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Statistical Thermodynamics (Mechanics) Pressure of Ideal Gas

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Statistical Thermodynamics (Mechanics) Pressure of Ideal Gas 1/16 [simolant -I0 -N100] 2/16 Statistical thermodynamics (mechanics) co05 Pressure of ideal gas from kinetic theory I co05 Molecule = mass point L Macroscopic quantities are a consequence ¨¨* N molecules of mass m in a cube of edge length L HY HH of averaged behavior of many particles HH Velocity of molecule = ~ , , 6 H = ( , ,y ,z) HH HH ¨¨* After reflection from the wall: , , y ¨ → − ¨ Next time it hits the wall after τ = 2L/, Force = change in momentum per unit time - Momentum P~ = m~ Change of momentum = ΔP = 2m, Averaged force by impacts of one molecule: 2 ΔP 2m, m, F, = = = τ 2L/, L → Pressure = force of all N molecules divided by the area N N 2 F, 1 m, p =1 = = 2 = 3 P L P L Kinetic energy of one molecule: 1 2 1 2 1 2 2 2 m ~ m = m(, + ,y + ,z) 2 | | ≡ 2 2 3/16 4/16 Pressure of ideal gas from kinetic theory II co05 Boltzmann constant co05 Kinetic energy = internal energy (monoatomic gas) 1 N 3 N 2 2 pV = nRT = NkBT Ekin = m = m, 2 1 2 1 X= X= N = nNA ⇒ N 2 1 m, 2 Ekin p = = 3 = P L 3 V R k 1.380649 10 23 JK 1 Or B = = − − NA × 2 ! pV = Ekin = nRT 3 Note: Summary: since May 20, 2019 it is defined: Temperature is a measure of the kinetic energy ( 0th Law) k 1.380649 10 23 JK 1, B = − − ∼ × 23 1 Pressure = averaged impacts of molecules NA = 6.02214076 10 mol , × − We needed the classical mechanics therefore, exactly: 1 1 R = 8.31446261815324 J mol K Ludwig Eduard Boltzmann (1844–1906) Once more: − − credit: scienceworld.wolfram.com/biography/Boltzmann.html N R 3n 3N 3 n = , kB = U Ekin = RT = kBT, CV,m = R NA NA ⇒ ≡ 2 2 2 5/16 [tchem/simolant1+2.sh] 6/16 Microstate, macrostate, ensemble, trajectory co05 Microcanonical ensemble and the ergodic hypothesis co05 microstate (state, configuration) = instantaneous Microcanonical ensemble = set of microstates in an isolated system quantum description: wave function, ψ Denoted: NVE (N = const, V = const, E = const) classical description: positions and velocities of all particles (better momenta...) 1 Ergodic hypothesis (quantum): πππ(ψ) = const = W ψ = (r~1,..., r~N, ~1 ..., ~N) (W = number of all states) macrostate = averaged microstates Ergodic hypothesis (classical): ensemble = set of all microstates with their respective probabilities πππ(ψ), trajectories fill the (phase) space uniformly In other words: Time averaged (mean) value = 1 t X t = lim X(t) dt 〈 〉 t ∞ t 0 → Z = ensemble mean value = microstate macrostate ensemble trajectory 1 X = X(ψ) 〈 〉 W ψ X for quantity X = X(ψ), where ψ = ψ(t) . but T = const is more practical 7/16 8/16 Canonical ensemble co05 Boltzmann probability co05 is the ensemble with a constant temperature. ... or the first half of the statistical thermodynamics. Denoted: NVT (N const, V const, T const) = = = Probability of finding a state of energy is proportional to Ergodic hypothesis: πππ ψ πππ ψ E ( ) = (E( )) E1 + E2 = E1 2 (small influence) (ψ) Em + πππ( ) = const exp E = const exp πππ E = probability of any state of energy E E k T RT ( ) · − B · − πππ(E1) πππ(E2) = πππ(E1+2) = πππ(E1 + E2) · Examples: E πππ(E) = const = exp(α βE) Ea a barrier (activation energy) Ea is overcome by exp RT molecules ⇒ − ∼ − 0th Law β is the empirical temperature Arrhenius formula ⇒ ⇒ E α is the normalization constant, so that πππ ψ 1, depends on the system a ψ ( ) = k = A exp − RT P 3 Determine β: monoatomic ideal gas, energy per atom = U1 = 2kBT energy needed to transfer a molecule from liquid to vapor is ΔvapH, probability 1 2 1 2 ψ (ψ)πππ( (ψ)) m~ πππ( m~ ) d~ 3 1 1 ΔvapH U E E 2 2 U β of finding a molecule in the vapor is proportional to exp Clausius– 1 = = 1 = = 1 = = RT 〈 〉 πππ( (ψ)) πππ m~2 d~ ··· 〈 〉 2 β ⇒ kBT ∼ − ⇒ P ψ E R (2 ) Clapeyron equation (integrated) P R ΔvapH 1 1 ΔvapH p = p0 exp = const exp R T T RT − − 0 · − [simolant -I2] 9/16 [cd ../simul/nmf/; blend -g butane]10/16 Barometric formula: particle in a gravitational field + co05 Boltzmann probability co05 ... once again. Example. Energy of the gauche conformation of butane is by ΔE = 0.9 kcal/mol higher than anti. Calculate the population of molecules which are in the gauche Potential energy of a molecule in a homogeneous gravitational field: state at temperature 272.6 K (boiling point). U = mgh Solution: 1 calth = 4.184 J Probability of finding a molecule at height h: πππ(gche) : πππ(nt) = exp[ ΔE/RT] = 0.190 U mgh Mgh − πππ ∝ exp = exp = exp Don’t forget that there are two gauche states! k T k T RT − B − B − Probability ∝ density ∝ pressure: 2πππ(gche) + πππ(nt) = 1 Mgh p = p0 exp ⇒ 2 exp ΔE/RT 2 0.190 − RT [ ] πππ = − = × = 0.275 The same formula can be obtained from the mechanical equilibrium + ideal gas 2 exp[ ΔE/RT] + 1 2 0.190 + 1 − × EOS: Note: we assumed that both minima are well separated and their shapes are iden- Mp dp = dhρg, ρ = tical. − RT (“ ” Boltzmann probability.) ⇒ 11/16 [jkv pic/BoltzmannTomb.jpg]12/16 Thermodynamics co05 Boltzmann equation for entropy co05 Internal energy . the second half of the statistical thermodynamics U = (ψ)πππ(ψ) β=1/kBT E πππ E exp α βE ψ k T α ln πππ ψ , dπππ ψ 0 ψ ( ) = ( ) E( ) = B [ ( )] ( ) = X − ⇒ − ψ Small change of U: X dU πππ ψ d ψ dπππ ψ ψ dπππ(ψ) (ψ) = dπππ(ψ)kBT[α ln πππ(ψ)] = kBT dπππ(ψ) ln πππ(ψ) = ( ) E( ) + ( ) E( ) E − − · ψ · ψ · ψ ψ ψ X X X X X d (ψ): energy level has changed E kBT d πππ ψ ln πππ ψ dπππ(ψ): probability of state has changed ψ = ( ) ( ) − ψ X 1st Law of Thermodynamics: dU = p dV + TdS − On comparison with TdS: p dV − S k πππ ψ ln πππ ψ “Piston” moved. Enegy changes by d (ψ) = mechanical work = Fd = F/ = B ( ) ( ) E − − A · − ψ d( ) = p(ψ) dV X A credit: schneider.ncifcrf.gov/images/ p ψ = “pressure− of state ψ”, (mean) pressure = p πππ ψ p ψ . 1/W for E = (ψ) boltzmann/boltzmann-tomb-8.html ( ) = ψ ( ) ( ) Microcanonical ensemble: πππ ψ E ( ) = 0 for E ψ = ( ) Transitions be- TdS P § 6 E Change of πππ ψ at constant V = change of probabilities of states of different tween states ( ) [ ] Boltzmann equation: S = kB ln W energies = heat dS/dt 0 ⇒ ≥ Property: S1+2 = S1 + S2 = kB ln(W1W2) = kB ln(W1+2) (H-theorem) 13/16 [traj/ice.sh]14/16 Ideal solution co05 Residual entropy of crystals at T 0 co05 → Energies of neighbors: – = – = – Crystal: 1 microstate S = k ln 1 = 0 (3rd Law) All configurations have• the• same• • energy• • ⇒ 3rd Law violation: CO, N2O, H2O. Not in the true equilibrium, but “frozen” Mix N1 molecules of 1 + N2 molecules of 2: because of high barriers N N! W = = N N !N ! Example 1: Entropy of a crystal of CO at 0 K 1 1 2 S k ln 2NA R ln 2 N1 N2 m = B = S = kB ln W kB N1 ln + N2 ln N N Example 2: Entropy of ice at 0 K ≈ − S k ln 1.507NA 3.41 JK 1 mol 1 Sm = R (1 ln 1 + 2 ln 2) m = B = − − − Pauling derivation: 4 We used the Stirling formula, ln N! N ln N N: 6 = 2 orientations of a water molecule ≈ − N N then an H-bond is wrong with prob.=1 by parts N 2 ln N! = ln ln d = [ ln ]1 = N ln N N + 1 N ln N N 1 ≈ 1 − − ≈ − 2NA bonds in a mole X= Z N asympt. 1 1 6 A 1 1 More accurate: ln N! = N ln N N + ln 2πN + + Sm = kB ln = 3.37 JK mol 3 22NA − − − 12N − 360N − · · · ⇒ p 15/16 16/16 Information entropy of DNA co05 Finishing the story + co05 Assuming random and equal distribution of base pairs. α = ? U S k πππ ψ α β ψ k α = B ( )[ E( )] = B Per one base pair: kB ln 4, per mole: R ln 4. − ψ − − − T X Corresponding Gibbs energy (at 37 C): Helmholtz energy: ◦ ΔG RT ln 4 3.6 kJ mol 1 = = − U TS F − − β (ψ) α − F kBT ln e E = = = − To be compared to: ATP ADP kBT kBT ⇒ − ψ X – standard: Δ G 31→kJ mol 1 r me = − − 1 – in usual conditions in a cell: ΔrGm = 57 kJ mol − − [...] = canonical partition function = statistical sum (Q or Z) Interpretation: number of “available” states (low-energy easily, high-energy difficult) All equilibrium quantites can be calculated from F (dF = pdV SdT): − − ∂F p = U = F + TS − ∂V H U pV ∂F = + S = G = F + pV credit: www.pbs.org/wgbh/nova/sciencenow/3214/01-coll-04.html − ∂T.
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