Lecture 7: Thermodynamic Potentials

Chapter II. Thermodynamic Quantities

A.G. Petukhov, PHYS 743

September 27, 2017

Chapter II. Thermodynamic Quantities Lecture 7: Thermodynamic Potentials A.G. Petukhov,September PHYS 27, 743 2017 1 / 11 Equation of State Parameters p, V, T (and additional external fields, if any) are thermodynamic parameters that specify a thermodynamic state. In equilibrium there is an equation f(p, V, T ) = 0, which relates three thermodynamic parameters p, V and T ⇒ p = P (V,T ). This equation is called ”the equation of state”. For example:

pV = RT

is the equation of state for one mole of an ideal gas. Energy (Natural variables S and V ). Consider the energy differential (first law of thermodynamics for quasi-static processes):

∂E  ∂E  dE = dS + dV = T dS − pdV (1) ∂S V ∂V S The energy differential dE is an exact differential and we can read off the partial derivatives:

Chapter II. Thermodynamic Quantities Lecture 7: Thermodynamic Potentials A.G. Petukhov,September PHYS 27, 743 2017 2 / 11 Conjugate Thermodynamic Variables

∂E  T = = T (S, V ) (2) ∂S V ∂E  p = − = p(S, V ) (3) ∂V S Therefore if we know E as a function of S and V we can use Eqs. (1)-(3) to relate four thermodynamic variables S,V ,T , and p by means of two equations (2) and (3). If we eliminate S we can find the equation of state: p = p(T,V )

There are only two independent thermodynamic variables It is convenient to treat S on the same footing with p, V and T . The pairs (S, T ) and (p, V ) are called conjugate thermodynamic parameters (variables). Each pair contains one extensive and one intensive variable Chapter II. Thermodynamic Quantities Lecture 7: Thermodynamic Potentials A.G. Petukhov,September PHYS 27, 743 2017 3 / 11 Characteristic Functions (Thermodynamic Potentials)

Knowledge of E(S, V ) allows us completely describe thermodynamics of the system. That is why E is called a characteristic function (thermodynamic potential) and S, V are called its natural variables. Note that if we knew E as a function of T,V we would not be able to obtain the equation of state. Example: from E = 3NT/2 for an ideal gas we cannot find p, i.e. it would be impossible to obtain the equation of state. It is possible to find three more thermodynamic potentials for which (T, p), (p, S), and (V,T ) are natural variables (arguments). It means that each thermodynamic potential is expressed as an exact differential of its natural arguments with partial derivatives equal to their conjugates. The procedure which allows us to generate other thermodynamic potentials is called Legendre Transformation

Chapter II. Thermodynamic Quantities Lecture 7: Thermodynamic Potentials A.G. Petukhov,September PHYS 27, 743 2017 4 / 11 Thermodynamic Potentials (TP) and Legendre Transformation Enthalpy (Natural Variables S and p). We need to perform Legendre transformation from (S, V ) to (S, p): dE = T dS − pdV = T dS − pdV − V dp + V dp = T dS + V dp − d(pV ) or d(E + pV ) = T dS + V dp ⇒ W = E + pV ← Enthalpy Thus ∂W  ∂W  dW = T dS + V dp, where = T ; = V ∂S p ∂p S Using E and W we can calculate specific heats:  ∂S  ∂E  cV = T = ∂T V ∂T V  ∂S  ∂W  cp = T = ∂T p ∂T p Chapter II. Thermodynamic Quantities Lecture 7: Thermodynamic Potentials A.G. Petukhov,September PHYS 27, 743 2017 5 / 11 Thermodynamic Potentials cont’d

