Phase Changes: Heating Curves, Vapor Pressure and Phase Diagrams

Why the temperature stalls during melting and boiling

Heat a block of ice at a steady rate and record its temperature, and you get a heating curve. The temperature climbs, stops at 0 °C while the ice melts, climbs again, then stops at 100 °C while the water boils. On the flat parts, the energy goes into separating molecules against their intermolecular attractions, not into making them move faster, so the temperature does not change until the phase change is complete.

The two kinds of segment need two different equations. Within one phase, q = m × c × ΔT, where c is the specific heat. During a phase change at constant temperature, q = n × ΔH, where ΔH is the molar enthalpy of fusion or vaporization. Melting, vaporization and sublimation are endothermic. Freezing, condensation and deposition release the same amounts of energy, so their ΔH values have the opposite sign.

Worked example: warming ice all the way to steam

How much heat turns 40.0 g of ice at −15.0 °C into water vapor at 100.0 °C? Data: specific heats of 2.093 J/(g·°C) for ice and 4.184 J/(g·°C) for liquid water (OpenStax Table 5.1); ΔHfus = 6.01 kJ/mol at 0 °C and ΔHvap = 40.67 kJ/mol at 100 °C (OpenStax section 10.3). The amount of water is n = 40.0 g ÷ 18.015 g/mol = 2.220 mol.

Step 1, warm the ice: 40.0 × 2.093 × 15.0 = 1256 J, or 1.26 kJ. Step 2, melt it: 2.220 mol × 6.01 kJ/mol = 13.3 kJ. Step 3, warm the liquid from 0 °C to 100 °C: 40.0 × 4.184 × 100.0 = 16 736 J, or 16.7 kJ. Step 4, boil it: 2.220 mol × 40.67 kJ/mol = 90.3 kJ. Total: 1.26 + 13.3 + 16.7 + 90.3 = 122 kJ, to three significant figures.

Boiling accounts for about 74 percent of the total, more than five times the heat needed to warm the liquid through 100 degrees. Use the value of ΔHvap at the temperature where the boiling happens. At 25 °C, water's ΔHvap is 44.01 kJ/mol, and using that number here would overstate step 4 by about 8 percent.

Vapor pressure and the Clausius–Clapeyron equation

In a closed container, a liquid evaporates until the rate of condensation matches it. The pressure of the vapor at that point is the vapor pressure, and it rises steeply with temperature. A liquid boils when its vapor pressure equals the pressure above it, so the normal boiling point is the temperature at which the vapor pressure reaches 1 atm.

The two-point form of the Clausius–Clapeyron equation links two vapor pressures to the enthalpy of vaporization: ln(P₂/P₁) = (ΔHvap/R)(1/T₁ − 1/T₂), with R = 8.314 J/(mol·K) and temperatures in kelvins. It assumes ΔHvap is constant over the range, so it works best over modest temperature spans.

Worked example 1. OpenStax Table 9.2 gives water's vapor pressure as 17.5 torr at 20 °C and 149.4 torr at 60 °C. ln(149.4/17.5) = 2.144, and 1/293.15 K − 1/333.15 K = 4.096 × 10⁻⁴ K⁻¹. ΔHvap = (8.314 × 2.144) ÷ 4.096 × 10⁻⁴ = 4.35 × 10⁴ J/mol, or 43.5 kJ/mol. This is an average over 20 °C to 60 °C. It lies between the values at 25 °C (44.01) and at 100 °C (40.67), as it should.

Worked example 2. At what temperature does water boil when the air pressure is 0.800 atm? Start from the normal boiling point, 373.15 K at 1.000 atm, and use ΔHvap = 40.67 kJ/mol. 1/T₂ = 1/373.15 − (8.314/40 670) × ln(0.800) = 2.7255 × 10⁻³ K⁻¹, so T₂ = 366.9 K, or 93.8 °C. As a check, 0.800 atm is 608 torr, and interpolating between 90 °C and 95 °C in Table 9.2 also gives 93.8 °C.

Reading the regions, lines and points of a phase diagram

A phase diagram plots pressure against temperature and shows which phase is stable under each set of conditions. Each region is one phase. Each line is a set of conditions where two phases coexist in equilibrium: solid and liquid along the melting curve, liquid and gas along the vapor-pressure curve, and solid and gas along the sublimation curve.

The three lines meet at the triple point, the single temperature and pressure where solid, liquid and gas all coexist. The vapor-pressure curve ends at the critical point. Above the critical temperature, no amount of pressure liquefies the gas, and the substance becomes a supercritical fluid with properties between those of a liquid and a gas.

To use the diagram, locate the starting conditions and move along the path the problem describes. Heating at constant pressure is a horizontal line from left to right, and every boundary it crosses is a phase change. Compressing at constant temperature is a vertical line moving upward. A horizontal line drawn below the triple-point pressure crosses only the sublimation curve, so the solid turns straight into gas.

Water's backward melting line and dry ice's high triple point

Water's triple point is at 0.01 °C and 0.0061 bar (NIST Chemistry WebBook), which is about 0.0060 atm. Its critical point is at 374 °C and 218 atm (OpenStax section 10.4). Unusually, its solid–liquid line slopes slightly to the left: raising the pressure lowers the melting point a little, because ice is less dense than liquid water and pressure favors the denser phase. Most substances have the opposite slope.

Carbon dioxide's triple point is at 216.58 K (−56.57 °C) and 5.185 bar, which is 5.12 atm (NIST Chemistry WebBook). Since that pressure is above 1 atm, liquid CO₂ cannot exist in open air. At 1 atm, solid CO₂ (dry ice) sublimes instead of melting. Its critical point, 31.0 °C and 73.8 bar, is low enough that supercritical CO₂ is used industrially as a solvent, for example to remove caffeine from coffee beans.

Worked example 3. Estimate the temperature at which dry ice sublimes at 1 atm (1.01325 bar), starting from the triple point and the enthalpy of sublimation, ΔHsub = 26.1 kJ/mol (NIST WebBook, measured near 207 K). The Clausius–Clapeyron equation works for the solid–gas line in the same way as for the liquid–gas line. ln(1.01325/5.185) = −1.633. 1/T₂ = 1/216.58 − (8.314/26 100) × (−1.633) = 5.137 × 10⁻³ K⁻¹, so T₂ = 194.7 K, or −78.5 °C. That matches the normal sublimation temperature of −78.5 °C quoted in LibreTexts.

Phase-change calculation mistakes

Most errors come from mixing the two kinds of heating-curve segment or from unit slips in the Clausius–Clapeyron equation. Sketch the heating curve and label each segment before calculating anything.

• Leaving out a phase-change step, or applying q = mcΔT across a plateau where ΔT is zero.
• Using the specific heat of liquid water for ice or for steam. Each phase has its own value.
• Mixing joules and kilojoules when adding the steps. Convert everything to one unit before summing.
• Multiplying ΔH by grams instead of moles, or c by moles instead of grams, unless you use molar heat capacities consistently.
• Using Celsius temperatures in the Clausius–Clapeyron equation. Only kelvins work, because the equation uses 1/T.
• Using R = 0.08206 L·atm/(mol·K) with ΔH in joules. With ΔH in J/mol, R must be 8.314 J/(mol·K).
• Reading a phase diagram's pressure axis as linear when it is logarithmic, as it often is for CO₂.