Thermodynamic Potentials and Maxwell's Relations

• Unit 10 explores the thermodynamic potentials, specifically free energy, Helmholtz free energy, and Gibbs free energy.

Introduction

• Entropy increases in all natural processes
• There is a need to supplement the first and second laws of thermodynamics
• Focus is on the condition of thermodynamic equilibrium of systems.
• Introducing Enthalpy (H), Helmholtz Free Energy (F) and Gibbs Free Energy (G)
• U, H, F, and G are collectively called thermodynamic potentials or free energies.
• Each free energy has its own pair of natural variables
• Thermodynamic potentials possess a great deal of information about a system

Maxwell's Relations

• Maxwell's relations are useful for obtaining thermodynamic relations
• They relate quantities which are apparently unrelated
• They enable to link experimental data obtained in different ways
• They help replace a difficult measurement by an easier one
• They can be used to obtain values of one property from calculations or measurement of another property
• Maxwell's relations simplify thermodynamic analysis

Thermodynamic Potentials

• The thermodynamic behavior of a gas can be described using two variables out of p, V, and T.
• A two-coordinate system needs several functions of state: p, V, T, S, U, and H.
• Only a few combinations of functions of state have physical significance
• Helmholtz and Gibbs free energies have dimensions of energy, like internal energy and enthalpy
• Knowledge of two-coordinate system behavior can be obtained from any of the four free energies

Four Free Energies

• Internal Energy: U
• Enthalpy: H = U + pV
• Helmholtz Energy: F = U – TS = H – pV – TS
• Gibbs Energy: G = U – TS + pV = F + pV
• U, H, F, and G are collectively referred to as thermodynamic potentials or free energies
• Helmholtz energy is imporant since it provides a crucial connection between thermodynamics and statistical mechanics
• It provides a bridge between macroscopic and microscopic viewpoints
• Gibbs free energy finds applications in the study of phase transitions
• The significance of thermodynamic potentials becomes clearer from their differential forms

Differentials of Potential Functions

• For a gaseous system undergoing an infinitesimal reversible process.
• Change in Internal Energy: dU = TdS – pdV
• Small change in enthalpy: dH = dU + pdV + Vdp = TdS + Vdp
• Helmholtz Free Energy: dF = dU – (TdS + SdT) = – SdT – pdV

Entropy and Pressure Equations

• Dependence of F on independent variations of T and V: F = F(T,V)
• Entropy of constant V and constant T systems: S = –(∂F/∂T)V
• Pressure of constant V and constant T systems: p = –(∂F/∂V)T
• These relations show once F is known for a system the thermal properties of the system can be determined
• Helmholtz energy decreases when temperature increases because entropy is positive definite
• The Helmholtz energy decreases when volume increases
• Gibbs energy (G = F + pV): dG = –SdT + Vdp
• T and p are the natural variables for Gibbs energy: G = G(T, p)
• Entropy from Gibbs Energy: S = –(∂G/∂T)p
• Volume from Gibbs Energy: V = (∂G/∂p)T

Gibbs-Helmholtz Equation

• Internal energy in terms of variables: U = F – T(∂F/∂T)V = F + T²[∂(F/T)/∂T]V = ∂(F/T)/∂(1/T)V
• By substituting for S and p from Eqs. (10.5a) and (10.5b) respectively
• Enthalpy: H= F-T(∂F/∂T)V -V(∂F/∂V)T
• G = F -V²[∂(F/V)/∂(1/V)]T
• Entire information about a thermodynamic system can be obtained given Helmholtz free energy

Thermodynamic Free Energies

• Behavior of any pVT system can be described by internal energy, enthalpy, Helmholtz energy, and Gibbs energy
• Each free energy is associated with a natural pair of variables
  • U = U(S,V)
  • H = H(S,p)
  • F = F(T,V)
  • G = G(T,p)

Maxwell's Relations

• Can be readily applied by determining change to simplify measuring
• z is a function of state which depends on two independent state variables x and y dz=(∂z/∂x)dx+(∂z/∂y)dy = Mdx+Ndy ∂M/∂y=∂²z/∂y∂x ∂N/∂x=∂²z/∂x∂y ∂M/∂y=∂N/∂x

Maxwell's Relations from Thermodynamic Potentials

• The use of thermodynamic free energies produces Maxwell's relations
• T and V as independent variables requires reference to Eq. (10.4)
• Replacing z with F, M with -S, N with -p, x with T, and y with V in Eq. (10.11b): (∂S/∂V)T=(∂p/∂T)V
• Replacing G with z, -S with M, and V with N; with T identified with x and p with y: (∂S/∂p)T=(∂V/∂T)p

Maxwell's Relations Notations:

