Measuring Partial Molar Volumes

Pour water into ethanol and the total volume shrinks below the sum of the two starting volumes. This surprising contraction reveals that molecules occupy different effective volumes once mixed than they do alone. Chemists capture that effect with the partial molar volume, a quantity that must be measured, not merely defined. This article shows how experimentalists actually extract partial molar volumes from real density data. You will see how to build a total-volume-versus-composition curve. You will learn why its slope, not the curve itself, is the physically meaningful quantity. You will also see the algebra that converts easily measured densities into rigorous partial molar volumes. Mastering this measurement matters because partial molar volumes feed directly into activity coefficients, mixing thermodynamics, and pressure-dependence calculations for solutions. The method here previews a deeper constraint, the Gibbs-Duhem equation, which links the partial molar volumes of different components to each other.

Key takeaways

The partial molar volume $\bar V_i$ is the slope of total volume $V$ against moles of component $i$ , at constant $T$ , $p$ , and other mole numbers.
Experimentally, chemists measure the density of solutions across a composition range and convert density to molar volume.
For a binary mixture, plotting the apparent molar volume against concentration and extrapolating gives $\bar V_i$ at any composition.
The method of intercepts uses a plot of molar volume versus mole fraction; tangent-line intercepts at $x=0$ and $x=1$ give both partial molar volumes simultaneously.
Partial molar volumes can be negative, as with concentrated aqueous MgSO $_4$ , because ordering effects around ions shrink the surrounding water.
Key idea: a partial molar volume is not measured directly; it is extracted as the slope of a measured total-volume curve, and that slope depends on composition.

The core insight

Because $V$ is a state function of $n_1, n_2, \dots$ at fixed $T$ and $p$ , the partial molar volume $\bar V_i = (\partial V/\partial n_i)_{T,p,n_{j\neq i}}$ is a local slope on the experimental $V$ versus composition surface, so measuring it means constructing that surface from precise density data and differentiating it graphically or analytically, not looking up a fixed per-mole volume for a substance.

The physical picture

Imagine a large beaker of a binary solution, made of $n_1$ moles of solvent and $n_2$ moles of solute, sitting at fixed temperature and pressure. The total volume $V$ depends on both mole numbers. If you add a tiny amount of component 2 while holding $n_1$ , $T$ , and $p$ fixed, the volume changes by an amount that depends on the local environment already present in solution — how crowded the solute molecules already are, how strongly they interact with solvent, and so on. That local, composition-dependent response is the partial molar volume of component 2. The experimental route to $\bar V_i$ begins with something easy to measure: density. For a series of solutions spanning a range of compositions, a chemist measures density $\rho$ with a vibrating-tube densitometer or precision pycnometer, converts to molar volume, and then builds a curve of total volume (per fixed total amount of solution, typically per mole of solvent or per kilogram of solvent) against composition. The partial molar volume of each component is then read off as a slope on that curve — either directly by numerical differentiation, or, more elegantly, using the method of intercepts, which exploits the fact that the tangent line to the molar volume curve at any composition intersects the pure-component axes at the two partial molar volumes simultaneously. A second, complementary picture is the apparent molar volume $\phi_V$ , defined for a solute in a large excess of solvent. This function is directly computable from density data without any curve-fitting, and it converges to the partial molar volume of the solute in the limit of infinite dilution, plus it can be differentiated to get $\bar V_2$ at any finite concentration.

Figure

Total volume of a water–ethanol mixture is not additive; the curve bows below the dashed line of ideal additivity

Figure

Apparent molar volume of MgSO4(aq) extrapolated to infinite dilution

The core relations

V = n_1 \bar V_1 + n_2 \bar V_2

where: $\bar V_i = (\partial V/\partial n_i)_{T,p,n_{j\neq i}}$ is the partial molar volume of component $i$ .

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

where: the derivative is taken with all other mole numbers, $T$ , and $p$ held fixed.

\phi_V = \frac{V - n_1 V_1^{*}}{n_2}

where: $V$ is total solution volume, $n_1 V_1^{*}$ is the volume the pure solvent alone would occupy, $n_2$ is moles of solute.

\bar V_2 = \phi_V + n_2\left(\frac{\partial \phi_V}{\partial n_2}\right)_{T,p}

where: this converts the directly measured apparent molar volume into the true partial molar volume of the solute.

