Chapter 5 Simple Applications of Macroscopic Thermodynamics

Preliminary DiscussionClassical, Macroscopic, ThermodynamicsDrop the statistical mechanics notation for average quantities. We know that All Variables are Averages Only! Well discuss relationships between macroscopic variables usingThe Laws of ThermodynamicsSome Thermodynamic Variables of Interest:Internal Energy = E, Entropy = S, Temperature = TFor Gases:External Parameter = V, Generalized Force = p(V = volume, p = pressure)For A General System:External Parameter = x, Generalized Force = X

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: Q = dE + pdV 2nd Law: Q = TdSThe Combined 1st & 2nd LawsTdS = dE + pdVNote that, in this relation, there are 5 Variables:T, S, E, p, VIt 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.

Now, A Brief, Pure Math DiscussionConsider 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)Suppose instead that we want to take y & z as independent variables. Then, There Must be a Function x = 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/xy) (2z/yx), etc.

Mathematics SummaryConsider a function of 2 independent variables: f = f(x1,x2).Its 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 theCombined 1st & 2nd Laws of Thermodynamics:TdS = dE + pdV

Some Methods & Useful Math Toolsfor Transforming Derivatives

Derivative Inversion

Triple Product (xyz1 rule)

Chain Rule Expansion to Add Another Variable

Maxwell Reciprocity Relationship

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

Jacobian TransformationsHave Several Useful Properties

TranspositionInversionChain Rule Expansion

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 and the inversion property

dE = TdS pdV (1)First, choose S & V as independent variables:E E(S,V)Properties of the Internal Energy EComparison of (1) & (2) clearly shows thatdEE(2)EEApplying the general result with 2nd cross derivatives gives:Maxwell Relation I!Eand

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

If T & V are chosen as independent variables, it is convenient to define the following energy:F F(T,V) E - TS Helmholtz Free EnergyUse 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 -pApplying the general result for the 2nd cross derivatives gives:Maxwell Relation III!

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 EnergyUse the combined 1st & 2nd Laws. Rewrite them in terms of dH: dE = TdS pdV = d(TS) - SdT [d(pV) Vdp] ordG = -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 VApplying the general result for the 2nd cross derivatives gives: Maxwell Relation IV!

1. Internal Energy:E E(S,V)2. Enthalpy: H = H(S,p) E + pV3. Helmholtz Free Energy: F = F (T,V) E TS4. Gibbs Free Energy: G = G(T,p) E TS + pVSummary: Energy FunctionsCombined 1st & 2nd Laws1. dE = TdS pdV2. dH = TdS + Vdp3. dF = - SdT pdV4. dG = - SdT + Vdp

Another Summary: Maxwell Relations(a) E = Q + W(b) S = (Qres/T)(c) H = E + pV(d) F = E TS(e) G = H - TS1. dE = TdS pdV2. dH = TdS + Vdp3. dF = -SdT - pdV4. dG = -SdT + Vdp

*Maxwell Relations: The Magic Square?VFTGPHSEEach side is labeled with anEnergy (E, H, F, G). The corners are labeled withThermodynamic Variables(p, V, T, S). Get theMaxwell Relationsby walking around thesquare. The partial derivatives are obtainedfrom the sides. TheMaxwell Relations are obtained from the corners.

*SummaryThe 4 Most CommonMaxwell Relations:

*Maxwell Relations: Table (E U)

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

Internal EnergyHelmholtz Free EnergyEnthalpyGibbs Free Energy

Some Common Measureable PropertiesHeat Capacity at Constant Volume:Heat Capacity at Constant Pressure:Volume Expansion Coefficient:Isothermal Compressibility:E Note!!Reifs notationfor this is The Bulk Modulus is inverse of theIsothermal Compressibility!! B ()-1

Some Sometimes Useful RelationshipsSummary of ResultsDerivations are in the text and/or left to the student!

Entropy:Enthalpy:Gibbs Free Energy:

A Typical ExampleGiven 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 aMaxwell Relation:

Combining these expressions gives:

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 V-1(V/T)pIsothermal Compressibility -V-1(V/p)TNote!! Reifsnotation forthis is

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:

Simplest Possible Example: The Ideal GasFor an Ideal Gas, its easily shown (Reif) that the Equation of State (relation between pressure P, volume V, temperature T) is (in per mole units!): P = RT. With this, it is simple to show that the volume expansion coefficient & the isothermal compressibility are: and

andSo, for an Ideal Gas, the volume expansion coefficient & the isothermal compressibility have the simple forms: 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)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. 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.

After a long enough time, the two objects are at the same temperature Tf. Since the 2 objects are identical, for this case, Object 1Initially at T1Object 2Initially at T2Q Heat FlowsFor some timeafter the initialcontact:

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

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

Internal Energy u(T,) Enthalpy h(T,P)

Entropy123

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