Unit 3: Maxwell’s Thermodynamic Relations and Phase Transitions

1. Maxwell’s Thermodynamic Relations and Derivations

Maxwell’s thermodynamic relations are a set of fundamental equations in thermodynamics derived from the definitions of thermodynamic potentials. They relate state variables such as temperature (T), pressure (P), volume (V), and entropy (S) to one another. These relations are essential because they allow us to express experimentally difficult-to-measure quantities (like entropy changes) in terms of easily measurable parameters (like temperature, pressure, and volume).

Thermodynamic Potentials and Exact Differentials

The derivations rely on Euler’s reciprocity theorem for exact differentials. If a thermodynamic state function Z is expressed in terms of two independent variables x and y as:

dZ = M dx + N dy

Where M and N are functions of x and y, then because Z is a state function, dZ is an exact differential. Therefore, the partial derivatives satisfy the condition:

(∂M/∂y)_x = (∂N/∂x)_y

Derivation of the First Maxwell Relation (From Internal Energy, U)

The first law of thermodynamics combined with the second law gives the change in internal energy (dU) for a reversible process:

dU = T dS - P dV

Here, U is a function of entropy (S) and volume (V), meaning U = U(S, V). The exact differential form is:

dU = (∂U/∂S)_V dS + (∂U/∂V)_S dV

Comparing coefficients, we get:

• T = (∂U/∂S)_V
• -P = (∂U/∂V)_S

Applying the reciprocity theorem:

(∂T/∂V)_S = -(∂P/∂S)_V

Physical Interpretation: This relation connects the temperature change during an isentropic (adiabatic) volume change to the pressure change during an isentropic entropy change.

Derivation of the Second Maxwell Relation (From Enthalpy, H)

Enthalpy is defined as:

H = U + PV

Differentiating both sides:

dH = dU + P dV + V dP

Substitute dU = T dS - P dV into the expression:

dH = (T dS - P dV) + P dV + V dP = T dS + V dP

Here, H is a function of S and P, meaning H = H(S, P). The exact differential form is:

dH = (∂H/∂S)_P dS + (∂H/∂P)_S dP

Comparing coefficients:

• T = (∂H/∂S)_P
• V = (∂H/∂P)_S

Applying the reciprocity theorem:

(∂T/∂P)_S = (∂V/∂S)_P

Derivation of the Third Maxwell Relation (From Helmholtz Free Energy, F)

Helmholtz Free Energy is defined as:

F = U - TS

Differentiating both sides:

dF = dU - T dS - S dT

Substitute dU = T dS - P dV:

dF = (T dS - P dV) - T dS - S dT = -S dT - P dV

Here, F is a function of T and V, meaning F = F(T, V). The exact differential form is:

dF = (∂F/∂T)_V dT + (∂F/∂V)_T dV

Comparing coefficients:

• -S = (∂F/∂T)_V
• -P = (∂F/∂V)_T

Applying the reciprocity theorem:

(∂S/∂V)_T = (∂P/∂T)_V

Derivation of the Fourth Maxwell Relation (From Gibbs Free Energy, G)

Gibbs Free Energy is defined as:

G = H - TS

Differentiating both sides:

dG = dH - T dS - S dT

Substitute dH = T dS + V dP:

dG = (T dS + V dP) - T dS - S dT = -S dT + V dP

Here, G is a function of T and P, meaning G = G(T, P). The exact differential form is:

dG = (∂G/∂T)_P dT + (∂G/∂P)_T dP

Comparing coefficients:

• -S = (∂G/∂T)_P
• V = (∂G/∂P)_T

Applying the reciprocity theorem:

(∂S/∂P)_T = -(∂V/∂T)_P

Summary Table of Maxwell’s Relations

2. Application: Clausius-Clapeyron Latent Heat Equation

The Clausius-Clapeyron equation governs the phase boundary between two phases of matter. It determines how the equilibrium pressure (vapor pressure, melting pressure) varies with temperature.

Derivation Using Maxwell’s Third Relation

Consider a system undergoing a first-order phase transition (e.g., liquid changing to vapor) at constant temperature T and pressure P. Maxwell's third thermodynamic relation states:

(∂S/∂V)_T = (∂P/∂T)_V

Since phase transitions occur at constant temperature and pressure, the change in entropy (dS) is directly related to the absorption or release of latent heat (L):

dS = dQ / T = L / T

For a change from Phase 1 to Phase 2, the finite change in entropy is:

S2 - S1 = L / T

The corresponding change in volume is:

V2 - V1

Substituting these values into the left-hand side of Maxwell's third relation:

(S2 - S1) / (V2 - V1) = dP / dT

Substitute the entropy change in terms of latent heat:

dP / dT = L / [ T (V2 - V1) ]

This is the Clausius-Clapeyron equation.

