Chapter 8: Free Energies Helmholtz Free Energy, F Gibbs Free Energy, G Enthalpy, H Table 8.1

I. Equilibrium – Ch. 7 In Ch. 7, we started with differences in T, p or μ between two subsystems. The total system was isolated from the surroundings but U, V or N were exchanged (the variables) between the subsystems. As a result, equilibrium was associated with maximum S(U, V, N).

Equilibrium – Ch. 8 Now consider a sealed test tube in which V and N are constant. This test tube is in a constant water bath, so heat (= energy) is exchanged to maintain constant T. (Confirm using First Law) When T, V and N are the variables, then equilibrium is associated with minimum F(T, V, N): dF ≤ 0. Eqns 8.1 – 8.6

Natural Variables There is a distinction between “variables” and “natural variables” of thermodynamic functions (TF). When a TF is expressed in terms of its natural variables, then the TF can be used to find equilibrium. I.e. the Extremum Principle is justified. –Max S(U, V, N) but not S(T, V, N) –Min G(T, p, N) but not G(U, V, N)

Thermodynamic Function, F F(T, V, N) = U – TS is called the Helmhotz Free Energy. Then dF = dU – T dS ≤ 0 and equilibrium is associated with min U and max S at constant T. dF = (δF/δT) V,N dT + (δF/δV) T,N dV + Σ(δF/δN j ) V,T,Ni dN j = - S dT – p dV + Σμ j dN j Examples 8.1, 8.2

Thermodynamic Function, H H(S, p, N) = U + pV is called Enthalpy. dH = T dS + V dp + Σμ j dN j Eqn 8.22 ΔH values are experimentally accessible (calorimetry).

Thermodynamic Function, G G(T, p, N) = H –TS is called the Gibbs Free Energy. dG = -S dT + V dp + Σμ j dN j Eqn 8.25 Then dG = dH – T dS ≤ 0 and equilibrium is associated with min H and max S at constant T. Note Table 8.1

II. More Applications Ch 7 Tools –In an adiabatic process, w  ΔU (dU = δw) –Using the First Law, q  ΔS (δq = dS/T) (Carnot Cycle) Ch 8 Tools –Heat Capacity –Cycles

Heat Capacity  ΔU, ΔS  ΔF q = heat α ΔT and the proportionality constant is specific heat (J/g-K) or heat capacity (C, J/mol-K) Note C may = f(T); Table 8.2 When ΔV = 0 (bomb calorimeter), dU = δq. Then C V = (δq/dT) V = (∂U/∂T) V = T(∂S/∂T) V dU = C V dT and dS = C V dT /TEqn 8.31 Ex 8.5, 8.6

Heat Capacity  ΔH, ΔS  ΔG When Δp = 0, then dH = δq. Then C p = (δq/dT) p = (∂H/∂T) p = T(∂S/∂T) p dH = C p dT and dS = C p dT /TEqn 8.34

Cycles (ΔTF = 0)  non- measurables Determine ΔH for a phase change not at equilibrium. Internal combustion engine –Adiabatic gas expansion –Otto Cycle (note compression ratio)