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Unformatted text preview: Mat E 510 Thermodynamics of Solids Mat E 510 Thermodynamics Thermodynamics of Solids
Based on Prof. Roger Doherty’s course Lecture #2 Yury Gogotsi
A.J. Drexel Nanotechnology Institute and gy Department of Materials Science & Engineering, Drexel University, Phil Philadelphia, Pennsylvania, USA USA Definitions
• • • • • • • • • Endothermic Exothermic Macroscopic state of the system Standard State Equation of state Enthalpy Enthalpy of fusion Enthalpy of vaporization Entropy First First Law of Thermodynamics
This is extension of the law of conservation of energy. Conservation Conservation of energy, U  Newton's Third Law. Equal and opposite forces. Work done by the body: dw = f dl f – force, l  distance Also conservation of momentum dm' = f dt. Potential Energy to Kinetic Energy: (Newton’s laws) work w = force x distance = mgh = mv2/2 Mechanical equivalent of heat. 4.154 J (Newtonmeter) = 1 Calorie 0.239 cal per joule – thermochemical calorie (rounded from 0.2389) 1 calorie (or 15º calorie) the heat to raise the temperature of 1 g of water from 14.5 to 15.5 °C Specific Heat, Cp, of water is 4.154 J/g 4.154 kJ/kg (at P = 1 atm). Originally 1 cal/g. Work transforms to heat by viscosity or friction. Temperature Temperature and Energy
• • • Two ways of raising "temperature": By heat (drop in a heated copper ball  copper at higher temperature") or or by mechanical work. Both raise the energy of the material. mechanical Both of Heat – Energy flow by virtue of difference in “temperature” Work – Energy flow capable of raising a weight For infinitesimal change of state: dU = δq  δw • • • Performance of the work decreases U Absorption of heat increases U This defines a change of internal energy, U q positive – endothermic process, heat flows out of a body q negative – exothermic process, heat flows into a body exothermic heat q zero – adiabatic process The The First law (Conservation of Energy)
We define Internal Energy, U, by: dU = δq  δw Can we measure the absolute value of the Internal Energy? How is it stored? • Specific heat  increased atomic vibration • Making or breaking of atomic bonds • Latent heat • Chemical Reaction Heat  breaking and remaking chemical bonds 2Mg + O2 > 2 MgO Statement of First Law: Internal Energy is a State Function: U = f (T,P,…) The same amount of work, however it is performed (motion, electrical current, friction, etc.) brings about the same change of the system (means, change of state state is path independent) Other Other Statements of the First Law
Equivalent statements: The change ΔUAB ΔUAB = UB  UA = q  w is path independent. In the case of a cyclic process which returns the system to its initial state, the change of U is zero B A or or cyclic integral ΔU = ∫dU + ∫dU = (UBUA) + (UAUB) = 0 A B Or: Energy is conserved. How do we know this? Empirical evidence. No one has ever made a perpetual motion machine = infinite source of energy. Another statement: The internal energy of an isolated system is constant Other Other Statements of the First Law
Numerous calculations, based on the 1st Law, give the right answer. Science must be predictive. Work = force x distance = f dl (J = newton meter, N.m) P = force /area (N/m2) V = area x length dV = area .dl (m3) Constant pressure process: dw = PdV = f dl Also: Charge x voltage. J = Coulombs x Volts – electrical energy Surface tension γ (N/m) J = γ dA ( N/m) x (m2)  surface energy dU = δq  δw = δq  δω'  PdV δw' is all other forms of work apart from PdV If dV > 0 work done by the system  so energy falls. Enthalpy Enthalpy
Also, heat, work are not state functions but volume is. So: dU + PdV = δq  δw'
dU  PdV = δq So specific heat, CV, is the slope of the graph of U vs T If (as is usual) dw' = 0, At constant volume, dV = 0: dU = δq = CV dT So (δq/dT)V =(dU/dT)V = CV Define a new function, H, "ENTHALPY" (greater than internal energy) "ENTHALPY" H = U + PV (a state function)
dH = dU + PdV + VdP At constant pressure dP = 0 (also with dw' = 0) dH = dU + PdV = δq ΔH shows the amount of energy released as heat, provided the system is free to expand. (accounting trick – easy to measure) Specific Specific Heat
Specific heat at constant pressure, CP. dH = dU + PdV = δq = CP dT So (δq/dT)P = (dH/dT)P = CP CP is often given as a power function: CP = a + bT + c/T2… Empirical equation, chosen to be easy to integrate. Note: V, U, H, C are EXTENSIVE State Properties  they increase (“extend”) with the amount of material. Can be normalized: V = volume/ mole, U = J/mole H = J /mole… Temperature, Pressure, also Electric, Gravitational and Magnetic fields (that also change U and H) are INTENSIVE State Properties (or Variables). They are independent of the amount of material. Temperature Temperature Dependence of Cp
CP = a + bT + c/T2… Temperature Dependence of Cv Enthalpy Enthalpy (cont.)
