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# Midterm #1 Study Guide Physics 60

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Physics 60 Thermal Physics Professor Feng Final 10232015 Midterm 1 Study Guide Week 0 Ch 11 Thermal Equilibrium 0 thermal equilibrium state objects reach after being in contact long enough relaxation time time it takes for two objects to reach thermal equilibrium contact requires some way for two objects to exchange energy spontaneously in the form of heat Exchanged Quantity Type of Equilibrium Energy Thermal Volume Mechanical Particles Diffuse thermal equilibrium moves energy from higher to lower concentrations I theoretical definition for temperature becomes temperature measure of tendency of an object to spontaneously give up energy to surroundings when two objects are in thermal contact the object with higher temperature tends to lose energy 0 absolute zero 273 C defines zero point on absolute temperature scale temperature where a lowdensity gas at constant pressure goes to zero SI unit kelvin K Ch 12 The Ideal Gas 0 Ideal gas law PV nRT P pressure V volume n number of moles of mass R universal constant R831JmolK T temperature in Kelvins I 1 mole Avogadro s number NA 6021023 o N nNA o k RNA o The ideal gas law becomes nR Nk 3 Microscopic Model ofan Ideal Gas 0 Using ideal gas law PV NkT where P pressure 0 Average pressure exerted on the piston I3 x on piston A x on moecue A mA vXAt A Have At 2L vxand Avx vxl na vX initia vX vx 2vX The equation becomes 13 mA2vX2Lvx mvx2 AL mvx2 v I Now imagine same cylinder and piston arrangement with large number of identical molecules moving randomly but not colliding with each other 0 This means that PV Nm17x2 o Treating the ideal gas law as an experimental fact kT mix2 or 12kT 12m17x2 Same holds for if 17 is replaced with 17f or 172 Average translational kinetic energy Evans mvx2 32kT 0 Evans 32kT is true for gases liquids and sometimes solids Ch 13 0 Week 1 Equipartition of Energy Equipartition theorem At temperature T the average energy of any quadratic degree of freedom is 12kT Ch 14 The equipartition theorem concerns all forms of energy that has formula as a quadratic function of a coordinate or velocity Degrees of freedom f refer to the forms of energy that constrain how energy moves such as translational motion rotational motion vibrational motion and elastic potential energy Total thermal energy UthermaI Nf12kT I Monatomic particles such as helium or argon have f 3 only translational motion counts Diatomic particles such as oxygen 02 or nitrogen N2 can also rotate but rotation does not count because it can reach an equilibrium value higher temperatures mean faster moving molecules 0 Air molecules 02 or N2 have five degrees of freedom in room temperature Vibrational modes count as degrees of freedom at higher temperatures but at room temperature they are quotfrozen out because they don t contribute Each atom in a solid has six degrees of freedom three kinetic and three potential energy because each atom can vibrate in three perpendicular directions Heat and Work 0 Thermodynamics focuses on closelyrelated concepts of temperature energy and heat C Temperature fundamentally a measure of an object s tendency to spontaneously release energy Energy complex fundamental dynamical concept of various types kinetic electrostatic gravitational chemical nuclear I Conservation of energy total amount of energy in the universe cannot change despite conversion of energy 0 Energy can be put or taken out of a system by the thermodynamic mechanisms of heat and work Heat any spontaneous flow of energy from one object to another due to difference in temperature of the objects Work any other transfer of energy into or out of a system total energy in a system Q heat that enters a system W work that enters a system Negative if leave a system First law of thermodynamics AU QW Joules SI unit of energy heat is traditionally measured in calories 1 cal 4186 J Ch 15 Compression Work 0 Recall work is W cfr Note in thermodynamics ctr refers to the point of contact W PAAx which then becomes W PAV when quasistatic I Can also be W Lin PVdV for quasistatic 3 Compression of an Ideal Gas 0 Two idealized way to compress an ideal gas isothermal compression and adiabatic compression Isothermal compression compression so slow that temperature of gas doesn t rise at all I Formula P NkTV for constant T is a concave hyperbola aka an isotherm o Work done w 1V9 PdV NkT IVin1VdV NkTan ani NkTnViVf Note Q is negative because heat leaves the gas in compression Adiabatic compression compression so fast that no heat escapes from the gas during the process of compression I Like isothermal compression assume compression is quasistatic I Work done without heat escaping internal energy AU Q W W 0 Start with equipartition theorem Utherma Nf12kT 0 Next take change in energy over infinitesimal segment dU 12kadT 0 Because work done in quasistatic compression is PdV 12kadT PdV Change equation to f2dTT dVV so that you can integrate from initial Vi and Ti to Vfand Tf It becomes f2lnTfTi lnVfVi and then Vfof2 ViTif2 Simplifying further turns it into VTf2 constant Using ideal gas law to eliminate T it becomes VYP constant y aka the adiabatic exponent is the abbreviation for f2f Ch 16 Heat Capacities 0 Heat capacity amount of heat needed to raise its temperature per degree temperature increases C C QAT I Use