Statistical thermodynamics mechanics) 1/15 Macroskopic quantities are a consequence of averaged behavior of many particles
[tchem/simplyn.sh] 2/15 Pressure of ideal gas from kinetic theory I Molecule = mass point L N molecules of mass m i in a cube of edge length L Velocity of molecule i = v i = v i,x, v i,y, v i,z ) After reflection from the wall: v i,x v i,x Next time it hits the wall after τ = 2L/v i,x Force = change in momentum per unit time Momentum P = m v Change of momentum = P x = 2m i v i,x Averaged force by impacts of one molecule: y x F i,x = P x τ = 2m iv i,x = m iv 2 i,x 2L/v i,x L Pressure = force of all N molecules divided by the area Ni=1 F i,x p = L 2 = Kinetic energy of one molecule: Ni=1 m i v 2 i,x L 3 1 2 m i v i 2 1 2 m iv 2 i = 1 2 m iv 2 i,x + v2 i,y + v2 i,z )
[tchem/simplyn.sh] 3/15 Pressure of ideal gas from kinetic theory II Kinetic energy = internal energy monoatomic gas) N N E kin = 1 2 i=1 m i v 2 i = 3 2 i=1 m i v 2 i,x Or p = Ni=1 m i v 2 i,x L 3 = 2 E kin 3 V pv = 2 3 E kin! = n Summary: Temperature is a measure of the kinetic energy 0th Law) Pressure = averaged impacts of molecules We needed the classical mechanics Once more: n = N N A, k B = R N A U E kin = 3n 2 = 3N 2 k BT, C V,m = 3 2 R
Boltzmann constant 4/15 pv = n = Nk B T N = nn A k B = R N A = 1.38065 10 23 J K 1 Ludwig Eduard Boltzmann 1844 1906) credit: scienceworld.wolfram.com/biography/boltzmann.html
Microstate, macrostate, ensemble 5/15 microstate state, configuration) = instantaneous quantum description: wave function, ψ classical description: positions and velocities of all particles ψ = r 1,..., r N, v 1..., v N ) better momenta...) macrostate = averaged microstates ensemble = set of all microstates with their respective probabilities πψ), microstate macrostate ensemble
[tchem/simolant1+2.sh] 6/15 Microcanonical ensemble and the ergodic hypothesis Microcanonical ensemble = set of microstates in an isolated system Denoted: NVE N = const, V = const, E = const) Ergodic hypothesis quantum): πψ i ) = const = 1 W W = number of all states) Ergodic hypothesis classical): trajectories fill the phase) space uniformly In other words: Time averaged mean) value = 1 X t = lim t t t 0 Xt) dt = ensemble mean value = X = 1 Xψ) W for quantity X = Xψ), where ψ = ψt)... but T = const is more practical ψ
Canonical ensemble 7/15 is the ensemble with a constant temperature. Denoted NVT N = const, V = const, T = const) ergodic hypothesis: πψ) = πeψ)) E 1 + E 2 = E 1+2 small influence) πe) = probability of any state of energy E πe 1 ) πe 2 ) = πe 1+2 ) = πe 1 + E 2 ) πe) = const E = expα i βe) 0th Law β is the empirical temperature α i is the normalization constant, so that ψ system πψ) = 1, depends on the Determine β: monoatomic ideal gas, energy per atom = U 1 = 3 2 k BT ψ U 1 = Eψ)πEψ)) 12 m v 2 π 1 ψ πeψ)) = 2 m v2 ) d v π 1 = = U 1 = 3 1 2 m v2 ) d v 2 β β = 1 k B T
Boltzmann probability 8/15... or the first half of the statistical thermodynamics. Probability of finding a state of energy E is proportional to [ πe) = const exp Eψ) ] = const exp E ) m k B T Examples: a barrier activation energy) E a is overcome by exp Arrhenius formula k = A exp E ) a E a ) molecules energy needed to transfer a molecule from liquid to vapor is vap H, probability of) finding a molecule in the vapor is proportional to exp Clausius Clapeyron equation integrated) vaph p = p 0 exp [ vaph R 1 T 1 T 0 )] = const exp ) vaph
[tchem/simplyn.sh] Barometric formula: particle in a gravitational field + 9/15... once again. Potential energy of a molecule in a homogeneous gravitational field: U = mgh Probability of finding a molecule at height h: π exp U ) = exp mgh ) = exp k B T k B T Probability density pressure: p = p 0 exp Mgh ) Mgh ) The same formula can be obtained from the mechanical equilibrium + ideal gas EOS: Boltzmann probability.) dp = dhρg, ρ = Mp
Boltzmann probability [cd tchem/molekuly; blend -g butane] 10/15 Example. Energy of the gauche conformation of butane is by E = 0.9 kcal/mol higher than anti. Calculate the population of molecules which are in the gauche state at temperature 272.6 K boiling point). Solution: 1 cal th = 4.184 J πgauche) : πanti) = exp[ E/] = 0.190 Don t forget that there are two gauche states! 2πgauche) + πanti) = 1 π = 2 exp[ E/] 2 exp[ E/] + 1 = 2 0.190 2 0.190 + 1 = 0.275 Note: we assumed that both minima are well separated and their shapes are identical.
