Unformatted text preview: Mat E 510 Thermodynamics of Solids Mat E 510 Thermodynamics Thermodynamics of Solids
Based on Prof. Roger Doherty’s course Lecture #3 Yury Gogotsi
A.J. Drexel Nanotechnology Institute and gy Department of Materials Science & Engineering, Drexel University, Phil Philadelphia, Pennsylvania, USA USA Entropy Entropy Scale: Third Law of Thermodynamics
(Nernst Heat Theorem  1906) As T tends to 0 K, the entropy of any system (in internal equilibrium) tends to a constant value So, independent of pressure and the state of aggregation (arrangement) of the atoms.
That is in any reaction A−>B : e.g., e.g., (A) 2 Na + Cl2 −> 2 NaCl (B) Cl or (A) White(metallic) Tin −> Gray(diamond) Tin (B) ΔSBA = 0 as T > 0K (An Experimental Result) This result can be demonstrated in several ways. (Non martensitic) Phase Changes (can be prevented by cooling quickly enough) Gray α−Tin > White β−Tin ΔV is large! Phase Phase Transition in Tin
Reversible transformation temperature in tine at 286 K (13.2°C), but white Sn can be supercooled to ≈ 0 K without transformation. (No martensitic shear transformation) The specific heat, CP, can be measured and, as always found, CP > 0 as T > 0 K.
298 Sα (298) = Sαo + ∫ CP(α) dT/T
0 298 Sβ (298) = Sβ + ∫ CP(β) dT/T
o 0 The transformation of gray αSn to white β−Sn occurs at 286 K reversibly with a latent heat heat, ΔHαβ. So Sβ (298) = Sα (298) + ΔHαβ/286 Measurements show that:
298 0 ∫ CP(α) dT/T + ΔHαβ/286 = ∫ CP(β) dT/T
0 298 That is SGo = Swo or ΔSGWo = ΔSαβo = 0 Phase Phase Transition in Tin
• Although the transformation temperature is 13.2 °C, the change does not take place unless the metal is of high purity, and only when the exposure temperature is well below 0 °C. This process is known as tin disease or tin pest. Tin pest was a particular problem in northern Europe in the 18th century as organ pipes made of tin alloy would sometimes be affected during long cold winters. Napoleon campaign. Some sources also say that during Napoleon's Russian campaign of 1812, the temperatures became so cold that the tin buttons on the soldiers' uniforms disintegrated, contributing to the defeat of the Grande Armée. Scott expedition to Antarctica. In 1910 British polar explorer Robert Scott hoped to be the first to reach the South Pole, but was beaten by Norwegian explorer Roald Amundsen. On foot, the expedition trudged th through the frozen deserts of the Antarctic, making for caches of food and kerosene deposited on the way in. In early 1912, at the first cache, there was no kerosene; the cans — soldered with tin — were empty. Members of the expedition later died in the cold and blizzards, only eleven miles from a massive depot of supplies. • • Definitions Definitions of the Third Law
• Definition 1: No finite sequence of cyclic processes can cool a body to absolute zero β= 1/kT, as Τ−> 0, β−> ∞ No finite ladder can be used to reach infinity • Definition 2: The entropy of every pure, perfectly crystalline substance approaches the same value as the temperature approaches zero • Definition 3: The entropy of all perfectly crystalline substances is zero at 0 K (we have simply chosen it to be 0 for convenience, thermodynamic laws don’t imply it) The 3rd law does not introduce a new TD function: it simply implies that S can be expressed on an absolute scale At 0.000 000 000 5 K, gas molecules travel at the speed of less than 1 mm/s Entropy Entropy of Fusion
• • The entropy of fusion is the increase in entropy when melting a substance. There is a proportionality between ΔSfusº for the elements and atomic number (or binding energy of crystals, which has the same trend). R.A. Swalin, Thermodynamics of Solids 4 Laws of Thermodynamics
0th Temperature is a state function. 1st  Failure to create energy 2nd  Failure to turn heat completely into work (in a cycle) 3rd  Failure to get to 0 K What is the importance / use of the Third Law? Allows prediction of the directions of all chemical reactions! A + B > C + D 2Hg + O2 > 2 HgO FeO + CO > Fe + CO2. α−Fe > γ−Fe We can calculate, using the 3rd Law, SAo = SBo = 0, T' o + ∫ C (A) dT/T At T = T' SA = SA P 0 Equivalently, for all elements or compounds: B, C, D.. So at T', ΔSR = Σ SProd  Σ SReact can be predicted, by calculations, using the 3rd law. Third Third Law
