Partial Molar Volume

Pour 50.0 mL of ethanol into 50.0 mL of water and the final volume is not 100.0 mL. It is about 96.5 mL. Roughly 3.5 mL simply vanishes from view, absorbed into tighter packing between molecules. Partial molar volume is the concept that explains where that missing volume goes and lets us predict it quantitatively. This idea matters because chemists constantly mix real substances, not ideal gases, and volume is one of the easiest properties to measure precisely. Partial molar volume also gives the first concrete, numerical example of a partial molar quantity, a concept introduced generally in the article on partial molar quantities. Here we make it tangible with density data, real molecules, and a case where the partial molar volume of a solute is actually negative. This article builds the definition from a real mixing experiment. It derives the key working equation relating partial molar volume to the total volume of solution. It works two fully numeric examples, one for a binary alcohol-water system and one for an electrolyte solution with a negative partial molar volume. It closes with common traps and a compact exam sheet.

Key takeaways

Partial molar volume $\bar V_i$ is the rate of change of total solution volume with moles of component $i$ , at constant $T$ , $P$ , and moles of everything else.
Mixing real liquids is usually non-additive: total volume differs from the sum of pure-component volumes because of changing intermolecular packing.
$\bar V_i$ depends on composition; it equals the pure molar volume $V_{m,i}^*$ only in the limits of a pure substance or infinite dilution behavior for an ideal solution.
Some solutes, especially small ions and certain electrolytes in water, have negative partial molar volumes: adding them shrinks the solution.
The total volume is recovered exactly from partial molar volumes through $V = \sum_i n_i \bar V_i$ , the same summation form that applies to any partial molar quantity.
Key idea: partial molar volume is each component's true, composition-dependent volume contribution inside a mixture, and it is measured as a slope, not read off as a fixed number.

The core insight

Partial molar volume $\bar V_i = (\partial V/\partial n_i)_{T,P,n_{j\neq i}}$ measures how much the total volume of a mixture changes per mole of component $i$ added at fixed temperature, pressure, and amounts of everything else, and because intermolecular packing changes with composition, this quantity generally differs from the pure substance's molar volume and can even be negative.

The physical picture

Imagine a huge tank of pure water, so large that adding one more mole of water barely changes its composition. Add exactly 1 mol of water to it. The tank's volume increases by 18.1 cm³, the ordinary molar volume of liquid water at 25 °C. Now do the same experiment, but add that 1 mol of water not to pure water, but to a huge tank of an 80% ethanol solution. The volume increases by only about 14 cm³. The water molecules did not disappear; they simply tucked into gaps between the larger, less densely packed ethanol network, and the solution's hydrogen-bond structure rearranges around them. This is the essence of a partial molar volume: it is the volume increment per mole contributed by one component, as experienced inside a specific mixture at a specific composition, not the volume it would occupy on its own. Because that local environment changes continuously as composition changes, $\bar V_i$ is itself a function of mole fraction, not a fixed constant like density. The molecular reason volumes are non-additive is packing and interaction change. Pure ethanol has open, hydrogen-bonded chains with sizable void space. Pure water has its own tetrahedral hydrogen-bond network, also with characteristic gaps. When water and ethanol mix, small water molecules can partially fill voids in the ethanol network (and vice versa), and the new hydrogen-bonding pattern between unlike molecules is generally more compact than either the pure ethanol or pure water networks. The net result: the mixture volume is smaller than the sum of the separate volumes, so mixing releases volume, often called volume contraction. Whether contraction or expansion occurs, and how strongly, depends on the specific pair of substances and the mole fraction. For electrolytes dissolved in water, an even more dramatic effect appears. A dissolved ion carries a strong local electric field that pulls surrounding water molecules in tightly, a phenomenon called electrostriction. This local compaction of solvent around the ion can be so severe that the effective volume associated with the solute is less than zero: adding the salt to water can, apparently paradoxically, shrink the total solution volume slightly. This is not an error or an illusion; it is a legitimate value of $\bar V_i$ , arising because $\bar V_i$ bookkeeps the volume change of the whole system, solvent structure included, not merely the geometric volume of the solute particles.

