Chapter 5 Simple Applications of Macroscopic Thermodynamics

Preliminary Discussion Classical, Macroscopic, Thermodynamics Now, we drop the statistical mechanics notation for average quantities. So that now, All Variables are Averages Only! We’ll discuss relationships between macroscopic variables using The Laws of Thermodynamics

Internal Energy = E, Entropy = S (V = volume, p = pressure) Some Thermodynamic Variables of Interest: Internal Energy = E, Entropy = S Temperature = T Mostly for Gases: (but also true for any substance): External Parameter = V Generalized Force = p (V = volume, p = pressure) For a General System: External Parameter = x Generalized Force = X

1st & 2nd Laws of Thermodynamics Assume that the External Parameter = Volume V in order to have a specific case to discuss. For systems with another external parameter x, the infinitesimal work done đW = Xdx. In this case, in what follows, replace p by X & dV by dx. For infinitesimal, quasi-static processes: 1st & 2nd Laws of Thermodynamics 1st Law: đQ = dE + pdV 2nd Law: đQ = TdS Combined 1st & 2nd Laws TdS = dE + pdV

Any 3 of these can always be expressed as functions of any 2 others. 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.

Brief, Pure Math Discussion dz  (∂z/∂x)ydx + (∂z/∂y)xdy (a) 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 Function z = z(x,y). From calculus, the total differential of z(x,y) has the form: dz  (∂z/∂x)ydx + (∂z/∂y)xdy (a)

dx  (∂x/∂y)zdy + (∂x/∂z)ydz (b) Suppose that, in this example of 3 variables: x, y, z, we want to take y & z as independent variables instead of x & y. Then, There Must be a Function x = x(y,z). From calculus, the total differential of x(y,z) is: dx  (∂x/∂y)zdy + (∂x/∂z)ydz (b) Using (a) from the previous slide [dz  (∂z/∂x)ydx + (∂z/∂y)xdy (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.

Combined 1st & 2nd Laws of Thermodynamics TdS = dE + pdV Mathematics Summary Consider a function of 2 independent variables: f = f(x1,x2). It’s exact differential is df  y1dx1 + y2dx2 & 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

Some Methods & Useful Math Tools for Transforming Derivatives Derivative Inversion Triple Product (xyz–1 rule) Chain Rule Expansion to Add Another Variable Maxwell Reciprocity Relationship 9

Pure Math: Jacobian Transformations A Jacobian Transformation is often used to transform from one set of independent variables to another. For functions of 2 variables f(x,y) & g(x,y) it is: Determinant! 10

Jacobian Transformations Have Several Useful Properties Transposition Inversion Chain Rule Expansion 11

Suppose that we are only interested in the first partial derivative of a function f(z,g) with respect to z at constant g: This expression can be simplified using the chain rule expansion & the inversion property 12

Properties of the Internal Energy E dE = TdS – pdV (1) First, choose S & V as independent variables: E  E(S,V) ∂E ∂E dE (2) Comparison of (1) & (2) clearly shows that ∂E ∂E and Applying the general result with 2nd cross derivatives gives: Maxwell Relation I! 13

H  H(S,p)  E + pV  Enthalpy 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! 14

F  F(T,V)  E - TS Helmholtz Free Energy 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! 15

G  G(T,p)  E –TS + pV  Gibbs Free Energy 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! 16

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 17

Another Summary: Maxwell’s Relations (a) ΔE = Q + W (b) ΔS = (Qres/T) (c) H = E + pV (d) F = E – TS (e) G = H - TS 1. 2. 3. 4. 1. dE = TdS – pdV 2. dH = TdS + Vdp 3. dF = -SdT - pdV 4. dG = -SdT + Vdp 18

Maxwell Relations: “The Magic Square”? Thermodynamic Variables 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. Partial derivatives are obtained from the sides. The Maxwell Relations are obtained from the corners. F V T E G S P H 19

The 4 Most Common Maxwell Relations: Summary The 4 Most Common Maxwell Relations: 20

Maxwell Relations: Table (E → U) 21

Maxwell Relations from dE, dF, dH, & dG Internal Energy Helmholtz Free Energy Enthalpy Gibbs Free Energy 22

Some Common Measureable Properties Heat Capacity at Constant Volume: ∂E Heat Capacity at Constant Pressure:

More Common Measureable Properties Volume Expansion Coefficient: Note!! Reif’s notation for this is α Isothermal Compressibility: The Bulk Modulus is the inverse of the Isothermal Compressibility! B  (κ)-1

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

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: Converting this result to a partial derivative gives:

This can be rewritten as: The triple product rule is: Substituting gives:

Note again the definitions: Volume Expansion Coefficient β V-1(V/T)p Isothermal Compressibility κ  -V-1(V/p)T Note again!! Reif’s notation for the Volume Expansion Coefficient is α

A GENERAL RELATIONSHIP Using these in the previous expression finally gives the desired result: Using this result as a starting point, A GENERAL RELATIONSHIP between the Heat Capacity at Constant Volume CV & the Heat Capacity at Constant Pressure Cp can be found as follows:

Using the definitions of the isothermal compressibility κ and the volume expansion coefficient , this becomes General Relationship between Cv & Cp

Simplest Possible Example: The Ideal Gas 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. ν = (V/n) With this, it is simple to show that the volume expansion coefficient β & the isothermal compressibility κ are: and

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:

Other, Sometimes Useful, Expressions

More Applications: Using the Combined 1st & 2nd Laws (“The TdS Equations”) Calorimetry Again! Consider Two Identical Objects, each of mass m, & specific heat per kilogram cP. See figure next page. Object 1 is at initial temperature T1. Object 2 is at initial temperature T2. 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.

Two Identical Objects, of mass m, & specific heat per kilogram cP Two Identical Objects, of mass m, & specific heat per kilogram cP. Object 1 is at initial temperature T1. Object 2 is at initial temperature T2. 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. Q  Heat Flows For some time after initial contact: Object 2 Initially at T2 Object 1 Initially at T1 After a long enough time, the two objects are at the same temperature Tf. Since the 2 objects are identical, for this case,

The Entropy Change ΔS for this process can also be easily calculated: Of course, by the 2nd Law, the entropy change ΔS must be positive!! This requires that the temperatures satisfy:

Some Useful “TdS Equations” 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.

Given these definitions, it can be shown that the Combined 1st & 2nd Laws (TdS) can be written in at least the following ways:

Internal Energy Enthalpy h(T,P): Student exercise to show that, starting with the previous expressions & using the definitions (per mole) of internal energy u & enthalpy h gives: Internal Energy u(T,ν): Enthalpy h(T,P):

Student exercise also to show that similar manipulations give at least the following different expressions for the molar entropy s: Entropy s(T,ν):