Physical Significance and Applications

• Boiling Point of Liquids: For vaporization, the volume of vapor (V2) is much greater than liquid (V1), so (V2 - V1) is positive. Since latent heat L is positive, dP/dT is positive. An increase in pressure increases the boiling point of a liquid.
• Melting Point of Solids:
  • For substances that expand upon melting (most solids, like wax): V2 > V1, making dP/dT positive. Increased pressure increases the melting point.
  • For substances that contract upon melting (like ice): V2 < V1, making dP/dT negative. Increased pressure decreases the melting point of ice (this explains glacier sliding and ice skating).

3. Application: Values of Cp - Cv (Difference in Heat Capacities)

The principal heat capacities of a substance are Cp (heat capacity at constant pressure) and Cv (heat capacity at constant volume). We use Maxwell's relations to find a general mathematical expression for their difference.

General Derivation

By definition, heat capacities are written in terms of entropy as:

Cp = T (∂S/∂T)_P
Cv = T (∂S/∂T)_V

Let entropy S be a function of temperature T and volume V: S = S(T, V). The total differential of S is:

dS = (∂S/∂T)_V dT + (∂S/∂V)_T dV

Divide both sides by dT at constant pressure P:

(∂S/∂T)_P = (∂S/∂T)_V + (∂S/∂V)_T (∂V/∂T)_P

Multiply the entire equation by T:

T (∂S/∂T)_P = T (∂S/∂T)_V + T (∂S/∂V)_T (∂V/∂T)_P

Substituting the definitions of Cp and Cv:

Cp - Cv = T (∂S/∂V)_T (∂V/∂T)_P

Using Maxwell’s third relation, substitute (∂S/∂V)_T = (∂P/∂T)_V:

Cp - Cv = T (∂P/∂T)_V (∂V/∂T)_P

This is the general equation for the difference in heat capacities. To express this in terms of measurable quantities, we use the triple product cyclic relation for P, V, T:

(∂P/∂T)_V (∂T/∂V)_P (∂V/∂P)_T = -1

Rearranging to solve for (∂P/∂T)_V:

(∂P/∂T)_V = -(∂V/∂T)_P / (∂V/∂P)_T

Substitute this back into the heat capacity equation:

Cp - Cv = -T [ (∂V/∂T)_P ]² / (∂V/∂P)_T

Let us define the coefficient of volume expansion (α) and isothermal compressibility (β):

α = (1/V) (∂V/∂T)_P
β = -(1/V) (∂V/∂P)_T

Substituting these definitions gives:

Cp - Cv = T V α² / β

Important Observation: Since T, V, α², and β are always positive for stable physical systems, Cp is always greater than Cv (Cp > Cv). At absolute zero (T = 0), Cp = Cv.

Case 1: Ideal Gas

For 1 mole of an ideal gas, the equation of state is:

PV = RT

Let us compute the required partial derivatives:

1. Differentiating PV = RT with respect to T at constant P: (∂V/∂T)_P = R/P
2. Differentiating PV = RT with respect to T at constant V: (∂P/∂T)_V = R/V

Substitute these into the general equation Cp - Cv = T (∂P/∂T)_V (∂V/∂T)_P:

Cp - Cv = T (R/V) (R/P) = T R² / (PV)

Since PV = RT, this simplifies to:

Cp - Cv = R

This is the familiar Mayer’s formula for an ideal gas.

Case 2: Van der Waals Gas

For 1 mole of a Van der Waals gas, the equation of state is:

(P + a/V²) (V - b) = RT

Expressing P explicitly:

P = RT / (V - b) - a/V²

Differentiating with respect to T at constant V:

(∂P/∂T)_V = R / (V - b)

To find (∂V/∂T)_P, we differentiate the pressure equation with respect to T at constant P:

0 = [ R / (V - b) ] - [ R T / (V - b)² ] (∂V/∂T)_P + [ 2a / V³ ] (∂V/∂T)_P

Rearranging terms to solve for (∂V/∂T)_P:

(∂V/∂T)_P [ R T / (V - b)² - 2a / V³ ] = R / (V - b)

(∂V/∂T)_P = [ R / (V - b) ] / [ R T / (V - b)² - 2a / V³ ]

Multiplying the numerator and denominator by (V-b)² / (RT):

(∂V/∂T)_P = [ (V - b) / T ] / [ 1 - 2a(V - b)² / (R T V³) ]

