Clausius-Clapeyron equation

Derivation and physical significance of the phase-equilibrium relation

The phase-equilibrium curve shown in a pressure–temperature (P–T) diagram can be described mathematically by a differential equation known as the Clausius–Clapeyron equation. It gives the relationship between pressure and temperature during a phase transition, such as melting, vaporisation or sublimation.

1. Derivation of the Clausius–Clapeyron Equation

Consider two points B and C lying very close to each other on a phase-equilibrium curve.

• At point B, the temperature and pressure are T and p, respectively.
• At point C, the temperature and pressure are T + dT and p + dp, respectively.

Since the two phases are in thermodynamic equilibrium at point B, their Gibbs free energies per molecule are equal:

g₁(T, p) = g₂(T, p) (1)

Similarly, at point C:

g₁(T + dT, p + dp) = g₂(T + dT, p + dp)

For small changes in temperature and pressure, the Gibbs free energies change by dg₁ and dg₂. Therefore,

g₁(T, p) + dg₁ = g₂(T, p) + dg₂ (2)

Using equation (1) in equation (2), we obtain

dg₁ = dg₂ (3)

2. Change in Gibbs Free Energy

The Gibbs free energy for phase i is defined as

G_i = E_i − TS_i + pV_i

Dividing by the number of molecules N, the Gibbs free energy per molecule is

g_i = e_i − Ts_i + pv_i (4)

Where:

• e_i = internal energy per molecule
• s_i = entropy per molecule
• v_i = volume per molecule
• T = absolute temperature
• p = pressure

Differentiating equation (4),

dg_i = de_i − T ds_i − s_i dT + p dv_i + v_i dp

According to the first and second laws of thermodynamics, for a reversible change involving only pressure–volume work,

de_i = T ds_i − p dv_i

or

T ds_i = de_i + p dv_i (5)

Substituting equation (5) into the differentiated expression and simplifying,

dg_i = −s_i dT + v_i dp dg₁ = dg₂

Using equation (6),

−s₁ dT + v₁ dp = −s₂ dT + v₂ dp

Rearranging,

(v₂ − v₁) dp = (s₂ − s₁) dT

Therefore,

dp/dT = (s₂ − s₁)/(v₂ − v₁) (7)

This is the general thermodynamic form of the Clausius–Clapeyron equation.

4. Expression in Terms of Latent Heat

During a phase transition, heat is either absorbed or released by a substance. This heat is called latent heat.

Let L₁₂ be the latent heat absorbed by one mole of a substance during its transition from phase 1 to phase 2. The change in molar entropy is

S₂ − S₁ = L₁₂/T

Since s₁ and s₂ are entropy per molecule,

N(s₂ − s₁) = L₁₂/T

The change in molar volume is

V₂ − V₁ = N(v₂ − v₁)

From equation (7), multiplying the numerator and denominator by N,

dp/dT = [N(s₂ − s₁)]/[N(v₂ − v₁)]

Substituting the expressions for entropy change and volume change,

dp/dT = [L₁₂/T]/(V₂ − V₁)

Clausius–Clapeyron Equation

dp/dT = L₁₂ / [T(V₂ − V₁)]

5. Physical Significance and Applications

The Clausius–Clapeyron equation describes how the equilibrium pressure changes with temperature during a phase transition. It is useful in studying:

1. Effect of pressure on melting point: It explains how the melting temperature changes with applied pressure.
2. Effect of pressure on boiling point: It describes the variation of boiling temperature with external pressure.
3. Phase equilibrium: It determines the slope of a phase-equilibrium curve in a pressure–temperature diagram.
4. Latent heat: It can be used to determine the latent heat associated with a phase transition when the other quantities are known.

Important formula for examination

dp/dT = ΔS/ΔV = L/(TΔV)

Where:

• L = latent heat per mole (J mol⁻¹)
• T = absolute temperature (K)
• ΔV = change in molar volume (m³ mol⁻¹)
• dp/dT = slope of the phase-equilibrium curve

Conclusion: The Clausius–Clapeyron equation establishes a quantitative relationship between pressure, temperature, latent heat and volume change during a phase transition.

Note: This derivation assumes two phases are in thermodynamic equilibrium and the phase transition is reversible. The equation applies to solid–liquid, liquid–vapour and solid–vapour equilibrium. Here, L and ΔV must refer to the same molar basis and the same direction of transition.