Why define Enthalpy H? It is USEFUL. Many changes occur at constant pressure so the energies measured include the work done against atmospheric pressure, ∫ PdV Avoids having to calculate the term ∫ PdV Definitions: Standard State: The most stable state of the material at P = 1atm and at the relevant temperature. So solid ice at 5°C ( 268K), but liquid water at +25°C (298K) (298K) 100+ elements and N ( > ∞) compounds. In MSE, we use the Chemistry definition. For elements H = 0 ( standard state) at P = 1 atm and T =298K Mech Eng (“Steam Tables”) H = 0 for H20. We can only measure changes, dU, dH, not absolute values of U or H. Calculating Calculating Enthalpy
Compound formation: 2H2 + O2 > 2H2O at 298K ΔHR = ? ΔHR = 2HH2O  2 HH2  HO2 ΔH298 = HH2O =  285 kJ/mole of H20. The enthalpy of the compound. This can be measured by calorimetry, usually at an elevated temperature but the value is easily corrected to 298K by use of specific heats and the first law. Find a calculable cyclic path: Use CP (dH=CpdT) At temperature T: A + B > AB ΔHT = ΔH1+ ΔH298 + ΔH2 ΔH1 = ∫ Cp(A+B)
T 298 ΔH2 = ∫ Cp(AB)
298 T At 298K: A + B > AB ΔH298 Change, ΔHT , is path independent (First Law). Coal CoaltoLiquid Process
• D. Hildebrandt et al. Producing Transportation Fuels with Less Work, Science, vol. 323, pp. 16801681, 27 March 2009 Improvements in efficiency of the Fischer‐Tropsch process can be achieved with a carbon dioxide and hydrogen route, rather than the traditional carbon monoxide and hydrogen route. The processes shown would produce 80,000 barrels of liquid fuel per day and have a theoretical minimum work of 350 MW; the work (via heat) inputs for each stage and for the overall processes are shown as red and green arrows. Catalysts: transition metals Co, Fe, Ru, etc. Applications Applications of First Law
Application of first law to a homogeneous phase. U = f (V,T) dU = (∂U/ ∂ T)V dT +(∂ U/ ∂ V)T dV But, δq = dU + PdV (dw' = 0) So, δq = (∂ U/ ∂ T)V dT +{(∂ U/ ∂ V)T + P}dV δq = CV dT +{(∂ U/ ∂ V)T + P}dV For an "ideal" gas PV = RT So, PdV + VdP = RdT, at constant pressure, PdV = R dT So: P(dV/dT) P = R and for Ideal Gas only(dU/dV)T = 0 No interactions between molecules in an Ideal Gas. A unique case. So: So: δq = CV dT +PdV (δq/dT)P = CP = CV + P(dV/dT)P = CV + R ONLY for Ideal gas Isothermal Isothermal Process
For an isothermal (dT = 0) change (an expansion) of ideal gas dU = δq  δw = 0 PV=RT
δq = CV dT + PdV = PdV = RT dV/V δq = PdV = δw
V2 V1 V2 V2 q = w = ∫ PdV = RT∫ dV/V = RT ln (V2/V1) = RT ln(P1/P2) RT RT Adiabatic Adiabatic Process
Adiabatic (constant heat) δq =0 dU = δq + δw = δq  PdV δq = 0 = dU + PdV = CVdT + PdV = CVdT + RT dV/V 0 = dT/T + (R/CV) dV/V (R/C 0 = dT/T + (R/CV) dV/V Integrate, at constant CV . K (a constant) = ln T +(R/CV) ln V Raise to power e: K' = exp K = T V(R/Cv) K' = T V(Cp Cv)/Cv But PV = RT so T = PV/R So, K' = (PV/R) V(Cp /Cv) 1) So, K'R ( a constant) = PV(Cp /Cv) =PV γ =PV Isothermal vs Adiabatic Process Second Second Law
Perhaps the most important discovery in 19th Century: Provides a foundation for understanding why ANY change occurs. Study of Heat engines: Sidi Carnot. French Engineer (1824) Developed “Carnot” cycle analysis of steam/heat engines engines ε=1 Tsink/Tsource Rudolph Clausius. German Physicist (1850): “Heat can never, of itself, flow from a colder to a hotter temperature” Lord Kelvin, (William Thomson) Scottish Physicist. (1852): “It is impossible to extract heat from a reservoir and convert it wholly into work, without causing other changes in in the universe” Both equivalent and more usefully expressed as: In every reversible cycle o dqrev/T = 0, while for any irreversible cycle o dqirrev/T < 0. ∫ ∫ Cycles with o dq/T > 0 are impossible. ∫ Heat Heat Flow