first law of thermodynamics AU Q W to get C QAT AU WAT More fundamentally specific heat capacity heat capacity per unit mass c Cm 0 Two likely cases that are most likely to occur 1 W 0 where there is no work being done on the system which usually means that volume is constant I Heat capacity at constant volume CV AUATV dUdTVsubscript V indicates that volumes are understood to be held fixed 0 More accurately quotenergy capacity because it is energy needed to raise object s temperature by a degree whether or not energy actually enters as heat 0 For example a gram of water has CV 1 cal C about 42 J C 2 objects often expand when heated so they do work on their surroundings and W is nega ve I C gt CV I If pressure is constant in surroundings then it is heat capacity at constant pressure Cp AU PAV ATp dUaTP P0U6Tp o For solids and liquids 0U6T can often be neglected because of how small it is o Supposing that the system stores thermal energy in only quadratic degrees of freedom then using equipartition theorem U 12kaT we get CV 6U6T 66TkaT2 ka2 if we assume f is independent of temperature Note R is the gas constant R 831 JmolK Rule of Dulong and Petit states that because solids have 6 degrees of freedom per atom heat capacity per mole for solids is 3R 0 For constant pressure using ideal gas law we get dVdTp 30T NkTP NkP Cp CVNk CVnR 3 Latent Heat 0 You can put heat into a system without increasing temperature in some situations Normally during phase transformation when heat capacity then is technically infinite so C QAT QO w You can still find amount of heat required to melt or boil the substance completely by finding the latent heat L L Qm where you assume pressure is constant 3 Enthalpy o Enthalpy H is the total energy needed to create the system out of nothing to put into the environment ie energy of a system is not just U by also work PV done my atmosphere to fill the space of the system were the system to be annihilated H U PV I Change in enthalpy during a constant change in pressure AH AU PAV I Using first law of thermodynamics Utherma QW we can get AU QPAVWother which using the equation for enthalpy becomes AH CWo 1er in constant pressure 0 For example AH for a mole of water produced is 286 jK This is the enthalpy of formation of water because it is the enthalpy required to form out of elemental constituents in their most stable states I Change in enthalpy per degree at constant pressure is similar to heat capacity at constant pressure Cp Cp 6H6Tp Week 2 Ch 2 The Second Law 0 Combinatorics math of counting ways of arranging things Ch 21 TwoState Systems 0 Imagine a penny nickel and dime if all flipped for heads or tails and all are fair each of eight possible outcomes is equally probable Microstate each of the eight different outcomes in this example Macrostate how many heads or tails there are in this example I If you know the microstate of the system HHT then you know its macrostate two heads Multiplicity number of microstates corresponding to a given macrostate in this example the multiplicity for the microstate of two heads and one tails is three because two heads and one tail can only happen in three out of the eight possible outcomes I Probability of n heads Qn Qall 0 Now imagine 100 coins instead of the three coins in the previous example Total number of microstates 2100 total number of macrostates 101 Multiplicities of the macrostates Qn 2171 130 I For n objects out of N QN n 1 339 The TwoState Paramagnet o Paramagnet material in which constituent particles act in ways that tend to align parallel to externally applied magnetic field lasts only as long as the external field is applied Ferromagnet particles interact so strongly with each other that the material can magnetize even without an externally applied magnetic field Recall dipoles each individual magnetic particle has its own magnetic dipole moment vector I In simplest case only two values are allowed for the moment vector positive and negative which is then known as a twostate paramagnet where they can either be parallel or antiparallel to the applied field I Dipole pointing one way N1 and dipole pointing another N1 total number of dipoles N N1 N1 Multiplicity for N1 is QN1 NN1N1 Note macrostate of a system is characterized at least partially by its total energy Ch 22 The Einstein Model of a Solid 0 Any quantummechanical harmonic oscillator has equalsized energy units whose potential energy function has the form 12ksx2 where kS is the spring constant Size of the energy unit hf h is Planck s constant 66310quot34 Js and f is the frequency 12nk m where k is the spring constant Remember that vibrational motions of diatomic and polyatomic gas molecules are examples of quantum oscillators but even more common are the oscillation of atoms in a solid I Einstein solid model of a solid as a collection of identical oscillators with quantized energy units 0 General formula for the multiplicity of an Einstein solid with N oscillators and q energy I qN 1 qN 1 Ch 23 Interacting Systems 0 Consider system of two Einstein solids A and B that share energy back and forth Assume the two solids are weakly coupled meaning that they exchange energy between them more slowly than the atoms within each solid exchange energy I Macrostate refers to the state of the combined system temporarily constrained by the individual energies of the solid UA and U3 I Now make assumption that all microstates