12 M uquot2 32 RT or uquot2 3RTM Gas diffusion and effusion o Diffusion is the gradual mixing of molecules of different gases 0 Effusion is the movement of molecules through a small hole into an evacuated space 0 Molecules effuse through holes in a rubber balloon for example at a rate molestime that is Proportional to T lnversely proportional to M 0 Therefore He effuses more rapidly than 02 at same T o Graham39s law governs effusion and diffusion of gas molecules 0 Studied relationship between effusion and gas density Rate of effusion 1sqrt d so 1 sqrt M 0 Rate of effusion is inversely proportional to the square root of its molar mass D PMRT 0 Can use graham39s law to determine M of unknown gas by comparing effusion rate with that of a known gas 0 Rate ofArate ofB sqrtM orBM ofA ldeaLgas equann 0 V 1P boyle39s law 0 V T charles39 law 0 V n avogadro39s law 0 Combining these we get V nTP o The relationship V nTP then becomes V R nTP or PVnRT o R is constant Using kinetic molecular theory to understand gas laws 0 Recall that KMT assumptions are Gases consist of molecules in constant random motion 0 P arises from collisions with container walls 0 No attractive or repulsive forces between molecules collisions elastic Volume of molecules is negligible Realgases o In the real world the behavior of gases only conforms to the idealgas equation at relatively high temperature and low pressure 0 Deviations from ideal behavior 0 The assumptions made in the kineticmolecular break down at high pressure andor low temperature Deviations from ideal gas law 0 Real molecules have volume 0 There are intermolecular forces Otherwise a gas couldn39t become a liquid 0 Account for the volume of molecules and intermolecular forces with van der Waal39s equation 0 Ideal gas equation PV nRT becomes 0 P nquot2 a Vquot2 V nb nRT P measured P V measured V Nquot2 aVquot2 intermolecular forces 0 Nb volume correction A and b are van der Waals constant 0 Cl2 gas has a 649 atmLquot2molquot2 b 0562 Lmol For 80 mol Cl2 in a 4 L tank at 27 degrees C what is the pressure as calculated from the ideal gas law and from the van der Waals equation N 80 mol V 40 L T 300 K R 08206 0 Pressure calculated from ideal gas equation p 80 mo08206300 K4 L o 492 atm o Pressure from the van der Waals P nRTVnb nquot2aVquot2 8008206300 4080562 8quot2 649 4quot2 o 295 0 Because both values differ C2 gas deviates from ideal gas behavior under these coniditions Corrections for nonideal behavior 0 The ideal gas equation can be adjusted to take these deviations from ideal behavior into account The correct ideal gas equation is known as the van der Waals equann Sample Problems 0 A sample of a gas 5 mol at 1 atm is expanded at constant temperature from 10 L to 15 L What is the nal pressure at atm PVnRT R 082057 n 5 mol T Ti Tf Vi 10 L Vf 15 L Pi1 atm niRTinfRTf gt PiViPfo Pf PiViNf Pf11015 67atm o If 321 mol of a gas occupies 562 L at 44 degrees C and 793 torr 529 mol of this gas occupies how many liters R 082057 P793 torr T 317K ni 321 mol nf 529 mol Vi 562L Vini Vfnf Vf nf Vini Vf 529 mol562L321 mol 0 926 L o The molar mass of a gas that has a density 09 575 gl at STP is how many gmol T 273K 1 bar R 082057 d575 gL 1 atm 1013 bar so 1 atm1013 987 atm M dRTP T 273 K R 082057 d 575 gL P 987 atm M 575 gL082057273 K 987 atm 1305 gmol


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