• Cross multiplication of the variables in the partial derivatives results in: (TS) = (pV)
  • Has dimensions of energy
• The independent variable of the partial differentiation on the left-hand side appears as a constant on the right-hand side and vice-versa
• The sign is positive if T appears with p in a partial derivative
• Pressure and volume variation of entropy in terms of partial derivatives can be studied involving extensive and intensive thermodynamic variables

Maxwell Relation Use Cases

• Explains the co-existence of two phases of a substance in equilibrium based on the first Maxwell relation
• The second relation explains the anomalous expansion of water when heated from 0°C to 4°C

Mnemonic Sentence for Maxwell's Relations

• Good Physicists Have Studied Under Very Fine Teachers: highlights a thermodynamic variable or free energy
• This is used to obtain the mnemonic diagram by placing the first letter of each word successively clockwise
• dG = ( ) dp – ( ) dT
• Gibbs Free Energy Equation: dG = Vdp – SdT

TdS Equations

• TdS-equations relate entropy of a substance to directly measurable quantities.
• These are provided that the equation of state and heat capacities are known.
• The independent variables dictate the TdS-equations.
• Expressing entropy of a substance as S = S(T,V) dS = (∂S/∂T)VdT+(∂S/∂V)TdV
  • Multiplying throughout by T TdS = T(∂S/∂T)VdT+T(∂S/∂V)TdV TdS = nCydT + T(∂p/∂T)vdV
• This equation is the first TdS-equation

Equations Involving Volume Expansivity

TdS = nCvdT + T (α/βT)dV

• (α/βT)dV = α dV α = 1/V(∂V/∂T)p and isothermal compressibility βT = -1/V(∂V/∂p)T
• In the system above the first equation expresses variation in entropy in terms of physically measurable quantities

Second TdS-Equation

• Using TdS=T(∂S/∂T)pdT + T(∂S/∂p)Tdp with p and T as independent variables: TdS = nCpdT –T(∂V/∂T)Tdp
• This is the second TdS-equation
• This equation can be rewritten in terms of volumne expansion: TdS = nCpdT -TVαdp

Third TdS-Equation

• By taking V and p as independent variables and writing S = S(p,V): TdS = T(∂S/∂p)Vdp+T(∂S/∂V)pdV = nCydp+nCpdV
• Hydrostatic System equations TdS = nCydT + T(∂p/∂V)dV TdS = nCpdT – T(∂V/∂p)dp

Energy Equations

• Energy equations provide a way to study how internal energy changes in volume, temperature, or pressure by using Maxwell's relations

The First Energy Equation

• Dividing dU = TdS – pdV by dV, derive: (∂U/∂V) =T(∂S/∂V) - p
• If T is held constant, this turns into partial derivatives: (∂U/∂V)T = T(∂S/∂V)T - p
• Apply Maxwell's relation (Eq. (10.12a)) (∂U/∂V)T = T(∂p/∂T)V - p
• The earlier result is the so-called first energy equation
• In an ideal gas U depends on only T

The Second Energy Equation

• Dividing Eq. (10.2) by dp and using Eq. (10.12b) results in (∂U/∂p) = T(∂V/∂T) - p((∂V/∂p)

Clausius-Clapeyron Equation

• (∂p/∂T) =(l/(T(vvap - V liq)
• This is one of the most important formulae in thermodynamics
• It gives the rate at which vapour pressure changes with temperature for two phases to coexist in equilibrium

Implications for Phase Changes

• Increase in pressure raises boiling point and vice-versa, since Vvap > Vliq
• This explains why food cooks faster in a pressure cooker, and is difficult to cook at high altitudes
• The arguments for Eq. (10.19) may be extended to phase change (solid-liquid, liquid-vapor and solid-vapor transition) involving latent heat
• First-order phase transitions involve the matter's change of phase with absorption or release of latent heat at constant temperature.
• Slope of the solid-liquid curve is positive implies materials expand on melting and_dp/dT is positive
• Water expands on freezing and its melting point decreases as pressure increases

Joule-Thomson Effect

• Gases have molecules that experience molecualr attraction in addition to having a finite size
• If intermolecular forces exist, gas expansion involves work done and lowers energy
• Through experiments Joule and Thompson found that if the temperature is below a certain inversion during adiabatic expansion, cooling occurs
• The Joule-Thomson effect is when gases experience cooling

Summaries of Joule-Thomson Effects

• All gases showed change in temperature on passing through porous plug
• All gases except helium and hydrogen showed cooling at ordinary temperatures
• All gases showed cooling at low enough temperatures
• The fall in temperatures varied with the pressure differential on each side of the porous plug.

Joule-Thompson Coefficient

• Defined as μ = ∆T/∆p
• Perfect gas (a=b=0) results in a zero joule-thompson coefficient