The Clausius–Clapeyron Equation

Core Concept

The Clausius–Clapeyron equation quantifies how a liquid's vapor pressure rises with temperature. Because vaporization is endothermic, more molecules can escape at higher temperature, and the relationship is exponential. In its two-point form, ln(P₂/P₁) = −(ΔHᵥₐₚ/R)(1/T₂ − 1/T₁), the equation lets you calculate the vapor pressure at one temperature if you know it at another, given the enthalpy of vaporization ΔHᵥₐₚ.

Key Ideas

The underlying physics

• Vaporization is endothermic (ΔHᵥₐₚ > 0). Raising temperature shifts the kinetic-energy distribution so more molecules exceed the escape threshold.
• Vapor pressure therefore increases exponentially, not linearly, with temperature.

The two forms

• Linear/graphical form: ln P = −(ΔHᵥₐₚ/R)(1/T) + C. A plot of ln P vs. 1/T is a straight line with slope −ΔHᵥₐₚ/R.
• Two-point form (most used in exams): ln(P₂/P₁) = −(ΔHᵥₐₚ/R)(1/T₂ − 1/T₁).

Assumptions

• ΔHᵥₐₚ is treated as constant over the temperature range (it actually varies slightly with T).
• Temperatures must be in kelvin; pressures may be in any consistent units.
• Valid below the critical point, away from extreme conditions.

Equations and Variables

• Clausius–Clapeyron (two-point): ln(P₂/P₁) = −(ΔHᵥₐₚ/R)(1/T₂ − 1/T₁)
  • P₁, P₂ = vapor pressures at temperatures T₁, T₂ (same units for P)
  • T₁, T₂ = absolute temperatures in kelvin
  • ΔHᵥₐₚ = molar enthalpy of vaporization (J/mol, to match R)
  • R = 8.314 J/(mol·K)
• Graphical form: ln P = −(ΔHᵥₐₚ/R)(1/T) + C; slope = −ΔHᵥₐₚ/R.

How It Works

1. Start with known (P₁, T₁). You typically know the normal boiling point (T where P = 1 atm) or a tabulated vapor pressure.
2. Convert temperatures to kelvin. T(K) = T(°C) + 273.15. Using °C gives nonsense results.
3. Make units consistent. Put ΔHᵥₐₚ in J/mol so it cancels with R = 8.314 J/mol·K.
4. Compute the exponent. Evaluate (1/T₂ − 1/T₁), multiply by −ΔHᵥₐₚ/R.
5. Solve. Exponentiate to find P₂/P₁, then P₂. To find a boiling point at a given pressure, solve for T₂ instead.

Worked Example

The normal boiling point of water is 100 °C (P₁ = 760 torr at T₁ = 373.15 K), and ΔHᵥₐₚ = 40.7 kJ/mol. Estimate the vapor pressure of water at 25 °C.

1. Convert. T₁ = 373.15 K, T₂ = 25 + 273.15 = 298.15 K, ΔHᵥₐₚ = 40,700 J/mol.
2. Compute 1/T₂ − 1/T₁. 1/298.15 − 1/373.15 = 0.003354 − 0.002680 = 0.000674 K⁻¹.
3. Compute the exponent. −(ΔHᵥₐₚ/R)(1/T₂ − 1/T₁) = −(40,700 / 8.314)(0.000674) = −(4895)(0.000674) = −3.30.
4. Exponentiate. ln(P₂/760) = −3.30 → P₂/760 = e⁻³·³⁰ = 0.0369.
5. Solve. P₂ = 760 × 0.0369 = 28 torr (measured value ≈ 23.8 torr; the small difference reflects ΔHᵥₐₚ varying with temperature).

How it works

1. Start with known (P₁, T₁). You typically know the normal boiling point (T where P = 1 atm) or a tabulated vapor pressure.
2. Convert temperatures to kelvin. T(K) = T(°C) + 273.15. Using °C gives nonsense results.
3. Make units consistent. Put ΔHᵥₐₚ in J/mol so it cancels with R = 8.314 J/mol·K.
4. Compute the exponent. Evaluate (1/T₂ − 1/T₁), multiply by −ΔHᵥₐₚ/R.
5. Solve. Exponentiate to find P₂/P₁, then P₂. To find a boiling point at a given pressure, solve for T₂ instead.

Common confusions

• "I can use Celsius in the equation." — Wrong. The 1/T terms must use absolute temperature (kelvin); Celsius gives incorrect ratios.
• "Any unit for ΔHᵥₐₚ works." — Wrong. Use J/mol to match R = 8.314 J/mol·K (or kJ/mol with R = 0.008314 kJ/mol·K).
• "The two pressures can be in different units." — Wrong. P₁ and P₂ must share the same units (they cancel in the ratio).
• "Vapor pressure rises linearly with temperature." — Wrong. It rises exponentially (hence the ln form).
• "The slope is +ΔHᵥₐₚ/R." — Wrong. The slope of ln P vs. 1/T is negative, −ΔHᵥₐₚ/R.

Quick review

• Two-point form: ln(P₂/P₁) = −(ΔHᵥₐₚ/R)(1/T₂ − 1/T₁).
• Use kelvin; use consistent pressure units; ΔHᵥₐₚ in J/mol.
• ln P vs. 1/T is linear with slope −ΔHᵥₐₚ/R.
• Worked: water, 760 torr at 373 K → ≈28 torr at 298 K.