Since S is state function: ΔSsyc = ΔSI + ΔSII + ΔSIII +ΔSIV = 0 Experimentally found: ΔSsyc = ΔSI + ΔSII + ΔSIII = 0 It means: ΔSIV = 0 Entropy of SiC is the same as entropy of Si + C at 0 K – 3rd law. The practical application – entropy change of a chemical reaction reaction can be calculated from heat capacity of pure substances R. De Hoff, Thermodynamics in Mater. Sci. Predicting Predicting Direction of a Reaction
In an isolated system the th equilibrium state is the state that has the maximum value of entropy that the system py can exhibit This statement applies only to the system isolated from its surrounding Enthalpy
We can also calculate ΔHR. Here we must measure ΔHR(T) and transform to T' using measured CP or by measuring ΔHF(298) calculate first ΔHR(298) then ΔHR(T') ΔHR = Σ HProd  Σ HReact Note: Heats can be measured for any reaction, reversible or not. We can only determine ΔSR from: ΔSR = ΔHR /T' if the reaction is reversible. Most chemical reactions are not reversible except at one temperature that may not be accessible. Can the reaction A + B > C + D occur at T‘ ? Yes if: ΔST = ΔSSys + ΔSSurr ≥ 0 ΔSSys = ΔSR Path independent. ΔHSurr =  ΔHR Note change of sign! So: ΔSSurr =  ΔHR /T' (The surroundings can always take up the heat reversibly, even if the heat is not generated generated reversible.) So ΔST = ΔSR  ΔHR /T' ≥ 0 Predicting Predicting Chemical Reactions
This then tells us if a chemical reaction CAN OCCUR. This remarkable result comes from SIMPLE THERMAL MEASUREMENTS  Specific heats over a range of temperatures, down to 0 K and Latent Heats and Heats of Reaction/Formation of all reactants and any known products. Note two limitations: The product must be a known compound. It only says if the reaction CAN occur  it may not occur if its kinetics are too slow. (Diamond > graphite) More accurately, TD only says if the opposite reaction cannot occur. If ΔST < 0, then the reaction cannot occur Gibbs Gibbs Free Energy
G = H  TS ( Josiah Willard Gibbs in 1886, Yale Univ. – founding father of chemical thermodynamics) This is a very clever mathematical trick to avoid worrying about the surroundings. 2nd Law Result: dST = dSSys + dSSurr ≥ 0 but SSurr =  dqSys/T =  dHSys/T H = U +PV, so dH = dU + PdV + VdP At constant pressure: dP = 0, dH = dU + PdV and, if dw' = 0, dU = δq  PdV, so then δq = dU + PdV and dH = δq (Note two restrictions: ΔP and δw = 0) So dST = dSSys  dHSys / T ≥ 0 Multiply thru by (T): T dST = TdSSys + dHSys ≤ 0 Gibbs Gibbs and Helmholtz Free Energy
Define G = H  TS Then Then dG = dH  TdS  SdT So at CONSTANT TEMPERATURE, dT = 0 dGSys = dHSys  TdSSys ≤ 0 So, at constant temperature and constant pressure and if no external work is done apart from PdV, a system will seek a minimum in (GIBBS) FREE ENERGY. ill G = H – TS or ΔG=ΔHTΔS An incredibly useful result! At constant volume δq = dU  PdV = dU So Helmholtz free energy, A = U – TS, will perform the same function for reactions at constant volume and temperature. Gibbs Energy Minimization
Criterion for spontaneity of process: Decrease in G (at P=const) (and increase in entropy of the universe) Criterion for equilibrium: dG=0 Most processes in biology, chemistry and many material synthesis/treatment processes occur at P, T=const P. Atkins, Four Laws Free Free Energy
Why FREE Energy ? It is the energy available in any reaction that can do useful work – e.g. in an ELECTROCHEMICAL CELL. Cu2+SO42 + Zn0 > Zn2+SO42 + Cu0 2H02 + O02 > 2H1+2O22Na0 + S0 > Na1+2S2Each of these reactions can be made into important EC Cells. (Primary, a "Fuel" Cell, or a “Rechargeable”) The cell can produce a reversible cell voltage  one with an equal opposing voltage so the reaction occurs essentially at zero current. We can measure the voltage in a potentiometer, with I > 0 amps. with Electrochemical Electrochemical Cell
Allow an infinitesimal amount of electrical work δw' (Coulombs x Volts) to be done (δw'< 0) by the reaction at constant pressure and temperature. dU = δq + δw = δq  PdV + δw' So dU+PdV = dH = δq+δw' δw'= n e V (J) n is the # of electrons, V the cell voltage. Thi This is reversible so: ΔH = δqrev + δw'rev G = H  TS ( T constant dT = 0) So: dG = dH  TdS = δqrev + δw'rev  TdS But dS = δqrev / T So: dG = δw' or dG =  δw' The decrease in free energy is the (electrical) work done by the cell (at constant P and T). Some Important Relationships of G and A
