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# Week 5 Book Notes - Thermal Physics Physics 60

UCI

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This 5 page Class Notes was uploaded by Hazel Medina on Sunday November 1, 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 118 views. For similar materials see Thermal Physics in Physics 2 at University of California - Irvine.

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Date Created: 11/01/15

Physics 60 10262015 Week 5 Ch 52 Free Energy as a Force toward Equilibrium o If system is not isolated consider the environment to be like a quotreservoirquot of energy that is large enough to release unlimited energy without its temperature changing dStota dS dSR the total entropy of the universe equals the sum of the entropy of the system plus the entropy of the reservoir environment I Using the thermodynamic identity dS dUT PTdV uTdN the equation for dStotal assuming VN and T are fixed becomes dStota dS 1TdU 1TdU TdS 1TdF recall F Helmholtz free energy I If P N and T are fixed dStota dS 1TdU PVdV 1TdU TdS PdV 1TdG recall G Gibbs free energy 0 S increases as constant energy and volume 0 F decreases at constant temperature and volume 0 G decreases at constant temperature and pressure 0 Remember that F E U TS As F decreases U decreases and S increase intuitively true because as a system loses energy the environment gains energy 0 Remember that G E U PV TS U and V leans towards decreasing while S leans towards increasing to maximize the total entropy of the universe 339 Extensive and Intensive Qualities 0 Key thermodynamic variables we have learned of U V N S T P u H F and G Extensive doubles with more quotth V N S U H F G mass Intensive does not change with more quotstuff T P 1 density I Multiplying intensive with extensive quantities result in extensive quantities for example volumedensity mass I Divining an extensive with another extensive quantity result in intensive quantities I Multiplying extensive quantities with each other result in neither I Adding similar types of quantities result in quantities of the same type eg extensive extensive extensive I You cannot add intensive with extensive or vice versa 339 Gibbs Free Energy and Chemical Potential 0 Remember 1 ac5N ie adding a particle under fixed temperature and pressure increases the Gibbs free energy of the system by u IfT and P are held fixed the value of 1 cannot change as you add more particles because G is an extensive quantity that grows in proportion to number of particles I G Np I u is the Gibbs free energy per particle 39 l1 dFGNHN 0 According to this equation 1 gradually increases as the system becomes more dense For more than one type of particle G N1p1 szz 2 Niui o For an ideal gas op6P 1N oGoP VN Integrating both sides from P0 to P gives uTP p T lenPP Ch 53 Phase Transformations of Pure Substances 0 Phase transformation discontinuous change in properties of a substance as its environment changes infinitesimally Phases different forms of a substance Phase diagram graph showing equilibrium phases as a function of temperature and pressure Line on this graph can show different forms of a substance coexisting stably at the same temperature and pressure eg water and ice at 0 degrees Celsius and 1 atm o Vapor pressure pressure at which a gas can coexist with its solid or liquid phase 0 Triple point where all three phases can coexist Most substances like carbon dioxide has a higher melting temperature when in more pressure Ice is different in how applying more pressure lowers its melting temperature Critical point point where there is no longer any discontinuous change from liquid to gas because the gas becomes more dense as pressure increases Liquid crystal phase where molecules move randomly like as liquids but tend to orient parallel to each other found in some materials made of long molecules Helium is the only element that remains liquid at absolute zero and only forms a solid phase at high pressures 0 Has two liquid phases a normal phase helium I and a superfluid phase helium II Superfluid many remarkable properties like zero viscosity and very high thermal conductivity Changing properties like composition and magnetic field strength can also affect phase transformations 0 TypeI superconductor such as lead tin or mercury zero electrical resistance when temperature and external magnetic field strength are low Ferromagnet such as iron magnetized phases up or down depending on the direction of applied field 0 Curie temperature temperature where magnetization completely disappears and the phase boundary ends at a critical point 1043 K for iron v Diamonds and Graphite 0 Two commonlyknown phases of elemental carbon diamond and graphite Graphite is more stable phase and process of diamonds converting to graphite is extremely slow at room temperature 0 Gibbs free energy of a mole of diamond is greater than that of a mole of graphite by 2900 J V 565th 0 Above a very high pressure diamonds should be more stable than graphite S oGoTpN v The ClausiusClapeyron