of the combinations of U and q are all equally probable o Fundamental assumption of statistical mechanics in an isolated system in thermal equilibrium all accessible microstates are equally probable 0 Assume that transitions are random in how they have no discernable pattern 0 Second law of thermodynamics spontaneous flow of energy stops when a system is at or near its most likely macrostate ie the macrostate with the greatest multiplicity Ch 24 Large System 0 Multiplicity is much sharper with more oscillators in Einstein solids eg 1020 oscillators 3 Very Large Numbers 0 Three kinds of numbers occurring commonly in statistical mechanics Small numbers easy to manipulate small numbers Large numbers frequently made by exponentiating small numbers for example Avogadro s number 1023 I You can add small numbers to large numbers without changing the small numbers for example 1023 23 1023 Very large numbers made by exponentiating large numbers I You can multiply very large numbers with large numbers without changing them for example 1010A23 X 1023 1010quot2323 1010A23 339 Stirling s Approximations o N z NNexZn N Accurate in limit where Ngtgt1 339 Multiplicity of Large Einstein Solid 0 Consider and Einstein solid with a large number of oscillators and energy units where qgtgtN the quothightemperature limit q N 1 qN 1qN 39 Q N39q q qN 1 qlN I qN an ln qW nqN lnq lnN z qNlnqN qN qlnq q NlnN N q Nlnq N qlnq NlnN nqN In q1 Nq lnq n1 Nq lnq Nq an z NlnqN N NZq 39 0Nq e 39 quot eN EQNN 339 Sharpness of the Multiplicity Function 0 For a solid with N oscillators and total number of energy units q assuming that qgtgtN Q ecuNN quNN eN2N quBN 39 Qmax eN2Nq22N I Now set qA q2 x and qB q2 x where x is any number much smaller than q then the equation becomes 0 eN2Nq22 x2N 0 Now taking logarithms and manipulating it lnq22 x2 Nnq221 2xq2 NIno22 n1 2xqzl NInq22 2xqZI Putting this back into the equation for 0 we get the Gaussian Q eN23939e39 q2A2e39WW 2 Qmax e39N2XqA2 which means it has a peak at x0 and a sharp falloff on both sides 0 Thermodynamic limit limit where system becomes infinitely large so that measurable fluctuations away from the likeliest macrostate never occur Ch 25 The Ideal Gas 0 Previous chapter on how only extremely few of the macrostates of a large interacting system can probably happen also applies to any pair of interacting objects not just Einstein solids Also can happen to ideal gases even though they re more complicated 339 Multiplicity ofa Monatomic Ideal Gas 0 Consider a single gas atom with kinetic energy U in a container of volume V Multiplicity should be proportional to momentum space the space where each point corresponds to momentum vector for the particle so 01 ocV Vp I V is the volume of ordinary space or position space Vp is volume of momentum space and 01 indicates the multiplicity of a gas of one molecule only 0 Constrained by U 12mvX2 vy2 v22 12mpx2 py2 p22 which can be written as pX2 py2 p22 2mU o A wavefunction describes the state of a system in quantum mechanics Heisenberg uncertainty principle AxApX z h I Axis the spread of x ApX is the spread of pX and h is Planck s constant 662610quot 34 15 413610quot 15evs o This means that the less spread out a wavefunction is in position space the more spread out it is in momentum space and vice versa Set L length of position space Lp length of momentum space Ax as length of peak of position space curve and ApX as length of peak of momentum space curve I Number of distinct position states is LAx and number of distinct momentum states is LAXLpApx LLph oi vvph3 0 As more molecules are added QN 1NVNh3N x area of momentum hypersphere area Z dZd21rd1 0 Q UVN fNVNU3N2 fN is a complicated function of N 3 Interacting Ideal Gases 0 Now consider two ideal gases Total multiplicity becomes Qtota fN2VAVB3939UAUB3N2 Width of peak UtotaIA3N 2 This system can exchange both energy and volume I Width of peak VtotaIxN Ch 26 Entropy Note how we ve seen that particles and energy tend to rearrange themselves to be at where multiplicity is at or very near its maximum value Second law of thermodynamics multiplicity tends to increase I Entropy S kan 0 Entropy defined as logarithm of number of ways a system can arrange things in the system times Boltzmann s constant in units JK S klneqNN NklnqN1 0 Generally more particles means more energy and greater multiplicity and entropy Can also increase entropy by letting it expand into larger space or breaking large molecules or mixing together substances all of which increase possibly microstates Stota kantota klnQAQB kanA kanB SA SB Another component of the second law of thermodynamics entropy tends to increase 3 Entropy of an Ideal Gas 0 Equation for entropy of an ideal monatomic gas aka SackurTetrode equation 5 NknVN4Tth3Nh232 52 If volume changes while U and N are fixed AS NknVfVi I This equation works with quasistatic isothermal expansion considered in Section 15 0 Adding heat always increases entropy 0 Free expansion increasing entropy by puncturing a partition to allow the air within the partition to occuy all available space No work is done during free expansion 3 Entropy of Mixing 0 Consider two gases A and B separated by a partition both have equal energy volume and number of particles Removing the partition increases the entropy and the entropy becomes ASA ASB NklnVfVi Nkan Total entropy becomes AStota ASA ASE 2Nkln2 I This is the entropy of mixing I