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Chapter 5 Simple Applications of Macroscopic Thermodynamics

Chapter 5 Simple Applications of Macroscopic Thermodynamics. Preliminary Discussion. Classical, Macroscopic, Thermodynamics Drop the statistical mechanics notation for average quantities. We know that All Variables are Averages Only !

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Chapter 5 Simple Applications of Macroscopic Thermodynamics

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  1. Chapter 5Simple Applications of Macroscopic Thermodynamics

  2. Preliminary Discussion Classical, Macroscopic, Thermodynamics • Drop the statistical mechanics notation for average quantities. We know that All Variables are Averages Only! • We’ll discuss relationships between macroscopic variables using The Laws of Thermodynamics • Some Thermodynamic Variables of Interest: Internal Energy = E, Entropy = S,Temperature = T For Gases: External Parameter = V, Generalized Force = p (V = volume, p = pressure) For A General System: External Parameter = x, Generalized Force = X

  3. Assume the relevant External Parameter = Volume V in order to have a specific case to discuss. For infinitesimal, quasi-static processes: The 1st & 2nd Laws of Thermodynamics 1st Law: đQ = dE + pdV 2nd Law: đQ = TdS The Combined 1st & 2nd Laws TdS = dE + pdV • Note that, in this relation, there are 5 Variables: T, S, E, p, V • It can be shown that: Any 3 of these can always be expressed as functions of any 2 others. • That is, there are always 2 independent variables & 3 dependent variables. Which 2 are chosen as independent is arbitrary.

  4. Now, A Brief, Pure Math Discussion • Consider 3 variables: x, y, z. Suppose we know that x & y are Independent Variables. Then, It Must Be Possible to express z as a function of x & y. That is, There Must be a Functionz = z(x,y). • From calculus, the total differential of z(x,y)has the form: dz  (∂z/∂x)ydx + (∂z/∂y)xdy (a) • Suppose instead that we want to take y & z as independent variables. Then, There Must be a Functionx = x(y,z). • From calculus, the total differential of x(y,z) has the form: dx  (∂x/∂y)zdy + (∂x/∂z)ydz (b) • Using (a) & (b) together, the partial derivatives in (a) & those in (b) can be related to each other. • We always assume that all functions are analytic. So, the 2nd cross derivatives are equal: Such as (∂2z/∂x∂y) (∂2z/∂y∂x),etc.

  5. Mathematics Summary • Consider a function of 2 independent variables:f = f(x1,x2). • It’s exact differential is df  y1dx1 + y2dx2, where, by definition: • Because f(x1,x2) is an analytic function, it is always true that Most Ch. 5 applications use this with the Combined 1st & 2nd Laws of Thermodynamics: TdS = dE + pdV

  6. Some Methods & Useful Math Tools for Transforming Derivatives

  7. Pure Math: Jacobian Transformations • A Jacobian Transformation is often used to • transform from one set of variables to another. • For functions of 2 variables f(x,y) & g(x,y) it is: Determinant!

  8. Jacobian Transformations Have Several Useful Properties

  9. Suppose that we are only interested in the first partial derivative of a function f(z,g) with respect toz at constant g: • This expression can be simplified using the chain rule expansion and the inversion property

  10. Properties of the Internal Energy E dE = TdS – pdV (1) First, choose S &Vas independent variables: E  E(S,V) ∂E ∂E (2) dE Comparison of (1) & (2) clearly shows that ∂E ∂E and Applying the general result with 2nd cross derivatives gives: Maxwell Relation I!

  11. If S &p are chosen as independent variables, it is convenient to define the following energy: H  H(S,p)  E + pVEnthalpy Use the combined 1st & 2nd Laws. Rewrite them in terms of dH:dE = TdS – pdV = TdS – [d(pV) – Vdp] or dH = TdS + Vdp (1) But, also: (2) Comparison of (1) & (2) clearly shows that and Applying the general result for the 2nd cross derivatives gives: Maxwell Relation II!

  12. If T &V are chosen as independent variables, it is convenient to define the following energy: F  F(T,V)  E - TSHelmholtz Free Energy Use the combined 1st & 2nd Laws. Rewrite them in terms of dF:dE = TdS – pdV = [d(TS) – SdT] – pdV or dF = -SdT – pdV (1) But, also: dF ≡ (F/T)VdT + (F/V)TdV (2) Comparison of (1) & (2) clearly shows that (F/T)V ≡ -S and (F/V)T ≡ -p Applying the general result for the 2nd cross derivatives gives: Maxwell Relation III!

  13. If T &p are chosen as independent variables, it is convenient to define the following energy: G  G(T,p)  E –TS + pVGibbs Free Energy Use the combined 1st & 2nd Laws. Rewrite them in terms of dH:dE = TdS – pdV = d(TS) - SdT – [d(pV) – Vdp] or dG = -SdT + Vdp (1) But, also: dG ≡ (G/T)pdT + (G/p)Tdp (2) Comparison of (1) & (2) clearly shows that (G/T)p ≡ -S and(G/p)T ≡ V Applying the general result for the 2nd cross derivatives gives: Maxwell Relation IV!