Helmholtz Free Energy (Natural Variable T and V ). We need Legendre transformation from (S, V ) to (T,V ):

dE = T dS−pdV = T dS−pdV +SdT −SdT = −SdT +pdV +d(TS)

or

d(E−TS) = −SdT −pdV ⇒ F = E − TS ← Helmholtz Free Energy

Thus ∂F  ∂F  dF = −SdT − pdV, where = −p; = −S ∂V T ∂T V

Chapter II. Thermodynamic Quantities Lecture 7: Thermodynamic Potentials A.G. Petukhov,September PHYS 27, 743 2017 6 / 11 Thermodynamic Potentials cont’d Gibbs Free Energy (Natural Variable T and p). We need Legendre transformation from (T,V ) to (T, p):

dF = −SdT −pdV = −SdT −pdV −V dp+V dp = −SdT +V dp−d(pV )

or

d(F + pV ) = −SdT + V dp ⇒ Φ = F + pV ← Gibbs Free Energy

Thus ∂Φ ∂Φ dΦ = −SdT + V dp, where = −S; = V ∂T p ∂p T

Φ = F + pV = E − TS + pV Remark: In the literature they often use G instead of Φ. Also sometimes they call Φ the Thermodynamic Potential while other three are called by the same name but in general sense.

Chapter II. Thermodynamic Quantities Lecture 7: Thermodynamic Potentials A.G. Petukhov,September PHYS 27, 743 2017 7 / 11 General Diagram We can relate W and Φ as: ∂Φ  ∂ Φ W = Φ + TS = Φ − T = T 2 ∂T p ∂T T p The following diagram is useful to generate eight Maxwell relations:

∂F  ∂E  ∂E  ∂W  p = − = − T = = ∂V T ∂V S ∂S V ∂S p ∂W  ∂Φ ∂Φ ∂F  V = = S = − = − ∂p S ∂p T ∂T p ∂T V

Chapter II. Thermodynamic Quantities Lecture 7: Thermodynamic Potentials A.G. Petukhov,September PHYS 27, 743 2017 8 / 11 Small Increment Theorem

One can introduce other variables and forces such as X dE = T dS − pdV − Λidλi i Since Legendre transformations for S, T, p, V do not affect λ, Λ we can obtain Λi using any thermodynamic potential: ∂E  ∂F  ∂W  ∂Φ −Λ = = = = ∂λ S,V ∂λ T,V ∂λ S,p ∂λ T,p Thus if λ changes slightly we obtain The small increment theorem:

−Λdλ = (δE)S,V = (δF )T,V = (δW )S,p = (δΦ)T,p ,

Chapter II. Thermodynamic Quantities Lecture 7: Thermodynamic Potentials A.G. Petukhov,September PHYS 27, 743 2017 9 / 11 Minimal Values of TP in Equilibrium We know that

T dS ≥ δQ = dE + pdV ⇒ d(TS) ≥ dE + pdV + SdT

V,T =const dF ⇒ d(E − TS) ≤ −pdV − SdT ⇒ ≤ 0 dt V,T =const Therefore at T,V = const this inequality always holds with ”=” sign corresponding to equilibrium. We can obtain similar inequalities E,W and Φ. Thus for all TP we have:

dE dW ≤ 0 ≤ 0 dt V,S=const dt p,S=const

dF dΦ ≤ 0 ≤ 0 dt V,T =const dt p,T =const Under different external conditions the thermodynamic potentials have minima in equilibrium, i.e. behave analogously to the potential energy in mechanical systems Chapter II. Thermodynamic Quantities Lecture 7: Thermodynamic Potentials A.G. Petukhov,September PHYS 27, 743 2017 10 / 11 Mixed Derivatives Using the properties of mixed derivatives:

∂f(x, y) ∂f(x, y) = ∂x∂y ∂y∂x or !  ∂ ∂f(x, y)  ∂ ∂f(x, y) = ∂x ∂y ∂y ∂x x y y x We can find four more Maxwell relations. For instance:  ∂ ∂E    ∂ ∂E   = ∂S ∂V S V ∂V ∂S V S Or  ∂p   ∂T  − = ∂S V ∂V S In the same fashion we can proceed with other relations (Read L&L !)

Chapter II. Thermodynamic Quantities Lecture 7: Thermodynamic Potentials A.G. Petukhov,September PHYS 27, 743 2017 11 / 11