The SackurTetrode equation becomes S NklnV4nmU3Nh232 32 for distinguishable molecules 0 Gibbs paradox says that if this formula were true then one can decrease entropy of gas in a container by half by inserting a partition to divide the total volume of the container by half The best resolution is to assume that all atoms are indistinguishable 3 Reversible and Irreversible Processes o Irreversible used to describe processes that create entropy Reversible used to describe processes that don t change entropy Slow compression of gas does not cause entropy to increase 0 Think of flow of heat from hot object to cold object Heat flow is reversible but quotreversible heat flow refers to very slow heat flow between two objects of nearly the same temperature Larger entropy increases are highly irreversible and are much more common eg wood burning sunlight warming the earth and even metabolism of nutrients in a body 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 oStotaloqA 0 or oStotaloUA 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 oSoUNV 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 NklnqN 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 0 v 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 o 2N1 N1 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 astotayavA 0 o Remembering that due to the fixed volumes dVA dVB 0 oStotaIVA asAavA asBavA asAavA 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 T39fTi 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 oSoUNVdU 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 o uchemistryE ToSonUV where n NNA 0 Therefore for diffusive equilibrium the chemical potentials are the same Physics 60 10172015 Week4 Ch 4 Engines and Refrigerators Ch 41 Heat Engines 0 Heat engine any device that absorbs heat and converts part of that energy into work For example steam turbines However heat engines can only convert part of energy absorbed as heat into work by the heat engine I Heat reservoir where heat is absorbed by the engine temperature Th I Cold reservoir where heat is dumped temperature T 0 Take Qh as heat absorbed from the heat reservoir in a certain period of time Qcas heat expelled to the cold reservoir and W as net work done by the engine Efficiency e E benefitcost WQh To get max efficiency use Qh QC W to replace the W in the cost for efficiency so that it becomes 9 Qh QCQh 1 QCah The second law reveals that QCTc 2 QhTh or QCQh 2 TcTh and the equation becomes e S 1 TcTh To make an engine less efficient than 1 TcTh produce additional entropy that must be disposed by dumping extra heat into the cold reservoir so that there is less energy to convert to work 3 The Carnot Cycle 0 Every engine has a quotworking substance the material that absorbs heat expels waste and does work oftentimes a gas In the process of the gas absorbing heat Q from the hot reservoir the entropy of the reservoir decreases QhTh and entropy of the gas increase QhTgas Since Tgas cannot equal Th imagine Tgas slightly less than Th the gas should be kept at Tgas by letting it expand as it absorbs heat while the gas expands isothermally I We need the heat to compress isothermally o Carnot cycle the cycle of the four steps where 1 isothermal expansion at Th while heat is absorbed 2 adiabatic expansion to Tc 3 isothermal compression at Tc while expelling heat and 4 adiabatic compression back to Th Although maximally efficient minimally probable because of how slow heat moves in the isothermal steps Ch 42 Refrigerators o Refrigerators considered heat engines in reverse Instead of quotefficiencyquot we find coefficient of performance COP benefitcost QcW 39 This time COP QCm 0 1QhQc 1 o This means that QhTh 2 CitTc or Qh QC 2 ThTc leads to COP S 1ThTc 1 TcTh Tc 0 You can get an ideal refrigerator by reversing steps for a maximum efficient heat engine ie reversing the Carnot cycle Ch 5 Free Energy and Chemical Thermodynamics Ch 51 Free Energy as Available Work 0 Enthalpy energy of a system plus energy required to make room for it in environment of constant pressure P H E U PV Have to relate enthalpy to heat I Can do this with Helmholtz free energy F E U TS o F is the energy that must be provided as work to create the system out of nothing therefore F is the available quotfreequot energy Have to relate enthalpy to work I Work done by surroundings to create system in constant pressure P and temperature T Gibbs free energy G E U TS PV Thermodynamic potentials four functions U H F and G I Change in F AF AU TAS Q W TAS where Q is heat added and W is work done on the system 0 Generally AF S W at constant T I Change in G AF AU TAS PAV Q W TAS PAV o W PAV Wother 0 AG S Wother at constant T and P AG AH TAS 339 Electrolysis Fuel Cells and Batteries 0 For electrolysis consider the chemical reaction H20 9 H2 1202 This is the electrolysis of liquid water into hydrogen and oxygen gas 0 Fuel cell that produces electrical work by efficiently expelling waste heat to get rid of excess entropy in the gases Battery similar to a fuel cell but with a fixed internal supply of fuel that is usually not gaseous 339 Thermodynamic Identities o Formulas used when given enthalpy or free energy resemble the thermodynamic identity dU TdS PdV udN H U PV shows that dH dU PdV VdP which becomes dH TdS VdP udN Similarly for F dF SdT PdV udN I S oFoTVN 0 Changing what s fixed P oFoVTN and u oFoN I For G dG SdT VdP udN 5 39dGdTP N V MS3mm l1 