GEN PHYSICS A STUDIO
GEN PHYSICS A STUDIO PHY 2048C
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This 13 page Class Notes was uploaded by Garett Kovacek on Thursday September 17, 2015. The Class Notes belongs to PHY 2048C at Florida State University taught by Simon Capstick in Fall. Since its upload, it has received 75 views. For similar materials see /class/205523/phy-2048c-florida-state-university in Physics 2 at Florida State University.
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Date Created: 09/17/15
Chapter 17 Temperature and the Kinetic Theory of Gases The three phases of water solid liquid and gas are in this flask This only occurs at a unique pressure and temperature Prof Chris Wiebe Prof Simon Capstick Thermal equilibrium Thermod namics study of heat and heat ow When two objects are in contact With each other and they exchange heat then they are in thermal contact with each other This will continue until no net heat flows between the two objects We call this thermal equilibrium In the next few chapters we will be exploring the three laws of thermodynamics There are actually four laws though so we need to start with the zeroth law of thermodynamics If two objects are in thermal equilibrium with a third object then they are in thermal equilibrium with each other i L ii a u 9 n D b If A and C are in thermal equilm and B and C are in thermal equilm then A and B have to be in thermal equilm Temperature What is temperature V qg lsludr39g gg trSP hgigaEgny To convert between the two scales kinetic energy of a system There are several units of 0 temperature tc 9tF 32 172 Celsius 0 degrees C freezing point of water or ice point 1 FAHRENHEIT CELSIUS CONVERSION atm pressure 100 degrees C boiling point of water 1atm pressure or steam pomt We can measure temperature for example by looking at how pressure changes in a fixed tube 900 of gas If we mark 100 divisions 39 o between the ice point and steam point then we have made a thermometer demo Fahrenheit 32 degrees F is the icepoint 212 is the steam point This thermometer measures how 5 much a metal expands inside of it as a function of temperature 350 0 1139 0 14 ll 0 09 gm 3 N3 Q 00 5 A more useful scale for temperature There is a more useful scale of temperature called the KeIVIn scale At zero Kelvin the pressure of a gas becomes zero as you cool down the pressure decreases Liquid N2 demo We can discover where this occurs by doing a pressure vs temperature ex eriment for a fixed amount 0 gas Where the x intercept O we have zero Kelvin O K Conversion T tC 27315 K Zero Kelvin 27315 degrees C 39 I v 27315 C t Temperature K min 1 01 lt Supernova IOS Hydrogen bomb 107 lntcrinr of the Sun 106 lt Solar corona 10j 104 Ken 103 lt Surtaco of the Sun 103 K Copper melts lt Water freezes Temp scale mu Liquid nitrogen 1071 7 Liquid hydrogen 1042 Liquid helium 104 lllt HeJ goes superfluid 10 8 J BoseEinstein condensate lt Iowest lemperalure achieved The Ideal Gas Law The graph we just made is an a n n o the ideal gas law P PV nRT or P nRVT a linear relationship P is the pressure V is the volume T is the temperature in Kelvin n is the number of moles of the gas quot massmolecular mass v39 1 mole 6022 x 1023 molecules 27315 C I t Avogadro s number R is the gas constant which depends upon our units R 8314 JSmol K 008206 L atmmO K Slope of graph This law only works for ideal gases no interactions between the molecules Note it also states that P 1V or PV a constant for an ideal gas This is Boyle s law demo P nRVT Ideal Gas Law experiment Pressure mm Temperature Hg 0 780 20 190 196 1000 100 1000 on o o Xintercept 265 degrees C close to 273 degrees C This is absolute zero 0 K m o 0 Pressure mm Hg Jz 8 N o o n I I I I I I I I 300 250 200 150 100 50 0 50 100 Temperature degrees C The result that we get based on data we took in class is 265 deg C This is close to the value of 273 deg C for 0 K Can you think of reasons why our result is off Isotherms o If you have a fixed amount of gas then n is a constant in youridealgas equann o In this situation P2V2T2 P1V1T1 o The PV curves are called isotherms since they each represent a single temperature Each one of these curves is an isotherm P 1V The Ideal Gas law again We can state PV nRT as PV NkBT where N is the fi TVIJmovx number of molecules you 860 8314JmolKR have and kB is 840 H2 Boltzmann s constant 8333 132 138 x 103923 JK 800 SO nR NkB 780 02 The ideal gas law is an example of an equation of 5 10 15 20 25 30 35 40 Platm state tells you the propertles of a system In At high pressures PVnT a certa39n State is not a constant Real ases more comp icated behavior at Why is this true high pressures need the Van der Waals equation Putting it all together with some examples Problem 42 Therefore V2V1 T2T1 100273 K50273K 115 Problem 48 a n massmolecular mass of He 10 g4003 gmol 250 mol 1 atm 1013 kPa 1013 X 103 Pa VnRTP 25 mol8314 JKmol42K1013x103 Pa v 862 x 104 m3 0862 L 103 m3 1 L b v nRTP 2583142931013X103 Pa 601 x 102 m3 601 L Kinetic Theory of Gases This is a theory that relates P V and T to the microsco ic pro erties of the gas ie the mo ecular mass and average kinetic energy Consider a volume V of gas with N molecules each has mass m with speed v Consider the molecules which hit a wall of area A The molecules which can hit this wall in an interval At must be within vXAt of the wall The number of molecules that hit the wall in time At is 12 WV vXAt A Numberunit volume x Volume The factor of 12 is because on the average only 12 of the molecules are moving to the right Z y 9 I 439 l Wall of area A Kinetic Theory of Gases o The magnitude of the total change in momentum of all the molecules during a time interval At is 2va elastic collisions speed is the same after as before no loss of energy times the number of molecules that hit the wall during that interval s IN N 2 A 2mv v AtA mv AAt plt xgtlt2Vx gtVx 0 So the pressure PFA is 3 luvxmu i XV I l lt I 7quot II A F1 N 2 mv A A At V x p PV 2 vax2 2Nmvx2 average Kinetic Theory of Gases We can rewrite this equation as PV NkBT 2N12va2average 1 1 imvav ikT 1719 THE AVERAGE ENERGY ASSOCIATED WITH MOTION IN THE x DIRECTION But we just picked the x direction there is nothin specia abzout It 2In fact v fav vy2av vZZaV and v2average VXZaV vy av v 3v 2 av X av The kinetic energy is then Kav 12va ng 17 21 21V AVERAGE KINETIC ENERGY OF A MOLECULE nd for the total amount of energy of N molecules we ave K Nmvzav gNkT gnRT 17 22 KINETIC ENERGY OF TRANSLATION FOR 71 MOLES OF A GAS Molecular speeds of gases Rearranging thizs equation we can get v 3kBTm 3RTMaverage Where M is the molar or molecular mass 602 x 1023 x massmolecule the molar mass is the mass per mole of molecules We can now define the root f v mean square speed vr S v2 x3RTM 3 x3kBt7er 39 This is not the average I We need to look at the dv distribution of as This shows the number of molecules to fin the root gas molecules fv mean square speed as a function of speed Vmax lt V lt Vrms averaqe
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