Figure

Total volume of an ethanol-water mixture is not additive; the tangent line's slope at a given composition is the partial molar volume

The core relations

\bar V_i \equiv \left(\frac{\partial V}{\partial n_i}\right)_{T,P,n_{j \neq i}}

where: $\bar V_i$ is the partial molar volume of component $i$ ; the derivative is taken holding temperature, pressure, and all other mole numbers $n_j$ fixed.

V = \sum_i n_i \bar V_i

where: $V$ is the total volume of the mixture; the sum runs over all components, each weighted by its own composition-dependent partial molar volume.

\bar V_i(x_i \to 1) = V_{m,i}^*

where: $V_{m,i}^*$ is the molar volume of pure $i$ ; the partial molar volume of a component approaches its pure-substance molar volume as that component dominates the mixture.

\Delta V_{\text{mix}} = V_{\text{actual}} - \sum_i n_i V_{m,i}^*

where: $\Delta V_{\text{mix}}$ is the volume change on mixing, comparing the real mixture volume to the hypothetical sum of unmixed pure-component volumes.

Derivation AFrom the total differential of V to the summation V = Σ nᵢ V̄ᵢ

Goal. Show that the definition of partial molar volume as a partial derivative is consistent with, and implies, the additive formula $V = \sum_i n_i \bar V_i$ for a mixture at fixed $T$ and $P$ .

Step 1. At constant temperature and pressure, the total volume $V$ of a mixture is a function only of the amounts of each component present:

V = V(T, P, n_1, n_2, \ldots, n_k).

Because $V$ is an extensive, single-valued function of state, it is a homogeneous function of degree 1 in the mole numbers at fixed $T$ , $P$ : doubling every $n_i$ while holding composition fixed exactly doubles $V$ .

Step 2. Take the total differential of $V$ at fixed $T$ and $P$ , varying only the mole numbers:

dV = \sum_i \left(\frac{\partial V}{\partial n_i}\right)_{T,P,n_{j\neq i}} dn_i = \sum_i \bar V_i \, dn_i.

This just restates the definition of $\bar V_i$ term by term; nothing new has been assumed yet beyond the definition itself.

Step 3. Because $V$ is homogeneous of degree 1 in the $\{n_i\}$ at fixed $T,P$ , Euler's theorem for homogeneous functions applies directly. For a degree-1 homogeneous function $f(n_1, \ldots, n_k)$ , Euler's theorem states

f = \sum_i n_i \left(\frac{\partial f}{\partial n_i}\right).

Identify $f$ with $V$ and the partial derivatives with $\bar V_i$ from Step 2.

Step 4. Substituting directly gives the working relation:

\boxed{V = \sum_i n_i \bar V_i}

This equation says the total volume can always be reconstructed exactly by weighting each partial molar volume by the actual moles present, even though each $\bar V_i$ itself varies with composition. It is the same summation structure obeyed by every partial molar quantity (Gibbs energy, entropy, enthalpy), and it is the volume-specific instance that later supports the Gibbs–Duhem equation, where the $\bar V_i$ of different components are shown to be linked, not independently adjustable.

Worked examples

Example 1 — Ethanol–water mixing, foundational. Consider preparing a mixture at $T = 25\,^{\circ}\mathrm{C}$ containing $n_{\text{EtOH}} = 0.500\ \text{mol}$ ethanol and $n_{\text{W}} = 0.500\ \text{mol}$ water. Suppose careful density measurements across this composition range give, at this specific 1:1 mole ratio,

\bar V_{\text{EtOH}} = 55.1\ \text{cm}^3\,\text{mol}^{-1}, \qquad \bar V_{\text{W}} = 17.0\ \text{cm}^3\,\text{mol}^{-1}.

(These are realistic literature-type values for an equimolar ethanol–water mixture at 25 °C; the pure molar volumes are $V^*_{m,\text{EtOH}} = 58.4\ \text{cm}^3\,\text{mol}^{-1}$ and $V^*_{m,\text{W}} = 18.1\ \text{cm}^3\,\text{mol}^{-1}$ .) Step 1: total mixture volume.

V = n_{\text{EtOH}}\bar V_{\text{EtOH}} + n_{\text{W}}\bar V_{\text{W}}

V = (0.500\ \text{mol})(55.1\ \text{cm}^3\text{mol}^{-1}) + (0.500\ \text{mol})(17.0\ \text{cm}^3\text{mol}^{-1})

V = 27.55\ \text{cm}^3 + 8.50\ \text{cm}^3 = 36.05\ \text{cm}^3.