Heat δq from hotter (T1) to colder body (T2), for example, down a temp gradient. T1 T2 δq> > δq +δq/T1  δq/T2 < 0 Since T1 > T2 δw > > δq Work in +δw > heat δq that then flows out dU = δw + δq = 0 δq = δw δq/T < 0 The opposite reactions: Heat from colder to hotter and heat completely to work never happens. Object (macroscopic: book, desk, building) jumping into the air. Why not? In macroscopic TD, there is no answer  the laws describe the world. They do not "explain" why it behaves this way. But atomic statistics can explain the second law. Entropy Entropy
Actually no fundamental explanations for fundamental physical laws: Newton’s Laws, Gravity, Coulombs Law, Schrödinger equation. Just very successful generalizations that predict how the world behaves. Thermodynamics is understandable in the light of those laws. Before we discuss Heat Engines and Carnot Cycles: Entropy, S We define dS = δqrev/T Heat/T – a useful idea! ∫ ∫ Why? 2nd Law: o dqrev/T = 0. Then o dS = 0 So this means, mathematically, that S is a STATE FUNCTION. So a mole of Copper, at a given T, P, etc., has a value of Entropy S that is unchanged unchanged after any cyclic path. That is: ΔSAB = SB  SA is Path Independent (reversible or not) B B However ΔSAB = ∫ dqrev/T But ΔSAB ≠ ∫ dqirrev/T A A Entropy Entropy Change in an Adiabatic Process
Require that the irreversible step (A>B) be adiabatic (as in the freezing of supercooled liquid in a Dewar) Then δqirrev = 0 Then Or: SA  SB < 0 or d (A>B) S A < SB dS ≥ 0 Back one step: B A o dqirrev/rev/T = ∫ dqirrev/T + ∫ dqrev/T ≤ 0 ∫ A B dS = 0 for a reversible step dS >0 if irreversible For an adiabatic system, the direction of any irreversible change is that which increases the entropy. Entropy Entropy Change in a NonAdiabatic Process
Great. But what about the many nonadibatic processes, with heat exchanged between the system and its surroundings? We must include the surroundings – and they may have to be the universe ( Heat radiating from the sun) Any Any heat exchanged by the system is done with thermal reservoirs each at the sample's current temperature. This allows heat to be received from or given to the reservoir reversibly, so we can calculate a heat exchange δq = CPdT This is an analytical (i.e. mathematical) idea to allow calculations of changes in entropy surroundings ↑ δq ↓ system δq = 0 By including the surroundings we make the total system adiabatic Entropy Entropy statement of the Second Law
We then retain: SA < SB But the entropy is that of the system and its surroundings. That is for any irreversible process: dST = dSsurr + dSsys > 0 For a reversible step: dST = dSsurr + dSsys = 0 So: dST = dSsurr + dSsys ≥ 0 • Entropy statement of 2nd Law: The entropy of the universe increases in the course of any spontaneous change Entropy Entropy
The whole thermodynamics can be expressed in terms of temperature, the internal energy and the entropy U – measure of quantity of energy S – measure of quality of energy. Lower entropy, higher Lower higher quality Review of reversible and non reversible heat engines (study yourself): All REVERSIBLE heat engines operating between the same two temperatures have the same efficiency. "Science owes more to the steam engine than the steam Science engine owes to science" Henderson (1917) Temperature Temperature Dependence of Entropy ...
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This note was uploaded on 02/12/2010 for the course MAT E 510 taught by Professor Yury during the Summer '09 term at Drexel.
 Summer '09
 Yury

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