Gibbs free energy: G = H TS H = U + PV So: So: dG = dH  TdS SdT dH = dU + PdV +VdP But dU = δq + δw = δq  PdV ( dw' = 0)  First Law of TD but dS = δqREV/T So: dU = TdS –PdV So δq = δqREV = TdS dH = TdS + VdP  Second Law of TD (combination of the 1st and 2nd Laws of TD) and dG = δq PdV + PdV + VdP  TdS  SdT So: Then dG = VdP  SdT (∂ G/ ∂ T)P = S and (∂G/ ∂ P)T = V Similarly, for Helmholtz free energy, from A = U TS dA = dU TdSSdT dU = δq + δw = δq – PdV (δw' = 0) = TdS PdV So So dA =  PdV  SdT PdV (∂ A/ ∂ T)V = S and (∂ A/ ∂ V)T = P Helmholtz free energy
A = U – T S ΔA = ΔU – T ΔS ‘A’ stands for ‘Arbeit’ Also called ‘work function’ ‐ energy left to do work (free to do useful work) Helmholtz Helmholtz free energy
Criterion for spontaneity of process: Decrease in A (at V=const) (and increase in entropy of the universe) Thermal Thermal Expansion and Compressibility
For ideal gases it is readily shown that: (∂ H/ ∂ T)P = CP = CV+{(∂ U/ ∂ V)T+P}(∂ V/ ∂ T)P All the terms except {(∂U/ ∂ V)T can be experimentally measured so that is how we can measure (∂ U/∂ V)T. The volume thermal expansion coefficient, α and the compressibility, β, are defined as: α = (1/V)(∂ V/∂ T)P β =  (1/V)(∂ V/∂ P)T (see Gaskell, Ch. 6.7 for derivation) (not thermodynamic β!) CP = CV + α2 V T / β (see Gaskell, Ch. 5.13 for derivation) Ch Gaskell (5.9) also quotes a large number of other useful thermodynamic, Maxwell, relationships. Not used in this course but useful in many applications. li Temperature Dependence of Bulk Modulus
• Bulk modulus (B) is the reciprocal of compressibility (β). The moduli (Y=3(12ν)B) of materials vs T can be predicted from thermodynamic parameters. Y Young’s modulus, v – Poisson’s ratio R.A. Swalin, Thermodynamics of Solids Heat of Formation vs Volume Change Stronger binding – smaller volume (larger ΔV) for reaction A+B ‐> AB R.A. Swalin, Thermodynamics of Solids Temperature Dependence of G,H, and S
Variation of H, S and G with Temperature – pure component: S
(∂H/ ∂T)P=  CP (∂ G/ ∂ T)P =  ΔS Both approach zero At 0 K dS =δq/T = CP dT/T G = H –TS so at T = 0 K G = H T dH = CP dT H = Hº + ∫ CP dT, Tº = 298 K 298 T dS = (CP /T) dT S = SO + ∫(CP /T) dT, TO = 0 K To (∂ S/ ∂ T)P = ‐ CP /T T Cp T CP ‐> 0 as T ‐> 0 CP / T remains finite as T ‐> 0 Phase Phase Change
Change of Phase examples: Solid state transformation α−Fe > γ−Fe (910°C) Melting AlS > AlL (660°C) Equilibrium / reversible at the transition temperature: ΔH = HL  HS > 0 ΔGLS = GL  GS = 0 at T = TEQ (equilibrium state when liquid and solid coexist) ΔGLS = GL  GS < 0 at T < TEQ ΔGLS = {HL  HS}  T (SL  SS ) = ΔHLS  T ΔSLS If CP(S) = CP(L), ΔHLS and ΔSLS = ΔHLS/ TEQ are constant with change of temperature: Then: ΔGLS = ΔHLS  T ΔHLS /TEQ = ΔHLS (TEQ  T) / TEQ = ΔHLS (ΔT / TEQ ), LS EQ LS LS where ΔT = TEQ  T Phase Phase Changes and Gibbs Free Energy P. Atkins, Four Laws Applications
Knowledge of Cp, S of substances, and heats of formation of compounds allows ΔH and ΔS calculation for any process. A few things to remember: • For pure elements H=0 at 298 K • S = 0 at 0 K (for substances in complete internal equilibrium) • The dependence of S and H of condensed phases on pressure at 01 atm is small enough to be ignored. The determination of ΔH and ΔS for any change of state at any T and P allows ΔG to be calculated as: ΔG= ΔH  ΤΔS This This is the criterion for equilibrium, which which can be determined from a knowledge of the thermochemical properties of the system.
Enthalpy, Helmholtz and Gibbs free energy are convenient accounting properties, not new concepts. No molecular description of H, A and G can be given. Example
Production of Uranium by the exothermal reaction: UF4 + 2 Mg −> 2 MgF2 + U Liquids must be produced for easy separation. TmMg = 922 K, TmU = 1405 K If the reactants are placed in an adiabatic container in the molar ratio Mg/UF4=2 and are allowed to react at 298K, is the heat released sufficient to increase T to 1773 K? ...
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 Summer '09
 Yury
 Thermodynamics, Entropy, Helmholtz free energy, Josiah Willard Gibbs

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