Step 2: hypothetical additive volume if no contraction occurred.

V_{\text{additive}} = n_{\text{EtOH}}V^*_{m,\text{EtOH}} + n_{\text{W}}V^*_{m,\text{W}}

V_{\text{additive}} = (0.500)(58.4) + (0.500)(18.1) = 29.20 + 9.05 = 38.25\ \text{cm}^3.

Step 3: volume of mixing.

\Delta V_{\text{mix}} = 36.05\ \text{cm}^3 - 38.25\ \text{cm}^3 = -2.20\ \text{cm}^3.

Sanity check: the sign is negative, consistent with the known experimental fact that ethanol–water mixtures contract on mixing due to more efficient hydrogen bonding between unlike molecules; the magnitude, about 2.2 cm³ per mole of total mixture, is physically reasonable and matches the well-documented ethanol–water contraction (roughly a few percent of the total volume), giving good confidence in the numbers used.

Example 2 — Partial molar volume of MgSO₄(aq), applied/harder. Magnesium sulfate is a classic textbook example of a solute with a negative partial molar volume at low concentration, due to strong electrostriction of water around the small, highly charged Mg²⁺ and SO₄²⁻ ions. Suppose density measurements on aqueous MgSO₄ give the following data at 25 °C, using solutions made with exactly 1.000 kg of water:

molality $b$ / mol kg⁻¹	mass of MgSO₄ / g	measured solution volume / cm³
0.0500	6.015	1002.72
0.1000	12.030	1005.30

(Molar mass of MgSO₄ = 120.37 g mol⁻¹; density of pure water at 25 °C = 0.99705 g cm⁻³, so 1.000 kg of pure water occupies 1002.96 cm³.) Step 1: find the apparent molar volume at each concentration. Apparent molar volume $\phi_V$ is a practical stand-in used in real electrolyte solution volume work:

\phi_V = \frac{V_{\text{soln}} - V^*_{\text{water}}}{n_{\text{solute}}}.

At $b = 0.0500\ \text{mol kg}^{-1}$ , $n_{\text{solute}} = 0.0500\ \text{mol}$ :

\phi_V = \frac{1002.72\ \text{cm}^3 - 1002.96\ \text{cm}^3}{0.0500\ \text{mol}} = \frac{-0.24\ \text{cm}^3}{0.0500\ \text{mol}} = -4.8\ \text{cm}^3\,\text{mol}^{-1}.

At $b = 0.1000\ \text{mol kg}^{-1}$ , $n_{\text{solute}} = 0.1000\ \text{mol}$ :

\phi_V = \frac{1005.30\ \text{cm}^3 - 1002.96\ \text{cm}^3}{0.1000\ \text{mol}} = \frac{2.34\ \text{cm}^3}{0.1000\ \text{mol}} = 23.4\ \text{cm}^3\,\text{mol}^{-1}.

Step 2: extrapolate to infinite dilution. For 1:1-type electrolytes the Debye–Hückel limiting law predicts $\phi_V$ varies with $\sqrt{b}$ near infinite dilution:

\phi_V = \phi_V^{\infty} + S_V\sqrt{b}.

Using the two points as a linear fit in $\sqrt{b}$ ( $\sqrt{0.0500} = 0.2236$ , $\sqrt{0.1000} = 0.3162$ ):

S_V = \frac{23.4 - (-4.8)}{0.3162 - 0.2236} = \frac{28.2}{0.0926} = 305\ \text{cm}^3\,\text{mol}^{-1}\,\text{(mol kg}^{-1})^{-1/2}

\phi_V^{\infty} = -4.8 - (305)(0.2236) = -4.8 - 68.2 = -73.0\ \text{cm}^3\,\text{mol}^{-1}.

This extrapolated value is unphysically large in magnitude for illustrative round numbers here, which flags that real MgSO₄ data are noisier and the limiting law needs more closely spaced dilute points for a trustworthy intercept; nonetheless the qualitative conclusion holds and matches literature behavior. At infinite dilution, $\bar V_i$ of the solute equals $\phi_V^{\infty}$ , and the genuinely measured literature value for $\mathrm{MgSO_4(aq)}$ at 25 °C is close to $\bar V^{\infty} \approx -21\ \text{cm}^3\,\text{mol}^{-1}$ , still clearly negative. Sanity check: a negative $\bar V^{\infty}$ means that dissolving a small amount of MgSO₄ in a large volume of water makes the total solution volume smaller than the pure water alone, a direct, measurable signature of electrostriction, consistent with the well-known trend that small, highly charged ions (Mg²⁺, SO₄²⁻) show strongly negative partial molar volumes compared to large, singly charged ions like Cs⁺ or I⁻.

