Partial Molar Quantities

Introduction

Adding one mole of liquid to a mixture need not increase its volume by the pure liquid's molar volume. Molecules reorganise, and the change depends on what is already present. Partial molar quantities capture this composition-dependent incremental change. They apply to volume, enthalpy, Gibbs energy and other extensive properties, providing the language needed for chemical potential and mixture thermodynamics.

Core explanation

For an extensive property X of a mixture at temperature T, pressure p and amounts n₁, n₂, …, define the partial molar property of component i as X̄ i = (∂X/∂n i) T,p,n j≠i. The derivative imagines adding a very small amount of i while holding T, p and every other component amount fixed. It is not generally the property of pure i and not the total X divided by n i. Its numerical value can change as composition changes.

For volume, V̄ i = (∂V/∂n i) T,p,n j≠i. An incremental addition dn i changes the mixture volume by approximately V̄ i dn i if the other amounts stay fixed. Because V is extensive, its total at a fixed state can be written V = Σ i n i V̄ i. This Euler relation does not say the partial molar volumes are constant across all compositions; their values are evaluated at the composition of that particular mixture. As composition changes, both n i and V̄ i can change.

Water and ethanol illustrate nonadditivity. Their molecules interact and pack differently in a mixture than in their separate liquids, so mixing measured portions can produce a final volume smaller than the sum of the unmixed volumes. The partial molar volume of ethanol in a water-rich solution is not necessarily its pure-liquid molar volume. This is not “missing matter”; the same atoms occupy a different total volume because intermolecular arrangement changes.

The same derivative definition applies to enthalpy H̄ i and Gibbs energy Ḡ i. The latter is the chemical potential μ i. Molar quantities and partial molar quantities coincide in a pure one-component phase under the relevant conditions, but a mixture introduces composition effects. A negative partial molar volume is mathematically possible for some solute descriptions under special conditions because adding solute can reorganise solvent enough to reduce total volume; it should not be interpreted as a molecule having negative physical size.

Step-by-step reasoning

Write X as a function of T, p and all independent amounts. Identify which amount changes and which variables stay fixed. Differentiate X with respect to that amount to obtain X̄ i, then evaluate at the stated composition. For a finite amount change, do not multiply by one constant X̄ i unless composition changes negligibly; otherwise integrate along the path or use final-state data.

Visual explanation

Plot total volume V against amount n B while holding T, p and n A fixed. The tangent slope at a selected composition is V̄ B. A secant line from the origin is not generally the same slope. Add a second curve for a different fixed n A to show that the incremental volume depends on background composition.

Real-world analogy

Adding another passenger to a bus changes occupied space differently when the bus is nearly empty than when seats and luggage must be rearranged. The incremental change depends on the existing mixture. The analogy captures context dependence, although molecules can attract and pack in ways passengers cannot.

Real-world example

When ethanol is mixed with water, the final volume can be less than the sum of the separately measured volumes. A chemist preparing an accurate volume percentage cannot simply add pure-liquid molar volumes. Composition-specific density or partial molar volume data are needed to predict the actual final volume.

Why?

Extensive properties depend on both how much material is present and how components interact. A derivative with respect to one amount isolates its marginal contribution at the current state. This is why partial molar values are useful in mixtures: they translate a small composition change into a measurable change in the total property without pretending every component retains its pure-substance behaviour.

Common misconception

The symbol V̄ B does not mean V/n B. The quotient allocates all mixture volume to B and ignores A, while the derivative measures the response to adding B. Nor can two partial molar volumes be varied independently at fixed T and p in a binary mixture without respecting the thermodynamic relationship between them.

Worked example

At fixed T and p, suppose a model binary mixture near a specified composition has volume V(n A,n B) = 18n A + 40n B − 4n An B/(n A+n B), with volumes in mL for mole amounts in mol. At n A = n B = 1, differentiate the last term with respect to n B while holding n A fixed: ∂[n An B/(n A+n B)]/∂n B = n A²/(n A+n B)² = 1/4. Thus V̄ B = 40 − 4(1/4) = 39 mL mol⁻¹. Its pure B value in this model is 40 mL mol⁻¹. The mixture's incremental B volume is lower because of the interaction term.

Quick check

1. What is held fixed when defining V̄ A in a binary mixture? Answer: Temperature, pressure and the amount n B are fixed while n A changes infinitesimally. The derivative is evaluated at the mixture's current composition.

Exam focus

Include all held-fixed subscripts in the definition. Use derivative rather than pure-molar values for a mixture. Check units: a partial molar volume has volume per mole. When using V = Σ n iV̄ i, take all V̄ i at the same final composition and state, not at unrelated pure-component states.

Advanced insight

Partial molar properties can be determined experimentally from composition-dependent total-property data. In a binary system, the tangent to a molar-property-versus-composition curve can be used to extract both components' partial molar values. The construction works because the extensive total obeys Euler's homogeneity relation.

Summary

A partial molar property X̄ i is the change in an extensive mixture property when an infinitesimal amount of component i is added at fixed T, p and other amounts. Partial molar volume explains why solution volumes may differ from sums of pure-liquid volumes. At a given composition, X = Σn iX̄ i, while the X̄ i themselves vary with composition.

Practice questions

1. For an idealised additive volume V = 20n A + 30n B in mL, find V̄ A and V̄ B. Answer: Differentiation gives V̄ A = 20 mL mol⁻¹ and V̄ B = 30 mL mol⁻¹, independent of composition in this idealised model. 2. Why may 1 mol of B added to an A-rich mixture change volume differently from 1 mol added to a B-rich mixture? Answer: The local intermolecular environment and packing differ, so V̄ B changes with composition. A finite one-mole addition may also move through a range of compositions, requiring integration for precision. 3. If a mixture has V = 100 mL, does V̄ B = 25 mL mol⁻¹ imply it contains exactly 4 mol of B? Answer: No. V̄ B is a marginal derivative, not V/n B. Other components contribute to total volume, and partial molar values depend on composition.