Apply molar phase energy only within the available phase inventory

Try: heat changes the amount of vapor

Begin with two moles of liquid ethanol at 351 K and supplied heat 38.7 kJ. The selected reference vaporization enthalpy is 38.7 kJ/mol. The nominal transferred amount is one mole, so the final vapor fraction is one-half, not one. Inspect the conserved liquid/vapor amounts before comparing a one-mole sample.

The activity assumes pure ethanol at phase equilibrium under its saturation pressure at the specified temperature. Volume can change; the sample is closed to matter and only pressure-volume work is included. The selected reference rows do not give a numerical pressure. We do not silently set it to one atmosphere or apply this model to a rigid sealed container.

Understand the molecular change

During vaporization, molecules move from the liquid phase to the vapor phase. The substance remains ethanol; this is not atomization or a reaction that breaks all its covalent bonds. Adding heat can change intermolecular arrangements and phase amounts without increasing the equilibrium temperature. Condensation reverses the energy transfer.

The amount scale shows moles rather than physical volume. A half-mole fraction of vapor does not imply that vapor occupies half the container. Amount conservation is n_liquid + n_vapor = n_total.

Calculate the phase fraction

Let Q be signed heat into the material in kJ, n total moles, L the molar vaporization enthalpy in kJ/mol, f₀ the initial vapor mole fraction and f the final fraction. Inside the phase interval, Q = nL(f − f₀), so f = f₀ + Q/(nL). This relation uses enthalpy at the specified temperature and pressure conditions, not a temperature-independent universal value.

At n = 2, f₀ = 0 and Q = 38.7, f = 0.5 using L = 38.7. The reported source uncertainty magnitude ±0.1 gives a conditional range from 38.7/(2×38.8) to 38.7/(2×38.6). We treat those endpoints as an envelope, not a probability distribution or confidence interval. Supplied n and Q are exact teaching inputs.

Experiment at and beyond the boundary

For one mole, the same 38.7 kJ produces nominal fraction one. But the source envelope gives fractions on both sides of one. The lesson therefore does not claim certain complete vaporization. A nominal endpoint is not automatically a robust phase classification.

Now start two moles with half already vapor. Removing 19.35 kJ condenses nominally half a mole, leaving vapor fraction 0.25. Removing more heat than can condense the initial vapor requires liquid cooling; adding more than can vaporize the liquid requires a vapor-temperature model. This phase-only activity supplies neither heat capacity nor a final temperature outside its domain. It refuses rather than clipping impossible fractions and discarding energy.

Compare the matched 320 K source value, 41.0±0.1 kJ/mol, with the 351 K value. The same heat transfers a different nominal number of moles. A deliberately unmatched reference packet withholds inference: the row temperature is part of the evidence. At zero total amount, no vapor fraction is defined.

Apply and continue

For three moles initially 20% vapor, heat 23.22 kJ transfers nominally 0.60 mol into vapor at the 351 K reference. The fraction increases by 0.20 to 0.40. Explain the initial inventory, the sign and why a constant temperature does not mean an unchanged state.

Reuse warming and melting for the basic distinction between sensible heat and phase heat, and vapor inventory for a different, fixed-volume equilibrium model. Those model boundaries are not interchangeable.

The NIST ethanol phase table lists 38.7±0.1 kJ/mol at 351 K and 41.0±0.1 at 320 K, attributed to Counsell, Fenwick and Lees (1970), with method C. The NIST method key identifies C as calorimetry. These empirical inputs are distinct from the original hypothetical inventories, transfers and diagrams. The activity supports AP topic 6.5 but does not establish a performed practical, a full uncertainty budget or independent learner mastery.

Transfer beyond vaporization: fusion and sublimation

Melting (fusion) transfers solid to liquid; freezing reverses it. Sublimation transfers solid directly to gas; deposition is the reverse. For an ordinary endothermic melting or sublimation step, signed heat into the material is positive; reversing the same step reverses its enthalpy. Molecular identity need not change. A plateau describes phase reorganization rather than an increase in average thermal kinetic energy.

Use an original stipulated substance, not the ethanol data above: molar fusion enthalpy is 6.0 kJ/mol at its melting conditions. Two moles begin 25% liquid. Predict the liquid fraction after adding 3.0 kJ while both phases remain. Write the changed amount and the conserved solid/liquid ledger before opening the comparison.

Compare the fusion calculation

Transferred amount is 3.0/6.0 = 0.50 mol. Initial liquid is 0.50 mol; final liquid is 1.00 mol, so liquid fraction is 0.50. Final solid is 1.00 mol. Temperature remaining constant does not mean that no energy entered.

For a separate stipulated common-temperature state-function packet, solid-to-liquid enthalpy is +6.0 kJ/mol and liquid-to-gas is +40.0 kJ/mol. The solid-to-gas difference is +46.0 kJ/mol by adding endpoints, whether or not a direct equilibrium path visits liquid. Depositing 0.20 mol releases 9.2 kJ. Predict how the sign changes for sublimating that amount. Do not add tabulated latent heats at different temperatures without accounting for intervening heating/cooling: those endpoints do not match.

These written transfers extend the lesson without pretending that the vapor-only slider simulates melting or sublimation. Concepts checked 30 October 2026 against OpenStax phase transitions. Values here are synthetic; no heating experiment is requested.