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# Week 3 Book Notes - Thermal Physics Physics 60

UCI

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This 5 page Class Notes was uploaded by Hazel Medina on Wednesday October 14, 2015. The Class Notes belongs to Physics 60 at University of California - Irvine taught by Feng, J. in Fall 2015. Since its upload, it has received 27 views. For similar materials see Thermal Physics in Physics 2 at University of California - Irvine.

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Date Created: 10/14/15

Physics 60 10122015 Week 3 Ch 3 Interactions and Implication 0 Even though the second law of thermodynamics is based on probability and math of large numbers alone we will treat it as fundamental Ch 31 Temperature 0 Recall that objects in equilibrium have the same temperature 0 Consider two Einstein solids A and B that are weakly coupled and can therefore exchange energy When ins at equilibrium oStotaIoqA 0 or oStotaIoUA 0 at equilibrium I Slope of Stota is slope of SA plus slope of SB so oSAoUA oSBoUB O 0 Because dUA dUB then oSAoUA oSBoUB at equilibrium This means that the slope of entropy vs energy graphs for both objects A and B are the same when they are in equilibrium Steep slopes correspond with quotlowquot temperature and shallower slopes correspond with high temperature 0 Thanks to Boltzmann s constant oSoU is in units of 1K This proposes that T oSoU391 which makes the definition of temperature 1T asauNv 339 A Silly Analogy 0 Think of a community of people where they exchange money to make themselves happier some are happier than others to give In this analogy money is energy happiness is entropy and generosity is temperature 39 For example more money tends to mean more generosity but some may get less generous with more money 0 Similar to how in physics there are no laws that say an object s temperature can t decrease with added energy I For example there are those who become happier upon losing energy 0 Comparable to an energy graph with negative slope 339 RealWorld Examples 0 Remember the equation S NknqN 1 Nkan NklneN Nk This means temperature is T oSoU391 NkU391 or U NkT Temperature then becomes T 32NkU391and U 32NkT Ch 32 Entropy and Heat 339 Predicting Heat Capacities 0 Remember CV oUoTNV for heat capacity at constant volume For an Einstein solid where qgtgtN the heat capacity is CV ooTNkT Nk For a monatomic ideal gas CV ooT32NkT 32Nk I Noticehow both an Einstein solid and a monatomic ideal gas heat capacity is independent of temperature and is equal to k2 times degrees of freedom 0 Steps to predict heat capacity 1 Find and expression for multiplicity Q in terms of UVN and other relevant variables by using quantum mechanics and combinatorics 2 Find entropy S by taking the logarithms of both sides of the function from Step 1 3 Get temperature T as a function of U and the other variables used by differentiating S with respect to U and taking its reciprocal 4 Solve for U as a function ofT and other variables used 5 Differentiate UT for the prediction of heat capacity C 3 Measuring Entropies 0 You can measure entropy by follow steps 35 listed above in reverse Entropy changes by dS dUT QT I When T is changing it s more convenient to use dS CVdTT As sf s 1 2 CVndT 0 At SO in principle equals zero which is where the system should settle into its unique lowest energy state meaning 0 1 and S 0 This is often referred to as the third law of thermodynamics Residual entropy difference in entropy between one object not in equilibrium and a substance in a crystal state close to absolute zero I Can also be from mixing different nuclear isotopes of an element I Another comes from multiplicity of alignments of nuclear spins which generally happens when the temperature is far below range of routine heat capacity measurements Note entropy is always positive and finite according to S kan unless CV or T approaches 0 I This result is also part of the third law of thermodynamics 339 The Macroscopic View of Entropy o Entropy traditionally defined as dS QT Entropy is half conserved it cannot by destroyed but is created often whenever heat flows between objects of different temperatures Ch 33 Paramagnetism 0 Remember twostate paramagnets from Ch 21 paramagnet material in which constituent particles are in a direction parallel to externallyapplied magnetic field 339 Notation and Microscopic Physics 0 System of a paramagnet is made up of N spin12 particles in constant magnetic field I in the 2 direction Dipoles particles that feels a torque to align its magnetic dipole moment with the magnetic field assume no interaction between dipoles I This is an ideal paramagnet I Particles are limited to certain discrete values meaning they are quantized o For spin12 particles there can only be two values quotupquot and quotdownquot along zaxis Amount