Clapeyron Equation: Melting Point and Pressure
The Clausius–Clapeyron equation is the famous one for liquid–vapour equilibrium. Its parent, the Clapeyron equation, is exact for any two-phase equilibrium of a pure substance, including solid–liquid. It explains why the melting line of water slopes backwards in the phase diagram and lets you calculate how far the melting point moves under pressure. This topic appears in physical chemistry for IIT-JAM, GATE and CSIR-NET, usually as a short numerical or a reasoning question.
The formula
Integrated (small range, constant ΔH and ΔV): P2 − P1 = (ΔfusH / ΔfusV) ln(T2/T1)
dP/dT is the slope of the solid–liquid line in the P–T phase diagram. ΔfusH is the molar enthalpy of fusion (J/mol) and is always positive. ΔfusV = Vm(liquid) − Vm(solid), in m3/mol. T is the melting temperature in kelvin. Keep SI units throughout: J/mol divided by (K × m3/mol) gives Pa/K.
Where it comes from
Along a phase boundary the two phases are in equilibrium, so their molar Gibbs energies are equal, and they stay equal when T and P change together. This gives dGm(s) = dGm(l), that is, −SsdT + VsdP = −SldT + VldP. Rearranging: dP/dT = (Sl − Ss)/(Vl − Vs) = ΔS/ΔV. At the melting point ΔfusS = ΔfusH/T, which gives the equation above. Because ΔfusH > 0 and T > 0, the sign of the slope is the sign of ΔfusV.
- Most substances expand on melting: ΔfusV > 0, so the slope is positive. Higher pressure raises the melting point.
- Water is unusual: ice is less dense than liquid water, ΔfusV < 0, so the slope is negative. Higher pressure lowers the melting point.
Worked example 1: melting point of ice under pressure
Question. Estimate dP/dT and the lowering of the melting point of ice for a pressure increase of 100 atm. Use ΔfusH = 6.01 kJ/mol, T = 273.15 K, and densities of ice 0.917 g/cm3 and liquid water 0.9998 g/cm3 near 0°C (M = 18.015 g/mol). Textbook values differ a little, so use the ones in your question.
Step 1: molar volumes. Vm(ice) = 18.015/0.917 = 19.65 cm3/mol. Vm(water) = 18.015/0.9998 = 18.02 cm3/mol.
Step 2: ΔfusV. 18.02 − 19.65 = −1.63 cm3/mol = −1.627 × 10−6 m3/mol.
Step 3: slope. dP/dT = 6010 / (273.15 × (−1.627 × 10−6)) = −1.35 × 107 Pa/K. In bar this is about −135 bar/K, or −133 atm/K.
Step 4: temperature change. dT/dP = 1/(−133.5 atm/K) = −0.00749 K/atm. For 100 atm, ΔT = −0.75 K. The ice now melts at about −0.75°C.
Check with the integrated form. T2 = T1 exp(ΔP · ΔV/ΔH) = 273.15 × exp(100 × 101325 × (−1.627 × 10−6)/6010) = 273.15 × exp(−0.002743) = 272.40 K. The change is −0.75 K. The two routes agree.
Worked example 2: very high pressure
Question. Using the same data, find the melting point of ice at 500 atm and at 1000 atm, and compare with the straight-line estimate.
Using T2 = 273.15 exp(ΔP·ΔV/ΔH): at 500 atm, ΔT = −3.72 K. At 1000 atm, ΔT = −7.39 K. The straight-line estimate with −0.00749 K/atm gives −3.75 K and −7.49 K. The straight line is close but slightly too large. This is because ΔH and ΔV are treated as constant. In reality they change with pressure, so treat results at very high pressure only as estimates.
Worked example 3: a substance that expands on melting
Question. A hypothetical solid melts at 400 K with ΔfusH = 5.00 kJ/mol and ΔfusV = +0.50 cm3/mol. These numbers are invented for practice. Find dP/dT and the melting point at +1000 bar.
Step 1. ΔV = 0.50 × 10−6 m3/mol. dP/dT = 5000/(400 × 0.50 × 10−6) = 2.5 × 107 Pa/K = 250 bar/K.
Step 2. ΔP = 1000 bar = 1.0 × 108 Pa. Linear estimate: ΔT = 1.0 × 108/2.5 × 107 = +4.0 K.
Step 3 (integrated check). T2 = 400 exp(1.0 × 108 × 0.50 × 10−6/5000) = 400 exp(0.0100) = 404.02 K. Rise = +4.02 K. Agrees with +4.0 K. The positive slope shows the melting point rises with pressure.
Clapeyron versus Clausius–Clapeyron
The Clausius–Clapeyron equation is a simplified form of the Clapeyron equation for liquid–vapour (or solid–vapour) equilibrium. There we assume Vgas is much larger than Vliquid, so ΔV is approximately Vgas = RT/P. That approximation does not work for melting, because the solid and liquid volumes are close. So for fusion always use the full Clapeyron equation with the real ΔfusV.
Common mistakes
- Using Clausius–Clapeyron for melting. The ideal-gas approximation does not apply; use ΔfusV.
- Mixed units. cm3 must be converted to m3 (× 10−6) when ΔH is in J. Otherwise the answer is wrong by a factor of a million.
- Wrong sign of ΔV. It is V(liquid) − V(solid). For water this is negative.
- Using °C for T. T in the denominator must be in kelvin.
- Over-explaining. Do not use this equation to claim that pressure melting is the main reason ice is slippery. Explanations of that are more complicated; follow what your syllabus states.
Exam relevance
| Question type | Approach |
|---|---|
| Why does the ice–water line have a negative slope? | ΔfusV < 0 in dP/dT = ΔH/(TΔV) |
| Change in melting point for a given ΔP | Compute dT/dP, or use the integrated form |
| Pressure needed for a given melting point shift | ΔP = (ΔH/ΔV) ln(T2/T1) |
| Which phase is denser | Read the sign of the slope |
Check the official syllabus and notification for your exam to see how deep this topic is expected to go. The reasoning above is the standard thermodynamic treatment.
This topic has no dedicated tool in the suite, but the Scientific Calculator handles the exponentials and unit conversions in these numericals.
Open the Chemistry Calculator Suite →