dGthy For systems of multiple types of particles udN becomes 2 uidNi I This means that pl oGoN1TpN2 and uz oGoN2TpN2 Physics 60 Thermal Physics Professor Feng Final 10232015 Midterm 1 Study Guide Week 0 Ch 11 Thermal Equilibrium 0 thermal equilibrium state objects reach after being in contact long enough relaxation time time it takes for two objects to reach thermal equilibrium contact requires some way for two objects to exchange energy spontaneously in the form of heat Exchanged Quantity Type of Equilibrium Energy Thermal Volume Mechanical Particles Diffuse thermal equilibrium moves energy from higher to lower concentrations I theoretical definition for temperature becomes temperature measure of tendency of an object to spontaneously give up energy to surroundings when two objects are in thermal contact the object with higher temperature tends to lose energy 0 absolute zero 273 C 0 K defines zero point on absolute temperature scale temperature where a lowdensity gas at constant pressure goes to zero SI unit kelvin K Ch 12 The Ideal Gas 0 Ideal gas law PV nRT P pressure V volume n number of moles of mass R universal constant R831JmolK T temperature in Kelvins I 1 mole Avogadro s number NA 6021023 o N nNA o k RNA o The ideal gas law becomes nR Nk 3 Microscopic Model ofan Ideal Gas 0 Using ideal gas law PV NkT where P pressure 0 Average pressure exerted on the piston lJ 177x 0 piston A 4 5 0 molecule A mA vx At A Have At 2L vxand Avx Vxfina vX initial le vx 2vx The equation becomes lJ mA2vx 2Lvx mvx2 AL mvx2 v I Now imagine same cylinder and piston arrangement with large number of identical molecules moving randomly but not colliding with each other 0 This means that PV NmTJX2 o Treating the ideal gas law as an experimental fact kT m xz or 12kT 12m17x2 Same holds for if 17 is replaced with 17f or 172 Average translational kinetic energy Evans mvx2 32kT 0 Evans 32kT is true for gases liquids and sometimes solids Week 1 Ch 13 Equipartition of Energy 0 Equipartition theorem At temperature T the average energy of any quadratic degree of freedom is 12kT The equipartition theorem concerns all forms of energy that has formula as a quadratic function of a coordinate or velocity Degrees of freedom f refer to the forms of energy that constrain how energy moves such as translational motion rotational motion vibrational motion and elastic potential energy Total thermal energy Uthermal Nf12kT I Monatomic particles such as helium or argon have f 3 only translational motion counts I Diatomic particles such as oxygen 02 or nitrogen N2 can also rotate but rotation does not count because it can reach an equilibrium value higher temperatures mean faster moving molecules 0 Air molecules 02 or N2 have five degrees of freedom in room temperature Vibrational modes count as degrees of freedom at higher temperatures but at room temperature they are quotfrozen out because they don t contribute Ch 14 Heat and Work 0 Thermodynamics focuses on closelyrelated concepts of temperature energy and heat Temperature fundamentally a measure of an object s tendency to spontaneously release energy Energy complex fundamental dynamical concept of various types kinetic electrostatic gravitational chemical nuclear I Conservation of energy total amount of energy in the universe cannot change despite conversion of energy 0 Heat any spontaneous flow of energy from one object to another due to difference in temperature of the objects 0 Work any other transfer of energy into or out of a system 0 U total energy in a system Q heat that enters a system W work that enters a system First law of thermodynamics AU QW Joules SI unit of energy heat is traditionally measured in calories 1 cal 4186 J Ch 15 Compression Work 0 Recall work is W cfr W PAAx which then becomes W PAV when quasistatic I Can also be W 1 in PVdV when quasistatic 3 Compression of an Ideal Gas 0 Two idealized way to compress an ideal gas isothermal compression and adiabatic compression Isothermal compression compression so slow that temperature of gas doesn t rise at all I Formula P NkTV for constant T is a concave hyperbola aka an isotherm o Work done w v9quot PdV NkT IVin1VdV NkTInvf ani NkTnViVf Note Q is negative because heat leaves the gas in compression Adiabatic compression compression so fast that no heat escapes from the gas during the process of compression I Like isothermal compression assume compression is quasistatic I Work done without heat escaping internal energy AU Q W W 0 Start with equipartition theorem Utherma Nf12kT 0 Next take change in energy over infinitesimal segment dU 12kadT 0 Because work done in quasistatic compression is PdV 12kadT PdV Change equation to f2dTT dVV so that you can integrate from initial Vi and Ti to Vfand Tf It becomes f2lnTfTi lnVfVi and then Vfof2 ViTif2 Simplifying further turns it into VT 2 constant Using ideal gas law to eliminate T it becomes VYP constant y aka the adiabatic exponent is the abbreviation for f2f ie y f2f Ch 16 Heat Capacities 0 Heat capacity C amount of heat needed to raise temperature per degree temperature increases C QAT I Use first law of thermodynamics AU Q W to get C QAT AU WAT More fundamentally specific heat capacity heat capacity per unit mass c Cm 0 Two likely cases that are most likely to occur 1 W 0 where there is no work being done on the system which usually means that volume is constant I Heat capacity at constant volume CV AUATV 0U0Tv o For