Using the general relation Cp - Cv = T (∂P/∂T)_V (∂V/∂T)_P:

Cp - Cv = T [ R / (V - b) ] * [ (V - b) / T ] / [ 1 - 2a(V - b)² / (R T V³) ]

Cp - Cv = R / [ 1 - 2a(V - b)² / (R T V³) ]

Since b is very small compared to V, we can approximate V - b as V:

Cp - Cv ≈ R / [ 1 - 2a / (R T V) ]

Using the binomial expansion (1 - x)^(-1) ≈ 1 + x for x << 1:

Cp - Cv ≈ R [ 1 + 2a / (R T V) ]

Observation: For a Van der Waals gas, the difference in heat capacities is greater than R, and it depends on both temperature and volume.

4. Application: TdS Equations

The TdS equations are useful for calculating heat changes (since dQ = TdS for reversible processes) when a system undergoes changes in its independent state variables.

First TdS Equation (Independent Variables T and V)

Let entropy S be a function of T and V: S = S(T, V). The total differential of S is:

dS = (∂S/∂T)_V dT + (∂S/∂V)_T dV

Multiply by T:

T dS = T (∂S/∂T)_V dT + T (∂S/∂V)_T dV

Using Cv = T (∂S/∂T)_V and Maxwell's third relation (∂S/∂V)_T = (∂P/∂T)_V:

T dS = Cv dT + T (∂P/∂T)_V dV

This is the First TdS Equation.

Second TdS Equation (Independent Variables T and P)

Let entropy S be a function of T and P: S = S(T, P). The total differential of S is:

dS = (∂S/∂T)_P dT + (∂S/∂P)_T dP

Multiply by T:

T dS = T (∂S/∂T)_P dT + T (∂S/∂P)_T dP

Using Cp = T (∂S/∂T)_P and Maxwell's fourth relation (∂S/∂P)_T = -(∂V/∂T)_P:

T dS = Cp dT - T (∂V/∂T)_P dP

This is the Second TdS Equation.

5. Application: Joule-Kelvin Coefficient (Ideal and Van der Waals Gases)

The Joule-Kelvin (Joule-Thomson) effect describes the temperature change of a real gas when it is forced through a multi-porous plug or throttling valve from a region of high pressure to low pressure under adiabatic conditions. This is an isenthalpic process (constant enthalpy, dH = 0).

Derivation of the Joule-Kelvin Coefficient (μ_JK)

The Joule-Kelvin coefficient is defined as the change in temperature with pressure at constant enthalpy:

μ_JK = (∂T/∂P)_H

To express this in terms of physical variables, we write the exact differential of enthalpy H(T, P):

dH = (∂H/∂T)_P dT + (∂H/∂P)_P dP

By definition, Cp = (∂H/∂T)_P. From dH = T dS + V dP, we can write:

(∂H/∂P)_T = T (∂S/∂P)_T + V

Using Maxwell’s fourth relation, (∂S/∂P)_T = -(∂V/∂T)_P:

(∂H/∂P)_T = -T (∂V/∂T)_P + V

Substitute these back into the differential for dH:

dH = Cp dT + [ V - T (∂V/∂T)_P ] dP

Since the process is isenthalpic, dH = 0:

0 = Cp dT + [ V - T (∂V/∂T)_P ] dP

Cp dT = [ T (∂V/∂T)_P - V ] dP

Solving for (∂T/∂P)_H:

μ_JK = (∂T/∂P)_H = (1/Cp) [ T (∂V/∂T)_P - V ]

This is the general equation for the Joule-Kelvin coefficient.

Case 1: Ideal Gas

For an ideal gas, PV = RT, which implies V = RT / P.

Differentiating with respect to T at constant P:

(∂V/∂T)_P = R / P

Substitute this into the expression for μ_JK:

μ_JK = (1/Cp) [ T (R/P) - V ]

Since TR/P = V, the bracketed term becomes:

μ_JK = (1/Cp) [ V - V ] = 0

Conclusion: An ideal gas experiences no temperature change (neither heating nor cooling) during a Joule-Kelvin expansion because there are no intermolecular forces to overcome.