of energy required to flip a dipole from up to down where the dipole prefers up is 2uB u constant related to the particle s magnetic moment Total energy of the system is U uBN1 N1 uBN 2N1 o Magnetization M the total magnetic moment of the system M uN1 N1 UB 0 2N1 NA NN1N1 3 Numerical Solution 0 Maximum multiplicity and entropy occur when UO ie when half of the dipoles point down and multiplicity and entropy decrease as more energy is added to the system Very different from Einstein solids we ve studied I Thinking of temperature as a function of energy if more than half of the dipoles face up total energy is negative temperature increases as energy is added 0 When UO the temperature is infinite meaning it will give up energy to any other system with a finite temperature 0 Negative temperatures can only occur for systems with limited energy which allows multiplicity to decrease as energy approaches its maximum For example nuclear paramagnets which have dipoles as their atomic nuclei can only have magnetic energy for short periods of time to give negative energy you start reverse the fields to make them antiparallel I Thinking of magnetization as a function of temperature means the zero positive system is saturated where all dipoles point up and have maximum magnetization o T w corresponds to maximum randomness rather than maximized with dipoles pointing down as would be expected when T 9 OO 0 Entropy is a fundamental quantity governed by the second law of thermodynamics 339 Analytic Solution 0 1T GS6U oN1oUoSoN1 12uBoSoN1 Differentiation 12uB oSoNt gives 1T kZuBlnNUuBNUuB U can be solved for U NuBtanhuBkT tanh is the hyperbolic tangent function tanh sinhxcoshx where sinhx 12eX ex and coshx 12eX e39X To find heat capacity differentiate U NuBtanhuBkT with respect to T to get CB cwam B Nk uBkT2cosh2uBkT I For an electronic twostate paramagnet the value of u is the Bohr magneton 13 eh4Ttme 9274 x 103924 JT 5788 x 10395 eVT e is the electron s charge and me is its mass o When 13 ltlt kT M z NuZBkT Curies law M oc 1T o For a nuclear paramagnet replace me with mass for a proton in Bohr magneton equa on Ch 34 Mechanical Equilibrium and Pressure 0 Now think of systems whose volumes can change as they interact exchange of volume governed by pressure Consider two systems that are free to exchange energy and volume separated by a movable partition with fixed total energy and volume I This means that astotayauA 0 and astotaavA 0 o Remembering that due to the fixed volumes dVA dVB 0 oStotaIVA oSAoVA asBavA oSAoVA asBavB which means that GSAWA asBavB at equilibrium 0 The systems are also in thermal equilibrium because while energy and volume can be exchanged heat does not meaning ToSoV is the same for both systems 0 To relate entropy and pressure P ToSoVUN Taking the log for the equation of multiplicity of a monatomic ideal gas 0 fNVNU get S Nkan 32Nkan klnfN I Then pressure becomes TooVNkan NkTV or PV NkT 339 The Thermodynamic Identity Imagine a system where volume and energy can be exchanged in small quantities AV and AU respectively I ts total energy changed by AS AS1 AS2 o Multiplying and dividing the first term by AU and doing the same with AV for the second term you get AS ASAUVAU ASAUUAV Because the changes are small the terms become derivatives and the equation becomes dS oSoUVdU oSoVUdV 1TdU PTdV It becomes the thermodynamic identity dU TdS PdV 3 Entropy and Heat Revisited 0 Recall the first law of thermodynamics dU Q W Using the first law of thermodynamics and the thermodynamic identity we get Q TdS for a quasistatic system meaning change in entropy is QT I When a process is both adiabatic and quasistatic it is isentropic I For constant pressure with temperature change the equation becomes ASp 1 cpTdT 3N2I we Ch 35 Diffusive Equilibrium and Chemical Potential 0 For two systems in thermal equilibrium temperatures are the same mechanical equilibrium pressures are the same 0 To find out what s the same for diffusive equilibrium consider two systems A and B that can exchange energy and particles Assume total number of particles and total energy is fixed note NANA UA UA VAVA At equilibrium total entropy is maximum so astotaauANAVA 0 and astota5NAUAVA 0 I astotayavA 0 as well I We can conclude that oSAoNA oSBoNB at equilibrium which means T oSAoNA T oSBoNB at equilibrium 0 Chemical potential u E ToSoNUV so uA uB at equilibrium We can conclude that particles tend to flow from the system of higher p into the one with lower u To include this in the thermodynamic identity with changes in N dS GSaumdu oSoVNUdV oSoNUVdN 1TdU PTdV uTdN I This becomes dU TdS PdV udN o udN is referred to as chemical work 39 I11 5 39TdS5N1UNN2 and H2 5 39TdS6N2UVN1 chemistryE TdSdnUVI Where n 0 Therefore for diffusive equilibrium the chemical potentials are the same

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