  14. Summary: Energy Functions 1. Internal Energy: E E(S,V) 2.Enthalpy: H = H(S,p)  E + pV 3.Helmholtz Free Energy: F = F (T,V)  E – TS 4.Gibbs Free Energy: G = G(T,p)  E – TS + pV 1. dE = TdS – pdV 2. dH = TdS + Vdp 3. dF = - SdT – pdV 4. dG = - SdT + Vdp Combined 1st & 2nd Laws

  15. 2. 1. 3. 4. Another Summary: Maxwell Relations (a) ΔE = Q + W (b) ΔS = (Qres/T) (c) H = E + pV (d) F = E – TS (e) G = H - TS 1. dE = TdS – pdV 2. dH = TdS + Vdp 3. dF = -SdT - pdV 4. dG = -SdT + Vdp

  16. Maxwell Relations:“The Magic Square”? Each side is labeled with an Energy (E, H, F, G). The corners are labeled with Thermodynamic Variables (p, V, T, S). Get the Maxwell Relations by “walking” around the square. The partial derivatives are obtained from the sides. The Maxwell Relations are obtained from the corners. F V T G E S P H 16

  17. Summary The 4 Most Common Maxwell Relations: 17

  18. Maxwell Relations: Table (E → U) 18

  19. Maxwell Relations Maxwell Relations from dE, dF, dH, & dG

  20. Some Common Measureable Properties ∂E Heat Capacity at Constant Pressure: Heat Capacity at Constant Volume: Volume Expansion Coefficient: Note!! Reif’s notation for this is α Isothermal Compressibility: The Bulk Modulus is inverse of the Isothermal Compressibility!! B  (κ)-1

  21. Some Sometimes Useful Relationships Summary of Results Derivations are in the text and/or left to the student! Entropy: Enthalpy: Gibbs Free Energy:

  22. A Typical Example • Given the entropy S as a function of temperature T & volume V, S = S(T,V), find a convenient expression for (S/T)P, in terms of some measureable properties. • Start with the exact differential: • Use the triple product rule & definitions: • Use a Maxwell Relation: • Combining these expressions gives:

  23. Converting this to a partial derivative gives an identity: • This can be rewritten as: • The triple product rule is: • Substituting gives: Note again the definitions: Volume Coefficient of Expansion βV-1(V/T)p Isothermal Compressibility κ -V-1(V/p)T Note!! Reif’s notation for this is α

  24. Using these in the previous expressionfinally gives the desired result: • Using this result as a starting point,A GENERAL RELATIONSHIPbetween The Heat Capacity at Constant Volume CV& The Heat Capacity at Constant Pressure Cpcan be found:

  25. For an Ideal Gas, it’s easily shown (Reif) that the Equation of State(relation between pressure P, volume V, temperature T) is (in per mole units!):Pν = RT. • With this, it is simple to show that the volume expansion coefficient β & the isothermal compressibility κare: Simplest Possible Example: The Ideal Gas and

  26. So, for an Ideal Gas, the volume expansion coefficient & the isothermal compressibility have the simple forms: and • We just found in general that the heat capacities at constant volume & at constant pressure are related as • So, for an Ideal Gas, the specific heats per mole have the very simple relationship:

  27. Other, Sometimes Useful, Expressions

  28. Consider Two Identical Objects, each of mass m, & specific heat per kilogram cP. See figure next page. Object 1 is at initial temperatureT1. Object 2 is at initial temperatureT2. Assume T2 > T1. When placed in contact, by the 2nd Law, heat Q flows from the hotter (Object 2) to the cooler (Object 1), until they come to a common temperature, Tf. More Applications: Using the Combined 1st & 2nd Laws (“The TdS Equations”)

  29. Two Identical Objects, of mass m, & specific heat per kilogram cP. Object 1 is at initial temperatureT1. Object 2 is at initial temperatureT2. T2 > T1. When placed in contact, by the 2nd Law, heat Q flows from the hotter (Object 2) to the cooler (Object 1), until they come to a common temperature, Tf. After a long enough time, the two objects are at the same temperature Tf. Since the 2 objects are identical, for this case, Q  Heat Flows Object 2 Initially at T2 Object 1 Initially at T1 For some time after the initial contact:

  30. The Entropy ChangeΔSfor this process can be easily calculated: • Of course, by the 2nd Law,the entropy changeΔSmust be positive!! This requires that the temperatures satisfy:

  31. NOTE:In the following, various quantities are written in per mole units! Work with the Combined 1st & 2nd Laws: • Definitions: • υ Number of moles of a substance. ν (V/υ)  Volume per mole. • u  (U/υ)  Internal energy per mole. h  (H/υ)  Enthalpy per mole. • s  (S/υ)  Entropy per mole. cv (Cv/υ)  const. volume specific heat per mole. • cP (CP/υ)  const. pressure specific heat per mole. Some Useful “TdS Equations”

  32. Internal Energyu(T,ν) Enthalpyh(T,P)

  33. Entropy 1 3 2

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