lecture25 - ∆S = T2 C ò T1 pdT T T1 = 0K Because there...

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Unformatted text preview: ∆S = T2 C ò T1 pdT T T1 = 0K Because there is an absolute zero for temperature, the entropy for any state can be calculated ... if we know S(0K). S(T) = T ò0 Cp(T) dT + S(0K) T We also defined entropy statistically: S(T) = k lnW(T) p The third law: Absolute entropies W(T) = number of accessible microstates at T W(T) depends on T through the Boltzmann distribution (fewer states accessible at low T): -∆E/kT Pn ∝ e Nils Walter: Chem 260 This definition gives us a useful S(T = 0K) “isoenergetic” Flip three coins HHH HHT HTH THH TTH THT HTT TTT A W = # of P= probability microstates S = k ln W 1/8 1 0k 3/8 3 1.10 k C 8 possible microstates 3/8 (distributions) 3 1.10 k D 1/8 1 0k B 4 possible macrostates If there is only one accessible microstate, the entropy of the system (macrostate) is given by: This is consistent with our view of entropy S = k ln W = k ln 1 = 0 as a measure of disorder. No disorder S = k ln W = k ln 1 = 0 (only one possible state) Þ S = 0 @ T = 0 K Nils Walter: Chem 260 The third law of thermodynamics S = 0 fforall perfectly crystalline materials at T = 0 K S = 0 or all perfectly crystalline materials at T = 0 K S = k ln W = k ln 1 = 0 S = k ln W = k ln 1 = 0 p If there is only one possible arrangement of atoms (i.e., in a perfect crystal): Since every substance has a finite (nonzero) heat capacity =0 T C (T) dT p and S(T) = + S(0K) T 0 ò Þ except for a perfect crystal at absolute zero, every substance has a finite positive entropy Standard absolute or “third law” molar entropies are tabulated, generally at 1 bar and 298 K Chem 260 Nils Walter: ( ) One can calculate the molar entropy of HCl 98.36 K Solid I 1.3 + 29.5 12.1 158.91 K Solid II 21.1 12.6 Liquid 9.9 188.07 K 85.9 Gas 13.5 Sum: S298.15 = 185.9 J K-1 mol-1 Sum: S298.15 = 185.9 J K-1 mol-1 Solid I S = ò Cp dlnT from 0 K to 16 K using Debye T3-law: Cp = aT3 S = ò Cp dlnT from 16 K to 98.36 K using experimental Cp S = ∆H/T for each phase transition Solid II S = ò Cp dlnT from 98.36 K to 158.91 K using experimental Cp Liquid S = ò Cp dlnT from 158.91 K to 188.07 K using experimental Cp Gas S = ò Cp dlnT from 188.07 K to 298.15Nilsusing experimental Cp K Walter: Chem 260 The exception: An imperfect crystal still has residual entropy a) Consider CO O C µ = 0.12 D If there were no preferred orientation W = 2N For one mole: W = 2NA S = k ln W Actual value: = k ln 2NA 4.2 J K-1 mol-1 = NAk ln 2 = R ln 2 = 5.8 J K-1 mol-1 b) Consider CH3D No dipole moment, same number of electrons No preferred orientation! W = 4N or for one mole W = 4NA S = R ln 4 = 11.5 J K-1 mol-1 Actual value: 11.7 J K-1 mol-1 Nils Walter: Chem 260 The three laws of thermodynamics Atkins Moore You can’t construct a perpetuum mobile that generates energy from nothing You can’t construct a machine that does nothing but convert heat into work 1. ∆U = q + w Energy Conservation The internal energy of an isolated system is constant 2. ∆S ≥ 0 everything tends toward increasing disorder The entropy of the universe tends to increase 3. S > 0 except for perfect crystals at T=0K The entropy of a perfectly crystalline substance is zero at T=0K You can never reach absolute zero Summary: The energy of the universe is constant; the entropy of the universe tends always toward a maximum. Nils Walter: Chem Rudolf Julius Clausius (1822-1888) 260 ...
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This note was uploaded on 12/26/2011 for the course CHEM 260 taught by Professor Staff during the Spring '08 term at University of Michigan.

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