Enter phase data on a consistent basis. Phase 1 and Phase 2 volumes must use the same unit and basis.
The Clapeyron equation links the slope of a phase boundary to entropy and volume changes: dP/dT = ΔS / ΔV.
If the latent heat ΔH is known at temperature T, then ΔS = ΔH / T and: dP/dT = ΔH / (T · ΔV).
Here ΔV = v₂ − v₁. Use consistent molar or mass basis throughout.
- Select a method: latent heat or entropy change.
- Enter temperature and choose K or °C.
- Choose a volume unit and enter v₁ and v₂.
- Provide ΔH or ΔS with a matching basis.
- Press Calculate to view slope and unit conversions.
Illustrative vaporization case near 373.15 K. Values are approximate for demonstration.
| Scenario | T (K) | ΔH (kJ/mol) | v₁ (L/mol) | v₂ (L/mol) | Computed dP/dT (Pa/K) | Computed dP/dT (bar/K) |
|---|---|---|---|---|---|---|
| Liquid → Vapor | 373.15 | 40.65 | 0.018 | 30.6 | 3562 | 0.0356 |
| Smaller ΔV case | 300.00 | 20.00 | 1.00 | 2.00 | 66667 | 0.6667 |
| Negative ΔV example | 273.15 | 6.00 | 20.0 | 18.0 | -1098 | -0.0110 |
1) What the slope represents
The Clapeyron slope dP/dT quantifies how equilibrium pressure changes with temperature along a phase boundary. A steep slope means small temperature shifts require large pressure adjustments to preserve coexistence. This is crucial in boiling, melting, and sublimation studies.
2) Core relationship and required inputs
The calculator uses dP/dT = ΔS/ΔV. You can enter entropy change directly, or supply latent heat ΔH and temperature to compute ΔS = ΔH/T. Accurate results depend on using a consistent basis: molar units with molar volumes, or mass units with specific volumes.
3) Typical magnitudes in real systems
For vaporization near ambient pressures, ΔV is large, so slopes are often in the range of 10³–10⁵ Pa/K, depending on temperature and latent heat. Near critical points, ΔV shrinks and the slope can grow rapidly, making numerical sensitivity more noticeable.
4) Sign of the slope and physical meaning
The sign is governed by ΔV = v₂ − v₁. Many melting curves have small or negative ΔV, which produces a negative slope. A classic example is ice, where the solid has larger specific volume than liquid water, allowing pressure to lower the melting temperature.
5) Unit handling and conversions
The output is computed internally in Pa/K, then converted to kPa/K, MPa/K, bar/K, and atm/K. Volume entries are converted to m³ per chosen basis. This approach minimizes rounding error when switching units for engineering checks.
6) Data quality checks you should apply
Use phase volumes at the same temperature and pressure as your latent heat or entropy data. For gases, volumes can vary strongly with pressure; a mismatch can distort ΔV and therefore dP/dT. If ΔV is extremely small, treat the computed slope as a sensitive indicator rather than a robust estimate.
7) Practical use cases
Engineers use slopes to estimate how saturation pressure changes in sealed vessels, to interpret phase diagrams, and to validate experimental measurements. In materials processing, the slope helps predict how pressure shifts melting lines, guiding sintering, crystal growth, and high-pressure experimentation workflows.
8) Interpreting results and exporting
After calculation, review the preferred unit and the conversion panel to ensure magnitudes are reasonable for your system. Use the CSV export for lab notebooks and the PDF export for reports. The cross-check lines help verify internal consistency between ΔH, ΔS, and T.
1) Should I use ΔH or ΔS?
Use ΔH when you have latent heat data and temperature. Use ΔS when entropy change is available directly. Both approaches are equivalent when the data are consistent and refer to the same state.
2) What basis should I choose?
Pick molar basis when you have kJ/mol and L/mol values. Pick mass basis when you have kJ/kg and m³/kg values. Mixing bases breaks the equation and will give incorrect slopes.
3) Why does my slope turn negative?
A negative slope happens when ΔV is negative, meaning the second phase has smaller volume than the first. This can occur in melting transitions where the liquid is denser than the solid.
4) What if ΔV is extremely small?
When ΔV is tiny, dP/dT becomes very sensitive to measurement noise. Use higher-precision data, verify states match, and interpret the result as an approximate trend rather than a single definitive value.
5) Can I use this for sublimation lines?
Yes. Enter solid and vapor volumes and the relevant latent heat or entropy change for sublimation. Because vapor volume is large, slopes can be moderate, but they still vary strongly with temperature.
6) Do the volumes need to be at equilibrium?
Ideally, yes. Volumes should correspond to coexisting phases at the same T and P. Using off-equilibrium values can bias ΔV, especially for compressible phases like gases.
7) How do I sanity-check the output?
Check that ΔH and ΔS obey ΔH ≈ T·ΔS. Confirm the slope magnitude matches your expectations, such as 10³–10⁵ Pa/K for many vaporization scenarios, unless near critical behavior.