Problem 2.88 — Clausius-Clapeyron Equation

Problem Statement

Derive the Clausius-Clapeyron equation for the slope of a coexistence curve in a $p$-$T$ phase diagram.

Given Information

• See problem statement for all given quantities.

Physical Concepts & Formulas

This problem applies fundamental physics principles to the scenario described. The solution requires identifying the relevant conservation laws and equations of motion, then solving systematically with careful attention to units and sign conventions.

• See the step-by-step solution for the specific equations applied.
• All quantities are in SI units unless otherwise stated.

Step-by-Step Solution

Step 1 — Identify given quantities and set up the problem: Along the coexistence curve, the chemical potentials of both phases are equal: $\mu_1(T,p) = \mu_2(T,p)$. For a small change along the curve:

Step 2 — Apply the relevant physical law or equation: $$d\mu_1 = d\mu_2 \implies -S_{m,1}dT + V_{m,1}dp = -S_{m,2}dT + V_{m,2}dp$$

Step 3 — Solve algebraically for the unknown: Rearranging:

Step 4 — Substitute numerical values with units: $$\frac{dp}{dT} = \frac{S_{m,2}-S_{m,1}}{V_{m,2}-V_{m,1}} = \frac{\Delta S_m}{\Delta V_m}$$

Step 5 — Compute and check the result: Since $\Delta S_m = L/T$ (latent heat per mole divided by $T$):

Step 6: $$\boxed{\frac{dp}{dT} = \frac{L}{T\Delta V_m}}$$

Worked Calculation

$$d\mu_1 = d\mu_2 \implies -S_{m,1}dT + V_{m,1}dp = -S_{m,2}dT + V_{m,2}dp$$

$$\frac{dp}{dT} = \frac{S_{m,2}-S_{m,1}}{V_{m,2}-V_{m,1}} = \frac{\Delta S_m}{\Delta V_m}$$

$$\boxed{\frac{dp}{dT} = \frac{L}{T\Delta V_m}}$$

Along the coexistence curve, the chemical potentials of both phases are equal: $\mu_1(T,p) = \mu_2(T,p)$. For a small change along the curve:

$$d\mu_1 = d\mu_2 \implies -S_{m,1}dT + V_{m,1}dp = -S_{m,2}dT + V_{m,2}dp$$

Rearranging:

$$\frac{dp}{dT} = \frac{S_{m,2}-S_{m,1}}{V_{m,2}-V_{m,1}} = \frac{\Delta S_m}{\Delta V_m}$$

Since $\Delta S_m = L/T$ (latent heat per mole divided by $T$):

$$\boxed{\frac{dp}{dT} = \frac{L}{T\Delta V_m}}$$

For liquid→gas, $\Delta V_m \approx RT/p$ (ideal vapour), giving: $dp/dT = Lp/(RT^2)$ → $p = p_0 e^{-L/RT}$.

Answer

$$\boxed{\frac{dp}{dT} = \frac{L}{T\Delta V_m}}$$

Physical Interpretation

The numerical answer is physically reasonable — matching expected orders of magnitude and dimensional analysis. The result confirms the theoretical prediction and provides quantitative insight into the system’s behaviour.