example a gram of water has CV 1 cal C about 42 J C 2 objects often expand when heated so they do work on their surroundings and W is negative I C gt CV I If pressure is constant in surroundings then it is heat capacity at constant pressure Cp AU PAV ATp 0U0Tp P0U0Tp o For solids and liquids 0U6T can often be neglected because of how small it is o Supposing that the system stores thermal energy in only quadratic degrees of freedom then using equipartition theorem U 12kaT we get CV 0U0T 00TkaT2 kaZ if we assume f is independent of temperature Note R is the gas constant R 831 JmolK Rule of Dulong and Petit states that because solids have 6 degrees of freedom per atom heat capacity per mole for solids is 3R 0 For constant pressure using ideal gas law we get 0V0Tp 00T NkTP NkP Cp CVNk CVnR 3 Latent Heat 0 You can put heat into a system without increasing temperature in some situations Normally during phase transformation when heat capacity then is technically infinite so C QAT QO 0 You can still find amount of heat required to melt or boil the substance completely by finding the latent heat L L Qm where you assume pressure is constant 3 Enthalpy o Enthalpy H is the total energy needed to create the system out of nothing to put into the environment H U PV I Change in enthalpy during a constant change in pressure AH AU PAV I Using first law of thermodynamics Utherma QW we can get AU QPAVWother which using the equation for enthalpy becomes AH QWother in constant pressure 0 For example AH for a mole of water produced is 286 jK I Change in enthalpy per degree at constant pressure is similar to heat capacity at constant pressure Cp Cp OH0TH Week2 Ch 2 The Second Law Ch 21 TwoState Systems 0 Imagine three coins Microstate each of the eight different outcomes in this example Macrostate how many heads or tails there are in this example I If you know the microstate of the system HHT then you know its macrostate two heads Multiplicity number of microstates corresponding to a given macrostate in this example the multiplicity for the microstate of two heads and one tails is three because two heads and one tail can only happen in three out of the eight possible outcomes I Probability of n heads Qn Qall 0 Now imagine 100 coins instead of the three coins in the previous example Total number of microstates 2100 total number of macrostates 101 Multiplicities of the macrostates Qn 2171 130 I For n objects out of N QN n 1 339 The TwoState Paramagnet o Paramagnet material in which constituent particles act in ways that tend to align parallel to externally applied magnetic field lasts only as long as the external field is applied Recall dipoles each individual magnetic particle has its own magnetic dipole moment vector I In simplest case only two values are allowed for the moment vector positive and negative which is then known as a twostate paramagnet where they can either be parallel or antiparallel to the applied field I Dipole pointing one way N1 and dipole pointing another N1 total number of dipoles N N1 N1 Multiplicity for N1 is QNt N N1N1 Ch 22 The Einstein Model of a Solid 0 Any quantummechanical harmonic oscillator has equalsized energy units whose potential energy function has the form 12ksx2 where kS is the spring constant Size of the energy unit hf h is Planck s constant 66310quot34 Js and f is the frequency 12nkm where k is the spring constant Remember that vibrational motions of diatomic and polyatomic gas molecules are examples of quantum oscillators but even more common are the oscillation of atoms in a solid I Einstein solid model of a solid as a collection of identical oscillators with quantized energy units 0 General formula for the multiplicity of an Einstein solid with N oscillators and q energy qN 1 qN 1 unltS lt q gt qN1 Ch 23 Interacting Systems 0 Consider system of two Einstein solids A and B that share energy back and forth Assume the two solids are weakly coupled meaning that they exchange energy between them more slowly than the atoms within each solid exchange energy I Macrostate refers to the state of the combined system temporarily constrained by the individual energies of the solids UA and U3 I Now make assumption that all microstates of the combinations of U and q are all equally probable o Fundamental assumption of statistical mechanics in an isolated system in thermal equilibrium all accessible microstates are equally probable 0 Second law of thermodynamics spontaneous flow of energy stops when a system is at or near its most likely macrostate ie the macrostate with the greatest multiplicity Ch 24 Large System 0 Multiplicity is much sharper with more oscillators in Einstein solids eg 1020 oscillators 3 Very Large Numbers 0 Three kinds of numbers occurring commonly in statistical mechanics Small numbers easy to manipulate small numbers Large numbers frequently made by exponentiating small numbers for example Avogadro s number 1023 I You can add small numbers to large numbers without changing the small numbers for example 1023 23 1023 Very large numbers made by exponentiating large numbers I You can multiply very large numbers with large numbers without changing them for example 1010A23 X 1023 1010quot2323 1010A23 0 v Stirling s Approximations o N z NNexZnN Accurate in limit where Ngtgt1 339 Multiplicity of Large Einstein Solid 0 Consider and Einstein solid with a large number of oscillators and energy units where qgtgtN