Partition Function - Thermodynamics and Statistical Mechanics - Lecture Slides, Slides of Thermodynamics

Download Partition Function - Thermodynamics and Statistical Mechanics - Lecture Slides and more Slides Thermodynamics in PDF only on Docsity! Thermodynamics and Statistical Mechanics Partition Function Docsity.com Free Expansion of a Gas Insulation” (h) Final state 7 Docsity.com Isothermal Expansion Reversible route between same states. ∫=∆ f i T QdS đQ = đW + dU Since T is constant, dU = 0. Then, đQ = đW. dV V nRTPdVWd == V dVnR V dV T nRT T Wd T QddS ==== Docsity.com Entropy Change 2ln2ln 2 nR V VnR V dVnRS V V =     ==∆ ∫ The entropy of the gas increased. For the isothermal expansion, the entropy of the Reservoir decreased by the same amount. So for the system plus reservoir, ∆S = 0 For the free expansion, there was no reservoir. Docsity.com Statistical Approach This image cannot currently be displayed. [ ]2)!2/( ! )!2/()!2/( ! 1 !0! ! lnlnln N N NN Nw N Nw w w kwkwkSSS f i i f ifif == ==       =−=−=∆ Docsity.com Boltzmann Distribution β β ε ε βε βε βε βε ∂ ∂ −=∂ ∂ − = == = ∑ ∑ ∑ ∑ = − = − = = − − Z Z Z N U eg eg NUN eg eg NN n j j n j jjn j jj n j j j j j j j j ln 1 1 1 1 Docsity.com Maxwell-Boltzmann Distribution •Correct classical limit of quantum statistics is Maxwell-Boltzmann distribution, not Boltzmann. •What is the difference? Docsity.com Maxwell-Boltzmann Probability ∏∏ ∏∏ == == == − = − −+ = n j j N j MB n j j N j B n j jjj j FD n j jj jj BE N g w N g Nw NgN g w gN gN w jj 11 11 ! ! ! )!(! ! )!1(! )!1( wB and wMB yield the same distribution. Docsity.com Chemical Potential •dU = TdS – PdV + µdN •In this equation, µ is the chemical energy per molecule, and dN is the change in the number of molecules. Docsity.com Chemical Potential VTN F dNPdVSdTdF SdTTdSdNPdVTdSdF TSUF dNPdVTdSdU ,       ∂ ∂ = +−−= −−+−= −= +−= µ µ µ µ Docsity.com Entropy                 −=       +−=       −= == ∑ ∑∑∑ ∑∑ ∏ = j j j j j j j jj j jj j j j jj n j j N j MBMB g N NNkS NNNgNkS NgNkS N g wwkS j ln lnln !lnln ! whereln 1 Docsity.com Chemical Potential      =−=      −−+−−=       ∂ ∂ = +−−= Z NkTZNkT N NkTNZkT N F NZNkTF VT ln)ln(ln 1)1ln(ln )1ln(ln , µ µ µ Docsity.com Chemical Potential αµ µµ µ −= =      =     = kT Z Ne Z N kTZ NkT kT So, Then ln so ln Docsity.com Boltzmann Distribution ∑ ∑ ∑∑ = − − = − − = −− = −− = = == = n j j j j n j j n j j n j j jj j j j j j eg eg NN eg Ne egeNN egeN 1 1 11 βε βε βε α βεα βεα Docsity.com Ideal Gas       +−      +−= +−−=       = 1ln2ln 2 3ln )1ln(ln 2 2 2/3 2 N h mkTVNkTF NZNkTF h mkTVZ π π Docsity.com Ideal Gas nRTNkTPV V NkT V FP N h mkTVNkTF NT == =      ∂ ∂ −=       +−      +−= 1 1ln2ln 2 3ln , 2 π Docsity.com Ideal Gas kTZ N U h mVZ h mV h mkTVZ 2 31 2 3ln ln 2 32lnln 22 2/3 2 2/3 2 2/3 2 == ∂ ∂ −= −      =       =      = ββ βπ β ππ Docsity.com Heat Capacity of Solids •Each atom has 6 degrees of freedom, so based on equipartition, each atom should have an average energy of 3kT. The energy per mole would be 3RT. The heat capacity at constant volume would be the derivative of this with respect to T, or 3R. That works at high enough temperatures, but approaches zero at low temperature. Docsity.com Heat Capacity •Einstein found a solution by treating the solid as a collection of harmonic oscillators all of the same frequency. The number of oscillators was equal to three times the number of atoms, and the frequency was chosen to fit experimental data for each solid. Your class assignment is to treat the problem as Einstein did. Docsity.com Heat Capacity Heat Capacity -5 0 5 10 15 20 25 30 0 100 200 300 400 T (K) C v (J /K ) Docsity.com