Relation 0 At phase boundary Gnquid G Ggas Gg and so dGnquid ngas SdT VdP Sngng P so dPdT Sg SVg V When L latent heat to convert the material from liquid to gas and AV Vg VI dPdT LTAV the ClausiusClapeyron relation 339 The van der Waals Model 0 For a liquidgas system van der Waals equation P aNZV2V Nb NkT Any relation among P V and T is called an equation of state Note how it modifies the ideal gas law PV NkT I V Nb fluids cannot have zero volume so here it has minimum volume Nb I P aNZV2 accounts for shortrange attractive forces between molecules not in contact with each other 0 Total potential energy aN2V Pdueto p ddvaN2v aN2v2 The van der Waals equation becomes P NkTVNb aNZVZ I Note that a and b have different values for different substances 0 Remember Gibbs free energy dG SdT VdP udN which becomes asawn VoPoNNT Using the van der Waals equation it becomes asawn NkTVV Nb2 2aN2V2 I Differentiating it becomes G NkTnV Nb NkTNbV Nb 2aN2V cT o Vapor pressure welldefined pressure at which liquidgas transformation takes place Critical point moment where Tc PC and Vccorrespond so that they give the substance similar properties as liquid and gas I Tc critical temperature Pc critical pressure Vc critical volume Ch 6 Boltzmann Statistics Ch 61 The Boltzmann Factor 0 In statistical mechanics it is important to know this simple formula encompassing the probability of finding a system in any particular microstate the system is in thermal equilibrium with a quotreservoirquot at a specific temperature Consider a single atom where the microstates of the system correspond to various energy levels of the atom I For example a hydrogen atom has only one ground state with energy 136 eV when spin is neglected but it has four independent states of energy 34 eV nine states with energy 15 eV etc etc I Each separate state is a microstate degenerate description for energy level corresponding to more than one independent state 0 Now think of atoms that exchange energy not isolated This means that the ratio of probabilities for any two states becomes 39Ps239Ps1 QRs2Qs1 s1 and s2 are the two states I Remembering that S kan the equation becomes 39Ps239Ps1 e entropy of the reservoir 0 Using the thermodynamic identity dSR 1TdUR PdV udNR and knowing that SRs2 SRs1 1TURs2 URs1 1TEs2 Es1 where E is the energy of the atom 1 sz1 sl e39E 52 quotTe39E SIW Boltzmann factor e39EWkT The equation can become 39Ps2e39E52kT 39Ps1e39E51kT where both sides are independent of 1 and s2 which means both sides equal a constant 12 and so any state s has 39Ps 1Ze39ESkTaka Boltzmann distribution or canonical distribution 552 Sslk where S is the 339 The Partition Function 0 Remember that the total probability of finding an atom in a state is 1 therefore 1 25135 2 et s 1z2 e39E SV This means 2 ZS e39ESkT sum of all Boltzmann factors I Z is the partition function which does not depend on state s but depends on temperature 0 Low temperature means 2 is approximately 1 and at high temperatures 2 is much larger 339 Thermal Excitation of Atoms 0 Consider a hydrogen atom in atmosphere of the sun about 5800 K What s the probability of finding it in one of its first excited states s2 compared to it being in ground state 1 I Remember 39Ps239Ps1 e39HSZV kTe39E51V kT e39E239E1VkT o The energy difference is 102 eV and kT is 86210quot5 eVK5800 K 050 eV The ratio is about e3920394 1410quot9 Notice that the units cancel Ch 62 Average Values 0 Probability of a system in a certain microstate s when in equilibrium with a reservoir at temperature T 39Ps 1Ze39BES where B is an abbreviation for 1kT The exponential factor is called the Boltzmann factor and Z is the partition function 2 2 e39BE S 0 Knowing this we can compute the average value of the energy of the atoms in state s I 25 ESNsN 25 ESNSN 25 ES1 s I Using 39Ps 1Ze39BES this becomes E 1ZZS ESe39BES I We can also use this to compute the average value of any other variable of interest X X 25Xs13s 12zs Ese39BES o The find the total average energy of the system from the average energy of one particle uNE 339 Pa ramagnetism 0 Remember that the ideal twostate paramagnet has two possible states up energy B and down energy uB For a single dipole Z ZS em eB B e39B B 2coshBuB I Probability of finding it in up state 39P1 eB BZ eB B2coshBuB and probability of finding it in down state 39P1 e39B BZ e39B B2coshBuB o This means that E ZS E39PS uB39P1 uB39P1 uB39P1 39P1 uBeB B e39B B2cosBuBl uBtanhBuB Total energy becomes U NuBtanhBuB By differentiating 2 with respect to B and multiplying by 1Z we find that E 1ZGZ6l3 The total magnetization is M NutanhBuB 339 Rotation of Diatomic Molecules o Rotational energies are quantized For diatomic molecules allowed rotational energies are Ej jj 1 I For level j the number of degenerate states is 2 1 o This means that Zrot 2de 1e39E kT 2de 1e39j 1ekT Zrot f0w2j 1e39j 1EkT kTe when kT gtgt 8 This means that Em 1ZoZoB BeooB1Be 18 kT when kT gtgt 8 Rotational partition function thz kTZe identical atoms kT gtgt 8

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