Clausius-Clapeyron Equation Calculator
Enter latent heat L, temperature T and volume change ΔV to compute the phase-transition slope dP/dT. Also supports the approximate form estimating vapor-pressure change with temperature, e.g. water 1 atm at 100°C → boiling point changes ~28°C/km altitude.
Input Data
Results
At a glance:The Clausius-Clapeyron equation (Rudolf Clausius, 1850; Clapeyron, 1834) describes the slope of a phase-equilibrium line in a P-T diagram: dP/dT=L/(T·ΔV), where L is the latent heat (J/mol or J/kg), T the absolute temperature, ΔV the molar (or specific) volume change. For a liquid-gas transition the approximate form assumes the gas is ideal and ΔV≈V_gas, giving ln(P₂/P₁)=−(L_m/R)·(1/T₂−1/T₁) (two-point form), which can estimate vapor pressure or boiling-point change with altitude. Physical meaning: the slope of the coexistence curve equals the ratio of latent heat to T·ΔV; at a phase transition the two phases have the same chemical potential, and the P-T slope is determined by entropy and volume changes. History: Clapeyron derived it in 1834 (Carnot cycle), Clausius re-derived and generalized it in 1850. Classic example: water at 100°C, 1 atm, L=2.26e6 J/kg, ΔV≈1.67 m³/kg → dP/dT=2.26e6/(373×1.67)=3629 Pa/K; rising 1 km (P drops ~9 kPa) → boiling point drops ~2.8°C (≈28°C per km, but strictly non-linear). Applications: (1) boiling-point change with altitude; (2) vapor-pressure curves; (3) sublimation (dry ice); (4) refrigeration cycles; (5) climate (water-vapor feedback).
Formula
Exact: dP/dT = L / (T·ΔV)
Approximate (liquid→gas, ideal gas): ln(P₂/P₁) = −(Lₘ/R)·(1/T₂ − 1/T₁)
Boiling-point change: ΔT ≈ (R·T²)/(Lₘ) · (ΔP/P) (linear approx)
Saturation pressure: P = P₀·exp(−Lₘ/R·(1/T − 1/T₀))
Sublimation: same form with L_sub = L_fus + L_vap
$$\frac{dP}{dT} = \frac{L}{T\Delta V}, \quad \ln\frac{P_2}{P_1} = -\frac{L_m}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right)$$How to Use
- For dP/dT: enter latent heat L, temperature T, volume change ΔV.
- For two-point vapor pressure: enter P1, T1, T2, enthalpy ΔHvap (J/mol), R.
- The tool outputs dP/dT or the vapor pressure P2 at T2.
Phase-Transition Parameters of Common Substances
| Substance | L_vap (J/kg) | T_b (K) | ΔV (m³/kg) | dP/dT (Pa/K) |
|---|---|---|---|---|
| Water | 2.26e6 | 373 | 1.67 | 3629 |
| Ethanol | 8.45e5 | 351 | 0.607 | 3960 |
| Ammonia | 1.37e6 | 240 | 1.25 | 4567 |
| CO₂ | 5.73e5 | 194.7 | 0.563 | 5224 |
| n-Octane | 3.06e5 | 399 | 0.226 | 3405 |
ΔV = V_gas − V_liquid ≈ V_gas (liquid volume neglected). dP/dT = L/(T·ΔV). Water dP/dT≈3630 Pa/K; 1 km altitude (ΔP≈−9 kPa) → boiling point drops ~2.8°C.
Case Studies
Boiling Point vs Altitude (Cooking)
Hong Kong Victoria Peak 552 m: P≈0.945 atm; water boils at ≈98.3°C (vs 100°C at sea level).
Lhasa 3650 m: P≈0.65 atm; water boils at ≈88°C. Rice and meat cook slower, needing pressure cookers.
Commercial pressure cookers raise internal P to ~2 atm → 120°C, cutting cooking time ~1/3.
Refrigeration Cycle Saturation Pressure
R134a at 5°C saturation pressure ~3.5 bar; at 45°C ~11.7 bar (compressor discharge). The Clausius-Clapeyron equation relates the two.
Designers use the equation to compute the condensing pressure at the hot-end temperature, ensuring the compressor can reach it.
Heat-pump COP depends on the evaporation/condensation temperature difference; the equation sets the working-pressure envelope.
FAQ
Why does the boiling point drop at high altitude?
Boiling occurs when vapor pressure equals ambient pressure. At high altitude atmospheric pressure is lower, so water vaporizes at a lower temperature. Using the approximate form with L≈2.26e6 J/kg, T=373 K, R=461 J/(kg·K): a 1 km drop in pressure (~9 kPa) lowers the boiling point by about 2.8°C. Lhasa (3650 m) boils at ~88°C. Pressure cookers restore high temperature for faster cooking.
When is the approximate form valid?
The approximate form ln(P₂/P₁)=−(Lₘ/R)(1/T₂−1/T₁) assumes: (1) the gas is ideal; (2) the liquid volume is negligible vs the gas volume (ΔV≈V_gas); (3) L is constant over the temperature range. It works well for liquid-vapor transitions away from the critical point. Near the critical point (water 374°C, 22 MPa) the gas-liquid distinction vanishes and the equation fails; use real-gas equations of state.
What is the difference between dP/dT and the two-point form?
dP/dT=L/(T·ΔV) is the differential slope at a point on the coexistence curve. The two-point form is the integrated version (assuming constant L and ideal gas) giving the exact P₂ at T₂. dP/dT tells you how steep the curve is locally; the two-point form predicts values over a range. For large temperature spans L varies, so integrate with temperature-dependent L or use Antoine parameters.
How does it apply to sublimation (dry ice)?
Dry ice (CO₂ solid) sublimates at −78.5°C at 1 atm. The sublimation curve also follows Clausius-Clapeyron with L_sub=L_fus+L_vap. At −78.5°C, P=1 atm is the triple-point-ish condition; below that CO₂ does not melt but sublimes. Cooling/venting keeps the pressure low. The equation gives the sublimation pressure at any temperature; it is used in freeze-drying and CO₂ fire extinguishers.
What is its role in climate science?
Water-vapor feedback: warmer air holds more water vapor (Clausius-Clapeyron predicts ~7%/°C increase in saturation vapor pressure). Water vapor is a greenhouse gas, so more of it amplifies warming — a positive feedback roughly doubling CO₂'s direct effect. Climate models use the equation to compute the saturation vapor-pressure profile and cloud formation. It is a core relation in atmospheric thermodynamics.
Related Tools
References
Content reviewed by the Calculatorism editorial team. Results are for reference only; please refer to the relevant authorities for the official figures.