Case 2: Van der Waals Gas

For a Van der Waals gas, we use the approximation for volume derived from its state equation at low to moderate pressures:

V ≈ RT/P + b - a/(RT)

Differentiating with respect to T at constant pressure:

(∂V/∂T)_P ≈ R/P + a/(R T²)

Multiply this derivative by T:

T (∂V/∂T)_P ≈ RT/P + a/(RT)

Substitute RT/P ≈ V - b + a/(RT) back into the equation:

T (∂V/∂T)_P ≈ V - b + 2a / (RT)

Substitute this into the general equation for μ_JK:

μ_JK = (1/Cp) [ V - b + 2a / (RT) - V ]

μ_JK = (1/Cp) [ 2a / (RT) - b ]

Discussion of Temperature of Inversion (Ti)

The behavior of a real gas depends on the term [ 2a / (RT) - b ]:

• Cooling (μ_JK > 0): If 2a / (RT) > b, then μ_JK is positive. A drop in pressure (dP < 0) leads to a drop in temperature (dT < 0).
• Heating (μ_JK < 0): If 2a / (RT) < b, then μ_JK is negative. A drop in pressure (dP < 0) leads to an increase in temperature (dT > 0).
• No Temperature Change (μ_JK = 0): If 2a / (RT) = b, there is no temperature change. The temperature at which this occurs is called the Temperature of Inversion (Ti):

Ti = 2a / (R b)

6. Application: Temperature Change during Adiabatic Processes

An adiabatic process occurs without heat exchange with the surroundings, meaning dQ = 0, and thus the process is isentropic (dS = 0). We can determine how temperature changes with pressure and volume during such processes using Maxwell’s relations.

Change of Temperature with Pressure: (∂T/∂P)_S

From Maxwell’s second relation:

(∂T/∂P)_S = (∂V/∂S)_P

Using the chain rule for the derivative on the right:

(∂V/∂S)_P = (∂V/∂T)_P (∂T/∂S)_P

Since Cp = T (∂S/∂T)_P, we have (∂T/∂S)_P = T / Cp. Substituting this yields:

(∂T/∂P)_S = (T / Cp) (∂V/∂T)_P

Physical Analysis: For most substances, the thermal expansion coefficient is positive, meaning (∂V/∂T)_P > 0. Since T and Cp are positive, (∂T/∂P)_S > 0. This means that adiabatic compression (increasing pressure) causes temperature to rise, while adiabatic expansion (decreasing pressure) causes temperature to fall.

Change of Temperature with Volume: (∂T/∂V)_S

From Maxwell’s first relation:

(∂T/∂V)_S = -(∂P/∂S)_V

Using the chain rule for the derivative on the right:

(∂P/∂S)_V = (∂P/∂T)_V (∂T/∂S)_V

Since Cv = T (∂S/∂T)_V, we have (∂T/∂S)_V = T / Cv. Substituting this yields:

(∂T/∂V)_S = -(T / Cv) (∂P/∂T)_V

Physical Analysis: For typical substances, pressure increases when heated at constant volume, so (∂P/∂T)_V > 0. Since T and Cv are positive, (∂T/∂V)_S < 0. This indicates that an increase in volume during an adiabatic process (adiabatic expansion) leads to a decrease in temperature.

7. Phase Transitions: First and Second Order

A phase transition is a physical process where a thermodynamic system changes from one phase or state of matter to another. Paul Ehrenfest classified phase transitions based on the behavior of the Gibbs free energy function and its derivatives at the transition temperature.

First-Order Phase Transitions

A first-order phase transition is characterized by a discontinuity in the first-order derivatives of the Gibbs free energy with respect to temperature and pressure. The Gibbs free energy itself remains continuous across the boundary.

• Entropy Discontinuity: Since S = -(∂G/∂T)_P, a jump in the first derivative of G means the entropy changes discontinuously. This discontinuity is associated with latent heat (L = T ΔS).
• Volume Discontinuity: Since V = (∂G/∂P)_T, the volume changes discontinuously at the transition point.

Examples:

1. Melting of ice to liquid water.
2. Boiling of water to steam.
3. Sublimation of dry ice.

Second-Order Phase Transitions

In a second-order phase transition, both the Gibbs free energy and its first-order derivatives (entropy and volume) are continuous. However, its second-order derivatives with respect to temperature and pressure are discontinuous or singular at the transition point.

• No Latent Heat: Since entropy is continuous (ΔS = 0), there is no latent heat associated with a second-order phase transition.
• Discontinuous Physical Properties: The second derivatives of G correspond to the heat capacity Cp, thermal expansion coefficient α, and isothermal compressibility β:
  • Cp = -T (∂²G/∂T²)_P
  • α = (1/V) (∂²G/∂P∂T)
  • β = -(1/V) (∂²G/∂P²)_T
  These properties change abruptly at the transition point.

Examples:

1. The transition of liquid Helium I to Helium II (the lambda transition).
2. The transition of a ferromagnetic material to a paramagnetic material at the Curie temperature.
3. The transition of a metal to a superconductor in the absence of a magnetic field.

Comparison of Phase Transitions