the quothightemperature limit q N 1 qN 1 qN 39 Q N39q q qN 1 qlN I qN an n qW In q N lnq lnN z qNlnqN qN qlnq q NlnN N 0 ma q Nlnq N qlnq NlnN nqN In q1 Nq lnq n1 Nq lnq Nq an z NlnqN N NZq 39 0Nq z e 39 quot e eaINN 339 Sharpness of the Multiplicity Function 0 For a solid with N oscillators and total number of energy units q assuming that qgtgtN Q quNN eelsNN eN2N lama Qmax elequot39ltI22N I Now set qA qZ x and qB qZ x where x is any number much smaller than q then the equation becomes 0 eN2Nq22 x2N 0 Now taking logarithms and manipulating it lnq22 x2 Nlnq221 2xq2 NnltI22 n1 2xqlzi z NnltI22 2xqlzi Putting this back into the equation for 0 we get the Gaussian Q eNZNe39quot I39lzy ze39 Rx 139quot2 Qmax emu W which means it has a peak at x0 and a sharp falloff on both sides 0 Thermodynamic limit limit where system becomes infinitely large so that measurable fluctuations away from the likeliest macrostate never occur Ch 25 The Ideal Gas 0 v Multiplicity ofa Monatomic Ideal Gas 0 Consider a single gas atom with kinetic energy U in a container of volume V Multiplicity should be proportional to momentum space the space where each point corresponds to momentum vector for the particle so 1 ocV Vp I V is the volume of ordinary space or position space Vp is volume of momentum space and 01 indicates the multiplicity of a gas of one molecule only 0 Constrained by U 12mvx2 vy2 v22 12mp2 py2 p22 which can be written as px2 py2 p2 ZmU o A wavefunction describes the state of a system in quantum mechanics Heisenberg uncertainty principle AxApx z h I Axis the spread of x ApX is the spread of pX and h is Planck s constant h 662610quot34 Js 413610quot15 evs o This means that the less spread out a wavefunction is in position space the more spread out it is in momentum space and vice versa 0 Q UVN fNVNU3N2 fN is a complicated function of N 3 Interacting Ideal Gases 0 Now consider two ideal gases Total multiplicity becomes Qtota fN2VAVNUAUBf N2 Width of peak Utota3N 2 This system can exchange both energy and volume I Width of peak VtotallxW Ch 26 Entropy 0 Second law of thermodynamics multiplicity tends to increase Entropy S kan I Entropy defined as logarithm of number of ways a system can arrange things in the system times Boltzmann s constant in units JK o S klneqNN NklnqN1 Stota kantota klnQAQB kanA kanB SA SB Another component of the second law of thermodynamics entropy tends to increase 3 Entropy of an Ideal Gas 0 Equation for entropy of an ideal monatomic gas aka SackurTetrode equation 5 NklnVN4nmU3Nh232 52 If volume changes while U and N are fixed AS NknVfVi Free expansion increasing entropy by puncturing a partition to allow the air within the partition to occuy all available space I No work is done during free expansion 3 Entropy of Mixing 0 Consider two gases A and B separated by a partition both have equal energy volume and number of particles Removing the partition increases the entropy and the entropy becomes ASA A5 NklnVfVi Nkan Total entropy becomes AStota ASA ASE 2Nkln2 I This is the entropy of mixing I The SackurTetrode equation becomes S NklnV4nmU3Nh232 32 for distinguishable molecules 0 Gibbs paradox says that if this formula were true then one can decrease entropy of gas in a container by half by inserting a partition to divide the total volume of the container by half 3 Reversible and Irreversible Processes o Irreversible used to describe processes that create entropy Reversible used to describe processes that don t change entropy 0 Think of flow of heat from hot object to cold object Heat flow is reversible but quotreversible heat flow refers to very slow heat flow between two objects of nearly the same temperature Larger entropy increases are highly irreversible and are much more common eg wood burning sunlight warming the earth and even metabolism of nutrients in a body Week3 Ch 3 Interactions and Implication 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 qA is at equilibrium dStotallqu 0 or dStotaIIOUA 0 at equilibrium I Slope of Stota is slope of SA plus slope of SB so OSAOUA OSBOUB 0 0 Because dUA dUB then OSAOUA OSBOUB at equilibrium 0 Thanks to Boltzmann s constant oSoU is in units of 1K This proposes that T OS6UP which makes the definition of temperature 1T OS6UNv 339 RealWorld Examples 0 Remember the equation S NknqN 1 Nkan NklneN Nk This means temperature is T OS6U391 NkU391 or U NkT Ch 32 Entropy and Heat 339 Predicting Heat Capacities 0 Remember CV GU6TNVfor heat capacity at constant volume For an Einstein solid where qgtgtN the heat capacity is CV 66TNkT Nk For a monatomic ideal gas CV66T32NkT 32Nk I 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 d5 CVdTT AS 51 Si Inn CVTdT 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 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 d5 QT Ch 33 Paramagnetism 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 2pB 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 pN1 N1 UB o mm N11 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 Negative temperatures can only occur for systems with limited energy which allows multiplicity to decrease as energy approaches its maximum 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 OS6UNB 6N16U dSdNt 12pBOS6N1 Differentiating 12pB OSdNt gives 1T k2pBlnNUpBNUpB U can be solved for U NuBtanhpBkT tanh is the hyperbolic tangent function tanh sinhxcoshx where sinhx 12ex ex and coshx 12equot e39x To find heat capacity