V_m(x_2) = x_1 \bar V_1 + x_2 \bar V_2

where: $V_m$ is molar volume of the mixture, $x_1, x_2$ are mole fractions; the tangent to $V_m(x_2)$ has intercepts $\bar V_1$ at $x_2=0$ and $\bar V_2$ at $x_2=1$ .

Derivation AFrom measured density to partial molar volume by the method of intercepts

Goal. Show that the tangent line to the molar volume curve $V_m(x_2)$ , drawn at any composition, meets the pure-component axes exactly at $\bar V_1$ and $\bar V_2$ , so that a single set of density measurements yields both partial molar volumes without separate experiments.

Step 1. Express molar volume in terms of the two partial molar volumes. For one mole of mixture with mole fractions $x_1 = 1-x_2$ and $x_2$ ,

V_m = x_1 \bar V_1 + x_2 \bar V_2 .

This follows from the Euler theorem for homogeneous functions of degree one, applied to the extensive volume $V = n_1\bar V_1 + n_2 \bar V_2$ divided through by total moles $n = n_1+n_2$ .

Step 2. Differentiate $V_m$ with respect to $x_2$ along the experimental curve. Since $x_1 = 1-x_2$ ,

\frac{dV_m}{dx_2} = -\bar V_1 + \bar V_2 + x_1\frac{d\bar V_1}{dx_2} + x_2\frac{d\bar V_2}{dx_2}.

The Gibbs-Duhem constraint at constant $T$ and $p$ requires

x_1\, d\bar V_1 + x_2\, d\bar V_2 = 0,

so the last two terms cancel exactly, leaving the clean result

\frac{dV_m}{dx_2} = \bar V_2 - \bar V_1 .

(This Gibbs-Duhem relation is the deeper constraint the next article develops in full; here it is the tool that makes the intercept method work.)

Step 3. Write the equation of the tangent line at a chosen composition $x_2^{*}$ . A line through the point $(x_2^{*}, V_m(x_2^{*}))$ with slope $\bar V_2 - \bar V_1$ is

y(x_2) = V_m(x_2^{*}) + (\bar V_2 - \bar V_1)(x_2 - x_2^{*}).

Step 4. Evaluate the tangent at $x_2 = 0$ . Using $V_m(x_2^{*}) = x_1^{*}\bar V_1 + x_2^{*}\bar V_2$ from Step 1,

y(0) = x_1^{*}\bar V_1 + x_2^{*}\bar V_2 - (\bar V_2-\bar V_1)x_2^{*} = x_1^{*}\bar V_1 + x_2^{*}\bar V_1 = \bar V_1,

because $x_1^{*}+x_2^{*}=1$ .

Step 5. Evaluate the tangent at $x_2 = 1$ .

y(1) = x_1^{*}\bar V_1 + x_2^{*}\bar V_2 + (\bar V_2-\bar V_1)(1-x_2^{*}) = \bar V_2,

by the same substitution.

Boxed result.

\boxed{\text{tangent intercept at } x_2=0 \text{ gives } \bar V_1;\quad \text{intercept at } x_2=1 \text{ gives } \bar V_2}

A single measured curve of molar volume versus composition, differentiated once, yields both partial molar volumes at any chosen composition.

Worked examples

Example 1 — Foundational: partial molar volume of NaCl in water from density data. Suppose density measurements on aqueous NaCl at 25 °C give the following apparent molar volumes of the solute, computed from

\phi_V = \frac{M_2}{\rho} - \frac{1000(\rho - \rho_0)}{m\,\rho\,\rho_0}

where $M_2 = 58.44$ g mol $^{-1}$ is the molar mass of NaCl, $\rho_0 = 0.99705$ g cm $^{-3}$ is the density of pure water at 25 °C, $m$ is molality in mol kg $^{-1}$ , and $\rho$ is the solution density. Take a solution of molality $m = 1.000$ mol kg $^{-1}$ with measured density $\rho = 1.0383$ g cm $^{-3}$ (a realistic literature value). Compute $\phi_V$ directly:

\frac{M_2}{\rho} = \frac{58.44 \text{ g mol}^{-1}}{1.0383 \text{ g cm}^{-3}} = 56.29 \text{ cm}^3\text{mol}^{-1}.