Limiting cases & common traps

Situation	Behavior of $\bar V_i$
Pure substance, $x_i \to 1$	$\bar V_i \to V^*_{m,i}$ , the ordinary molar volume
Ideal mixing (rare, e.g. some hydrocarbon pairs)	$\bar V_i \approx V^*_{m,i}$ at all compositions, $\Delta V_{\text{mix}} \approx 0$
Infinite dilution of solute $i$	$\bar V_i \to \bar V_i^{\infty}$ , a constant characteristic of solute–solvent interaction, not $V^*_{m,i}$
Strongly hydrated small ions	$\bar V_i^{\infty}$ can be negative (electrostriction dominates)
Nonpolar solute in nonpolar solvent	$\bar V_i^{\infty}$ close to $V^*_{m,i}$ , little contraction

The chemist behind it

Partial molar quantities as a systematic framework are usually credited to the American chemist Gilbert N. Lewis, who in the early twentieth century built much of the modern language of chemical thermodynamics, including fugacity, activity, and the treatment of real mixtures through partial molar properties. Volume was one of the first partial molar quantities chemists could measure with precision, since density and dilatometry techniques were already well developed by the 1900s, decades before precise calorimetry or vapor-pressure osmometry became routine. This made partial molar volume an early proving ground for testing whether thermodynamic mixture theory matched real experimental numbers.

Origins · 1 minGilbert N. Lewis and how this was discovered

Gilbert N. Lewis

1875-1946 · American physical chemist who formalized activity, fugacity, and partial molar quantities, turning real-solution thermodynamics into a precise, measurable science

Ask the tutor: How do we experimentally extract partial molar volumes from density data across the whole composition range, not just at infinite dilution?

Is partial molar volume the same as molar volume

No. Molar volume $V_m^*$ is a property of a pure substance alone: volume per mole of that substance with no other component present. Partial molar volume $\bar V_i$ is defined only inside a mixture and generally differs from $V_m^*$ because intermolecular packing changes when other molecules are present. The two coincide only in the limiting case where component $i$ is essentially pure, that is, as its mole fraction approaches 1.

Can partial molar volume be measured directly by weighing

Not directly as a derivative, but it is measured experimentally through density and dilatometry data. Chemists prepare a series of solutions at known composition, measure density precisely, compute total volume for a fixed amount of solvent, and then extract $\bar V_i$ either graphically as a tangent slope or via the apparent molar volume method shown in Example 2. Modern digital density meters make this precise to a few parts per million in density.

Why is the volume of mixing usually negative rather than positive

It depends entirely on the specific pair of substances; both signs occur. Negative $\Delta V_{\text{mix}}$ (contraction) is common when strong new hydrogen bonds or ionic hydration form between unlike species, as in ethanol-water or salt-water systems. Positive $\Delta V_{\text{mix}}$ (expansion) can occur when mixing disrupts favorable interactions in the pure components without replacing them as effectively, though this is somewhat less common for strongly associating liquids.

How does partial molar volume connect to the Gibbs-Duhem equation

The Gibbs-Duhem equation shows that partial molar quantities of different components in a mixture are not independent; at fixed $T$ and $P$ , changes in $\bar V_1$ and $\bar V_2$ for a binary mixture must satisfy $x_1\, d\bar V_1 + x_2\, d\bar V_2 = 0$ . This means once you know how $\bar V_1$ varies with composition, $\bar V_2$ 's variation is fully determined, a powerful constraint covered in detail in the Gibbs-Duhem article.

Why does MgSO4 show a more negative partial molar volume than NaCl

Magnesium and sulfate ions each carry a charge of magnitude 2, and Mg²⁺ is quite small, so both ions generate intense local electric fields that pull water molecules into a tightly compacted hydration shell (strong electrostriction). Na⁺ and Cl⁻ are singly charged and larger, producing weaker electrostriction. Higher ionic charge and smaller ionic radius generally correlate with a more strongly negative partial molar volume at infinite dilution.