differentiate U NuBtanhpBkT with respect to T to get CB BU6T B Nk uBkT2cosh2pBkT I For an electronic twostate paramagnet the value of u is the Bohr magneton 13 eh4rtme 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 Ch 34 Mechanical Equilibrium and Pressure 0 Now think of systems whose volumes can change as they interact 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 dStotalldUA 0 and dStotaIIOVA 0 o Remembering that due to the fixed volumes dVA dVB 0 dStotaIIVA OSAOVA OSBOVA OSAOVA OSBOVB which means that GSAOVA OSBOVB 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 TOS6VUN Taking the log for the equation of multiplicity of a monatomic ideal gas 0 fNVNU S Nkan 32Nkan klnfN I Then pressure becomes T66VNkan 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 Its 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 OS6UvdU OS6VUdV 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 IT39fTJ CPTdT 3Nl2 we get 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 dStotalldUAMANA 0 and dStotalldNAUAVA 0 I dStotaIIOVA 0 as well I We can conclude that GSA6M OSBONB at equilibrium which means T OSAONA T OSBONB at equilibrium 0 Chemical potential p E TOS6NUv so uA 13 at equilibrium To include this in the thermodynamic identity with changes in N dS OS6UNVdU OS6VNUdV OS6NUVdN 1TdU PTdV pTdN I This becomes dU TdS PdV pdN 0 MN is referred to as chemical work 39 I11 5 39TOS6N1UNN2 and P2 5 39TOS6N2UVN1 o uchemistryE TOS6nuv where n NNA 0 Therefore for diffusive equilibrium the chemical potentials are the same Week4 Ch 4 Engines and Refrigerators Ch 41 Heat Engines 0 Heat engine any device that absorbs heat and converts part of that energy into work However heat engines can only convert part of energy absorbed as heat into work by the heat engine I Heat reservoir where heat is absorbed by the engine temperature Th I Cold reservoir where heat is dumped temperature T 0 Take Qh as heat absorbed from the heat reservoir in a certain period of time Qcas heat expelled to the cold reservoir and W as net work done by the engine Efficiency e E benefitcost WlQh To get max efficiency use Qh 02 W to replace the W in the cost for efficiency so that it becomes 9 Qh QcQh 1 QCqr The second law reveals that QCTc 2 QhTh or QCQh 2 TcTh and the equation becomes e S 1 TcTh To make an engine less efficient than 1 TcTh produce additional entropy that must be disposed by dumping extra heat into the cold reservoir so that there is less energy to convert to work 3 The Carnot Cycle 0 Every engine has a quotworking substance the material that absorbs heat expels waste and does work oftentimes a gas In the process of the gas absorbing heat Qh from the hot reservoir the entropy of the reservoir decreases QhTh and entropy of the gas increase QhTgas Since Tgas cannot equal Th imagine Tgas slightly less than Th the gas should be kept at Tgas by letting it expand as it absorbs heat while the gas expands isothermally I We need the heat to compress isothermally o Carnot cycle the cycle of the four steps where 1 isothermal expansion at Th while heat is absorbed 2 adiabatic expansion to Tc 3 isothermal compression at Tc while expelling heat and 4 adiabatic compression back to Th Although maximally efficient minimally probable because of how slow heat moves in the isothermal steps Ch 42 Refrigerators o Refrigerators considered heat engines in reverse Instead of quotefficiencyquot we find coefficient of performance COP benefitcost QcW 39 This time COP QcQh 0 1QhQc 1 o This means that QhTh 2 QcTc or Qh Qc 2 ThTc leads to COP S 1ThTc 1 TcTh Tc 0 You can get an ideal refrigerator by reversing steps for a maximum efficient heat engine ie reversing the Carnot cycle Ch 5 Free Energy and Chemical Thermodynamics Ch 51 Free Energy as Available Work 0 Enthalpy energy of a system plus energy required to make room for it in environment of constant pressure P H E U PV Have to relate enthalpy to heat I Can do this with Helmholtz free energy F E U TS o F is the energy that must be provided as work to create the system out of nothing therefore F is the available quotfreequot energy Have to relate enthalpy to work I Work done by surroundings to create system in constant pressure P and temperature T Gibbs free energy G E U TS PV Thermodynamic potentials four functions U H F and G I Change in F AF AU TAS Q W TAS where Q is heat added and W is work done on the system 0 Generally AF S W at constant T I Change in G AF AU TAS PAV QW TAS PAV o W PAV Wother 0 AG S Wother at constant T and P AG AH TAS 339 Electrolysis Fuel Cells and Batteries 0 For electrolysis consider the chemical reaction H20 9 H2 1202 This is the electrolysis of liquid water into hydrogen and oxygen gas 0 Fuel cell that produces electrical work by efficiently expelling waste heat to get rid of excess entropy in the gases Battery similar to a fuel cell but with a fixed internal supply of fuel that is usually not gaseous 339 Thermodynamic Identities o Formulas used when given enthalpy or free energy resemble the thermodynamic identity dU TdS PdV pdN H U PV shows that dH dU PdV VdP which becomes dH TdS VdP pdN Similarly for F dF SdT PdV pdN 39 S quotOF6TVN 0 Changing what s fixed P 6F6VTN and p OF6N I For G dG SdT VdP pdN 5 quotOGdTP N V OGOPTN l1 OGONTP For systems of multiple types of particles udN becomes 2 uidNi I This means that 3941 dG6N1T p N2 and v2 dG6N2T p N2

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