5.60 Thermodynamics & Kinetics Spring 2008

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1 MIT OpenCourseWare Thermodynamics & Kinetics Spring 2008 For information about citing these materials or our Terms of Use, visit:

2 5.60 Spring 2008 Lecture #26-27 page 1 STTISTICL THERMODYMICS Calculation of macroscopic thermodynamic results Entropically driven examples: Free expansion of a gas vacuum V 2 gas gas Lattice model for ideal gas translation: Molecular volume v, Total volume V ll molecular positions have equal energy ε trans = 0 ll system microstates have equal energy E trans = 0 Calculate S = k ln Ω in each state Molecular degeneracy g = V/v V 2 System degeneracy Ω = g /! = (V/v) /! For expansion from volume to V 2, V v! 2 Δ S = kln Ω2 kln Ω 1 = kln = kln Ω V v! V Δ S= kln 2 V = nrln 2 Ω ( 2 ) 1 ( ) 1 V2 Should look familiar! nd Δ G = ΔH TΔ S = nrtln Entropy change is positive, free energy change is negative, as we expect.

3 5.60 Spring 2008 Lecture #26-27 page 2 Ideal gas mixture V V = + V = V + V ssume same initial (p,t) for & same (p,t) for mixture ssume equal molecular volumes & lattice cell sizes. Then initially (V v) S = kln Ω + kln Ω = kln Ω Ω = kln 1 (V v)!! fter mixing: Count how many ways to distribute molecules of and molecules of among the (V/v) lattice sites s before, the number of ways to distribute molecules among (V/v) sites is (V/v). To correct for indistinguishability, divide by!! So the final state entropy is ( Vv) S 2 = kln Ω = kln!! ( ) ( ) ( v ) ( Vv ) kln = kln ( ) ( ) V ( + ) Δ S = S = kln V V 2 S 1 = kln Vv V v!!!! V v V v V V Since the initial pressures are the same, the initial volumes must be in the ratio of the number of molecules, i.e. V /V = / = X and V /V = X, so Δ = ( ) S kln V V = klnx klnx = k X lnx + X lnx (> 0) V V With a simple microscopic model we can derive the macroscopic entropy change!

4 5.60 Spring 2008 Lecture #26-27 page 3 Ideal liquid mixture Lattice model is different from gas because all the cells are occupied. Then in the pure liquid there is no disorder at all! S = kln Ω = kln1 = 0 + S = kln Ω = kln1 = 0 Mixture: molecules for sites. First molecule has choices, second ( 1), etc. # ways to put the molecules into sites =! Correct for overcounting by dividing by!! ΔS mix = S mix (S + S ) = S mix = kln Ω mix = kln!!! Stirling s approximation ln! ln ΔS mix = kln k ( kln k + kln k ) = ( + )kln kln kln = kln + kln = k ( X lnx + X lnx ) Real liquid has additional states - positional disorder, molecular rotation, etc. but these occur in both the pure and mixed liquids, so ΔS mix is dominated by the disorder in molecular positions that the lattice model describes reasonably well. *********************************************************************** Combinatorics: Simple example Mix 2 molecules + 3 molecules How many distinct igurations Ω? Ω = 10 = 5!/2!3!

5 5.60 Spring 2008 Lecture #26-27 page 4 Energy & entropy changes We saw one example earlier, with 4-segment polymers. Molecular state: Energy ε: 0 ε int ε int ε int Degeneracy g: 1 3 We ve redefined the zero of energy as the ground state energy. Configurational molecular partition function is ε kt 0kT ε int kt ε int kt ε int kt q = e i, = e + e + e + e states i ε int kt ε kt = ge ε i kt 0kT = e + 3e = 1 + 3e int energy levels ε i ε i For a solution of noninteracting polymer molecules, Q = q = ( 1 + 3e ε int kt ) We can determine the thermodynamic properties: ( + 3e ) = ktlnq = ktln ( 1 + 3e ε int kt ) = ktln 1 ln 1 + 3e ε U = = int = β V, β 1 + 3e V, lnq ( ) 3 e Energy scales with : molecules are not interacting with each other so total energy is just a sum of individual molecule energies.

6 5.60 Spring 2008 Lecture #26-27 page 5 verage energy per molecule is U 3ε int e ε = = 1 3e + ε i e βε i + ε e ut we also know ε = ε i P i = i = 0 3 int i q 1+ 3e - same result S = + U T T = klnq 1 lnq T β = kln ( 1 + 3e V, ) 3ε int e + T 1+3e lso scales with sum over individual molecule entropy contributions verage molecular igurational entropy is s = kln ( 1 + 3e 1 3 e ) + T 1 + 3e ε int In high-t (low-β) limit, it s kln(4) as expected. In low-t limit, it s kln(1) = 0. μ = = kt lnq = ktln 1 + 3e T,V ( ) T,V Chemical potential is just per molecule, and scales with so it s just /. U 1 U 3ε int e C V = = 2 = 2 βε kt β V, kt β + int T V, 1 3e βε 3ε + int βε ( int βε int βε (1 3e ) ε ( 3 int inte ) e ε int e ) int = kt 2 (1 3e βε + int ) 2 Scales with, so we can think of a igurational heat capacity per molecule. Complicated function, but its limits are understandable:

7 5.60 Spring 2008 Lecture #26-27 page 6 C V 0 as T 0 ε int Low-T t low T, all molecules are in the lowest state. If kt increases infinitesimally, all the molecules are still in the lowest state! So the igurational 0 energy U doesn t change! limit kt C V 0 as T t high T, the molecules are equally distributed among all the states. If kt increases, they are still equally distributed among all the states! So 0 U doesn t change. kt High-T limit ε int The low-t limit C V 0 is common to almost every degree of freedom since ultimately a temperature is reached at which only the lowest level is occupied. The high-t limit C V 0 is characteristic of systems or degrees of freedom with a finite number of states, i.e. a maximum possible energy. In that case, ultimately a temperature is reached at which the equilibrium distribution is ~ equal probability of all the levels being occupied. This is the case for molecular igurations as discussed here and for spin states of nuclei and electrons.