\frac{1000(\rho-\rho_0)}{m\,\rho\,\rho_0} = \frac{1000(1.0383-0.99705)}{1.000 \times 1.0383 \times 0.99705} = \frac{41.25}{1.0352} = 39.85 \text{ cm}^3\text{mol}^{-1}.

\phi_V = 56.29 - 39.85 = 16.44 \text{ cm}^3\text{mol}^{-1}.

Masterton's classic tabulated value for NaCl(aq) at $m=1.0$ mol kg $^{-1}$ , 25 °C is close to $16.6$ cm $^3$ mol $^{-1}$ , confirming the arithmetic is physically reasonable. Repeating this calculation at several molalities ( $m = 0.1, 0.5, 1.0, 2.0$ mol kg $^{-1}$ ) and plotting $\phi_V$ against $\sqrt m$ (the natural variable for the Debye-Hückel limiting law) gives a near-straight line. Extrapolating that line to $m \to 0$ gives the partial molar volume of NaCl at infinite dilution, experimentally about $\bar V_2^{\infty} = 16.6$ cm $^3$ mol $^{-1}$ , essentially unchanged here because NaCl's apparent molar volume is only weakly concentration-dependent in this range. The partial molar volume of water itself follows from the intercept method or from

\bar V_1 = \frac{M_1}{\rho_1^{*}} \quad \text{(pure solvent limit)}, \qquad M_1 = 18.02 \text{ g mol}^{-1},\ \rho_1^{*}=0.99705 \text{ g cm}^{-3}

giving $\bar V_1 \approx 18.07$ cm $^3$ mol $^{-1}$ at infinite dilution of NaCl, dropping only slightly as NaCl concentration increases.

Example 2 — Applied: locating partial molar volumes in ethanol–water at $x_{\text{EtOH}} = 0.40$ . Real density data for ethanol-water mixtures at 25 °C (a well-known contraction system) give molar volumes (per mole of mixture) approximately:

x_{\text{EtOH}} = 0.20:\ V_m = 21.95 \text{ cm}^3\text{mol}^{-1}; \qquad x_{\text{EtOH}} = 0.40:\ V_m = 26.20 \text{ cm}^3\text{mol}^{-1}; \qquad x_{\text{EtOH}} = 0.60:\ V_m = 31.40 \text{ cm}^3\text{mol}^{-1}

(these are representative values consistent with published ethanol-water excess volume curves, in which the mixture volume lies below the ideal straight line joining $V_m^{*}(\text{H}_2\text{O}) = 18.07$ cm $^3$ mol $^{-1}$ and $V_m^{*}(\text{EtOH}) = 58.4$ cm $^3$ mol $^{-1}$ ).

Step 1: estimate the local slope at $x_2=0.40$ by central difference.

\left(\frac{dV_m}{dx_2}\right)_{x_2=0.40} \approx \frac{31.40 - 21.95}{0.60-0.20} = \frac{9.45}{0.40} = 23.6 \text{ cm}^3\text{mol}^{-1}.

Step 2: use the tangent-intercept algebra derived above. From the derivation, slope $= \bar V_2 - \bar V_1$ , and

\bar V_1 = V_m(x_2^{*}) - x_2^{*}(\bar V_2 - \bar V_1) = 26.20 - 0.40(23.6) = 26.20 - 9.44 = 16.76 \text{ cm}^3\text{mol}^{-1}.

Step 3: solve for $\bar V_2$ .

\bar V_2 = \bar V_1 + 23.6 = 16.76 + 23.6 = 40.36 \text{ cm}^3\text{mol}^{-1}.

Sanity check. Pure water's molar volume is 18.07 cm $^3$ mol $^{-1}$ and pure ethanol's is 58.4 cm $^3$ mol $^{-1}$ . The extracted $\bar V_1 = 16.76$ cm $^3$ mol $^{-1}$ is below pure water's molar volume, consistent with water being compressed by ethanol's presence (volume contraction on mixing). The extracted $\bar V_2 = 40.36$ cm $^3$ mol $^{-1}$ is well below pure ethanol's 58.4 cm $^3$ mol $^{-1}$ , showing ethanol also occupies less volume once surrounded by water. Both values fall physically between zero and the pure-component molar volumes, and the weighted sum reproduces $V_m$ :

0.60(16.76) + 0.40(40.36) = 10.06 + 16.14 = 26.20 \text{ cm}^3\text{mol}^{-1},

matching the input value exactly, confirming internal consistency of the intercept algebra.