Thermodynamics 11/15 Internal energy U = ψ Eψ)πψ) Small change of U: du = ψ πψ) deψ) + ψ dπψ) Eψ) deψ): energy level has changed dπψ): probability of state has changed ψ 1st Law of Thermodynamics: du = p dv + TdS p dv Piston moved. Enegy changes by deψ) = mechanical work = Fdx = F/A dax) = pψ) dv pψ) = pressure of state ψ, mean) pressure = p = ψ πψ)pψ). TdS Change of πψ) at constant [V] = change of probabilities of states of different energies = heat
Boltzmann equation for entropy [jkv BoltzmannTomb.jpg] 12/15... the second half of the statistical thermodynamics πe) = expα i βe) Eψ) = 1 β [α i ln πψ)], dπψ) = 0 dπψ)eψ) = ψ ψ = k B T d ψ dπψ) 1 β [α i ln πψ)] = 1 β πψ) ln πψ) ψ dπψ) ln πψ) ψ On comparison with TdS: Microcanonical ensemble: πψ) = S = k B πψ) ln πψ) ψ Boltzmann equation: { 1/W pro E = Eψ) 0 pro E Eψ) S = k B ln W Property: S 1+2 = S 1 + S 2 = k B lnw 1 W 2 ) = k lnw 1+2 ) If transitions between states are considered, ds dt 0 H-theorem) can be derived
Ideal solution Energies of neighbors: = = All configurations have the same energy 13/15 Mix N 1 molecules of 1 + N 2 molecules of 2: ) N W = = N! N 1 N 1!N 2! S = k B ln W k B N 1 ln N 1 N + N 2 ln N ) 2 N S m = R x 1 ln x 1 + x 2 ln x 2 ) We used the Stirling formula, ln N! N ln N N: N ln N! = ln i i=1 More accurate: N 1 ln x dx by parts = [ x ln x x ] N 1 = N ln N N + 1 N ln N N ln N! asympt. = N ln N N + ln 2πN + 1 12N 1 360N 3 +
Residual entropy of crystals at T 0 [traj/ice.sh] 14/15 Crystal: 1 microstate S = k ln 1 = 0 3rd Law) 3rd Law violation: CO, N 2 O, H 2 O. Not in the true equilibrium, but frozen because of high barriers Example 1: Entropy of a crystal of CO at 0 K S m = k B ln 2 N A = R ln 2 Example 2: Entropy of ice at 0 K S m = k B ln 1.507 N A = 3.41 J K 1 mol 1 Pauling derivation: 6 = 4 2 ) orientations of a water molecule then an H-bond is wrong with prob.= 1 2 2N A bonds in a mole S m = k B ln 6 N ) A 2 2N = 3.37 J K 1 mol 1 A
Information entropy of DNA 15/15 Assuming random and equal distribution of base pairs. Per one base pair: k B ln 4, per mole: R ln 4. Corresponding Gibbs energy at 37 C): G = ln 4 = 3.6 kj mol 1 To be compared to: ATP ADP standard: r G m = 31 kj mol 1 in usual conditions in a cell: r G m = 57 kj mol 1 credit: www.pbs.org/wgbh/nova/sciencenow/3214/01-coll-04.html