The Clausius-Clapeyron Equation & P-T Diagrams

Estimate how vapor pressure changes with temperature using ΔHvap, and interpret phase boundaries on a P-T diagram.

Material & Energy BalancesChemical Engineering Year 1Free preview
⏱️ About 16 min

If you know a liquid's vapor pressure at one temperature and its heat of vaporization, can you estimate vapor pressure at another temperature without a table?

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The big idea: Clausius-Clapeyron links vapor pressure's temperature sensitivity to ΔHvap; the integrated form gives a practical two-point estimate.
🎯 By the end, you'll be able to
  • Write the differential and integrated Clausius-Clapeyron relations
  • Use consistent units (ln, Kelvin, J/mol) in vapor-pressure estimates
  • Interpret P-T phase diagrams: triple point, critical point, and phase boundaries
  • Compute an estimated water vapor pressure at a new temperature with verified arithmetic
📎 Helpful to know first

Clausius-Clapeyron (vaporization approximation)

Along the liquid-vapor equilibrium line, an approximate form of the Clausius-Clapeyron equation is:

d(ln P*)/dT = ΔHvap/(R T²)

This comes from the Clapeyron equation with the additional approximation that the vapor behaves ideally and the liquid molar volume is negligible compared to the vapor molar volume.

\[ \frac{d(\ln P^*)}{dT}=\frac{\Delta H_{\text{vap}}}{R\,T^2} \]
Differential Clausius-Clapeyron form (ideal vapor approximation). Use T in K, ΔHvap in J/mol, and ln (natural log).
\[ \ln\!\left(\frac{P_2^*}{P_1^*}\right)= -\frac{\Delta H_{\text{vap}}}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right) \]
Integrated two-point form (assuming ΔHvap is approximately constant over the temperature interval).

P-T Phase Diagrams (conceptual map)

A P-T phase diagram shows boundaries where two phases coexist in equilibrium:

  • Sublimation curve (solid-vapor)
  • Fusion curve (solid-liquid)
  • Vaporization curve (liquid-vapor), which ends at the critical point

All three curves meet at the triple point, where solid, liquid, and vapor coexist.

⚠️ Pitfall: Use Kelvin and natural log (ln), not °C and log10

In Clausius-Clapeyron, T must be in K and the equation uses ln. Mixing °C or log10 with this form is a common source of large errors.

📝 Worked example: Estimate the vapor pressure of water at 90.0°C using Clausius-Clapeyron. Use the known point at 100.0°C: P1* = 760 mmHg = 1.000 atm, ΔHvap ≈ 40,660 J/mol, and R = 8.314 J/(mol·K). Assume ΔHvap is constant between 90-100°C.
  1. Convert temperatures to Kelvin: T₁ = 100.0°C = 373.15 K, T₂ = 90.0°C = 363.15 K
  2. Use the integrated form: ln(P₂*/P₁*) = −(ΔHvap/R)(1/T₂ − 1/T₁)
  3. Compute ΔHvap/R: 40,660 / 8.314 = 4890.55
  4. Compute the reciprocal-temperature difference exactly via (T₁−T₂)/(T₁T₂):
  5. T₁ − T₂ = 373.15 − 363.15 = 10.00 K
  6. T₁ × T₂ = 373.15 × 363.15 = 135,509.42
  7. (1/T₂ − 1/T₁) = 10.00 / 135,509.42 = 7.3795×10⁻⁵ K⁻¹
  8. ln(P₂*/P₁*) = −(4890.55)(7.3795×10⁻⁵) = −0.36090
  9. Exponentiate: P₂*/P₁* = exp(−0.36090) = 0.69705
  10. P₂* = (0.69705)(1.000 atm) = 0.697 atm
  11. In mmHg (760 mmHg = 1 atm): P₂* = 0.69705 × 760 = 529.8 mmHg
✓ P*(90.0°C) ≈ 0.697 atm ≈ 530 mmHg (two-point Clausius-Clapeyron estimate)
✏️ Practice: Using the same data (P*(100.0°C)=1.000 atm, ΔHvap=40,660 J/mol, R=8.314 J/(mol·K)), estimate P* of water at 95.0°C. Give the answer in atm.
atm
Solution
  1. T₁ = 373.15 K, T₂ = 95.0°C = 368.15 K
  2. ΔHvap/R = 40,660/8.314 = 4890.55
  3. T₁ − T₂ = 5.00 K; T₁ × T₂ = 373.15 × 368.15 = 137,375.17
  4. (1/T₂ − 1/T₁) = 5.00 / 137,375.17 = 3.6397×10⁻⁵ K⁻¹
  5. ln(P₂*/P₁*) = −(4890.55)(3.6397×10⁻⁵) = −0.17800
  6. P₂*/P₁* = exp(−0.17800) = 0.83695
  7. P₂* = 0.83695×1.000 atm = 0.837 atm

Check your understanding

1. In the integrated Clausius-Clapeyron equation, which combination is correct?
The standard integrated form uses natural log and absolute temperature (K).
2. On a P-T phase diagram, the liquid-vapor coexistence curve terminates at the:
The liquid-vapor boundary ends at the critical point, beyond which liquid and vapor are indistinguishable.
🔎 Related lessons
Raoult's Law & Relative Volatility · soon
✅ Key takeaways
  • Clausius-Clapeyron relates vapor pressure changes to ΔHvap: d(ln P*)/dT = ΔHvap/(RT²)
  • Integrated form: ln(P2*/P1*) = −(ΔHvap/R)(1/T2 − 1/T1) with T in K
  • P-T diagrams: three coexistence curves meet at the triple point; vaporization curve ends at the critical point
  • Two-point estimates are useful but assume ΔHvap is roughly constant over the interval
➡️ Now that you can compute or estimate pure-component vapor pressures, we can predict vapor-liquid equilibrium compositions for mixtures using Raoult's law and relative volatility.
Want to test yourself on this? Try the Chemical Aptitude test →