Limiting cases & common traps

Situation	What actually happens	Common error
Infinite dilution ( $x_2 \to 0$ )	$\bar V_2 \to \bar V_2^{\infty}$ , a well-defined finite limit even though the solute has "no volume of its own" to measure directly	Assuming $\bar V_2^\infty$ equals the pure-solute molar volume; it usually does not
Ideal mixing	$\bar V_i = V_{m,i}^{*}$ for all compositions, and $V_m(x_2)$ is a straight line	Assuming all real mixtures behave this way; most show curvature
Concentrated electrolyte solutions	$\bar V_2$ can be negative (e.g., MgSO $_4$ (aq) near infinite dilution), because electrostriction shrinks the surrounding water more than the ion's own volume adds	Treating a negative partial molar volume as a measurement error
Single-density-value shortcut	One density measurement gives only $V_m$ at one composition, never a partial molar volume	Reporting $V_m$ or $\phi_V$ itself as "the partial molar volume"

The chemist behind it

Careful density-based determination of partial molar volumes grew out of late nineteenth- and early twentieth-century physical chemistry's drive to quantify non-ideal mixing, following foundational work by Josiah Willard Gibbs on the thermodynamics of multicomponent systems and later systematized by Gilbert N. Lewis, who popularized partial molar quantities as practical laboratory tools rather than abstract thermodynamic constructs. Lewis championed the view that every extensive thermodynamic property of a mixture could be decomposed into partial molar contributions, measurable in principle from composition-dependent experiments exactly like the volume measurements described here.

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

Gilbert N. Lewis

1875-1946 · American physical chemist who formalized partial molar quantities and activity, turning Gibbs's abstract thermodynamics into a practical experimental toolkit for solution chemistry

Ask the tutor: How does the Gibbs-Duhem equation let you calculate one component's partial molar volume if you already know the other's across the full composition range?

Why can't I just measure the partial molar volume directly with one experiment

Partial molar volume is defined as a derivative, $(\partial V/\partial n_i)_{T,p,n_j}$ , so it is inherently a rate of change, not a single measurable quantity like mass or length. You cannot isolate "the volume belonging to component $i$ " in a mixture because molecular volumes are not simply additive once molecules interact. Instead, chemists measure total volume (via density) across a range of compositions and extract the slope numerically or graphically, which is why the method described here requires a series of measurements rather than one reading.

What is the difference between apparent molar volume and partial molar volume

Apparent molar volume $\phi_V$ pretends all non-ideality is carried by the solute while the solvent behaves as if pure, which makes it directly calculable from one solution's density plus the pure solvent's density. Partial molar volume $\bar V_2$ is the true thermodynamic derivative and generally differs from $\phi_V$ except in the special case of infinite dilution, where the two coincide because the solvent genuinely is behaving as pure solvent there.

Why do we plot apparent molar volume against the square root of concentration rather than concentration itself

For dilute electrolyte solutions, the Debye-Hückel limiting law predicts that apparent molar volume varies linearly with $\sqrt c$ at low concentration, not linearly with $c$ itself. Plotting against $\sqrt c$ straightens the low-concentration data, making the extrapolation to infinite dilution ( $c \to 0$ ) far more reliable and reducing the risk of extrapolation bias from curve-fitting a nonlinear function.

Can partial molar volumes be negative, and if so what does that mean physically

Yes; a classic example is MgSO $_4$ in dilute aqueous solution, where the partial molar volume of the salt extrapolates to a negative value near infinite dilution. The strong electric field around the small, highly charged Mg $^{2+}$ and SO $_4^{2-}$ ions pulls surrounding water molecules into a more compact arrangement than they had in bulk water, a phenomenon called electrostriction. This solvent compaction outweighs the intrinsic volume the ions themselves would occupy, making the net partial molar volume of the solute negative.

How does this measurement connect to the Gibbs-Duhem equation

The tangent-intercept method relies on a hidden simplification: differentiating the molar-volume equation naturally produces extra terms in $d\bar V_1$ and $d\bar V_2$ , and these vanish only because of the Gibbs-Duhem constraint at constant $T$ and $p$ . That constraint says the two partial molar volumes cannot vary independently of each other across composition; they are coupled. This coupling is explored fully as its own topic, since it also governs partial molar entropy, enthalpy, and Gibbs energy in the same way.